text:
\section{Introduction}\label{intro}

An $n$-dimensional Riemannian manifold $(M^n,g)$ is called a gradient Yamabe soliton if there exists a smooth function $F$ on $M$ and a constant $\rho\in \mathbb{R}$, such that
$$\nabla \nabla F=(R-\rho)g,$$
where, $R$ is the scalar curvature on $M$.
If $F$ is constant, $M$ is called trivial.

One of the most interesting problem of the Yamabe soliton is the Yamabe soliton version of Perelman's conjecture, that is, ``Any complete gradient Yamabe soliton with positive scalar curvature under some natural assumption is rotationally symmetric".
The problem was first considered by P. Daskalopoulos and N. Sesum \cite{DS13}.
They showed that any locally conformally flat complete gradient Yamabe soliton with positive sectional curvature is rotationally symmetric.
Later, G. Catino, C. Mantegazza and L. Mazzieri \cite{CMM12} and H.-D. Cao, X. Sun and Y. Zhang \cite{CSZ12} also considered the same problem. In particular, Cao, Sun and Zhang showed that any locally conformally flat complete gradient Yamabe soliton with positive scalar curvature is rotationally symmetric.

To understand the gradient Yamabe soliton, many generalizations of it have been introduced.

\begin{enumerate}
\item
Almost gradient Yamabe solitons \cite{BB13}:

For smooth functions $F$ and $\rho$ on $M$,
$$(R-\rho)g=\nabla\nabla F.$$

\item
Gradient $k$-Yamabe solitons \cite{CMM12}:

For a smooth function $F$ on $M$ and $\rho\in\mathbb{R}$,
$$2(n-1)(\sigma_k-\rho)g=\nabla\nabla F,$$
where, $\sigma_k$ denotes the $\sigma_k$-curvature of $g$, that is,
$$\sigma_k=\sigma_k(g^{-1}A)=\sum_{i_1<\cdots<i_{k}}\mu_{i_1}\cdots\mu_{i_k}~~~~~(\text{for}~1\leq k\leq n),$$
where, $A=\frac{1}{n-2}({\rm Ric}-\frac{1}{2(n-1)}Rg)$ is the Schouten tensor and $\mu_1,\cdots,\mu_n$ are the eigenvalues of the symmetric endomorphism $g^{-1}A$. Here, ${\rm Ric}$ is the Ricci tensor of $M$.\\

\item
$h$-almost gradient Yamabe solitons \cite{Zeng20}:

For smooth functions $F$, $\rho$ and $h~(h>0~\text{or}~h<0)$ on $M$,
$$(R-\rho)g=h\nabla\nabla F.$$

\end{enumerate}

To consider all these solitons, we consider the gradient conformal soliton defined by G. Catino, C. Mantegazza and L. Mazzieri \cite{CMM12}:

\begin{definition}[\cite{CMM12}]
Let $(M,g)$ be an $n$-dimensional Riemannian manifold. For smooth functions $F$ and $\varphi$ on $M$, $(M,g,F,\varphi)$ is called a {\it gradient conformal soliton} if it satisfies
\begin{equation}\label{GCS}
\varphi g=\nabla\nabla F.
\end{equation}
If $F$ is constant, $M$ is called trivial.
The function $F$ is called the potential function.
\end{definition}

\begin{remark}
By the definition, all results in this paper can be applied to gradient Yamabe solitons, gradient $k$-Yamabe solitons, almost gradient Yamabe solitons and $h$-almost gradient Yamabe solitons.

We also remark that gradient conformal solitons were studied by Cheeger-Colding $($\cite{CC96}, see also \cite{Tashiro65}$)$.
\end{remark}

Recently, A. W. Cunha and E. L. Lima \cite{CL22} showed the following.

\begin{theorem}[\cite{CL22}]\label{CL22main}
Let $(M,g)$ be a complete gradient steady $k$-Yamabe soliton.
If $\sigma_k\geq0$, $F\geq0$, and
$$\int_{M\setminus B(x_0,R_0)}\frac{F}{d(x_0,x)^2}<+\infty,$$
for some $x_0\in M$ and $R_0>0,$
then $F$ is harmonic.
\end{theorem}

In this paper, under the same assumption, we completely classify gradient conformal solitons. In particular, we give an affirmative partial answer to the Yamabe soliton version of Perelman's conjecture.

\begin{theorem}\label{main}
Any nontrivial complete gradient conformal soliton $(M^n,g,F,\varphi)$ with $\varphi,F\geq0$ and
$$\int_{M\setminus B(x_0,R_0)}\frac{F}{d(x_0,x)^2}<+\infty,$$
for some $x_0\in M$ and $R_0>0$ is
$$([0,\infty)\times\mathbb{S}^{n-1},ds^2+a^2{\bar g}_{S},as+b,0),$$
where, $\bar g_{S}$ is the round metric on $\mathbb{S}^{n-1},$ and $a>0$ and $b\geq0$.
\end{theorem}

The remaining sections are organized as follows.
Section $\ref{proofofmain}$ is devoted to the proof of Theorem~$\ref{main}$.
In Section~$\ref{trivial}$, we show some triviality results under the assumption in Cunha and Lima \cite{CL22}. In Appendix, we give some classifications of complete gradient conformal solitons.

\section{Proof of Theorem \ref{main}}\label{proofofmain}

In this section, we prove Theorem \ref{main}. To show Theorem \ref{main}, we use the following result shown by the author \cite{Maeta24}.

\begin{theorem}[\cite{Maeta24}]\label{main of Maeta24}
A nontrivial complete gradient conformal soliton $(M^n,g,F,\varphi)$ is either

$(1)$ compact and rotationally symmetric, or

$(2)$ the warped product
$$(\mathbb{R},ds^2)\times_{|\nabla F|} \left(N^{n-1},\bar g\right),$$
where, the scalar curvature $\bar R$ of $N$ satisfies
$$|\nabla F|^2R=\bar R-(n-1)(n-2)\varphi^2-2(n-1)g(\nabla F,\nabla\varphi),$$

or

$(3)$ rotationally symmetric and equal to the warped product
$$([0,\infty),ds^2)\times_{|\nabla F|}(\mathbb{S}^{n-1},{\bar g}_{S}),$$
where, $\bar g_{S}$ is the round metric on $\mathbb{S}^{n-1}.$

Furthermore, the potential function $F$ depends only on $s$.
\end{theorem}

Therefore, to consider the Yamabe soliton version of Perelman's conjecture,  we only have to consider $(2)$ of Theorem \ref{main of Maeta24}.

\if0
Why???

\begin{align*}
\int_{B_r}\varphi^2
=&\frac{1}{n}\int_{B_r}\varphi\Delta F\\
=&-\frac{1}{n}\left\{\int_{B_r}\langle\nabla \varphi ,\nabla F +\int_{\partial B_r}\varphi\langle \nu,\nabla  F\rangle \right\}\\
=&-\frac{1}{n(n-1)}\left\{\int_{B_r} {\rm Ric}(\nabla F,\nabla F)+\int_{\partial B_r}\varphi\langle \nu,\nabla  F\rangle \right\}\\
\end{align*}
Since $|\nabla F|\leq $

Why???
\fi

\begin{proof}[Proof of Theorem $\ref{main}$]

We first take a cut off function $\eta$ on $M$ satisfying that
 \begin{equation}
\left\{
 \begin{aligned}
&0\leq\eta(x)\leq1\ \ \ (x\in M),\\
&\eta(x)=1\ \ \ \ \ \ \ \ \ (x\in B(x_0,R)),\\
&\eta(x)=0\ \ \ \ \ \ \ \ \ (x\not\in B(x_0,2R)),\\
&|\nabla\eta|\leq\frac{C}{R}\ \ \ \ \ \ \ (x\in M),\ \ \ \text{for some constant $C$ independent of $R$},\\
&\Delta\eta \leq\frac{C}{R^2}\ \ \ \ \ \ \ (x\in M),\ \ \ \text{for some constant $C$ independent of $R$},
\end{aligned}
\right.
\end{equation}
where, $B(x_0,R)$ and $B(x_0,2R)$ are the balls centered at a fixed point $x_0\in M$ with radius $R$ and $2R$, respectively. By $\Delta F=n\varphi$, one has
\begin{align}\label{intev}
0\leq
&\int_{B(x_0,2R)}\eta\varphi\\
=&\frac{1}{n}\int_{B(x_0,2R)}\eta\Delta F\notag\\
=&\frac{1}{n}\int_{B(x_0,2R)\setminus B(x_0,R)}\Delta\eta F\notag\\
\leq&\frac{C}{nR^2}\int_{B(x_0,2R)\setminus B(x_0,R)} F.\notag
\end{align}
Take $R\nearrow +\infty$. By the assumption, we have that the right hand side of \eqref{intev} goes to $0$.
Therefore, we have
$$\int_M\varphi=0.$$
Since $\varphi\geq0$, one has $\varphi=0$, hence
\begin{equation}\label{nnF}
\nabla\nabla F=0.
\end{equation}
By Theorem \ref{main of Maeta24}, we have 3 types of gradient conformal solitons.
\\

\noindent
Case 1.  $M$ is compact and rotationally symmetric.

