text:
\section{Introduction}
This article aims to initiate the study of quotients of smooth Poisson manifolds $(M,\pi) $ by an \underline{infinite} discrete group $G$ of Poisson automorphisms of $(M,\pi)$.
In the sequel, we call \emph{quotient of $M$ by $G$} and denote by $ M\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ (with an abuse of notation) the quotient of $M$ under the closure of the equivalence relation induced by $G$, \emph{i.e.} $ x \sim y $ if and only if there exists sequences $ (x_n)_{n \in \mathbb N},(y_n)_{n \in \mathbb N}$ with limits $x$ and $y$ respectively such that $ x_n$ and $y_n$ are in the same $G$-orbit for all $n \in \mathbb N$. Such quotients are neither varieties
but they are always "Poisson sets" \cite{Hichem} in the following sense: a real valued function on $M\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ is said to be \emph{smooth} if its pull-back to $M$ is smooth on $M$. In other words, we define the algebra of smooth functions on $M\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ by  $ C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G) := C^\infty(M)^G$. Since $G$ preserves the Poisson structure $\pi $, the algebra $ C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$ comes equipped with a Poisson bracket (see \cite{c3}, Section 5.4).

\begin{defi} \cite{Hichem1} \label{def:Lassoued}
Let $(M,\pi_M)$ be a Poisson manifold, and $C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G) := C^\infty(M)^G$.
A \emph{symplectic realization} of $M\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ is a triplet $(\Sigma, \pi_{\Sigma}, \varphi)$ where $(\Sigma, \pi_{\Sigma})$ is a symplectic manifold (maybe with boundary), $\varphi: \Sigma \to M\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ is a map such that
\begin{enumerate}
    \item[i)] $\varphi \colon \Sigma \to M\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ is smooth, \emph{i.e.} $\varphi^*(F)$ is a smooth function on $\Sigma $  for all $F \in C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$,
    \item[ii)] the algebra morphism $\varphi^*\colon C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G) \to C^{\infty}(\Sigma)$ is a Poisson morphism.
\end{enumerate}
Such a symplectic realization $(\Sigma,\pi_\Sigma,\varphi) $ is called a \emph{symplectic resolution} of $M\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ when:
\begin{enumerate}
    \item[iii)] there exists an open dense $G$-saturated subset $U \in M$ on which the $G$-action is discrete, proper and free\footnote{This implies that $U\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ is a manifold (and makes sense of Condition \emph{(iiii)}).},
    \item[iiii)] $ \varphi : \varphi^{-1}(U\hspace*{-1.5pt} / \hspace*{-1.5pt}G) \to U\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ is a diffeomorphism.
\end{enumerate}
\end{defi}

\noindent
Symplectic realizations of $M\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ defined as in
Definition \ref{def:Lassoued} echoes the nowadays classical notion of symplectic realizations of smooth Poisson manifolds \cite{BX16}-\cite{DCW}-\cite{DZ}-\cite{Karazev}.
Also, for symplectic resolutions of $M\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ defined as in Definition \ref{def:Lassoued}, the manifolds $ \Sigma$ and $M$ have the same dimension while in the latter theory, the dimension is doubled. Definition \ref{def:Lassoued} is therefore more similar to that of symplectic resolutions of affine Poisson varieties as in \cite{sym}-\cite{BaoHua}-\cite{c1234}: indeed, it is an extension of this notion to the context of real differential geometry (as in \cite{c1234}).

\noindent
In algebraic geometry, examples of symplectic resolutions  include quotients of symplectic vector spaces by \textit{finite} subgroup of the symplectic group, see \cite{BellamySchedler}. The most basic examples of such quotients are the very classical Du Val singularities $ (A_n, D_n,E_6,E_7,E_8)$, see \cite{Du Val}.
It is known that the minimal desingularization of these singularities  is a symplectic resolution \cite{sym}-\cite{res}. This implies that those (more precisely, their real parts) are also symplectic resolutions in the sense of Definition~\ref{def:Lassoued}.

\noindent
We wish to investigate quotients of Poisson vector spaces under infinite discrete groups and their symplectic resolutions.
Notice that for $G$ a discrete infinite subgroup of $GL(V) $, there may be very few non-constant algebraic $G$-invariant functions, while $G$-invariant smooth functions $ C^{\infty}(V \hspace*{-1.5pt} / \hspace*{-1.5pt} G ) $ may be still large enough to be worth of study. Our problem can not be addressed within algebraic geometry: We have to work with smooth functions.

\noindent
But this is certainly not an easy matter. Our present purpose is to understand in detail two examples that we think to be "generic", as least for b-manifolds. Both will turn to be surprisingly subtle, although they are simply the quotients of $\mathbb R^2$ by the groups
 \begin{equation}\label{eq:GG'}
 G = \left\{ \left(  \begin{array}{cc} 1 & n\\ 0 & 1 \end{array}
 \right) \,  \middle| n \in \mathbb Z \right\} \hbox{ and }  G' = \left\{ \left(  \begin{array}{cc} 1 & n\\ 0 & \epsilon \end{array}
 \right) \,  \middle|  \epsilon \in \{\pm1\}, n \in \mathbb Z   \right\} \end{equation}
which are isomorphic to $\mathbb{Z} $ and $\mathbb{Z}\hspace*{-1.5pt} / \hspace*{-1.5pt}2\mathbb{Z} \rtimes \mathbb{Z}$ respectively.
These infinite but discrete groups act on $\mathbb R^2 $ by Poisson automorphisms for the   Poisson structure
 \begin{equation}\label{eq:Poisson1} \pi_{\mathbb R^2} := \frac{1}{2 \pi} \, q  \, \frac{\partial}{ \partial p} \wedge \frac{\partial}{ \partial q}  \end{equation}
where $(p,q) $ stands for the canonical coordinates on $\mathbb R^2 $.  This Poisson bracket is log-symplectic, making $\mathbb R^2 $ a b-manifold \cite{Miranda}.

\noindent
It was proven in \cite{Hichem1} that the previous Poisson structure does not admit symplectic resolutions.  In contrast, the conclusion of the present article is that the quotients $\mathbb R^2 \hspace*{-1.5pt} / \hspace*{-1.5pt}G$ and $\mathbb R^2 \hspace*{-1.5pt} / \hspace*{-1.5pt}G'$ do admit symplectic resolutions.

\vspace{.5cm}

\noindent Let us describe the content of the article. In Section \ref{sec:quotient}, we study the algebras $C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G):= C^{\infty}(\mathbb{R}^2)^G$ and  $C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'):= C^{\infty}(\mathbb{R}^2)^{G'}$.
In the process, we will explain why, unlike in the  finite case, real analytic or holomorphic contexts are not relevant here because they admit too little invariant functions.
For smooth invariant functions, we give a useful Fourier series expansion.

\noindent Section \ref{sec:DuVal} is dedicated to reminders on Du Val's symplectic resolutions (in the real case) of $ A_{2k}$ and $ D_{2k+1}$.

\noindent In section \ref{sec:Quotient2}, we show that for all $k \geq 2$, the quotient space $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ is in bijection (as a set) with the quotient singularity of Du Val $$ A_{2k}=\left\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=z^{2k}\right\}.$$
We then show that this bijection $\underline{\phi_k}$ behaves well with respect to Poisson structures. By "behaves well", we mean that $\underline{\phi_k}^*:C^\infty(\mathbb R^3)  \to C^{\floor{k/2}} (\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$ is a Poisson algebra morphism.

\noindent We also show similar results for the second group $G' \simeq  \mathbb{Z} \rtimes \mathbb{Z}\hspace*{-1.5pt} / \hspace*{-1.5pt}2\mathbb{Z}$, but it is then the Du Val singularities of type $ D_{2k+1}=\left\{(x,y,z)\in\mathbb{R}^3\mid zx^2+y^2=z^{2k+1}\right\} $ which intervenes (more precisely, $D_{2k+1} \cap \{ z \geq 0 \}$).

\noindent
 Section \ref{sec:resolution} concludes this construction be an extremely surprising result: the (usual) symplectic resolution $\varphi:Z_k \to A_{2k} $ of the Du Val singularity $A_{2k}$ induces a symplectic resolution of the Poisson quotient $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G $, when composed with $\underline{\phi_k}^{-1}: A_{2k} \to \mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G   $. In other words, $ \underline{\phi_k}^{-1} \circ \varphi : Z_k \to  \mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ satisfies the requirements of Definition \ref{def:Lassoued}. This is not an easy result, and we have to deploy various techniques to achieve this, including Fourier analysis, and subtle considerations on the Du Val resolution.

\noindent We also show similar results for the second group $G':=\mathbb{Z} \rtimes \mathbb{Z}\hspace*{-1.5pt} / \hspace*{-1.5pt}2\mathbb{Z}$.
The usual symplectic resolution $\varphi:Z_k \to D_{2k+1} $, restricted to $\varphi^{-1} ( D_{2k+1} \cap \{z \geq 0\})  $, composed with $\overline{\phi_k}^{-1} :  D_{2k+1} \cap \{z \geq 0\}) \to \mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' $  gives a smooth symplectic resolution of the Poisson quotient $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$ in this case.

\noindent
 It is of course interesting to see that there are infinitely many symplectic resolutions for both quotient spaces. It is unexpected to find $A_{2k}$ in one case and $D_{2k+1}$ in the other.

\section{Quotient of $\mathbb{R}^2$ by two infinite groups}\label{sec:quotient}

We equip $ \mathbb{R}^2 $ with the coordinates $(p,q)$. Consider the infinite groups $G \simeq \mathbb{Z}$ and  $G' \simeq \mathbb{Z}\rtimes\mathbb{Z}\hspace*{-1.5pt} / \hspace*{-1.5pt}2\mathbb{Z}$ as in (\ref{eq:GG'}) and their actions on $\mathbb{R}^2$ by

\begin{equation*}
\begin{array}{ccccc}
 & & \mathbb{Z}\times\mathbb{R}^2 & \to & \mathbb{R}^2 \\
 & & n \cdot (p,q) & \mapsto & (p+nq,q)
\end{array} \hbox{ \hspace{.5cm} and } \begin{array}{ccccc}
 & & (\mathbb{Z}\rtimes\mathbb{Z}\hspace*{-1.5pt} / \hspace*{-1.5pt}2\mathbb{Z}) \times\mathbb{R}^2 & \to & \mathbb{R}^2\\
 & & (n,\epsilon) \cdot (p,q) & \mapsto & (p+nq,\epsilon q)
\end{array}
\end{equation*}
for all $(n,\epsilon)\in\mathbb{Z}\times\{-1,1\}$ and $(p,q) \in \mathbb{R}^2$.

\noindent
The closure  $\sim $ of the equivalence relation induced by the action of $G = \mathbb{Z}$ above is given by
\begin{equation}\label{hna} (p,q) \sim (p+nq,q),
\end{equation}
for any pair $(p,q) \in \mathbb{R}^{2}$, $n\in\mathbb{Z}$ and $(p,0) \sim (p',0)$ for all $p,p' \in \mathbb{R}$. This equivalence relation is closed \emph{i.e.}, $\left\{(p_1,q_1)\times(p_2,q_2)\in\mathbb{R}^2 \times \mathbb{R}^2  \mid (p_1,q_1)\sim(p_2,q_2)\right\}$ is a closed set of $\mathbb{R}^2\times\mathbb{R}^2$.
For $G'=\mathbb{Z}\rtimes\mathbb{Z}\hspace*{-1.5pt} / \hspace*{-1.5pt}2\mathbb{Z}$, we still have to identify $(p,q)$ and $(p,-q)$.
We denote by $ \sim'$ this relation.

\begin{convention}
We denote (with a slight abuse of notation)  the equivalence classes of the closed relations $\sim $ and $\sim' $ by $ \mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ and $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$.
\end{convention}

In general, for $\sim$ be a closed equivalence relation on a manifold $M$, we call \emph{smooth function, real analytic function or function of class $C^k $ on  $M\hspace*{-1.5pt} / \hspace*{-1.5pt}\sim$} a function whose pull-back on $M $ through the natural projection $M \mapsto M\hspace*{-1.5pt} / \hspace*{-1.5pt}\sim $ is  smooth, respectively real analytic or of class $C^k $. Equivalently, functions of on of these given types on $M\hspace*{-1.5pt} / \hspace*{-1.5pt}\sim$ are functions on $M$ constant on the equivalences classes.