By $\eqref{nnF}$, $\Delta F=0$. Since $M$ is compact, by the standard Maximum principle, we have that $F$ is constant.
\\

\noindent
Case 2. $M$ is the warped product
$$(\mathbb{R},ds^2)\times_{|\nabla F|} \left(N^{n-1},\bar g\right).$$

By $\eqref{nnF}$, we have
$$\nabla |\nabla F|^2=2\nabla_j\nabla_iF\nabla_iF=0.$$
Hence,  $\nabla F$ is a constant vector field.
Since $F$ depends only on $s\in \mathbb{R}$, one can get that
$$\nabla F=F'(s)\partial _s,$$
where, $F'(s)$ is a constant, say $a$.
If $a\not=0$, one has
$F(s)=as+b\geq0 $ on $\mathbb{R}$, which cannot happen.
Hence, $F$ is constant. Therefore, $M$ is trivial.
\\

\noindent
Case 3. $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),ds^2)\times_{|\nabla F|}(\mathbb{S}^{n-1},{\bar g}_{S}).$$

By the same argument as in Case 2, we have $F(s)=as+b\geq0$ on $[0,\infty)$. Therefore, one has $a>0$ and $b\geq0$.

\end{proof}

\if0
\begin{remark}
In Case $3$ of the proof of Theorem $\ref{main}$, we can observe the following.

One can get that the volume of the ball $B(x_0,s)$ of $[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})$ is $C(a,n,\omega_n) s$, where $C(a,n,\omega_n)$ is a constant, which depends only on $a$, $n$ and $\omega_n$.
From this and $s\leq d(x_0,x)$, one has
$$\int_{B(x_0,s)\setminus B(x_0,R_0)}\frac{s}{d(x_0,x)^2}<+\infty.$$
Therefore, the assumption of Theorem $\ref{main}$
\begin{align*}
\int_{\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,R_0)}
\frac{as+b}{d(x_0,x)^2}
<+\infty
\end{align*}
is satisfied.
\if0
\begin{align*}
\int&_{\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,r_0)}
\frac{as+b}{d(x_0,x)^2}\\
=&a\int_{\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,r_0)}\frac{s}{d(x_0,x)^2}
+b\int_{\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,r_0)}\frac{1}{d(x_0,x)^2}\\
<&a\int_{\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,r_0)}\frac{1}{d(x_0,x)}
+b\int_{\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,r_0)}\frac{1}{d(x_0,x)^2}\\
<&\infty.
\end{align*}

Therefore, we have
$${\mathrm Vol}(\{[0,\infty)\times_a(\mathbb{S}^{n-1},\bar{g}_{S})\}\setminus B(x_0,r_0))\leq O(d(x_0,x)),$$
which cannot happen. In fact,
\fi

\end{remark}
~\\
\fi

\section{Some triviality results of gradient conformal solitons}\label{trivial}

Cunha and Lima \cite{CL22} also showed that the $\sigma_k$-curvature is constant under some assumptions.
We first recall parabolicity of $M$: A Riemannian manifold $M$ is parabolic, if every subharmonic function on $M$ which is bounded from above is constant (see \cite{Grigoryan99}).

\begin{theorem}[\cite{CL22}]\label{CLmain1}
Let $(M,g)$ be a complete gradient $k$-Yamabe soliton.
If one of the following is satisfied, then $\sigma_k=\lambda.$

$(A)$ The potential function satisfies that $F\geq K>0$ for some $K\in \mathbb{R}$, $\sigma_k\leq \lambda$, and one of the following is satisfied $(i)$ $M$ is parabolic, $(ii)$ $|\nabla F|\in L^1(M)$, $(iii)$ $F^{-1}\in L^p(M)$ for some $p>1$, or $(iv)$ $M^n$ has linear volume growth.

$(B)$ The soliton $M$ is nontrivial, ${\rm Ric} (\nabla F,\nabla F)\leq0$, and
$|\nabla F|\in L^{p}(M)$ for $p>1$.

\end{theorem}

In this section, we give triviality results of gradient conformal solitons under the same assumption. As a corollary, one can also get triviality results of gradient $k$-Yamabe solitons under the assumption of Theorem \ref{CLmain1}.

\begin{proposition}\label{sub}
There exists no nontrivial complete gradient conformal soliton $(M^n,g,F,\varphi)$ with one of the following is satisfied.

$(A)$ The potential function satisfies that $F\geq K>0$ for some $K\in \mathbb{R}$, $\varphi\leq0$, and one of the following is satisfied $(i)$ $M$ is parabolic, $(ii)$ $|\nabla F|\in L^1(M)$, $(iii)$ $F^{-1}\in L^p(M)$ for some $p>1$, or $(iv)$ $M^n$ has linear volume growth.

$(B)$ The Ricci curvature satisfies that ${\rm Ric} (\nabla F,\nabla F)\leq0$, and
$|\nabla F|\in L^{p}(M)$ for $p>1$.
\end{proposition}

We recall formulas which will be used later. For any gradient conformal soliton, we have
\begin{equation*}
\Delta {\nabla}_iF={\nabla}_i\Delta F+R_{ij}{\nabla}_jF,
\end{equation*}
\begin{equation*}
\Delta {\nabla}_iF={\nabla}_k{\nabla}_k{\nabla}_iF={\nabla}_k(\varphi g_{ki})={\nabla}_i\varphi,
\end{equation*}
and
\begin{equation*}
{\nabla}_i\Delta F={\nabla}_i(n\varphi)=n{\nabla}_i\varphi.
\end{equation*}
Hence, we have
\begin{equation}\label{f1.1}
(n-1){\nabla}_i\varphi+R_{ij}{\nabla}_jF=0,
\end{equation}
where, $R_{ij}$ is the Ricci tensor of $M$.
Therefore, one has
\begin{equation}\label{RicnF}
\langle \nabla \varphi,\nabla F\rangle=-\frac{1}{n-1}{\rm Ric} (\nabla F,\nabla F).
\end{equation}
By applying ${\nabla}_l$ to the both sides of $(\ref{f1.1})$, we obtain
\begin{equation}\label{f1.2}
(n-1){\nabla}_l{\nabla}_i\varphi+{\nabla}_lR_{ij} \cdot {\nabla}_jF+R_{ij}{\nabla}_l{\nabla}_jF=0.
\end{equation}
Taking the trace, we obtain
\begin{equation}\label{f1}
(n-1)\Delta{\varphi}+\frac{1}{2}~g(\nabla R,\nabla F)+R \varphi =0.
\end{equation}
~\\

\begin{proof}[Proof of Proposition $\ref{sub}$]~

\noindent
(A) If $M$ is compact, by $\Delta (-F)=-n\varphi \geq0$, the standard Maximum principle shows that $F$ is constant.
Therefore, we assume that $M$ is noncompact.
~\\

\noindent
$(i)$ Since $-F\leq-K$, and $\varphi\leq0$, we have
$$\Delta(-F)=-n\varphi\geq0.$$
Since $M$ is parabolic, $-F$ is constant. Therefore, $M$ is trivial.
\\

\noindent
$(ii)$
By a direct computation,
\begin{equation}\label{deltaF}
{\rm div}\nabla (-F)=\Delta (-F)=-n\varphi\geq0.
\end{equation}
By Theorem \ref{main of Maeta24}, we have 3 types of gradient conformal solitons.
\\

\noindent
Case 1.  $M$ is compact and rotationally symmetric.

This case cannot happen.

To consider Cases 2 and 3, let us recall the following:

\begin{lemma}[\cite{CSC10}]\label{divX}
Let $X$ be a smooth vector field on a complete noncompact Riemannian manifold, such that, ${\rm div} X$ does not change the sign on $M$. If $|X|\in L^1({M}),$ then ${\rm div}_MX=0$.
\end{lemma}

By Lemma \ref{divX}, we have
$\Delta F=0.$
Hence, we have $\varphi=0$, and $\nabla\nabla F=0.$
\\

\noindent
Case 2. $M$ is the warped product
$$(\mathbb{R},ds^2)\times_{|\nabla F|} \left(N^{n-1},\bar g\right).$$

Since $\nabla\nabla F=0$, we have
$$\nabla |\nabla F|^2=2\nabla_j\nabla_iF\nabla_iF=0.$$
Hence,  $\nabla F$ is a constant vector field.
Set $|\nabla F|=a$.
If $a\not=0$,
$$a{\rm Vol}\,(\mathbb{R}\times N^{n-1})=+\infty.$$
From this and the assumption, we have $a=0$. Therefore, $F$ is constant, and $M$ is trivial.
\\

\noindent
Case 3. $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),ds^2)\times_{|\nabla F|}(\mathbb{S}^{n-1},{\bar g}_{S}).$$
By the same argument as in Case 2, we have that $F$ is constant.
\\

\noindent
$(iii)$ Since $\Delta (-F)\geq0,$ one has
$$\Delta F^{-1}=2F^{-3}|\nabla F|^2-\Delta F F^{-2}\geq0.$$
By the Yau's Maximum principle, one has that $F^{-1}$ is constant.
\\

\noindent
$(iv)$
By Lemma 6.3 in \cite{SY94} and the assumption, we have
\begin{align}\label{2(4)}
\int_{B(x_0,R)}|\nabla F^{-1}|^2
\leq&\frac{C}{R^2}\int_{B(x_0,2R)}F^{-2}\\
\leq&\frac{C}{R^2K^2}{\rm Vol}(B(x_0,2R))\notag\\
\leq&\frac{\bar C}{RK^2}.\notag
\end{align}
Take $R\nearrow +\infty$. The righthand side of \eqref{2(4)} goes to $0$.
Therefore, we have that $F$ is constant.
\\

\noindent
$(B)$ By $\Delta F=n\varphi$ and \eqref{RicnF},
we have
\begin{align}\label{6-1}
\frac{1}{2}\Delta |\nabla F|^2
=&|\nabla\nabla F|^2+{\rm Ric}(\nabla F,\nabla F)+\langle \nabla F,\nabla \Delta F\rangle\\
=&|\nabla\nabla F|^2-\frac{1}{n-1}{\rm Ric}(\nabla F,\nabla F).\notag
\end{align}
From this and ${\rm Ric}(\nabla F,\nabla F)\leq 0$, we have
$$\Delta |\nabla F|^2\geq0.$$

If $M$ is compact, by the standard Maximum principle, we have that $|\nabla F|$ is constant.

Assume that $M$ is noncompact. By the Yau's Maximum principle, $|\nabla F|$ is constant.

By $\eqref{6-1}$,
$$|\nabla\nabla F|^2-\frac{1}{n-1}{\rm Ric}(\nabla F,\nabla F)=0.$$
Therefore, we have $\nabla\nabla F=0.$

By Theorem \ref{main of Maeta24}, we have 3 types of gradient conformal solitons.
\\

\noindent
Case 1.  $M$ is compact and rotationally symmetric.

Since $\Delta F=0$ and $M$ is compact, by the standard Maximum principle, we have that $F$ is constant.
\\

Cases 2 and 3 are considered by the same argument as in (A)-(ii).
\if0
\noindent
Case 2. $M$ is the warped product
$$(\mathbb{R},ds^2)\times_{|\nabla F|} \left(N^{n-1},\bar g\right).$$

Since $\nabla F$ is a constant vector field.
Set $|\nabla F|=a$.
By the assumption
$$a{\mathrm Vol}(\mathbb{R}\times N^{n-1})<+\infty.$$
Hence, we have $a=0$, and $F$ is constant.
\\

\noindent
Case 3. $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),ds^2)\times_{|\nabla F|}(\mathbb{S}^{n-1},{\bar g}_{S}).$$
By the same argument as in Case 2, we have that $F$ is constant.
\fi
\\

\end{proof}

\section{Appendix}\label{appendix}

In this section, we also give classification results under the similar assumption of Cunha and Lima's paper \cite{CL22}.
Cunha and Lima also showed the following.
\begin{theorem}[\cite{CL22}]\label{CLmain}
Let $(M,g)$ be a complete gradient $k$-Yamabe soliton.
If one of the following is satisfied, then $\sigma_k=\lambda.$

$(A)$ The manifold $M^n$ is noncompact, $|\sigma_k-\lambda|\in L^1(M)$, $\int_M{\rm Ric} (\nabla F,\nabla F)\leq0$,
and $F$ has at most quadratic growth on $M$.

$(B)$ The soliton $M^n$ $(n\geq3)$ is nontrivial, $\sigma_k\geq \lambda$, and $|\nabla F|$ has at most linear growth on $M$, that is, for some $x_0\in M$, it holds $|\nabla F(x)|\leq Cd(x,x_0)$ near infinity, where $C$ is some uniform constant.  Let $u$ be a non-constant solution of
\begin{equation*}
\left\{
\begin{aligned}
\Delta u+h(u)=0,\\
\int_Mh(u)\langle\nabla u,\nabla F\rangle \geq0,
\end{aligned}
\right.
\end{equation*}
for $h\in C^1(\mathbb{R})$, and
the function $|\nabla u|$ satisfies
$$\int_{B(x_0,R)}|\nabla u|^2=o(\log R),~~~\text{as}~~~R\rightarrow +\infty.$$

$(C)$ The manifold $M^n$ is parabolic and nontrivial with ${\rm Ric} (\nabla F,\nabla F)\leq0$ and $|\nabla F|\in  L^\infty(M)$.
\end{theorem}

\begin{proposition}\label{main2}
Let $(M^n,g,F,\varphi)$ be a nontrivial complete gradient conformal soliton.
Assume that $M$ satisfies the following $(A)$, $(B)$ or $(C)$.

\noindent
$(A)$ The function $\varphi$ satisfies that $|\varphi|\in L^1(M)$, $\int_M{\rm Ric} (\nabla F,\nabla F)\leq0$, and $|\nabla F|$ has at most linear growth on $M$.

\noindent
$(B)$ The function $\varphi$ is nonnegative, and $|\nabla F|$ has at most linear growth on $M$.
Let $u$ be a non-constant solution of
\begin{equation*}
\left\{
\begin{aligned}
\Delta u+h(u)=0,\\
\int_Mh(u)\langle\nabla u,\nabla F\rangle \geq0,
\end{aligned}
\right.
\end{equation*}
for $h\in C^1(\mathbb{R})$, and
the function $|\nabla u|$ satisfies
$$\int_{B(x_0,R)}|\nabla u|^2=o(\log R),~~~\text{as}~~~R\rightarrow +\infty.$$

\noindent
$(C)$ The manifold $M$ is parabolic and nontrivial with ${\rm Ric} (\nabla F,\nabla F)\leq0$ and $|\nabla F|\in L^\infty(M)$.

Then, $M$ is either

$(1)$
$$(\mathbb{R}\times N^{n-1},ds^2+a^2\bar g,as+b,0),$$
where, $a,b\in\mathbb{R}$, or

$(2)$
$$([0,\infty)\times\mathbb{S}^{n-1},ds^2+a^2{\bar g}_{S},as+b,0),$$
where, $\bar g_{S}$ is the round metric on $\mathbb{S}^{n-1}$, and $a,b\in\mathbb{R}.$
\end{proposition}

\if0

\begin{remark}
Proposition $\ref{main2}$ is also a generalization of Theorem $3$ of \cite{MM12}.
\end{remark}
\fi

We observe the following lemma.
\begin{lemma}\label{cptRicnF}
Any compact gradient conformal soliton with $$\int_M{\rm Ric} (\nabla F,\nabla F)\leq 0$$ is trivial.
\end{lemma}
\begin{proof}
By $\Delta F=n\varphi$ and \eqref{RicnF}, we have
\begin{align*}
\int_{M}\varphi^2
=&~\frac{1}{n}\int_{M}\varphi\Delta F\\
=&~-\frac{1}{n}\int_{M}\langle \nabla \varphi, \nabla F\rangle\\
=&\frac{1}{n(n-1)}\int_M{\rm Ric}(\nabla F,\nabla F)\leq0.
\end{align*}
Thus, one has $\varphi=0$ and $\nabla\nabla F=0.$ Hence, we have $\Delta F=0$. By the standard maximum principle, we have that $M$ is trivial.
\end{proof}

\begin{remark}
From this lemma, the assumption that $M$ is noncompact is not needed in $(A)$ of Theorem $\ref{CLmain}$.
\end{remark}
~\\

\begin{proof}[Proof of Proposition $\ref{main2}$]
~\\

\noindent
$(A)$ If $M$ is compact, by Lemma \ref{cptRicnF}, $M$ is trivial.