When the equivalence relation is the closure of a relation given by the action of a group $G$, smooth functions (resp. real analytic or $C^k$-functions) on the quotient are exactly $G$-invariant smooth functions (resp. real analytic or $C^k$-functions) on $M$.
In the case we are interested with, we therefore have $\mathcal{F}[\mathbb{R}^2]^G\simeq \mathcal{F}[\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G]$ and $\mathcal{F}[\mathbb{R}^2]^{G'}\simeq \mathcal{F}[\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G']$ for $\mathcal F = C^\infty$ or $\mathcal C^k $.

Let us study this algebra.
 finite groups,  the algebra of invariant functions $\mathcal{F}[\mathbb{R}^2]^G$ or $\mathcal{F}[\mathbb{R}^2]^{G'}$ may not be finitely generated. We take $G= \mathbb{Z} $ and we start with the algebra $C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)= C^\infty (\mathbb{R}^2)^{G}$.

\begin{lem}\label{didii}
Any function $f \in C^{\infty}(\mathbb{R}^2)^G $ satisfies for all $i \geqslant 1, j \geqslant 0$
   $$\frac{\partial^{i+j}f}{\partial p^i\partial q^j}(p,0)=0, \hspace{1cm} \forall p\in \mathbb{R} .$$
\end{lem}
\begin{proof}[Proof]
The $ k^{th} $ derivative of the relation $f(p,q)=f(p+nq,q)$ with respect to $q$ gives:
\begin{equation*}
    \frac{\partial^kf}{\partial q^k}(p,q)=\sum_{i=0}^{k} {{k}\choose{i}} n^{k-i}\frac{\partial^kf}{\partial p^i\partial q^{k-i}}(p+nq,q).\end{equation*}
For $q=0$, we have $$\sum_{i=0}^{k} {{k}\choose{i}} n^{i}\frac{\partial^kf}{\partial p^i\partial q^{k-i}}(p,0)=\frac{\partial^kf}{\partial q^k}(p,0).$$
The right hand term coincides with the $i=0$ term on left hand. We therefore obtain for all $n \in \mathbb Z$:
 $$\sum_{i=1}^{k} {{k}\choose{i}} n^{i}\frac{\partial^kf}{\partial p^i\partial q^{k-i}}(p,0)=0.$$
The coefficients of a polynomial function that admits infinity many roots are zero. In particular, the coefficients of the polynomial
 $$P_{k} (X):=\sum_{i=1}^{k}  {{k}\choose{i}} \frac{\partial^kf}{\partial p^i\partial q^{k-i}}(p,0) \, X^{i}$$
are zero for all $k \in {\mathbb N}^*$ and $ p \in {\mathbb R}$. This
completes the proof.
\end{proof}

\begin{prop}\label{honi}
The algebra $C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$ of smooth functions on $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ breaks down as follows $$C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)=C^{\infty}(q)+\widetilde{C},$$
where $C^{\infty}(q)$ is the algebra of smooth functions which depends only on the variable $q$, and $\widetilde{C} \subset C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$ is the subalgebra of $G$-invariant smooth functions vanishing with all their partial derivatives along the line $q=0$.
\end{prop}

\begin{proof}[Proof]
\noindent Consider a $G$-invariant smooth function $f(p,q)$. Define smooth functions $g,h$ on  $\mathbb{R} $ and $\mathbb{R}^2 $ respectively by $h(q):=g(0,q)$ and $g(p,q):=f(p,q)-h(q)$. We have $ f=h+g$ by construction. It therefore remains to show that $ g $ is an element of $\widetilde{C}$.

\noindent
Since the function $g$ is obviously $G$-invariant, Lemma \ref{didii} implies that for all $i \geq 1, j \geq 0$:
$$\frac{\partial^{i+j} g}{\partial p^i\partial q^j }(p,0)=0, \ \forall p\in \mathbb{R}.$$
By integration,
$$\frac{\partial^j g}{\partial q^j}(p,0)=\frac{\partial^j g}{\partial q^j}(0,0)+\int_{t=0}^p\frac{\partial^{1+j}g}{\partial p\partial q^j}(t,0)dt.$$
Since $g(0,q)=0$ by construction, we have $\frac{\partial^jg}{\partial q^j}(0,0)=0$, and the first term on the right vanishes.
Lemma \ref{didii} implies the vanishing of the second one. This completes the proof.
\end{proof}

There is a similar result for the $G'$-action on $ \mathbb R^2$.
\begin{prop}\label{prop:honi2}
The algebra $C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G')$ of smooth functions on $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$ breaks down as follows $$C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)=C^{\infty}(q^2)+\widetilde{C},$$
where $C^{\infty}(q^2)$ is the algebra of smooth even functions depending only on the variable $q$, and $\widetilde{C'} \subset C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G')$ is the subalgebra of $G'$-invariant smooth functions vanishing with all their partial derivatives along the line $q=0$.
\end{prop}

\noindent The following corollary shows that the real analytic case is not interesting: there are too little $G$-invariant functions.

\begin{coro}
Real analytic functions on  $\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G $  (on  $\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' $) are the functions (resp. even function) which depend only on the variable $q$.
\end{coro}
\begin{proof}[Proof]
A real analytic function on $ \mathbb{R}^2 $ which vanishes along a line with all its partial derivatives is zero on $ \mathbb{R}^2 $.
The result now follows from Propositions \ref{honi} and \ref{prop:honi2}.
\end{proof}

\noindent This corollary also means that it is not possible to make such a study in the holomorphic context and study  $\mathbb{C}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ instead. There is not enough $G$-invariant holomorphic functions.
Let us complete our description of smooth functions on  $\mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G$:

\begin{prop}\label{refr}
Any function $ f \in C^{\infty}(\mathbb{R}^2)^G \simeq C^{\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G) $ reads
\begin{equation}\label{seriesf}
f(p,q)=\alpha_{0}(q)+\sum_{n\geq 1}\alpha_{n}(q)\cos\left(2n\pi\frac{p}{q}\right)+\sum_{n\geq1}\beta_{n}(q)\sin\left(2n\pi\frac{p}{q}\right),
\end{equation}
where $ \alpha_0$ is a smooth function on $\mathbb R $ and where   $\left(\alpha_{n}(q),\beta_{n}(q)\right)_{n\geq1}$ are smooth functions on $\mathbb R $ that vanish together with all their derivatives at $ q = 0$.
Both above series, moreover, are absolutely convergent on any open ball.
\end{prop}

\begin{proof}[Proof of Proposition \ref{refr}]
\noindent {\it{Step 1.}} Let us construct functions $\alpha_n, \beta_n $ that satisfy (\ref{seriesf}).
For $q\in \mathbb{R}^*$ fixed, the function $p \mapsto f(p,q)$ is smooth and $q$-periodic. Its Fourier transform with respect to the variable $ p $ reads
$$f(p,q)=\alpha_{0}(q)+\sum_{n\geq 1}\alpha_{n}(q)\cos\left(2n\pi\frac{p}{q}\right)+\sum_{n\geq1}\beta_{n}(q)\sin\left(2n\pi\frac{p}{q}\right) $$
\noindent Let us use the change of variables $u=\frac{p}{q}$ to find a friendly expression of the Fourier coefficients:
$$
\begin{array}{rcll}
\alpha_{0}(q)&=& \frac{1}{q}\int_{-q\hspace*{-1.5pt} / \hspace*{-1.5pt}2}^{q\hspace*{-1.5pt} / \hspace*{-1.5pt}2}f(p,q)dp,  & \\
& =& \int_{-1/2}^{1/2}f(uq,q)du & (*_0), \\
& \hbox{and} & & \\
\alpha_n(q)&=&\frac{2}{q}\int_{-q/2}^{q/2}f(p,q)\cos\left(2n\pi\frac{p}{q}\right)dp,& \\
&=& 2 \int_{-1/2}^{1/2}f(uq,q)\cos\left(2n\pi u \right)du, &  (*_n) \\
& \hbox{and} & & \\ \beta_n(q)&=&\frac{2}{q}\int_{-q/2}^{q/2}f(p,q)\sin\left(2n\pi\frac{p}{q}\right)dp,& \\
 &=&  2\int_{-1/2}^{1/2}f(uq,q)\sin\left(2n\pi u \right)du.& (**_n)
\end{array}
$$
Formulas $(*_0,*_n,**_n) $ still make sense for $q=0$, so that the functions $\alpha_n, \beta_n $ are defined on $ \mathbb R$. They are  smooth as integrals of smooth functions on a compact interval.

\noindent {\it{Step 2.}}  Let us show that for all $ n \geq 1 $, the Fourier coefficients $\alpha_n(q),\beta_n(q)\in C^{\infty}(\mathbb{R},\mathbb{R})$ are smooth functions which are zero with all their derivatives at $q=0$.
\begin{enumerate}
    \item[A.] If the function $f(p,q)$ only depends on $q$, then $\alpha_n=\beta_n=0$ for all $n \geqslant 1$, while $\alpha_0(q)=f(p,q) $ and the result is obvious.
    \item[B.] If $f(p,q) $ belongs to $ \tilde{C}$, it vanishes with all its partial derivative along the line $q=0 $. A simple computation using ($*_n,**_n$) then gives the result.
\end{enumerate}
By Proposition \ref{honi}, any function in $C^\infty(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G) $ is a sum of a function of the  type A with a function of the second type B. The result therefore holds.

\noindent {\it{Step 3.}} Let us show that the series in (\ref{seriesf}) are absolutely convergent on any relatively compact subset $\mathcal U$ of $\mathbb{R}^2 $.  Let us integrate by part the right hand side of ($*_n$)  twice. Since $u \mapsto f(uq,q) $ is periodic of period $1$, all boundary term disappear, and we obtain:
\begin{equation}\label{eq:alphaqk}  \alpha_n(q) = \frac{-1}{n^2} \frac{q^2}{4 \pi^2} \int_{-1/2}^{1/2} \frac{\partial^2 f(uq,q)}{ \partial q^2} {\rm cos}(2n \pi u) du. \end{equation}
Let  $K= \frac{B^2}{2 \pi^2} N$, where $B,N$ are the greatest possible value of $|q|$ and $|\frac{\partial^2 F(p,q)}{ \partial p^2}|$ on $\mathcal U $. We have $|\alpha_n(q)| < \tfrac{K}{n^2} $. The same applies to $(**_n) $.
Since the series $ \sum 1/n^2$ converges, both series in \eqref{seriesf} are absolutely convergent. \end{proof}

In fact, the inequality \eqref{eq:alphaqk} can be made much stronger - and it will be needed later.

\begin{lem}
\label{lem:alphank}
For every $i \geq 1$, $k \geq 1 $ and $M >0$,  there exists a contant $K $ that does not depend on $n$ such that:
$$ \left|  \frac{\alpha_n^{(i)}(q)}{q^k}  \right|  \leq \frac{K}{n^k} \hbox{ and } \left|  \frac{\beta_n^{(i)}(q)}{q^k}  \right|  \leq \frac{K}{n^k}  $$
for all $-M<q<M $
\end{lem}
\begin{proof}
 Using successive integration by part, we obtain:
\begin{equation}\label{eq:alphaqk2}  \alpha_n^{(i)}(q) = \frac{1}{n^k} \frac{q^k}{(2 \pi)^k} \int_{-1/2}^{1/2} \frac{\partial^{k+i} f(uq,q)}{ \partial q^{k+i}} {\mathrm cossin_k}(2n \pi u) du. \end{equation}
where $ {\rm cossin_k}$ is ${\mathrm{ cos}},{\mathrm{ sin}}, -{\mathrm{ cos}}, -{\mathrm{ sin}} $ for $k \equiv 0,1,2,3  $ modulo $4$ respectively. Since $|{\rm cossin_k} | \leq 1$,
$$ K = \frac{ 1 }{(2\pi)^k} {\mathrm{Max}} \left\{ \left|  \frac{\partial^{k+i} f(uq,q)}{ \partial q^{k+i}}\right|  , {(u,q) \in [0,1] \times [-M,M]} \right\} $$
This maximum exists by compactness: This concludes the proof for the functions $\alpha_n^{(i)} $. The proof is similar for the functions $\beta_n^{(i)} $.
\end{proof}

Here is an immediate consequence of Lemma \ref{lem:alphank}:

\begin{prop}
Both series in Equation \eqref{seriesf} are uniformly convergent on any compact set, together with all their partial derivatives.
\end{prop}

\noindent For the group action $ G'$, a similar argument leads to:

\begin{prop}\label{refr1}
A function $f\in C^{\infty}(\mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G')$ reads $$f(p,q)=\alpha_{0}(q)+\sum_{n\geq 1}\alpha_{n}(q)\cos\left(2n\pi\frac{p}{q}\right)+\sum_{n\geq1} \beta_{n}(q)\sin\left(2n\pi\frac{p}{q}\right),$$
where the functions $\left(\alpha_{n},\beta_{n}\right)_{n\geq1}$ are smooth and vanish with all their derivatives at $ q = 0 $. Moreover the functions $ \alpha_{n} $ are even functions and the functions $ \beta_{n} $ are odd functions for all $n\in \mathbb{N}$.
The above series, moreover, is absolutely convergent on any relatively compact open set, and so are any of its partial derivatives.
\end{prop}

\section{Real Du Val symplectic resolutions}\label{sec:DuVal}

\noindent In this section, we mainly recall from \cite{sym} and adapt to the real case the construction of the symplectic resolution of the Du Val singularities $ A_{2k} $ and $ D_{2k+1} $.