Therefore, we assume that $M$ is noncompact.
By $\Delta F=n\varphi$ and \eqref{RicnF}, we have
\begin{align*}
\int_{B(x_0,R)}\varphi^2
=&~\frac{1}{n}\int_{B(x_0,R)}\varphi\Delta F\\
=&~-\frac{1}{n}\left\{\int_{B(x_0,R)}\langle \nabla \varphi, \nabla F\rangle +\int_{\partial B(x_0,R)}\varphi\langle \nu,\nabla F\rangle \right\}\\
=&~-\frac{1}{n}\left\{\int_{B(x_0,R)}-\frac{1}{n-1}{\rm Ric}(\nabla F,\nabla F) +\int_{\partial B(x_0,R)}\varphi\langle \nu,\nabla F\rangle \right\}\\
\leq&~\frac{1}{n(n-1)}\int_{B(x_0,R)}{\rm Ric}(\nabla F,\nabla F)
+CR\int_{\partial B(x_0,R)}|\varphi|,
\end{align*}
where, $\nu$ is the outward unit normal to the boundary $\partial B(x_0,R)$.
Since $|\varphi|\in L^{1}(M)$, by taking $R\nearrow +\infty,$ one has
$$CR\int_{\partial B(x_0,R)}|\varphi|\rightarrow 0.$$
Therefore, by the assumption, we have
$$\int_M\varphi^2=0,$$
hence, $\varphi=0.$

By Theorem \ref{main of Maeta24}, we have 3 types of gradient conformal solitons.
\\

\noindent
Case 1. $M$ is compact. This case cannot happen.

\noindent
Case 2. $M$ is the warped product
$$(\mathbb{R},ds^2)\times_{|\nabla F|} \left(N^{n-1},\bar g\right).$$

Since $\nabla \nabla F=0,$ we have
$$\nabla |\nabla F|^2=2\nabla_j\nabla_iF\nabla_iF=0.$$
Hence,  $\nabla F$ is a constant vector field.
Therefore, we have $F(s)=as+b.$
\\

\noindent
Case 3. $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),ds^2)\times_{|\nabla F|}(\mathbb{S}^{n-1},{\bar g}_{S}).$$

By the same argument as in Case 2, we have that $M$ is
$$([0,\infty)\times\mathbb{S}^{n-1},ds^2+a^2{\bar g}_{S},as+b,0).$$
~\\

\noindent
$(B)$ We take a cut off function $\eta$ on $M$ satisfying that
 \begin{equation}
 \eta(r)=
\left\{
 \begin{aligned}
&1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r\leq R,\\
&2-\frac{\log r}{\log R}\ \ \ \ r\in [R,R^2],\\
&0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r\geq R^2.
\end{aligned}
\right.
\end{equation}
By the soliton equation, we have
\begin{align}\label{7-1}
\int_M\varphi|\nabla u|^2\eta^2
=&\int_{B(x_0,R^2)} \nabla_i\nabla_jF\nabla_iu\nabla_ju\eta^2\\
=&-\int_{B(x_0,R^2)} \nabla_iF\nabla_j\nabla_iu\nabla_ju\eta^2
 +\nabla_iF\Delta u\nabla_ju\eta^2\notag\\
 &-\int_{B(x_0,R^2)}\nabla_iF\nabla_iu\nabla_ju\nabla_j\eta^2.\notag
\end{align}
Substituting
$$-\int_{B(x_0,R^2)} \nabla_iF\nabla_j\nabla_iu\nabla_ju\eta^2
=\frac{1}{2}\int_{B(x_0,R^2)} n\varphi|\nabla u|^2\eta^2+|\nabla u|^2\langle \nabla F,\nabla \eta^2\rangle,$$
into $\eqref{7-1}$, we have
\begin{align*}
\int_M\varphi|\nabla u|^2\eta^2
=&\frac{1}{2}\int_{B(x_0,R^2)} n\varphi|\nabla u|^2\eta^2
 +2\eta|\nabla u|^2\langle \nabla F,\nabla \eta\rangle\\
 &+\int_{B(x_0,R^2)} \langle \nabla F,\nabla u\rangle h\eta^2
 -2\eta\langle \nabla F,\nabla u\rangle \langle \nabla u,\nabla \eta\rangle.
\end{align*}
Therefore, we have
\begin{align*}
\frac{n-2}{2}\int_M\varphi|\nabla u|^2\eta^2
=&-\int_{B(x_0,R^2)} \eta|\nabla u|^2\langle \nabla F,\nabla \eta\rangle\\
 &-\int_{B(x_0,R^2)} \langle \nabla F,\nabla u\rangle h\eta^2
 +2\int_{B(x_0,R^2)} \eta\langle \nabla F,\nabla u\rangle \langle \nabla u,\nabla \eta\rangle\\
\leq&~3\int_{B(x_0,R^2)} \eta|\nabla u|^2 |\nabla F| |\nabla \eta|
 -\int_{B(x_0,R^2)} \langle \nabla F,\nabla u\rangle h\eta^2\\
\leq&~\frac{C}{\log R}\int_{B(x_0,R^2)\setminus B(x_0,R)} |\nabla u|^2
 -\int_{B(x_0,R^2)} \langle \nabla F,\nabla u\rangle h\eta^2,
\end{align*}
where, the last inequality follows from $|\nabla F|\leq Cr$ near infinity and the definition of the cut off function $\eta$.
Take $R\nearrow +\infty.$ From this and the assumption, we have
\begin{align*}
\frac{n-2}{2}\int_M\varphi|\nabla u|^2\eta^2
 \leq -\int_M \langle \nabla F,\nabla u\rangle h\eta^2\leq0.
\end{align*}
Since $u$ is a non-constant solution, we have $\varphi=0.$
Therefore, we have $\nabla\nabla F=0$.

By Theorem \ref{main of Maeta24}, we have 3 types of gradient conformal solitons.
\\

\noindent
Case 1.  $M$ is compact and rotationally symmetric.

Since $M$ is compact, by the standard Maximum principle, we have that $F$ is constant.
Therefore, $M$ is trivial.
\\

Cases 2 and 3 are considered by the same argument as in $(A)$.
~\\

\noindent
$(C)$
 Since \eqref{RicnF} and $\Delta F=n\varphi$, by a direct computation,
 \begin{align*}
 \frac{1}{2}\Delta |\nabla F|^2
 =&~|\nabla\nabla F|^2+{\rm Ric} (\nabla F,\nabla F)+\langle \nabla F,\nabla \Delta F\rangle\\
 =&~|\nabla\nabla F|^2-\frac{1}{n-1}{\rm Ric} (\nabla F,\nabla F)\\
 \geq&~0,
 \end{align*}
 where, the last inequality follows from the assumption.
 Since $M$ is parabolic and ${\rm sup}|\nabla F|<+\infty$, we have that $|\nabla F|$ is constant.
 Therefore, we have $\nabla \nabla F=0$.
 By the same argument as in $(B)$, we complete the proof.
\end{proof}

\noindent
{\bf Conflict of interest}~

There is no conflict of interest in the manuscript.

\bibliographystyle{amsbook}

redpajama_set_name: RedPajamaArXiv


Swimmers from the Let's Swim Bahamas program were treated to an afternoon of snorkeling at Athol Island on Saturday, October 20th. We couldn't have had a better day – calm seas and the sun was shining. Eleven students from 8 different schools were selected to go on the trip based on having completed their Swim America Award of Achievement. For the majority of these students this was their very first-time snorkeling and seeing underwater marine life. Most of them were a little hesitant at first, but with the help of the chaperones, they soon were all snorkeling independently. Let's Swim Bahamas wants to thank Avery Wilson from Stuart Cove for arranging the trip and all the other staff members on the boat.


LG Display saw its shares surge to a one-month high on Wednesday following a report that the company will begin supplying OLED panels to Apple's new range of iPhones this year LG is already a supplier of LCD panels, currently used in the iPhone 8 and iPhone 8 Plus, but Samsung dominates supply of OLED displays for Apple's flagship iPhone X. Indeed, last week LG Display revealed in a regulatory filing that it didn't supply any of the OLED panels currently used in the iPhone X. However, that could change this year, since Apple is expected to launch two new iPhones with OLED displays in 2018. According to today's report, LG Display will supply Apple with the 6.5-inch OLED panels, while Samsung will supply the 5.8-inch or 6-inch OLED panels. So i guess we'll just have to wait and see what Apple has planned up there sleeve.


This paper reports the results from a laboratory experiment designed to study political distortions in the accumulation of public debt. A legislature bargains over the levels of a public good and of district specific transfers in two periods. The legislature can issue or purchase risk-free bonds in the first period and the level of public debt creates a dynamic linkage across policymaking periods. In line with the theoretical predictions, we find that public policies are inefficient and efficiency is increasing in the size of the majority requirement, with higher investment in public goods and lower debt associated with larger majority requirements. Debt is lower when the probability of a negative shock to the economy in the second period is higher indicating that even in a political equilibrium debt is used to smooth consumption and to insure against political uncertainty. Finally, also in line with the theoretical predictions, we find that dynamic distortions are eliminated independently of the voting rule when the first period majority can commit to a policy for the second period. The experiment however highlights two phenomena that are surprising in terms of standard theory and have not been previously documented. First, balancing the budget in each period is a focal point, leading to lower distortions than predicted. Second, higher majority requirements induce significant delays in reaching an agreement.


text:
Breakout Profile: O.J. Howard
I'm not one to be conservative, but O.J. Howard is a top 5 talent at the Tight End position! He marries a well-rounded game with an athletic profile only surpassed by his metrics. Now! I'm not one to be hyperbolic, But Howard is the clearest breakout TE to spot in years!!!!
At age 24, O.J. Howard's draft capital, talent, and efficiency have left us fantasy skids salivating. His usage however, be-it staying in to block as opposed to running routes, or his overall priority in the offence, have left us all banging our heads against… the fire Dirk Koetter wall.
Entering the offseason, Koetter wasn't the only body gone. We also saw DeSean Jackson and Adam Humphries walk in free agency. When Tampa Bay hired Bruce Arians as their new Head Coach, our fantasy ears perked up! The famous fantasy football community immediately put the Buccaneers under a microscope. With the masses paying close attention, they proceeded to use free agency and the draft to add… absolutely nobody on offence! (Sorry Andre Ellington/ Sorry not sorry Breshad Perriman).
Buccaneers management, and Arians alike, recognized they have THREE weapons on offence, all through the receiving game. If we polled 30 analysts asking them who they think Jameis Winston's 4th read in the passing game is, I think we would hear several names bandied about. O.J. Howard would not be one of them. To my point, after being fifth in the pecking order to date, O.J. Howard has moved up to third on the totem pole.
Through his two combined seasons in the league Howard is 3rd behind only Travis Kelce and Rob Gronkowski in PFF's Yards Per Route Run metric. He's put up 11 TDs through 2 seasons on only 87 targets (12.6%- compared to Hunter Henry's 10% and Evan Engram's 5%). Howard's 16.6 Yards per Reception usually reflects a Wide Receiver who sees his targets deep down the field! Last season O.J. Howard was the only TE in the NFL with both an average depth of target above 10 yards AND a catch rate of at least 70%.
So why do we still see O.J. Howard with a draft price near the 6th round? In my opinion, it's mostly people in high places driving us down narrative street. They may need to turn around though, I think they're lost! And we all know, those voices can be drowning. Not to fear though, as we always carry with us our trusty "How to do Fantasy Football right Map". So instead of getting lost, we're at a truckstop re-fueling while plotting our checkpoints to the Fantasy promise land.
In 13 seasons as either Head Coach or coordinator, Bruce Arians offences have produced only ONE Tight End who performed as a top 20 fantasy option. Steelers fan favourite Heath Miller's TE 7, 9, and 15 finishes are the 3 best fantasy seasons of any Tight End under the Kangol Killer.
That sounds damning, but it isn't damning to O.J. Howard. In those 13 years Heaaaaath- was the best Tight End Arians has had at his disposal. Out of the other 9 TEs he's coached, most football fans would be hard pressed to remember 3 of them. Let me jog your memory, because they DID exist. Coby Fleener in 2012 was TE23 with Bruce Arians as OC. Remember that hair! (Referring to Coby Fleener of course, not Arians!) The only other two pseudo household names would be Darren Fells, and Jermaine Gresham who finished between TE31 and TE33 under Arians in Arizona.
Now we can truly appreciate the lack of weaponry Arians has had at the position. You may still argue his offence and schemes devalue the Tight End position overall. But a no risk-it- no biscuit offence can always use an athletic freak (40 Yard Dash, Speed Score, and Burst Score all at least in the 97th percentile). O.J. Howard can stretch the seam and run routes like a WR. He operates in all areas of the field and has shown historic efficiency doing so.
Showing Howard's talent, and Arians' motley crew of Tight Ends just serves to dismiss the circumstantial evidence against my client! It's the concrete numbers that show O.J. Howard is likely to Blow!
The Tampa Bay Buccaneers have the 4th most vacated targets in the NFL going into 2019. With Adam Humphries and DeSean Jackson going to Tennessee and Philadelphia respectively, Tampa Bay has 234 targets to distribute. Those 234 available targets is 38% of the Buccaneers passing game. With little to none in the way of offseason additions, the trio of Mike Evans, Chris Godwin, and O.J. Howard should all outkick their ADP coverage this season. My thoughts on the Running back group (especially their pass catching abilities) The lack of receiving ability in the Running Back core really enforces that statement.
When doing projections, one of the first steps is always to analyse the coaching staff changes. We all know Bruce Arians is as pass heavy as it gets. When we begin to look at pass to run ratios and historical data we find that there shouldn't be much in the way of a regression to the mean in Tampa Bay. We know how much Arians loves to pass, so the play calling should be Brucie through the sky with Jameis! (Sing as Lucy in the sky with diamonds, c'mon guys).
With the volume available, and a pass heavy offence, O.J. Howard's target share will skyrocket! In Howard's rookie year his infinitesimal 6% share of the passing game was equal to how many touchdowns he scored! Last year, in Howard's sophomore season, his market share % only crept up to 7.75. A head scratcher to most of us, it's compounded by Cameron Brate out targeting him 126 to 87 over the last two seasons. Brate has also seen more work in the red zone over that time, although the margin is much thinner.
Another benefit the hiring of Arians provides to O.J. Howard is low usage of 2 Tight End sets. This should help us fantasy gamers to know that Cam Brate will be on the field less. Anyway you slice it Tampa Bay becomes a more condensed offence. This can also combat Jameis Wintson's tendency to spread the ball around.Not only should Brate's role continue to dissipate, there are a lot of high leverage fantasy points available for this team.
Along with having the 4th most passing work up for grabs, the Buccaneers also have the 5th most Red Zone targets available in 2019. Surprisingly, the red zone is where O.J. Howard really dominates. His criminally low Market Share % increases dramatically the closer Tampa Bay gets to the goaline, and his efficiency is undeniable. In his rookie campaign he caught 4 of his 6 targets inside the 20 yard line for 3 TDs. Inside the 10 yard line he converted 4 targets into 3 receptions, all 3 of which went for scores.
We hoped last season to see that role expand, and while we did see Cam Brate's role diminish, Adam Humphries, Desean Jackson, Chris Godwin and Mike Evans continued to be prioritized ahead of Howard. Despite their brutal efficiency in relation to Howard's. In 2018, O.J. still managed 4 TDs on 5 catches in the red zone, and identical to his rookie season, his 4 targets inside the 10 yard line turned into 3 TDs on 3 snags.
O.J Howard has an ADP in early August in the back half of the 5th round, with Evan Engram and Hunter Henry both going early 6th. These guys are being drafted as TE 4 through 6 and over the last 3 seasons, 25% of Tight Ends drafted in the top 6 at their position have busted outright, finishing outside the top 18. It's likely that 2 out of those top 6 will break your heart this season! Adding Eric Ebron, Jared Cook, David Njoku and Vance McDonald, we have 7 TEs being drafted between rounds 5 and 8. If we do draft a Tight End in this range, personally, I'm expecting to pick the player who will perform as that TE 4, if not creep into that elusive elite tier behind Travis Kelce. Not unlike George Kittle in 2018. To bring it home, out of those 7 Tight Ends falling into our ADP bucket, only 2 can vault up to top 4 at their position based on historical trends.
2019 may have a cheat code in projecting O.J. Howard's rise into elite status. When you look at the 3 offences in question, as mentioned, Tampa Bay has a ton of targets up for grabs and no new comers or first timers! The Chargers and Giants either have new or returning players to inherit targets, or very few vacated targets available at all. Most importantly, both the Chargers and Giants are likely to be passing 25-30% of the time to their Running Backs! For the Buccaneers, unless Ronald Jones miraculously… I couldn't even write it.
My arguments for O.J. Howard have been laid out to leave you riddled with bias in reaching a conclusion, so I'd be remiss to not pose this question: If you had an ultimatum where you had to pick ONE of those Tight Ends to crack the top 4 at the position, who's your pick!? As you can see by the True North Projections, our minds are made up. In 2019 we just hope that when Jameis Winston is eating those W's, he's washing them down with some O.J.!!
Tyrell Maclachlan
@TNFFTyrell