First, we define a Poisson structure on the varieties $ A_{2k} $ and $ D_{2k+1} $.
For any polynomial function $ F(x,y,z)$ on $\mathbb{R}^3 $, a Poisson bracket on $\mathbb{R}^3 $ is given by \cite{Pichereau2}:
\begin{equation}
 \label{eq:Pichereau} \{x,y\} = \frac{\partial F}{\partial z}, \{y,z\} = \frac{\partial F}{\partial x}, \{z,x\} = \frac{\partial F}{\partial y} .\end{equation}
This Poisson bracket admits $F$ as a Casimir, and therefore goes to the quotient to $\tfrac{\mathbb R[x,y,z]}{(F)} $. Unlike in the complex case, the latter may not be easily identified to polynomial functions on the set $F=0$ (for instance, for $ F= x^2+y^2+z^2$, the latter set is a point).
It only makes sense when the quotient $\tfrac{\mathbb R[x,y,z]}{(F)} $ injects itself into the algebra of real valued functions on $\{F =0\} $. This is the case, in particular, for $F= x^2 + y^2 - z^{2k}$: We denote by $ A_{2k}$ the set
$$A_{2k} =  \{(x,y,z) \in \mathbb{R}^3 \,  \mid \,  x^2 + y^2 = z^{2k}\} .$$
Formula \eqref{eq:Pichereau} equips $A_{2k} $ with the Poisson structure:
\begin{equation}\label{hichem2}
\left\{x,y\right\}=-2kz^{2k-1},\quad \left\{y,z\right\}=2x,\quad \left\{z,x\right\}=2y.
\end{equation}

Let us study the successive blowing up at the origin (the only singular point) of $A_{2k} $ and the behavior of the above Poisson structure.

Let us first describe the minimal resolution of $ A_{2k}$.
\begin{enumerate}
    \item The blowup $\varphi:Z\to A_{2k}$ at the origin $(0,0,0) \in A_{2k}$ is covered by $3$  pieces.
     \begin{enumerate}
         \item In two of them, the strict transform  of $A_{2k}$ is smooth.
         \item In the last one, the $z$-chart, with coordinates $u, v, z$, the morphism is given by
$\varphi_1:  Z_1 \to \mathbb{R}^3 $ defined by $x = uz, y = vz, z = z$, and the inverse image of $A_{2k}$
is defined by $u^2z^2+v^2z^2=z^{2k}$ so that  the strict transform is $ u^2+v^2 =z^{2(k-1)} $.
As a consequence, the strict transform  on this chart is isomorphic to~$ A_{2k-2} $.
     \end{enumerate}
    \item We then blow-up $ A_{2k-2}$, and repeat the procedure
    \item[]
    \item[$\vdots$]
    \item[]
    \item[ (k)]  Therefore, after $k$-successive blowups at $(0,0,0)$: $\varphi_k:  Z_k \to Z_{k-1}$, the
strict transform is given by $x^2+y^2=1$ with $z$ a free variable.
\end{enumerate}
At this point, we obtain a smooth projective variety, called the \emph{(real) Du Val resolution of $A_{2k} $}.

Let us describe the behaviour of the Poisson structure under the successive blow-up leading to the Du Val resolution.
\begin{enumerate}
    \item The Poisson structure \eqref{hichem2} pulls-back to a Poisson structure on the strict transform of $ A_{2k}$ under the first blow-up $ A_{2k} $
    \begin{enumerate}
        \item In the two first affine charts, the pulled-back Poisson structure is symplectic
        \item In the third one, (where the strict transform is $\simeq A_{2k-2}$), the pulled-back Poisson structure is of the form \eqref{hichem2} with $k$ being replaced by $ k-1$.
    \end{enumerate}
     \item The same applies then to the blow-up of $ A_{2k-2}$
    \item[]
    \item[$\vdots$]
    \item[]
    \item[ (k)] Therefore, after $k$-successive blowups at $(0,0,0)$, the pulled-back Poisson structure, which is still well-defined and has no singularity, is given by
\begin{equation}\label{hic23}
\left\{x,y\right\}=0,\quad \left\{y,z\right\}=2x,\quad \left\{z,x\right\}=2y.
\end{equation}
which is symplectic.
\end{enumerate}

We have therefore obtained by successive blowing a symplectic resolution of $A_{2k}$, which we denote by $(Z_k,\pi_k)$. Let us recapitulate.

\begin{prop}
\label{prop:sympresolA2k}
For all $k \geq 2$:
\begin{enumerate}
    \item
The map $\phi: Z_k $ is onto, and restricts to a diffeomorphism $ Z_k \backslash \phi^{-1}(\{0\}) \simeq A_{2k} \backslash \{ 0\} $
    \item
The pull-back of the Poisson structure \eqref{hichem2} to the Du Val resolution of $ \varphi : Z_k \to A_{2k}$ is a symplectic structure.
\end{enumerate}
Equivalently, $(Z_k,\varphi,\pi_k)$ is a symplectic resolution of $ A_{2k}$.
 \end{prop}

\noindent The next point is easy to verify and will be important. Notice that, now, it is only true on $\mathbb{R}$, and not valid on $\mathbb{C}$.

\begin{prop}\label{coco3}
The pull back of the rational functions $\frac{x}{z^{k}}$ and $\frac{y}{z^{k}}$ by the symplectic Du Val resolution $Z_k\rightarrow A_{2k}$ are smooth functions on $Z_k$.
\end{prop}

\begin{proof}
 The unique point where the rational function $\frac{x}{z^{k}}$ is not defined is  $x=y=z=0$.
Let us see the behaviour under the successive blowups that lead to the Du Val resolution:
\begin{enumerate}
    \item
For the blowing-up of the singularity $ A_{2k} $ at the origin, the following phenomenon occurs:
    \begin{enumerate}
        \item in the two first affine charts (the $x$-chart and the $y$-chart), the pull-back is a smooth function, because the pull-back of $z$ can not be equal to $0$,
        \item in the last one, the restriction to the strict transform (which is isomorphic to $A_{2k-2} $) of the pull back of $\frac{x}{z^{k}}$ is the function $\frac{x}{z^{k-1}}$.
    \end{enumerate}
    \item[] We are therefore brought back to the previous function, upon changing $ k $ by $ k-1 $. We can then repeat the procedure.
    \item[$\vdots$]
    \item[]
    \item[ (k)] At the last step, the strict transform is given by $ x^{2}+y^{2}-1=0$  with $z$ being free, and the  back pull of the function $x/z^k $ is simply the function $x$, which is smooth.
\end{enumerate}
This completes the proof.
\end{proof}

\noindent Let us define two sequences $(x_n)_{n \geq 1}$ and $ (y_n)_{n \geq 1}$ of rational functions on $A_{2k} \subset \mathbb{R}^3$ by $x_1(x,y,z):=x$, $y_1(x,y,z):=y$ for all $(x,y,z) \in A_{2k}$ and
\begin{equation}
\label{eq:xnyn}
  \left\{\begin{array}{rcl}
              x_{n}=\frac{x_{1}}{z^{k}}x_{n-1}-\frac{y_{1}}{z^{k}}y_{n-1}\\
						  y_{n}=\frac{x_{1}}{z^{k}}y_{n-1} + \frac{y_{1}}{z^{k}}x_{n-1}.
 \end{array}\right.
\end{equation}
From Proposition \ref{coco3}, it follows that

\begin{coro}\label{coronYn}
For all $n\in \mathbb{N} $ the pull back $X_n := \varphi^* x_n$ and $Y_n = \varphi^* y_n $ of the functions $x_n,y_n$ defined in (\ref{eq:xnyn}) to the Du Val resolution $ \varphi: Z_k \to A_{2k}$ are smooth functions on $Z_k$.
\end{coro}

\noindent By same procedure, we equip the variety defined by
$$  D_{2k+1} := \left\{ (x, y, z) \in \mathbb{R}^3 \middle| zx^2 + y^2 = z^{2k+1} =0  \right\}$$
with the Poisson structure defined as in \eqref{eq:Pichereau} with the help of the function $ zx^2 + y^2 - z^{2k+1}$:
 \begin{equation}\label{PoissonDk} \{x,y\}= -(2k+1)z^{2k}+x^2 , \{y,z\}= 2zx, \{z,x\}= 2y .\end{equation}
Let us now describe a symplectic resolution of the latter. The procedure is at first very similar.
\begin{enumerate}
    \item
The strict transform of the blowing up at zero is covered by the three natural charts.
\begin{enumerate}
    \item In the $x$-chart the strict transform is the variety $  zx-z^{2k+1}x^{2k-1} = - y^2$, whose unique singular point is $(0,0,0) $.
    \begin{enumerate}
    \item The strict transform of the pull-back at the origin of the latter variety is smooth, and the pull-back of the Poisson structure \eqref{PoissonDk} is symplectic.
    \end{enumerate}
    \item  In the $y$-chart the strict transform is smooth and the pull-back of the Poisson structure \eqref{PoissonDk} is symplectic.
     \item  In the $z$-chart the strict transform is $ D_{2(k-1)}$, and the pull-back of the Poisson structure \eqref{PoissonDk} is the Poisson structure  \eqref{PoissonDk}, with $k$ being replaced by $ k-1$.
\end{enumerate}
   We are therefore brought back to the previous problem in changing $ k $ by $ k-1 $ and then repeat the operation.
    \item[$\vdots$]
    \item[]
    \item[ (k)] At the $k^{th}$-step, the situation is quite different from the one of $ A_{2k}$. The strict transform is the variety:
    \begin{equation*}
            zx^{2} + y^{2}-z=0
\end{equation*}
     which admits two singular points: $(1,0,0)$ and $(- 1,0,0)$.
     The pull-back of the Poisson structure \ref{PoissonDk} is given (as in \eqref{eq:Pichereau}) by
       \begin{equation}\label{PoissonD0} \{x,y\}= x^2 -1, \{y,z\}= 2x z, \{z,x\}= 2y .\end{equation}
     \begin{enumerate}
         \item The blow up at $(1,0,0)$  admits a singular point.  Blowing-up at this point, we obtain at last a smooth variety, and the pull-back of the Poisson structure \eqref{PoissonD0} is symplectic.
         \item The blow up at $(-1,0,0)$  admits a singular point.  Blowing-up at this point, we obtain at last a smooth variety, and the pull-back of the Poisson structure \eqref{PoissonD0} is symplectic.
         We denote by $(Z_{k+2}, \pi_{k+2}) $ this symplectic manifold.
     \end{enumerate}
\end{enumerate}

Let $D_{2k+1}^+=D_{2k+1} \cap \{z>0\}$ and  $Z_{k+2}^+ :=\phi^{-1}(\{z>0\}) \subset Z_{k+2} $.  $Z_{k+2}^+$ is  manifold (with boundary). We have:

\begin{prop}
\label{prop:sympresolA2k1}
For all $k \geq 2$:
\begin{enumerate}
    \item
The map $\varphi: Z_{k+2}^+\to D_{2k+1}^+ $ is onto, and restricts to a diffeomorphism $ Z_{k+2}^+ \backslash \phi^{-1}(\{0\}) \simeq D_{2k+1}^+ \backslash \{ 0\} $
    \item
The pull-back of the Poisson structure \eqref{PoissonDk} to the Du Val resolution of $ \varphi : Z_{k+2}^{+} \to D_{2k+1}^+$ is a symplectic structure.
\end{enumerate}
Equivalently, $(Z_{k+2}^+,\varphi,\pi_k)$ is a symplectic resolution of $ D_{2k+1}^+$.
 \end{prop}

\begin{prop}
The pull back $\frac{x}{z^{k+\nicefrac{1}{2}}}$ and $\frac{y}{z^{k+\nicefrac{1}{2}}}$ by the map $Z_{k+2}\rightarrow D_{2k+1}^+$ are smooth functions on $Z_{k+2}$.
\end{prop}

\begin{proof}[Proof]
The proof is similar to that of the Proposition \ref{coco3}. We leave it to the reader.
\end{proof}

\noindent We shall consider the following functions, similar to those in (\ref{eq:xnyn}):

\begin{equation}
\label{eq:xnyn1}
  \left\{\begin{array}{rcl}
              x_{n}=\frac{x_{1}}{z^{k+\nicefrac{1}{2}}}x_{n-1}-\frac{y_{1}}{z^{k+\nicefrac{1}{2}}}y_{n-1}\\
						  y_{n}=\frac{x_{1}}{z^{k+\nicefrac{1}{2}}}y_{n-1} + \frac{y_{1}}{z^{k+\nicefrac{1}{2}}}x_{n-1}.
 \end{array}\right.
\end{equation}
It follows from all of the above that:

\begin{coro}\label{coronYn23}
The pull back $X_n := \varphi^* x_n$ and $Y_n := \varphi^* y_n $ of functions $x_n,y_n$ defined in (\ref{eq:xnyn1}), by the Du Val resolution $ \varphi : Z_{k+2} \to D_{2k+1}$ are polynomial functions on $Z_{k+2}$.
\end{coro}

\section{Quotient Poisson structure}\label{sec:Quotient2}

\noindent We now introduce the Poisson bracket on the quotient space $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ (resp $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$), with $G$ and $G'$ acting on $M = \mathbb R^2 $ as in Section \ref{sec:quotient}.
We introduce for $ k \geq 2 $ the algebra of functions noted $C^{k,\infty}(\mathbb{R}^2)$ which are:
\begin{enumerate}
    \item[a)] differentiable  $\floor{k/2}$-times on $\mathbb R^2 $,
    \item[b)] and smooth outside the straight line $q=0 $.
\end{enumerate}

\noindent
 We now  equip $M=\mathbb{R}^2$ with the following Poisson bracket
\begin{equation} \label{eq:Poisson}\{p,q\}=\frac{q}{2\pi}. \end{equation}

\noindent
The following lemmas are easily verified.
\begin{lem} \label{lem:stable} For every $k\in\left\{2,3,\ldots,+\infty\right\}$, the algebra $C^{k,\infty}(\mathbb{R}^{2})$ is stable under the Poisson bracket.
\end{lem}

\noindent
and

\begin{lem}
\label{lem:invariant}
The Poisson bracket \eqref{eq:Poisson} is invariant under the actions of $ G = \mathbb{Z} $ and $G'$.
\end{lem}

\noindent
In view of Lemmas \ref{lem:stable} and \ref{lem:invariant}, the following algebras come equipped with the induced Poisson brackets:
 $C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G) := C^\infty(M)^G$, $C^\infty(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G'):=C^\infty(M)^{G'}$, $ C^{k,\infty}(M\hspace*{-1.5pt} / \hspace*{-1.5pt}G):= C^{k,\infty}(M)^G$, and $C^{k,\infty}(M)^{G'} $ (see Section \ref{sec:quotient} for notations).

\begin{rema}
\normalfont
The symplectic Poisson bracket $ \left\{p,q\right\}=1 $ is also invariant under the actions of $G$ and $G'$, but does not satisfy Lemma \ref{lem:stable}.
Also, since the singular locus of both group actions (\emph{i.e.} the straight line $q=0$)  projects to a single  point in the quotient space, it makes more sense to choose a Poisson structure on $\mathbb R^2 $ that vanishes on this straight line.
\end{rema}

From now on, we  deal with the $G$-action only: results for the $G'$-action will be stated at the end of the section.

\noindent Let $\mathcal{A}$ be the subalgebra of $C^{k,\infty}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$ generated by $$\left(x(p,q):=q^{k}\cos\left(2\pi\frac{p}{q}\right), \, \, y(p,q):=q^{k}\sin\left(2\pi\frac{p}{q}\right), \, \, z(p,q):=q\right).$$

\begin{prop}
The algebra $\mathcal A $ is stable under the Poisson bracket \eqref{eq:Poisson}.
\end{prop}
\begin{proof}
 A direct computation with this bracket gives:
\begin{equation}
\label{eq:bracketsinA}
\left\{\begin{array}{rcll}
\{x,y\}&=&-2k \, z^{2k-1}  &\in \mathcal{A}\\
          \{y,z\}&=&2 \, x& \in \mathcal{A}\\
           \{z,x\}&=&2 \, y & \in \mathcal{A}.
 \end{array}\right.\end{equation}
 This proves the result.
\end{proof}

Let us consider the map $ \mathbb R^2 \to \mathbb R^3$
defined by
 $$ \phi_k \colon \big(p,q\big) \mapsto \big(x(p,q),y(p,q),z(p,q)\big) .$$
 Since $x,y,z$ are $G$-invariant functions, and are equal to $0$ on the straight line $ q=0$, this map goes to the quotient to define a map:
 $\underline{\phi_k} \colon \mathbb R^2 \hspace*{-1.5pt} / \hspace*{-1.5pt} G \to \mathbb R^3$.
Since $x^2 + y ^2 = z^{2k} $, this map takes values in the Du Val singular surface $A_{2k} $ (see Section \ref{sec:quotient}).

\noindent
\begin{lem}
\label{afair}
The map $ \underline{\phi_k}  \colon \mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G \to A_{2k}$
 is one-to-one.
Its restriction to $ \mathbb R^* \times \mathbb R \hspace*{-1.5pt} / \hspace*{-1.5pt} G \simeq A_{2k} \backslash \{0\} $ is a diffeomorphism.
 \end{lem}
 \begin{proof}
The relation \begin{equation}\label{presquefin}
\begin{split}
x^{2}+y^{2}&=\left(q^{k}\cos\left(2\pi\frac{p}{q}\right)\right)^{2}+\left(q^{k}\sin\left(2\pi k\frac{p}{q}\right)\right)^{2}\\
           &=q^{2k} =z^{2k}.
\end{split}
\end{equation}
proves that $\phi_{k} $ is valued in $A_{2k} $. Hence so is $\underline{\phi_{k}}$.

Injectivity follows from the following two points:
\begin{itemize}
\item For any two $(p,q),(p',q')$ with $q,q'\ne0$,  $x(p,q)=y(p,q)$ if and only if $q=q'$ and $p=p'+nq$ for some $n\in\mathbb{N}$.
\item $x(p,q)=y(p,q)=z(p,q)=0 $ implies $q=0$
\end{itemize}
Hence $\underline{\phi_k} (p,q) = \underline{\phi_k}(p',q')$ if and only if $(p,q) \sim (p',q') $.

\noindent
 Surjectivity follows easily from the surjectivity of $u\to(\cos{u},\sin{u})$ from $\mathbb{R}$~to~$S^1$.
\end{proof}

\noindent
Both $\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ and $A_{2k}$ come equipped with Poisson structures. We would like to state that $\underline{\phi_k} : \mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G \simeq A_{2k} $ is a Poisson isomorphism.
But we have to be careful because there are several algebras of functions on $\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$: the one that allows to make this statement is $\mathcal A $, as we now state
(here $\mathcal{F}(A_{2k})= \frac{\mathbb R[x,y,z]}{x^2+y^2-z^{2k}}$ stands for the algebra of polynomial functions on $A_{2k} $):

\begin{prop}\label{jti}
The pull-back map of $\underline{\phi_{k}}:{\mathbb R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G \to A_{2k}$ is a Poisson morphism:
 $$C^\infty(\mathbb R^3)  \to C^{\floor{k/2}} (\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G).$$
 It restricts to a Poisson
algebra isomorphism
$\underline{\phi_k}^* :\mathcal{F}(A_{2k})\simeq \mathcal A$.
\end{prop}

\begin{proof}[Proof]
By Equation \eqref{presquefin}, the functions $x,y,z \in C^{k,\infty}(\mathbb R^2) $ satisfy  $x^2+y^2 = z^{2k} $.
Since there is no other relations between them, save for multiples of this relation, and since the generators $x,y,z $ of $\mathcal{F}(A_{2k}) $ are also subject to the relation  $x^2+y^2 = z^{2k} $, the algebra morphism $\underline{\phi_k}^*:\mathcal{F}(A_{2k})\hookrightarrow \mathcal A$ is an algebra isomorphism.
It is still necessary to show that it is a Poisson morphism.

\noindent
A comparison of the Poisson bracket on $\mathcal{F}(A_{2k}) $ defined in \eqref{hichem2}
with the Poisson brackets of $x,y,z \in C^{k,\infty}(M)$ as computed in (\ref{eq:bracketsinA})
shows that $\phi_k: \mathbb R^2  \to \mathbb R^3  $ is a Poisson map.
In view of the following commutative diagram
\begin{equation}\label{prsq1}
\xymatrix{ \mathbb{R}^{2} \ar[r]^{\phi_k} \ar@{->>}[d]^{P} & \mathbb{R}^{3} \\ \mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G \ar[r]^{\underline{\phi_k}}& A_{2k} \ar@{^{(}->}[u]^{i} }
\hbox{ \begin{tabular}{c} \\  \\ \\  hence  \\ \end{tabular}}
\xymatrix{
    \mathbb R[x,y,z] \ar[r]^{{\phi_k}^{*}} \ar@{->>}[d]^{i^{*}} & C^{\floor{k/2}}(\mathbb{R}^{2}) \\
    \mathcal F (A_{2k}) \ar[r]^{\underline{\phi_k}^{*}} & C^{\floor{k/2}}(\mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G) \ar@{^{(}->}[u]^{P^{*}}}
\end{equation}
this implies that $\underline{\phi_k}^* $ is a Poisson morphism.

\end{proof}

\noindent For the quotient space ${\mathbb R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$, we consider the subalgebra $\mathcal{A}'\subset C^{k}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G')$ generated by
\begin{equation*}
\left(x=q^{2k}\cos\left(2\pi\frac{p}{q}\right),  \,  y=q^{2k+1}\sin\left(2\pi\frac{p}{q}\right), \,  z=q^2\right), \, \mbox{for any integer } k\geq2.
\end{equation*}

\begin{prop}
The algebra $\mathcal{A}'$ is stable under the Poisson bracket \eqref{eq:Poisson}.
\end{prop}

\begin{proof}[Proof]
A direct computation with this bracket gives:
\begin{equation*}
\left\{\begin{array}{rcll}
\{x,y\}&=&-(2k+1) z^{2k}+x^2 &\in \mathcal{A}'\\
          \{y,z\}&=&2z x&\in \mathcal{A}'\\
           \{z,x\}&=&2y &\in \mathcal{A}'.
 \end{array}\right.\end{equation*}
 This proves the result.
\end{proof}

\noindent Let us consider the map $ \mathbb R^2 \to \mathbb R^3$
defined by
 $$ \phi_k \colon \big(p,q\big) \mapsto \big(x(p,q),y(p,q),z(p,q)\big) .$$
 Since $x,y,z$ are $G'$-invariant functions, and are equal to $0$ on the straight line $ q=0$, this map goes to the quotient to define a map:
 $\underline{\phi_k} \colon \mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' \to \mathbb R^3$.
Since $zx^2 + y ^2 = z^{2k+1} $, this map takes values in the Du Val singular surface $D_{2k+1} $ (see Section \ref{sec:quotient}). More precisely, it takes values in $D_{2k+1}^+ = D_{2k+1} \cap \{z \geq 0\} $

\noindent
\begin{lem}
\label{afair2}
The map $ \bar{\phi_k}  \colon \mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' \to D_{2k+1}^+$
 is one-to-one
 \end{lem}
 \begin{proof}
The relation \begin{equation}\label{presquefin1}
\begin{split}
zx^{2}+y^{2}&=\left(q^{2k+1}\cos\left(2\pi\frac{p}{q}\right)\right)^{2}+\left(q^{2k+1}\sin\left(2\pi k\frac{p}{q}\right)\right)^{2}\\
           &=q^{4k+2} =z^{2k+1}
\end{split}
\end{equation}
proves that $\phi_{k} $ is valued in $D_{2k+1} $. Since $ q^2 \geq 0$, it takes indeed values in $D_{2k+1}^+$. Hence so is $\bar{\phi_k}$. It remains to prove that $\bar{\phi_k}$ is bijective: The proof is identical to that of the Lemma \ref{afair}.
\end{proof}

\noindent As for $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ and $A_{2k} $, since $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$ and $D_{2k+1}^+$ are in bijection, and since they are also Poisson sets, we would like to assert that $\bar{\phi_k}$ is a Poisson isomorphism. As for $\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ and $A_{2k} $, the precise statement will use the subalgebra $\mathcal{A}'\subset C^{k}(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G')$.

\begin{prop}\label{jti1}
The pull-back map of $\bar{\phi_{k}}:{\mathbb R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' \to D_{2k+1}^+$ is a Poisson morphism:
 $$C^\infty(\mathbb R^3) \to C^{\floor{k\hspace*{-1.5pt} / \hspace*{-1.5pt}2}} (\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G). $$
It restrict to a Poisson algebra isomorphism
$\bar{\phi_k}^* :\mathcal{F}(D_{2k+1})\simeq \mathcal A'$.
\end{prop}

\noindent The proof of this Proposition is identical to that of the Proposition \ref{jti}.