redpajama_set_name: RedPajamaCommonCrawl


A small black wolf, with Russian blue tinted fur, walked along the caves. She was the shortest of the pack, and wasn't treated to well. Even if she was a scout, they treated her like the omega. She closed her azure eyes. The feathers on her body were beautiful, but very uncommon for a wolf. A purple ribbon was tied to her left fore paw. It smelled of willow trees, daisies, and sunflowers. The same scent of the girl she tried so hard to save. A sad look took over her face as she remembered the girl. Azura shook her head and walked forward.
The bush didn't hold much, not enough to cover him for today. He'd have to go and find a hare or something. That fridge down in town was tempting, but definitely not for him. He might live with dogs, but it didn't mean he had to eat like them.
Jaeger scented the air again, checking for any sign of prey. So far, none beside deer. But he wasn't going after a deer, not today. Maybe a fawn, but he didn't smell any of those. They were far sweeter than regular deer scent.
He dropped his head, picking up the rest of the berries he'd found and swallowing them whole, not bothering to chew them. He didn't like berries, but if he just swallowed them, he wouldn't have to taste their incredibly sweet juices.
Azura lifted her head to the scent of loner. She broke out into a run after the scent. Out of the caves and into the dark forest. The trees closed in on her, but she expertly weaved around them. She stopped at a small bush, and squeezed under it. There was a wolf there. She sniffed the air, and found the scent of berries. Azura curled into a tight ball and watched him.
They didn't last long. Maybe a good thirty seconds? By the time he'd finished swallowing the berries, he wanted more. Jaeger looked down with a unhappy face, wishing there was more to be feasting upon, only to find a face looking up at him.
Jumping back, the wolf felt his neck bristle up with surprise. It was just a she-wolf, but she held the scent of the pack. Nothing but danger brushed down his spine, reminding him that he had to watch his step. He couldn't be dumb enough some how tick her off, if that was even possible. He was on pack territory, wasn't he?
Azura jumped back, getting scratched by the twigs. "I'm sorry I didn't scare you did I?" Azura said creeping forward again. She really hadn't meant to put him on edge she was just wondering what he was doing.
"What do you want?" The wolf didn't mean to sound hostile, but after years of dealing with other loners and predators, it was hard to be nice when someone popped out of no where. He could feel his heart speeding up slightly, each pound faster than the one before it, a drum thrumming in his ears constantly.
"I was just out and I smelled a somebody else. I wanted to see who it was." Azura replied. She still didn't dare leave the bush. She was much smaller than him, and was afraid he would attack. She watched every movement, every flicker of his eye. Listened to his breath, his heartbeat. She didn't even dare to blink. It was a natural reaction, she was trained as a scout. Making her very cautious.
"And you found me." The wolf replied somewhat calmer, but not too much. Looks didn't prove anything. Even if the pack wolf before him was small, skinny, and appeared weaker than him, he wasn't about to believe she couldn't prove a threat. He'd fallen for that too many times. Her ears were giant, far bigger than any regular wolf's would be. How did she get them? "Now what?"
Azura flicked one of her long ears. She crawled out from under the bush, and when she sat down t was easy to see she was less than half his height. In the dim sunlight the ribbon on her paw and the feathers in her fur had a shine to them. She looked up at him not really knowing what to say.
The feathers weren't from around her, and that purple thing was not from wolves. This wolf was a human's pup, or was anyway. He sneered, disgusted with the new fact. "Pet." Jaeger stood, backing a step away, his fangs unintentionally baring in the presence of a threat. He hated humans.
Azura flicked her ear. "I never lived with humans. I came from far away. There was a human girl lost and dyeing. So I protected her. She wore this on her fur on her head." Azura said picking her paw up. "She placed it on me when she died." Azura said. She was hardly a pet. She was just taking care of the girl. "The feathers came from the bird that lived in the moutians. I liked them and placed them in my fur."


text:
Naive Bayes
What Does Naive Bayes Mean?
Techopedia Explains Naive Bayes
A naive Bayes classifier is an algorithm that uses Bayes' theorem to classify objects. Naive Bayes classifiers assume strong, or naive, independence between attributes of data points. Popular uses of naive Bayes classifiers include spam filters, text analysis and medical diagnosis. These classifiers are widely used for machine learning because they are simple to implement.
Naive Bayes is also known as simple Bayes or independence Bayes.
A naive Bayes classifier uses probability theory to classify data. Naive Bayes classifier algorithms make use of Bayes' theorem. The key insight of Bayes' theorem is that the probability of an event can be adjusted as new data is introduced.
What makes a naive Bayes classifier naive is its assumption that all attributes of a data point under consideration are independent of each other. A classifier sorting fruits into apples and oranges would know that apples are red, round and are a certain size, but would not assume all these things at once. Oranges are round too, after all.
A naive Bayes classifier is not a single algorithm, but a family of machine learning algorithms that make uses of statistical independence. These algorithms are relatively easy to write and run more efficiently than more complex Bayes algorithms.
The most popular application is spam filters. A spam filter looks at email messages for certain key words and puts them in a spam folder if they match.
Despite the name, the more data it gets, the more accurate a naive Bayes classifier becomes, such as from a user flagging email messages in an inbox for spam.
Independence Bayes, Simple Bayes
Bayesian Filter
INFOGRAPHIC: International Internet Scam Hotspots
Email Spam: What Goes Around Comes Around?
Too Much Spam? 5 Technologies Designed to Block It
Autonomic Systems and Elevating Humans from Being Middleware: Q&A with Ben Nye, CEO of Turbonomic
Machine Learning 101
The Promises and Pitfalls of Machine Learning
Software DevelopmentData Management


text:
Maiden is a  aluminium ocean racing yacht built in 1979, designed by Bruce Farr and raced by Pierre Fehlmann, Bertie Reed and Tracy Edwards and John Bankart. Edwards bought the yacht in 1987 to compete in the 1989–90 Whitbread Round the World Race with an all-female crew. The yacht achieved good results and broke records, leading to Edwards becoming the first female winner of the Yachtsman of the Year Trophy, changing the perception of women in ocean racing.

Edwards sold Maiden after the race; the yacht subsequently had several different owners. The last of these abandoned her in the Seychelles. She was brought back to England by Edwards in 2017 after a fundraising drive. After a major refit, the boat set off on a global voyage in 2018 to raise funds and awareness for girls' education under the foundation "The Maiden Factor". A second world tour began in 2021.

In 2018, a documentary titled Maiden was made about Edwards and the crew competing in the race; its first showing was at the 2018 Toronto International Film Festival.

Racing career

1979–1987
The  aluminium-hulled racing yacht was commissioned by the Swiss Ocean Racing Club. Designed by Bruce Farr and named Disque D'Or 3 (or Disque D'Or III) in 1979,  she came fourth in the 1981–82 Whitbread Round the World Race, skippered by Swiss sailor Pierre Fehlmann. She was renamed Stabilo Boss for the  1986–87 BOC single-handed challenge, and came seventh, skippered by South African Bertie Reed.

Whitbread 1989–90

The boat, then named Prestige, was bought by Tracy Edwards in 1987 with the intention of entering the first all-female Whitbread crew in the 1989–90 Whitbread Round the World Race; in the previous race, only five women had been among the 200 sailors. Edwards remortgaged her house to buy the boat, and was supported by King Hussein of Jordan and sponsorship from Royal Jordanian Airlines. Edwards recruited a 12-woman crew; the boat nearly sank in the River Hamble on the way to the boatyard, but they rebuilt the boat from scratch in 1988. She was christened Maiden by Sarah, Duchess of York. The 1989 Fastnet race was intended to be a warm-up for Maiden and crew, but was abandoned after the start owing to an injury to crewperson Jo Gooding. In the Whitbread race, Maiden finished second in her class (which included Rucanor Sport (BEL), Esprit de Liberté (FRA) and La Poste (FRA)), and was winner in Division D on two out of six individual legs of the race (including the longest leg, the iceberg strewn 7,260 nmi leg in the Southern Ocean), the best finish by a British boat for 17 years. Edwards was awarded the Yachtsman of the Year Trophy, the first woman to receive it, and made an MBE.