\section{Symplectic resolutions of the quotients $\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ and $\mathbb R^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' $}\label{sec:resolution}

\noindent We first deal with the case of $G$. We use the notations of the previous section.

\noindent The map $ \underline{\phi_k}: \mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G \to A_{2k}$ is bijective by Lemma \ref{afair}. Let us denote by $\psi_k :A_{2k} \to \mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G $ its inverse map. We denote by $\varphi:Z_{k} \to A_{2k}$ the usual symplectic resolution of $A_{2k}$, see Section \ref{sec:DuVal}. Corollary \ref{coronYn} will now be used to show the very surprising following result:

\begin{prop}\label{jiti}
The map $\psi_k  \, \circ \, \varphi$ is a smooth map from $(Z_{k},\{\cdot, \cdot\}_{Z_{k}})$ to $(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G, \{\cdot,\cdot\}) $, \emph{i.e.} for every smooth function $F \in C^\infty(\mathbb{R} ^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$, its pull-back $(\psi_k\circ\varphi)^*F$ is a smooth function on $Z_k$.
\end{prop}

\begin{proof}[Proof]
 Let $F \in C^\infty(\mathbb{R} ^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G)$. By Proposition \ref{refr}, we have
$$ F(p,q) = \alpha_{0}(q)+\sum_{n\geq 1}\alpha_{n}(q)\cos\left(2n\pi\frac{p}{q}\right)+\sum_{n\geq1}\beta_{n}(q)\sin\left(2n\pi\frac{p}{q}\right) $$
for some smooth functions $(\alpha_{n})_{n\geq 1}$ and $(\beta_{n})_{n\geq 1}$ on $ \mathbb R$ that vanish at $0$ together with all their derivatives, and some smooth function $\alpha_{0}$ (all unique).

\noindent We wish to define a function $ H $ on $ Z_{k} $ by:
\begin{equation}\label{eq:defH} H := \alpha_{0}(Z)+\sum_{n\geq 1} \frac{\alpha_{n}(Z)}{Z^k}   X_n +\sum_{n\geq1}
\frac{\beta_{n}(Z)}{Z^k}  Y_n\end{equation}
where the functions $ X_n,Y_n \in C^\infty(Z_k)$ are as in Corollary \ref{coronYn} and $ Z= \varphi^* z$.
This needs some justification.
\begin{enumerate}
\item All terms of both series defining $H $ (see \eqref{eq:defH}) are smooth:
\begin{enumerate}
\item Corollary \ref{coronYn} states that $X_n,Y_n $ are smooth functions for all $n \geq 1$.
\item $Z = \varphi^* z$ is also a smooth function.
\item Last, since $\alpha_n, \beta_n $ vanish at $0$ together with all their derivatives, the quotients
$$ \alpha_n(z) / z^n, \beta_n (z) / z^n $$
are smooth functions on $\mathbb R $. In view of (b),  $\alpha_n(Z) / Z^n, \beta_n (Z) / Z^n $ are also smooth functions.
\end{enumerate}
\item Both series are convergent, together with all their partial derivatives.
We will come back to this point at the end the proof.
\end{enumerate}

Now that the existence and the smoothness of the function $H$ is well-established, let us check that
\begin{equation} \label{eq:tirearriere} (\psi_k \circ \varphi)^* F = H  .\end{equation}
It suffices to check that the pull-back through $(\psi_k \circ \varphi)^*$ of each one of the terms of the series defining $F$ are the terms of both series defining $H$.
Consider the following smooth functions on $\mathbb R^3 $:
$$f_n(x,y,z):=\frac{\alpha_{n}(z)}{z^k} x_n  \hbox{ and } g_n(x,y,z) := \frac{\beta_{n}(z)}{z^k} y_n  $$
where the functions $x_n,y_n$ are as in (\ref{eq:xnyn}).
We have:
$$ \left\{ \begin{array}{rcll}
 \varphi^* f_n &= &    \frac{\alpha_{n}(Z)}{Z^n} X_n & \hbox{by definition of $X_n$ (see Cor. \ref{coronYn}) } \\
  \underline{\phi_k}^* f_n & = & \alpha_{n}(q)\cos\left(2n\pi\frac{p}{q}\right)  & \hbox{by definition of $\underline{\phi_k} $}
\end{array} \right. $$
and
$$ \left\{ \begin{array}{rcll}
 \varphi^* g_n &= &   \frac{\beta_{n}(Z)}{Z^n} Y_n & \hbox{by definition of $Y_n$ (see Cor. \ref{coronYn}) } \\
  \underline{\phi_k}^* g_n & = & \beta_{n}(q)\sin \left(2n\pi\frac{p}{q}\right)  & \hbox{by definition of $\underline{\phi_k} $}
\end{array} \right. $$
Hence
 \begin{align*}
 \begin{array}{rcl}
     \varphi^* \circ \psi_k^* \left( \alpha_{n}(q)\cos \left(2n\pi\frac{p}{q}\right)\right) &=&\alpha_{n}(Z) X_n  \\ \varphi^* \circ \psi_k^* \left(  \beta_{n}(q)\, \sin \left(2n\pi\frac{p}{q}\right)\right) &=& \beta_n (Z) \, Y_n\end{array}
 \end{align*}
 \noindent Summing up these identities, we obtain \eqref{eq:tirearriere}.

\noindent To complete the proof, we need to prove uniform convergence of the series \eqref{eq:defH} defining $ H $, and uniform convergence of all the series obtained by applying a given partial differential operator to all its terms.

 \noindent
Let $C \subset Z_k $ be a compact neighborhood of $\phi^{-1}(0) $.
Pulling-back the relation:
 $$ x_n^2+y_n^2=z^{2k} $$
 we obtain that
 \begin{equation}\label{eqXn} X_n^2 + Y_n^2 = Z^{2k} \end{equation}
 so that $X_n,Y_n $ is bounded on $C$ by some constant $k = {\mathrm{max}}_C |Z| $ that does not depend on $n$.

 \noindent
Now let us consider $\varphi_k^{-1} ( \phi(C)) \subset \mathbb R^2  $.
This is a neighborhood of the horizontal line in $\mathbb R^2 $, contained into a band $\mathcal B $ of the form $ - M <q< M $ for some $M \in \mathbb R $.
In view of Lemma \ref{lem:alphank}, there exists a constant $K$ such that
$| \alpha_n(q)/q^k |< K / n^k $ and $| \beta_n(q)/q^k |< K / n^k $ for all $ -M<q<M $, so that
$$ \left| \frac{\alpha_n(Z)}{Z^k} \right| < \frac{K}{n^k} \hbox{ and } \left| \frac{\alpha_n(Z)}{Z^k} \right| < \frac{K}{n^k}  $$
uniformly on $C$. This proves the uniform convergence on $C$ of the series \eqref{eq:defH} defining~$H$.

 \noindent
To prove that $H$ is smooth, it suffices to prove that the series that define $D(H)$ converge for every partial derivative operator $D$ on $Z$.

 \noindent
 For every partial derivative operator $D$ of order $d$, we derive from form the recursion relation \eqref{eq:xnyn} that:
  $$ \begin{pmatrix} D(X_{n+1})\\ D (Y_{n+1}) \end{pmatrix} = \begin{pmatrix} X_1& -Y_1\\ Y_1 & X_1 \end{pmatrix} \begin{pmatrix} D (X_n) \\ D(Y_n) \end{pmatrix}  +  \begin{pmatrix} L(X_n,Y_n) \\ M(X_n,Y_n) \end{pmatrix} $$
where $L(X,Y),M(X,Y)$ depend linearly on a finite number of partial derivatives of order $\leq d-1 $ of $X,Y $.
Hence
$$ U_{n}^D :=  (  D(X_{n}))^2 + ( D(Y_{n}))^2$$
satisfies
\begin{equation}\label{eq:Un} U_{n+1}^D = U_n^D + S^D(X_n,Y_n) \end{equation}
 where $S^D(X,Y)$ depend linearly on a finite number of partial derivatives of order $\leq d-1 $ of $X,Y $.

 \noindent
 We claim that for any partial derivative operator $D$ of degree $d$, there exists constants $A,B $ such that
 $ U_n^D \leq  A +  B n^D $.
 For $d=0$, Equation \eqref{eqXn} implies the desired assumption.
 If this is true for all $D$ of degree $\leq d-1 $, then in
 Equation \eqref{eq:Un}, there exists $A,B$ such that
  $$ |S^D(X_n,Y_n)| \leq A + B n^{d-1} $$
Equation \eqref{eq:Un}
implies then that there exists constants $A',B' $ such that
 $ U_n^D \leq  A' +  B' n^d $.

 \noindent
The convergence to zero of the $d$-first derivatives of $|\alpha_n(Z)/Z^k|, |\beta_n(Z)/Z^k| $ can be
 made smaller that $\frac{K}{n^r} $ for an arbitrary $r$, uniformly on $\phi^{-1}(C) $. This proves the uniform convergence of  the series obtained by applying  $D$ to all the terms in the series in \eqref{eq:defH}. This proves the smoothness of its limits, and therefore the smoothness of $H$.

\end{proof}

\noindent We conclude with the main result of this article:

\begin{thm}\label{resultatp}
The map $\psi_k \circ \varphi: Z_{k} \rightarrow \mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G$ is a symplectic resolution of  $(\mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G,\{\cdot,\cdot \}) $.
\end{thm}

\begin{proof}[Proof]
Let us check that all four conditions in Definition \ref{def:Lassoued} are satisfied.
Condition \emph{(i)} has been established in Proposition \ref{jti}.
Condition \emph{(ii)} follows from Proposition \ref{jti} and the first item in Proposition \ref{prop:sympresolA2k}.
Conditions \emph{(iii)} and  \emph{(iiii)} hold in view of Lemma \ref{afair} and the second item in Proposition~\ref{prop:sympresolA2k}.
\end{proof}

\noindent For the quotient space ${\mathbb R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$, the map $ \bar{\phi_k}: \mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' \to D_{2k+1}^+$ is bijective by Lemma \ref{afair2}. Let us call $\underline{\psi_k} :D^+_{2k+1} \to \mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G' $ its inverse map. We denote by $\varphi:Z_{k+2} \to D_{2k+1}^+$ the usual symplectic resolution of $D_{2k+1}^+$, see Section \ref{sec:DuVal}.

\begin{prop}
The map $\underline{\psi_k} \circ \varphi$ is a smooth morphism of $(Z_{k+2},\{\cdot, \cdot\}_{Z_{k+2}})$ to $(\mathbb{R}^2\hspace*{-1.5pt} / \hspace*{-1.5pt}G', \{\cdot,\cdot\}) $.
\end{prop}

\noindent
\begin{proof}
The proof of this result is identical to that of Proposition \ref{jiti}. We simply use Proposition \ref{refr1} instead of Proposition \ref{refr}, and Corollary \ref{coronYn23} instead of Corollary~\ref{coronYn}.
\end{proof}

\begin{thm}
The map $\underline{\psi_k} \circ \varphi : Z_{k+2} \rightarrow \mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G'$ is a smooth symplectic resolution of the quotient Poisson $(\mathbb{R}^{2}\hspace*{-1.5pt} / \hspace*{-1.5pt}G',\{\cdot,\cdot \} ) $.

\end{thm}

\begin{proof}[Proof]
Let us check that all four conditions in Definition \ref{def:Lassoued} are satisfied.
Condition \emph{(i)} has been established in Proposition \ref{jti1}.
Condition \emph{(ii)} follows from Proposition \ref{jti1} and the first item in Proposition \ref{prop:sympresolA2k1}.
Conditions \emph{(iii)} and \emph{(iiii)} hold in view of Lemma \ref{afair2} and the second item in Proposition~\ref{prop:sympresolA2k1}.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Tag: Archeology
China and the Cincinnati Bengals: Being tough, winning and losing.
When you talk to just about anybody about sports they are quick to declare what their favorite team should do in order to win. "Get rid of T.O. He costs too much and is a pain in the ass!" Or, "get rid of Chad, he runs his mouth too much, he's too expensive and they can't even win with him." I am refereeing to a couple of players for the Cincinnati Bengals, and I hear comments to that effect all the time.
But speak to those same people about how to deal with Social Security, or Education, or any number of social programs, and people clam up and refuse to commit an opinion. I suppose that's because the game under which politics is played is just too complicated for many of them, or they are taking something out of the systems in question, and lack the courage to assert an opinion.
And that's the beauty of sports. Sports allow people to become arm-chair coaches because they don't have anything invested in the team other than committing to an occasional game or a sport jersey. So they can be objective as to the possible problems with the team they're watching.
People like Doc Thompson, and myself, can be objective about social issues, because we aren't expecting government to do anything for us. I wrote off Social Security a long time ago, along with all the other entitlements that are floating around out there. So I particularly enjoyed Doc's show on January 18, 2011 where he laid it on the line as to what the real problems are. Listen to that here.
Hey, he's not exaggerating. The issue truly is whether or not the United States will stay on top of the heap in world affairs. We won't do it complaining about silly issues as to whether or not Native American bones are returned to their graves, or whether or not the entire Constitution can be read because of our internal guilt over slavery. The rest of the world is not hindered by that type of restrictive guilt, and we have to compete with them economically.
My team, the Tampa Bay Buccaneers are not in the playoffs, but I am proud of how they played over the 2010 season. I watched how management approached the off-season last year and I believe they are on the march to winning ways going forward. But the team in my home town, the Cincinnati Bengals continue to be a bad team no matter how much money they spend.
Now you can go to any sports bar in America and even a drunken fool could tell you why the Bengals can't win. And the same holds true for our county. Everybody knows how to fix the problems. But we won't win if we don't toughen up. It's that simple.
What Doc talks about in that clip is a perfectly articulated synopsis of our counties problem. It sounds easy to hear him say it, but he has the luxury of seeing things clearly, because he doesn't want anything from government. People like Thompson rely on themselves first to do most things, so the problems are easy to see.
So America, you better get tough quick. Because being tough is how you win.
overmanwarrior	Budget problems, Congress, Conservative politics, Creative Writing, Doc Thompson, Feminisim, Finance, Football, Football fans, Guilt, Justice, Literature, National Debt, NFL, Organized Labor, Policial Science, Political Theroy, Politics, Progressive, Regulations, Rich Hoffman, School levy, Social Networking, Socialists, Sports, Taxes	14 Comments January 19, 2011 January 19, 2011 2 Minutes
Net Neutrality is the Villian from Tron
The following line is NOT a bit of dialogue from the new Disney film Tron.
"We must take action to protect consumers against price hikes and closed access to the Internet—and our proposed framework is designed to do just that: to guard against these risks while recognizing the legitimate needs and interests of broadband providers." FCC Chairman Julius Genachowski
Who do these people think they are? That's almost as audacious as trying to regulate the air. But that statement almost sounded like the uttering's of the character Clu as he addresses a sea of program soldiers about to be unleashed upon society in the new Tron Legacy film.
This is clearly a power grab by the FCC to further attempt to justify their existence. It must be terribly frustrating for them to have a part of the media that they don't have any control over, and this is an attempt by a government agency to get control.
If you don't understand what NET NEUTRALITY is, listen to Bill Cunningham of 700 WLW discuss it here:
As far as the platform the FCC is using to justify this move, the prices on the internet have regulated themselves. Prices too high are beat out by companies that offer better rates. It's pretty simple really.
I can remember when the internet was new, and charged by the hour, then service for the month for a flat fee. There were many different combinations of price packages that were created along the way, but ultimately cheaper prices in a competitive fashion created the internet we have today, where a single user can talk on their Smart Phone, while talking on a headset over their XBOX and answering their email on a laptop, all done on the internet. The bandwidth drop that the FCC is talking about is a complete fabrication.
It has regulated itself and is one of the only aspects of society that is free. The internet is the model for what society should mimic.
But the FCC under the disguise of "fairness" is attempting to get their little feet in the door, and once there they will unleash many, many, many more regulations, just as they have for television and radio. It won't take long for everyone that wants a website to have apply for one to the FCC just as a radio broadcaster does. And if you violate their terms, you'll lose your license.
You have to consider what we're talking about here, and that is access to the internet. Since the internet is just a bunch of computers talking independently of each other, it might as well be air. It has no central location and does not stop at a nation's border. Therefore, the only way a government can get control of the internet is to control the on-ramp to the World Wide Web. In this case the FCC is using "fairness" to begin a relationship with internet providers that will become a much more parental relationship down the road.
This power move is so audacious that they by-passed congress to do it. We can't even blame the incompetence of our congressional politicians on this one. This is a raw power grab from a branch of government functioning without any checks or balances, and in the context of things, is very serious.
If you dear reader scanning over these words don't find this whole NET NEUTRALITY issue to be EVIL, and doesn't provoke you into an outrage of forced accountability of the FCC, you can count the days that you soon won't be able to read these words, because I will, along with this entire Blog Site will be found in violation of the FCC regulations, and will be shut down. And you will have to get thoughtful comments from an "orthodox" source that the FCC has given an approved license to.
That's what's at stake. As always, government sells itself as having your best interest in mind. It's all about fairness, according to them. But history says otherwise.
The internet has thrived without any government regulation. And now, those public employees that have their socialist tendencies toward securing their own wallets in a collective utopia desire to hang themselves like parasites from internet commerce.
As always, it's about the money.
While I write this there is a court case against WLW for setting up a Christmas manger at their tower in Mason, Ohio. The reason for the law suit is because they have a FCC license and that the argument is any government entity should have a separation of church and state.
Also, the FCC is using AL Sharpten to crack down on conservative talk show hosts like Rush Limbaugh, because the bottom line is people like Rush need to be shut down in order for government to expand, which is the goal of these types of regulators. They desire to secure employment for themselves from tax dollars. That's all there is to it. The whole concept is incredibly short-sighted and selfish. But that's what you usually find are the quality of people that desire to make a living off government work.
Once this line is crossed, these parasites of government regulation will embed themselves like a tick on a dog to suck the life-force of the creature.
Without Net-Neutrality, an entire industry, which young people are quite aware of has risen. These sad politicians wouldn't know much about that industry because they are too busy pursuing lobby supplied prostitutes on K-street, or exotic dinners where they believe their discussion will pave the way for the nation.
You can see the future by simply turning on a Nintendo Wii. The interface of either the Wii, the XBOX or the PlayStation 3, which is all internet based, provides at a glance Netflex, weather updates, news from around the world, and a portal to play Black Ops with a friend from half-way around the world in real time. That is the world of tomorrow, and it was created without governmental control.
Before I went with my family to see the new Tron film, I watched my kids swordfight with each other on the Wii, and many of those images came to me while I was watching that very interesting film. Tron is all about the world of cyberspace and the empires that can rise and fall within that world. And every new game created is a new experience for each and every human being on this planet to be a part of. But the overall theme of Tron was the perilous course of pursuing perfection.
In Tron, Jeff Bridges created an exact opposite of himself called Clu to implement the perfect world, a utopia. What it turned out to be was a hellish digital fascism that Bridges was unable to control and that society completely turned on him. It was heady stuff, not intended to be watched with just a light eye for entertainment.
Net Neutrality is in our world the fictional villain of the Tron film. Initiated with good intentions, but resulting in a hellish existence.
Art becomes reality, most of the time. "Your focus will determine your reality," a line from another great science fiction film proclaims, and it's true. Too bad human beings don't think through things and learn from their art, instead of repeating the same mistakes over, and over, and over, and over, and over, and over…………………………………infinity.
We don't remember what life was like before social security, or Medicare, or any number of government programs. But every one of us knows what the internet was like before the government decided to stick its nose in it. And it has thrived.
Don't believe the lies. Encourage the new congress to escort the FCC to court for this gross imposition on our freedoms and liberty, and instruct them to stay out of our nation's way.
Guess who is the main man behind this push………………..George Soros. Listen to Rush Limbaugh discuss it here.
overmanwarrior	Action, Art, Congress, Conservative politics, Creative Writing, Education, George Soros, Glenn Beck, Literature, Movie Review, Net Neutrality, Policial Science, Political Theroy, Politics, Progressive, Taxes, Technology, Video Games	8 Comments December 23, 2010 December 23, 2010 6 Minutes
WHY DO SOME CRITICS DISLIKE 'DARK KNIGHT RISES?' CLICK HERE TO FIND OUT WHY.
I found a unique treasure at the 2009 Mothman Festival at Point Pleasant in, West Virginia while on a trip with my wife. It's a map of paranormal activity in the State of Ohio called Hidden Ohio. It features Ancient American sites, haunted locations, sacred geography, scenic byways, strange creatures, unexplained mysteries, and unique sites.
It was on this map that I noticed two sites where giant human beings were discovered. One location was just outside of Cleveland, and the other was just south of Cincinnati. The below picture I don't believe is real, but it gives you an idea. I am currently looking for the private citizens that have these discovered bones. In both cases the bones of the giant humans were 8 to 10 feet tall. Prior to seeing those excavation sites on that map, I had never heard of giant humans in an actual historical sense.
Below is a history of the giant bones discovered in North America. Some of the bones are without question stuck in drawers at museums. But most are in the hands of private collectors. I would love to see them for myself.
Historical North American Giants
1792 New York, Buffalo: Turner's History of the Holland Purchase reports that 7 and 8 foot skeletons were found at an earthen fort in Orleans county with broad flat topped skulls.
1800 Ohio, Conneaut: Among the normal size skeletons found in the remains of mounds were found gigantic bones. Some of the skulls and jaws were large enough to fit over the head and face of a normal man
1821 Tennessee, White County: An ancient fortification contained skeletons of gigantic stature averaging at least 7 feet in length.
1825 Ohio Valley: David Cusick, a Tuscorora by birth, wrote that among the legends of the ancient people of the stock, there was a powerful tribe called Ronnongwetowanca. They were giants, and had a "considerable habitation." When the Great Spirit made the people, some of them became giants. After a time, and having endured the outrages of these giants, it is said that the people banded together, and through the final force of about 800 warriors, successfully annihilated the abhorrent Ronnongwetowanca. After that, it was said that there were no giants anywhere. This was supposed to have happened around 2,500 winters before Columbus discovered America, around 1000 BC.
1829 Ohio, Chesterville: In digging away a mound where a hotel was to be built, a large human skeleton was found, but no measurements were made. It is related that the jawbone was found to fit easily over that of a citizen of the village. The local physicians examined the cranium and found it proportionately large, with more teeth than the white race of today. The skeleton was taken to Mansfield, and has been lost sight of entirely.
1833 California, Lompock Rancho: Soldiers digging at Giant Lompock Rancho, California, discovered a male skeleton 12 feet tall. Carved shells, stone axes, and other artifacts surrounded the skeleton. The skeleton had double rows of upper and lower teeth. Unfortunately, this body was secretly buried when local Indians became upset about the remains.
1835 Illinois, Lake County: In the numerous mounds in the county, skeletons ranging between 7 and 8 feet are discovered.
1845 Virginia: A human jaw bone of great size was uncovered in a burial mound on which the teeth stood transversely in the jawbone.
1849 New York: From "Illustrations of the Ancient Monuments of Western New York" comes the report that an elliptical mound above near the Conewango Valley held eight big skeletons. A thigh bone was found to be 28" long. Exquisite stone points, enamelwork, and jewelry were found. Also discovered in the area were a number of other large skeletons one almost 9 feet in height.
1850 New York: From the History of Allegany County in 1879 a report that very large human bones were uncovered during excavation for the railroad
1851 New York: A skull rib bone, and shinbone were found that indicated the height to be over 8 feet tall.
1856 West Virginia, Wheeling: A human skeleton was discovered by labourers while ploughing a vineyard measuring almost 11 feet tall.
1858 Ohio, Vermillion Township: Skeletons of a race of beings much larger than the local inhabitants were discovered.
1870 Ohio: In Brush Creek Township a large mound contained skeltal remains of several humans up to nine feet tall. A large stone tablet with unknown insriptions similiar to Greek writing was also found.
1872 Ohio, Seneca Township: When the "Bates" mound was opened the remains of three skeletons, whose size would indicate they measured in life, at least, eight feet in height, were found. A remarkable feature of these remains was they had double teeth in front as well as in back of mouth and in both upper and lower jaws.
1873 Ohio, Seville: An Ohio Bicentennial Commission historical marker serves as a reminder that the Giants of Seville, Captain Martin Van Buren Bates and his wife, Anna Swan Bates, lived in the village of Seville in Medina County. Anna stood 7 feet 11 1/2 inches tall and weighed 413 pounds. Martin was 7 feet 9 inches tall and weighed 480 pounds.
1873 Washington DC: "The objects here collected which have not been given, or acquired by exchange, have been purchased for the use of the museum by order of the surgeon-general… There is a skeleton of a giant, who, in life, measured seven feet, prepared by Auzoux and mounted by Blanchêne's method, which, if I may use that term, is really a beauty. It is as white and clean as new fallen snow, and the brass joints and screws which keep it together are bright, and of the latest style and finish." From the article "The Army Medical Museum in Washington" by Louis Bagger, Appletons' Journal: A Magazine Of General LiteratureVolume 9, Issue 206
1875 West Virginia, Rivesville: Workmen constructing a bridge near the mouth of Paw Paw Creek uncovered three giant skeletons with strands of reddish hair clinging to the skulls. The skeletons had supported people approximately 8 feet tall.
1876 Wisconsin: Mounds were excavated containing a giant skull and vertebrae.
1877 Missouri, Kansas City: A giant skull was unearthed when mounds wore opened and giant man tracks belonging to humans 25 to 30 feet tall were discovered.
1877 Nevada, Eureka: Prospectors found a human leg bone and kneecap sticking out of solid rock. Doctors examined the remains and determined they were from a human being, and one that stood over 12 feet tall. The rock in which the bones were found was dated geologically to the Jurassic Period, over 185 million years old.
1878 Ohio, Ashtabula County: While excavating the ground for graves, bones were exhumed, which seemed to have belonged to a race of giants. A skull and jaw were found, which were of such size that the skull would fit easily over a large man's head like a loose fitting helmet, even with the jaw in place. The number of these graves has been estimated to be between two and three thousand.
1879 Indiana, Brewersville: A skeleton almost ten feet tall was excavated from a mound.
1880 Ohio, Zanesville: A skeleton was reported to have been of enormous dimensions found in a clay coffin, with a sandstone slab containing hieroglyphics.
1880 Minnesota, Clearwater: Several giant skeletons were found with double rows of teeth.
1881 Ohio, Medina County: A jawbone of great size belonging to a human being was discovered, which contained eight jaw-teeth in each side, of enormous size; and the teeth stood transversely in the jawbone. It would pass over any man's face with entire ease.
Why don't we learn about this in school? That's another story that I've covered elsewhere.
This is what people are saying about my new book–Tail of the Dragon
"Just finished the book and am sweating profusely. Wow, what a ride !!! Fasten your seat belts for one of the most thrilling rides ever in print."
While you wait for Tail of the Dragon, read my first book at Barnes and Nobel.com as they are now offering The Symposium of Justice at a discount which is the current lowest price available.
overmanwarrior	Archeology, Drama, Mystery, Rich Hoffman	54 Comments December 20, 2010 August 11, 2012 5 Minutes
My Love of Real Science and the Adventure of Discovery!
It's certainly not that I'm against education. Quite the contrary, I believe an education never stops. However, since I have continued my own education well into my adulthood, there are some glaring issues that emerge regarding traditional education.
About a decade ago, I read Forbidden Archeology: The Hidden History of the Human Race, a very meticulously written book. In short it is a catalog of anthropological and archeological artifacts that don't fit nicely into the authentic endorsement of historians. The book demonstrates that relics that don't fit in the scheme of things are routinely ignored. At first that sounds conspiratorial, but my experience says this is quite possible. Education institutions are routinely provided with facts that they ignore. Namely funding issues, where the unions insulate their membership from the evolution that is occurring in the outside world.
The new way of learning is quickly turning to computers; Rosetta Stone Software for foreign language comes to mind. And if the free market was opened up, and unions didn't resist the change, similar programs would emerge for mathematics, reading, history, literature, just about every academic endeavor. The role of the teacher is changing, yet they are being paid more highly than ever, in a time that education is on the precipice of change, for the better. Do we still need a teacher in the front of the room teaching like an authority figure? But because of the protectionism that goes on, nobody can even ask the question.
And the same thing happens to relics. When evidence is presented that doesn't fit the endorsed university theories, there is a lot of resistance, because that puts the credibility of the university at risk who published the original theories. It shouldn't, but in the world of big business education, change threatens the income stream. So the result is that education is stagnate, and not constantly evolving as quickly as our technology allows.
Below, I'll let Micheal Cremo explain for himself.
Because of this resistance from authentic education, emerging sciences, such as cryptozoology, and paranormal research is being ignored. Those sciences are at a phase similar to where archeology was at the turn of the century. Universities and their published results were what gave legitimacy to archeology and anthropology in the first place. And because universities are slow to accept new and larger ideas, they are missing the boat. The History Channel, and Travel Channel, and these kinds of cable entertainment are doing actual scientific investigation that will soon be considered legitimate. A show like Ghost Hunters is doing what universities should be doing, but they aren't because at some point in the past, so-called legitimate scientists with university backing proclaimed paranormal research as hokey.
Yet all across the country, there are thousands of paranormal reports each year, and this phenomenon has a direct correlation to our own history.
For instance, how can you study the motives of North American Indians without investigating the source of their beliefs where they routinely reported visits from the Gods, and had a strong belief that the rulers of the spirit world dwelled in animals?
A few years ago I went to the Mothman Festival in Point Pleasant, West Virginia. In case you don't know the story, in 1967, to 1968, Point Pleasant was ravaged with UFO reports, strange government agents, and a creature called by the press, a Mothman. The Mothman is a 6 foot tall creature that flies around with red eyes terrifying people, and supposedly shows up before major disasters.
I don't want to get into all that here, but I couldn't help but compare the stories of the Mothman in Point Pleasant to the civilization of Cahokia just outside of St. Louis.
Cahokia was an ancient city of Native American habitation around 1200 AD and contained at least 20,000 residents. It wasn't discovered until developers built a neighborhood over it in East St. Louis. Developers thought the ancient North American city was just a bunch of hills. As it turned out, Monks Mound at Cahokia is the third largest pyramid in the world by volume.
At this site they worshipped a god called "Birdman." Its description was almost exactly like that of the Mothman, which was just up the river about 400 miles in Point Pleasant some 700 years later. This in itself is a mystery. And while at this archeological site, you can't help but think of the world without buildings. You can see St. Louis from the top of Monks Mound. The primary highway runs right by it, yet most of America doesn't know much about Cahokia because it's not taught in school. It doesn't fit nicely into the scheme of our educated belief system.
But the evidence is right in front of all of us. And I'm the type of person that will try and connect the dots even further. I'd be inclined to ask why such a city was built at that particular location in the first place. I mean there are lots of locations along the Mississippi River.
An interesting aspect of St. Louis is that it is built on a series of caves. Caves used to be very important to Native American cultures. Most agriculturally based cultures, and especially hunting and gathering cultures have ceremonies for the youth that involve going into caves and letting the spirits guide them to some sort of revelation.
One of the most hau nted spots in St Louis is the Lemp Mansion, which was a family of early beer brewers that had a tendency to commit suicide. The Lemp Mansion had access to these caves where a fantastic pool was build down in the cave system, under the mansion. It is hard for me to not connect some of these mysteries that are hidden to us by the busy contemporary existence that hurries itself on the surface of things. I can't help but think that the caves of St. Louis were important aspects to life in the city of Cahokia, and that the same god that residents of that city worshipped is possibly the same creature that hunted Point Pleasant, West Virginia. And the same "spirits" that resided in the caves, that ancient Cahokian's sought out in their rituals were the same ones that haunted the residents of the Lemp Mansion and inspired them to decrypted lives of misery and suicide. It's just a thought, you never know.
However, modern science doesn't have anyway to deal with all that. But that doesn't mean they don't exist or aren't connected in some way.
In the end, it all comes back to money. Money is the god of modern institutions. And because they are always primarily concerned about how they will get it, or how they will preserve their union endeavors, they will overlook obvious mysteries that are all around them, and actually take steps to suppress the information to protect their budgets for fear of explaining how scientific theories of the past are suddenly wrong.
Basically, we are changing as a human civilization. And we're changing faster than orthodox systems can keep up with. And that makes just about everything they are protecting irrelevant.
We know now that there are over 10 known dimensions. Quantum physicists can even name them to some extent, even if we can't yet understand how to relate to something that is not part of our 4 dimensional realities. Much like radio waves, these other dimensions exist even if we don't have a device that can detect them. There are many such revelations that are coming at us at a furious pace. And yet orthodox education and politics is always in the way of understanding.
So when I speak out about institutionalism, it's for reasons such as those mentioned above. I look at people who proclaim the above as "hokey," "crazy," or even "silly," as being a big part of the problem. They are the same fools that proclaimed the world was flat, or that Hell was under the earth because they were aware that volcano's spewed lava. Science is expanding and institutions cleaving to the old ways will soon be as relevant as a dinosaur fossil in a museum and we'll laugh at how old and archaic we used to learn information truly was.
When it came to Forbidden Archeology, the evidence of suppression was so obvious it started me thinking of how much of that suppression actually goes on. And it is clear that there are many scientific factors that are not yet discovered, and not part of our human existence by understanding. They may be felt, or suspected, but not yet proven. So to cleave to the old ways is to cleave to what is false.
overmanwarrior	Archeology, Education, Rich Hoffman	20 Comments October 12, 2010 October 12, 2010 7 Minutes