The crew (not all for the whole race) were Tracy Edwards (skipper), Mandi Swan (Amanda Swan Neal), Mikaela Von Koskull, Claire Warren, Michele Paret, Tanja Visser, Sally Creaser, Dawn Riley, Nancy Hill, Jeni Mundy, Jo Gooding (from Leg 2), Sarah Davies (reserve), Kristin Harris (Leg 1) and Angela Farrell.

After Whitbread
Having taken out loans to buy the boat, Edwards sold her after the race. In 1990 Edwards published an account in Maiden, co-authored with Tim Madge.  There were several subsequent owners. In 1996/97 she was chartered by Stuart Bowen Davies and skippered by John Bankart and was successfully campaigned in the Hong Kong Round the World yacht Race which was organized by Jimmy Cornell.
The last of this series of owners abandoned the yacht in a marina in the Seychelles. The marina seized the boat and offered it to Edwards, who was reunited with the boat after a break of 27 years and was shocked at her condition. Nevertheless, the intention, after fundraising and some restoration work in Cape Town, was to sail her back to England with original crew members and use her for fundraising to promote girls' education under the banner "Maiden Rescue". In the event, the yacht, considered unfit to be sailed, left Cape Town on 26 March 2017 by cargo ship. The shipping was arranged by Peters and May, who had worked with the Maiden team in the Whitbread race.

Return to England
Maiden arrived in Southampton in April 2017 and the original crew reunited. The yacht was refitted by Hamble Yacht Services in Hamble, near Southampton, in the same shed where she had been refitted in 1988. She was re-rigged by Ocean Yacht Systems, the same firm that had rigged her in the 1980s.

World Tour 2018-20

Maiden was re-launched in the summer of 2018 and in September 2018 departed from Southampton Boat Show with a new crew of women, to prepare for a three-year world tour on behalf of the charity "The Maiden Factor Foundation", which was set up by Edwards with the support of the late King Hussein of Jordan's daughter, Princess Haya. A principal project of the charity is to use Maiden to raise funds and awareness for girls' education globally. The yacht left Hamble in November, and called into Plymouth for repairs. Their first scheduled stop was Kerala, India. They were headed for the Strait of Gibraltar on 26 November. On the voyage, crews and skippers were rotated; among the guest skippers planned were Dee Caffari, Sharon Ferris and Wendy Tuck; the first is Nikki Henderson.

The yacht arrived in Malta on 13 December 2018, and was passing through the Suez Canal in February 2019. Skippered by Wendy Tuck, Maiden reached Auckland, New Zealand, on 1 May 2019.

In November 2019, having crossed the Pacific Ocean via Hawaii to North America, she visited Vancouver, Seattle, San Francisco, Los Angeles and San Diego, all centres for yachting and opportunities for fundraising. She passed through the Panama Canal, skippered by Liz Wardley, bound for Antigua. At each stop-off, hundreds of schoolgirls visited the boat and crew members gave talks in schools. Edwards commented: " We can't keep up with the number of schools that want us. Teachers love it."

World Tour 2021-24
After a shakedown voyage from Dubai and passing through the Suez Canel, in March 2022 Maiden was in Palma, Majorca, a maintenance stop on the first leg of the new tour, this time for three years, first crossing the Atlantic to Miami, Florida. Numerous other Atlantic stops would see Maiden arrive in Cape Town later in 2022. In partnership with DP World, the Tour sees a resumption of fundraising and an emphasis on education for females, particularly in STEM subjects. In June 2022 she was in Brooklyn and in August was in Newport, Rhode Island.

See also
 List of Volvo Ocean Race sailors

References

External links
 Photo-diary of 2017/18 refit at HYS
 The Maiden Factor
 The Maiden Factor: world tour blog 2018-2020
 Tracking Maiden (MMSI 232016680) on MarineTraffic

Individual sailing vessels
1970s sailing yachts
Volvo Ocean Race yachts
Sailing yachts built in the United Kingdom
Sailing yachts of the United Kingdom
Sailboat type designs by Bruce Farr

redpajama_set_name: RedPajamaWikipedia


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Starting questions - any food?
Hi everyone! This is my first post on this board.
I'm a 23 year old student studying in Oxford. I'm planning on starting CD or LL next week, but I have some questions that need answering first before I can go ahead with it.
1) I have been doing Slim.Fast and only taking in 500 calories a day for 3 days now, in preparation for cutting down completely, but I feel awful. I'm absolutely shattered, I ache and I am actually fantasising about food. After two days. I caved in and ate a pitta bread just now, and now I'm annoyed, because if I was on CD or LL, it would have ruined my ketosis. Do the cravings and hunger disappear for real, and do you feel OK on such few calories?
2) Is any food allowed on the diets, even like celery?
3) How do the shakes/ soups/ tetras/ bars taste? The powders are made with water, right? Are they watery or are they palatable?
4) I know alcohol isn't allowed on the diet, but what are the consequences if you were to get drunk? You go out of ketosis? Massive hangover?
5) I'm in two musicals this term and have an exam in a month. I'm going to be busy and having to study a lot. On such low calories so far (on my preparation Slim.Fast plan) I am unable to concentrate and I'm getting woozy if I'm too active. Basically this term is going to be crazy busy, will I survive?
6) What happens if you actually do end up eating something?
7) I've been in touch with LL and CD counsellors in Oxford. I'm not sure if I need the CBT/TA of LL, and it is a lot more expensive (for a student, this is important). Does anyone have any views on the neccessity of the CBT/TA in LL vs. CD?
OK. Well, I've left enough questions there! I'm quite excited about this, but I'm worried I'm getting into something I won't be able to finish, or something that will make me suffer.
Thanks for all your help, I'm hoping to get a lot out of this.
Well I can only answer for CD, but I think they're really good, much nicer than SFast, and there's a lovely range of sweet & savoury, there really is something for everyone. The soups can be a bit watery, but they're still very flavoured as are the shakes - they do actually taste of what they're meant to taste of.
You could make yourself seriously ill, alcohol is a big no-no & can be dangerous whilst SSing. Your body will not cope with alcohol at all & really should be avoided.
Yes you will, honestly there is no comparision between a proper VLCD and Slimfast, you really will have loads of energy - it will amaze you.
You'll knock yourself out of ketosis, and it can lead to cravings, plus you'll have to go through the process of getting back into ketosis again. If you can do the diet & stick to it 100%, it is so much easier & you'll get to the end a whole lot quicker too.
It's really down to personal choice, if you feel you need the counselling element then LL may be better for you, but if you think you can get by with your CDCs assistance, and there's always Minimins of course, then CD is a lot cheaper.
Hope that helps John and if you need to know anything else, just holler.
I think Dotty has answered all your ?'s now. Well done Dotty!
But just wanted to say that if you have lasted 3 days on Slimfast and only caved in for a pitta you will absolutely love and thrive on Cambridge or Lighter Life. If the flavour of the shakes and soups is most important to you I would suggest CD. If the counselling and group support will be good for you I would suggest LL. CD is also cheaper and you see your counsellor at a time to suit you so may fit in with your commitments better. Good Luck whatever you decide.
Dotty has answered your questions admirably John.
You'll be shocked by the amount of energy you have on a VLCD but you HAVE to stick to it 100% for it to work at its best. It's only when people start customising a VLCD that it all goes to pot.
John.. I started my MA the same week as I started LL. It was really good timing. I'm so busy with studies and exams and stuff that I really forget about food. Also, not having all those food worries has really allowed me to focus on my studies! So all in all it's been fantastic! Good luck with whatever you decide.
Isn't this only if you are ssing? I'm on 790 which means you can eat some chicken/turkey/tuna/white fish and some of the green and white vegs.
You're right fiojam - small amounts of food (specified in the yellow CD booklet) are allowed on the 790 plan and also on AAM week.
LL don't have alternative plans - they SS straight through the first 100 days and in that case (if John decides on LL) absolutely no food is allowed.
I have basically ruled out LL because of the expense and lack of AAM week (although AAM week could potentially derail me, so if I was coping on SS I might stay on it, but it'll be CD cos of the flavours/options and price).
I am worried about one thing. I'm in uni, I do a lot of theatre and I socialise a lot. Which involves drinking alcohol. Obviously alcohol is the scourge of SS, but I know (and I am fully prepared to admit this) I would find it virtually impossible to abstain from alcohol for however long SS takes. I think if I was allowed alcohol once a week (and we're talking student amounts here), I could handle it. I know that drinking alcohol on SS is dangerous, unpleasant and really sets your weight loss back.
I'm thinking that maybe now isn't the right time to start CD. How long do you have to abstain for, and even when you no longer have to abstain, is your intake limited (i.e. in the later steps)?
Are we all completely sure alcohol is a no-no on SS? Even one night a week?
I guess this all makes me sound completely alcohol-dependent.
Alcohol is a complete no-no. Your liver is working hard as it is - adding alcohol to the mix whilst doing SS is just too much.
I had a client who was convinced giving up alcohol would be the straw that broke the camel's back, but she actually found that as the weight came off, the keto-high, etc that she didn't miss alcohol at all. She went to birthday parties, a Robbie Williams concert, etc and didn't touch a drop.
It's about priorities I guess. The first time you do a VLCD is like a golden time - are you prepared to put in what it takes to make this work for you? Weight loss for men is fantastic - for example, a male client of mine lost 8 stone in 15 weeks - don't you think it's worth it for the amount of time it would take?
The last three stages are for when you are coming off the packs and returning to food. It takes approx 6 weeks to go through maintenance and stabilisation, so you need to factor that into the time it will take you to lose your weight.
No one can make the decision for you - we can tell you the facts, but only you can decide whether you are ready to give this your best shot, or wait til later!
That has put my worries at ease about feeling too drainned to do things as I am working 40 hours a week and studying CIMA (Management Accountancy) to. But I suppose on an up side you do not have to food shop (O/H might be joing me, if not can do the food), and you can make a shake and study it is a lot less time out of an evening.
Can I ask did any of you feel drainned/tired the first few weeks?
-First weight in - lost ?? & now weigh - ??
Every time I do this diet, I feel totally rough!
Im on day 11 now so I'm hoping that I will start to feel alot better soon!
Questions - Starting in couple of days.