redpajama_set_name: RedPajamaCommonCrawl


Many pet owners are curious about what is involved when their pet is placed under anesthesia. At Star Meadow Animal Clinic your pet's safety and comfort are our top priority so you can be sure that your pet will receive only the best and safest anesthetic and surgical care.
Safe Anesthesia — a very safe anesthetic gas which is also used in human pediatric medicine.
Experienced Monitoring Support — our trained technicians use state-of-the-art anesthetic monitors, pulse oximetry, electrocardiograms (ECG), capnography, and blood pressure monitors (Oscillometric and Doppler) to continuously check your pet's pulse rate, oxygen levels and blood pressure. We also ensure that your pet's temperature remains stable with a warming unit.
IV Catheter Placement — fluids are given during surgery to maintain blood pressure and to help your pet recover quickly from the anesthesia.
Induction Agents — necessary treatments through the intravenous line.
Pain Medication — multi-modal approach to pain control before, during and after your pet's surgical procedure.
Star Meadow Animal Clinic provides surgical services for dogs and cats. We offer a clean and well-equipped facility and experienced team to provide your pet with high quality surgical care in a stress-free and relaxing environment.
Our team of veterinarians and technicians are experienced with a range of surgeries. All of our procedures include a thorough pre-surgical physical examination by a veterinarian, surgical monitoring and lots of care and attention throughout the day.
Our veterinarians work closely with veterinary specialists in the area. A visiting board-certified radiologist and surgeon allow us to offer ultrasonography, echocardiography, orthopedic surgery, and other specialized surgeries without your pet having to travel to an unfamiliar hospital.


This bearing is already a cut above the competition before you get it out of the box. It's ability to withstand relatively high temperatures and loads, gives you the benefits of keeping your speed in the most critical of situations. the high level of race surface finish will keep it's speed longer. this is an advantage you cannot afford to be without. so when you're looking at stepping up a level, you're already there.


text:
Q: How to include Moq in VS 2017 xUnit Project (.NET Core) I was trying to create some unit tests in xUnit using .NET Core and using Moq. I created the project then added Moq using NuGet. I see in the dependencies of the xUnit project under NuGet that Moq (4.7.99) is there but I am not able to use it in code.
When I add:
using Moq;

I get an error saying Moq could not be found. What am I missing?

redpajama_set_name: RedPajamaStackExchange


text:
Q: how to lock device but prevent the screen to be turned off in android? I'm using DevicePolicyManager to lock the device while needed. what I want is to bring up the lock screen but using DevicePolicyManager and calling it's method lockNow() will lock the device and turn the screen off. how to prevent turning off the screen?
UPDATE: I have tried this:
public class LockMeNowActivity extends Activity {
protected PowerManager.WakeLock mWakeLock;
private DevicePolicyManager mgr = null;
private ComponentName cn = null;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    setContentView(com.commonsware.android.lockme.R.layout.main);
    PowerManager pm = (PowerManager) getSystemService(Context.POWER_SERVICE);
    mWakeLock = pm.newWakeLock(PowerManager.SCREEN_DIM_WAKE_LOCK, "My Tag");
    cn = new ComponentName(this, AdminReceiver.class);
    mgr = (DevicePolicyManager) getSystemService(DEVICE_POLICY_SERVICE);
}

public void lockMeNow(View v) {
    if (mgr.isAdminActive(cn)) {

        mgr.lockNow();
    } else {
        Intent intent = new Intent(
                DevicePolicyManager.ACTION_ADD_DEVICE_ADMIN);
        intent.putExtra(DevicePolicyManager.EXTRA_DEVICE_ADMIN, cn);
        intent.putExtra(DevicePolicyManager.EXTRA_ADD_EXPLANATION,
                "For experimentation purposes only");
        startActivity(intent);
    }
}

public void pullWakeLock() {
    mWakeLock.acquire();
}

public void onPause() {
    super.onPause();
    pullWakeLock();
}

public void onResume() {
    super.onResume();
    releaseWakeLock();
}

public void releaseWakeLock() {
    mWakeLock.release();
}

}

A: Along with your lockNow(), add PowerManager.WakeLock:
Import
import android.os.PowerManager;

Global Variable
protected PowerManager.WakeLock mWakeLock;

Public Methods
    public void pullWakeLock() {
        final PowerManager pm = (PowerManager) getSystemService(Context.POWER_SERVICE);
        this.mWakeLock = pm.newWakeLock(PowerManager.SCREEN_DIM_WAKE_LOCK, "My Tag");
        this.mWakeLock.acquire();
    }

    public void releaseWakeLock() {
        this.mWakeLock.release();
    }

Use the follwing permission in manifest file :
<uses-permission android:name="android.permission.WAKE_LOCK" />

A: I finally found a way for this problem. as I want to lock the device when onStop() happens I did:
    public void onStop() {
    mgr.lockNow();
    super.onStop();
    WakeLock screenLock = ((PowerManager) getSystemService(POWER_SERVICE))
            .newWakeLock(PowerManager.SCREEN_BRIGHT_WAKE_LOCK
                    | PowerManager.ACQUIRE_CAUSES_WAKEUP, "TAG");
    screenLock.acquire();
}

and it worked well. turning on the screen after locking it...


International network marketing leader Qplus unveiled its latest products and services designed to improve health and promote overall wellness. These were the revolutionary Bio-Resonance Hair Analysis, the Ole Leaf Extract nutritional supplement and the Amezcua Lifestyle Set. The presentation took place at one of Qplus' signature Super Saturday events, a gathering attended by industry practitioners and media alike.
Bio-Resonance Hair Analysis is a new wellness concept available from Amezcua Vital Life Center. The state-of-the-art procedure, developed over several decades is a non-invasive procedure that examines trace elements in the hair, enabling training health professionals to determine one's overall health and nutritional state by examining the cells in one's hair. Once the levels of toxins, pathogens, food sensitivities and nutritional requirements have been examined, Amezcua staff can then recommend a specifically-tailored regimen to assist the patient on the road to wellness, correcting existing problems and preventing future ones. The Bio-Resonance Hair Analysis is exclusively available at the Amezcua Vital Life Center.
The Amezcua Lifestyle Set consists of three discs; Amezcua EAT, Amezcua DRINK and Amezcua MOVE. Each discs serves a specific dietary purpose, specifically designed to re-energize the user in his daily life.
Ole Olive Leaf Extract is a nutritional supplement with five times the antioxidants potential of Vitamin C. Developed from years of study that began with the isolation of a class of chemical compound called phenols, it has been consistently proven that olive leaf extract contains a wealth of nutrients and vitamins beneficial to the human immune system. An excellent antioxidant, Ole not only slows oxidative damage, but is also powerful combatant against free radicals when combined with a balanced diet and proper exercise.
Qplus is a trailblazing, exceptionally-positioned network marketing company in the Philippines. Part of the Qi Group of Companies, which now has presence in over 100 countries, Qplus offers extraordinary business opportunities, outstanding health and wellness products as well as unwavering near and long-term support to our stakeholders.
Rooting from Amezcua Wellness Center's main advocacy of Preventive Health through Natural Medicine, the Vital Life Center's approach to health care revolves around the power and importance of Prevention. It seeks to detect weakness in the body before symptoms of disease manifest themselves and provide natural therapies that strengthen the immune system.
Through its variety of modern screening methods which range from the analysis of body fluids to the measurement of electrical energy, and cell analysis, the VLC's licensed medical practitioners are able to address the root of most health problems, not just their symptoms.
can i get the contact number of wow direct sales outlet in sn. fdo pampanga?just send your reply in my email add splendidruby@ymail.com thanks!
Ole is a very good product as i read from this article. I think we should try taking this, more powerful antioxidant than other antioxidants out in the market right now..
Really a good product as I reviewed it! Thanks for sharing us..
I realy like your blog and this product seam to be a very good suppliment for your hair and body. I also found that natures own vitamines and nutritions also have a udge impact and I believe that in combine with your product will be a very good solution.
Nice site and I like your Article and how you present this!
Thanks you for sharing and indeed keep us informed on your progress.


text:
Q: React - How to get an id and use it to delete So in react I made a button to delete some data so I need to get teh car by id and remove it by using react js axios so the response I got is an empty array so can some one help me please.
Here is the code :
in service data file :
get(id) {

  return http.get(`/get/${id}`);
}
delete(id) {

  return http.delete(`/delete/${id}`);
}

Component.jsx
 this.state = {
        idCars: null,
        carName: "",
        carModel: "",
        cars: [],
        submitted: false
      };

    }
    getCar(idCars) {
      DataService.get(idCars)
        .then(response => {
          this.setState({
            cars: response.data

          });
          console.log(response.data);

        })
        .catch(e => {
          console.log(e);
        });
    }

    componentDidMount() {

      this.getCar(this.props.match.params.idCars);
      this.retrieveCars()
    }
 deleteC() {

      DataService.delete(this.state.cars.idCars)
        .then(response => {

          this.props.history.push('/Classement');
          this.refreshList()
        })
        .catch(e => {
          console.log(e);
        });
    }
 render() {
      const { cars } = this.state;
           return (

                    {
                      cars.map(data => (

                        {data.idCars}

                      {data.carName}
                      {data.carModel}

                      <button  className="btn btn-danger"  onClick={this.deleteC}>
                        Remove
                      </button>

                      ))
                    }

And doesn't remove nothing, how can I fix this guys

A: At the moment your DeleteC function tries to read:
this.state.cars.idCars

But this.state.cars is an array so idCars is only defined for a given index such as in:
this.state.cars[0].idCars

What you could do is modify your onClick behavior for this:
onClick={() => deleteC(data.idCars)}

This way the deleteC function will be called with the idCars of the selected row.


Filed under: Bar Exam, Best Practices & Curriculum, Uncategorized |	Comments Off on Are the Students Failing the Bar Exam Today Canaries in the Coal Mine warning us of a More General Need to Change Legal Education?
In addition to her post on Law School Café about alleged correlations between bar exam scores and lawyer discipline (discussed on this blog here), Professor Deborah Merritt recently offered another bar exam-related post. This one provides intriguing historical perspective on the current need to expand the range of skill sets tested on the bar exam. Following up on points made by Professor Derek Muller, Professor Merritt discusses a 1980 study by the California Committee of Bar Examiners, cosponsored by the National Conference of Bar Examiners (NCBE), on adding a clinical component to the bar exam. Several hundred applicants who had taken the July 1980 California Bar Exam volunteered to complete an additional clinical evaluation requiring them, among other things, to interview a client and examine witnesses. Professional actors played the role of clients, akin to the standard patient role that actors perform for clinical evaluations in medicine. The applicants were scored based on an elaborate protocol.
Written performance testing was the last meaningful innovation in bar exam testing. In thinking about who might lead an effort toward the next one that introduces greater clinical evaluation, including possibly of oral skills, I think not of a top-down effort from the resolutely conservative NCBE. It is focused on getting as many jurisdictions as possible to adopt its Uniform Bar Exam (26 and counting as of today). Rather, I think of a bottom-up effort by individual states—perhaps with California in the lead—serving as laboratories for testing methods that could ultimately spread to other jurisdictions, thereby persuading or forcing the NCBE to join.
The history of written performance testing is illustrative of my point. Long before the NCBE went forward with the MPT in 1997, not just California but also Alaska and Colorado devised performance tests of their own and administered them on the bar exam. Indeed, those three states were administering performance tests in the early 1980s, playing an important initial role in advancing the cause of a needed bar exam reform. Here, for example, is a follow-up study of the 1983 California Bar Exam, discussing its two performance tests.
The biggest barrier to innovation at the state level is the NCBE's influence, which increases with each state that adopts the UBE and thereby constrains itself to offer the conventional bar exam that the NCBE requires it to. Indeed, both Alaska and Colorado, two of the original performance test states from the 1980s, have adopted the UBE, meaning neither of those states will be doing any more bar exam innovation. That leaves California (and any of the other 23 states that have yet to join the UBE, none of which matches the influential profile of California).
But I also come back to the states, and to California specifically. There is a vigorous debate going on about whether California should lower its bar exam cut score. That's an important discussion to have. But I might suggest another discussion to have about the California Bar Exam: Shouldn't California resist the UBE and instead conduct a new study of alternative methods for assessing today's relevant lawyering skills that are not encompassed by the UBE?


text:
from flask.views import MethodView
from flask import request
from flask import jsonify
from app.core.db.sql.models import User
from ..  import api_v1_blueprint as api

# REST API方法实现如下面格式
class UserAPI(MethodView):
    """
    URL	HTTP 方法	描述
    /users/	GET	获得全部用户的列表
    /users/	POST	创建一个新用户
    /users/<id>	GET	显示某个用户
    /users/<id>	PUT	更新某个用户
    /users/<id>	DELETE	删除某个用户
    """
    def get(self, user_id):
        if user_id is None:
            # 处理列表
            user_list = []
            users = User.query.all()
            for user in users:
                user_list.append(user.to_json())
            return jsonify({'status':True, 'result':user_list})
        else:
            # 处理单个用户
            user = User.query.get_or_404(user_id)
            return jsonify(user.to_json())

    def post(self):
        print request.form
        return jsonify({'status': True, 'msg':"post request"})

    def delete(self, user_id):
        # delete a single user
        print 'delete user: ', user_id
        return jsonify({'status': True})

    def put(self, user_id):
        # update a single user
        print 'update user: ', user_id
        return jsonify({'status': True, 'msg':"put request"})

# 注册路由
user_view = UserAPI.as_view('user_api')
api.add_url_rule('/users', defaults={'user_id': None}, view_func=user_view, methods=['GET',])
api.add_url_rule('/users', view_func=user_view, methods=['POST',])
api.add_url_rule('/users/<int:user_id>', view_func=user_view, methods=['GET', 'PUT', 'DELETE'])

# 多个类似API可以使用下面的方法
# def register_api(view, endpoint, url, pk='id', pk_type='int'):
#     view_func = view.as_view(endpoint)
#     app.add_url_rule(url, defaults={pk: None},
#                      view_func=view_func, methods=['GET',])
#     app.add_url_rule(url, view_func=view_func, methods=['POST',])
#     app.add_url_rule('%s<%s:%s>' % (url, pk_type, pk), view_func=view_func,
#                      methods=['GET', 'PUT', 'DELETE'])
# register_api(UserAPI, 'user_api', '/users/', pk='user_id')

redpajama_set_name: RedPajamaGithub


A range of fun workshops for young people will be brought to Louth in the October half term - with the three day event costing just £20 per person.
The exciting new 'Unleashed' project, for ages 8 - 18, is run by the organisers of Young Americans 2014 and the three day event will take place from Wednesday, October 28 until Friday, October 30.
Workshops include street dance, ballet, tribal dance, and a workshop run by a professional graffiti artist which will give youngsters the unique opportunity to spray their graffiti art on to a car which has been kindly donated.
The project will feature two days of workshops, followed by a show rehearsal on the Friday afternoon with the final show taking place on Friday evening at St James' Church in Louth.
The icing on the cake is that, thanks to a number of generous donations, the three-day course costs just £20 per youngster, including a hot meal before the final show and a show t-shirt which can be kept forever.
Places are strictly limited to 150 youngsters, but there are still some available - if you would like to take part, call 01507 604174 or 07990 639805, or pop in to the Renew shop in Queen Street.
Payment is due in full upon booking, either by card or cash payments.


Size: 18 cm Packing: 100pcs/pack A safe way to attach tags to your garments and luggages Ea..
Size: 1.5 mm thick x 100 m Ideal for securing parcels, books & many other packaging items ..
Size: 19 mm x 100 yds ..
Width: 1 inch Length: 22 m Ideal for tying and gift wrapping Comes in two available c..