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Archive for the 'AMC: Hell On Wheels' Category
AMC Just Broke the Mold by Scheduling Hell on Wheels Season 3… on Saturdays
April 16, 2013 DStanley	Leave a comment
There's always been one universally accepted and unwritten rule of television scheduling that's not in the Network Executiving for Dummies book: Don't schedule new primetime programming on Saturday. Just don't! Saturday is the day when Americans go out and spend all night being Americans and celebrating the freedom of the weekend, according to conventional wisdom. No one is even home to watch television, so just air infomercials and aging sketch shows (and Rules of Engagement).
But conventional wisdom doesn't set trends, conventional wisdom isn't a maverick, and conventional wisdom can suck it, says AMC. The basic cable network announced today that it's moving its Western drama Hell on Wheels to Saturdays for the show's third season, which premieres Saturday, August 3. Though banishment to Saturday is usually seen as a death sentence, this is is in no way a demotion for Hell on Wheels. According to AMC, the idea is to make Hells on Wheels the capper on the network's longstanding tradition of airing Western movies on Saturdays.
"A new episode of Hell on Wheels on Saturday night after a full day of Western fan favorites is going to be like the saddle on top of the horse," AMC president Charlie Collier said in the network's official statement. Oh sure, but what's going to be the cowboy, Mr. Collier?
AMC's move could be about more than just keeping a consistent theme, however. Television's exponential growth in the last five years has meant a more crowded schedule from Sunday to Friday. Sunday night, where AMC normally programs its scripted originals, is a murderer's row of quality television, and there isn't a different day of the work week that's particularly inviting for new shows. Logically, branching out to Saturdays makes sense with the economy keeping people inside more often on the weekends, DVRs time-shifting shows on viewers' schedules, and viewers' growing preference of online streaming. And Hell on Wheels, one of AMC's lower-profile scripted dramas, is the perfect show to experiment with.
Categories: AMC: Hell On Wheels
Hell on Wheels: Jennifer Ferrin Joining Season 3 as Series Regular
Jennifer Ferrin, who's currently pursuing Kevin Bacon on The Following, is joining the third season of Hell on Wheels as a series regular, TVLine has learned exclusively.
The actress will play Louise Ellison, a smart, witty and flirtatious journalist hired by the New York Sun to cover the "story of the century" — the building of the Transcontinental Railroad. Louise imbues her work with an element of human emotion as well as thorough, blunt, truthful fact-finding. She is strong and independent and demands to be treated with the same respect that any man in her position would be given.
The gig reunites Ferrin with her former Cape EP John Wirth, who was recently named Wheels' new showrunner.
In addition to The Following and The Cape, Ferrin's resume includes stints on CBS' late soap As the World Turns and ABC's Life on Mars.
AMC's period Western is slated to kick off its 10-episode third season this summer.
'Hell On Wheels' Showrunner John Shiban Departs, Series' Renewal On Hold
November 2, 2012 DStanley	Leave a comment
There is some behind-the-scenes drama on AMC's Hell On Wheels. The Western drama on Monday was given a 10-episode third-season renewal, but it was picked up without its creators/executive producers Joe and Tony Gayton, who left their day-to-day duties on the show when their contract was not renewed. (They may consult.) At the time, AMC said John Shiban would continue as executive producer/showrunner. Now I hear that veteran showrunner Shiban has informed the network that he won't be coming back after two years on the show. As a result, I hear Hell On Wheels' pickup has been put on hold until a new showrunner is found.
This is the latest AMC series hit by major creator/showrunner departures. The Walking Dead's Frank Darabont left in Season 2, Rubicon creator/exec producer Jason Horwitch exited during production on its first and only season, and Hell On Wheels now is looking to continue without its creators and showrunner. Before joining Hell On Wheels, The X Files alum Shiban worked as a consulting producer on another AMC series from a former X Files writer-producer, Vince Gilligan's Breaking Bad.
Hell On Wheels wrapped its second season earlier this month with a solid average of 2.4 million viewers. That was down from the average of 3 million viewers for the show's freshman season — but this time around, Hell On Wheels did not have blockbuster The Walking Dead as a lead-in and was impacted by the lack of AMC carriage on Dish Network. Hell On Wheels, from Entertainment One and Endemol USA, is exec produced by Jeremy Gold and David Von Ancken.
'Hell on Wheels' Renewed by AMC for Season 3
October 30, 2012 DStanley	Leave a comment
AMC has renewed Hell On Wheels for a 10 episode third season, Deadline reports.
Hell on Wheels – Season 2 – This Season Promo
Hell on Wheels – Season 2 – Promotional Photos
Categories: AMC, AMC: Hell On Wheels
Hell on Wheels – Season 2 – Searching Promo


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Canal Café Theatre
Situated in the heart of Little Venice, the Canal Café Theatre has been a popular venue for British comedy since 1979. Featuring a 60-seat performance space, the theatre plays host to over 14 shows per week, including sketch, improv, stand-up, and plays.
The theatre is the home for NewsRevue, a politically themed sketch show that runs weekly and transfers to the Edinburgh Fringe Festival annually. The show currently holds the Guinness World Record for longest running live comedy.
The Canal Café is available for hire and rehearsal space and is always looking out for new comic writing.
Get social with Canal Café Theatre:
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Seven Lives, One Building, Many Stories as Peepshow comes to Canal Cafe
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Atlanta selects UNCW's Goodrum in MLS draft
Phillip Goodrum UNCW (Source: wect)
January 13, 2020 at 5:07 PM EST - Updated January 13 at 5:07 PM
WILMINGTON, N.C. (UNCW release) - UNCW senior forward Phillip Goodrum was selected by Atlanta United FC in the third round of the Major League Soccer SuperDraft on Monday afternoon.
Goodrum, the 75th overall pick, was the second UNCW player selected in the draft after Danny Reynolds was taken by Seattle in the second round.
The duo joins Colin Bonner and Daniel Roberts as former UNCW standouts to be selected in a Major League Soccer draft.
"I am so happy to be a part of Atlanta United," Goodrum said. "It's a dream come true to get to continue playing soccer for such a great club. My time at UNCW was the best four years of my life and the soccer program shaped me into a better player and person. I'm so excited for the future and can't thank [Head Coach] Aidan Heaney and the UNCW soccer staff enough."
Goodrum, a three-time First-Team United Soccer Coaches All-Atlantic Region honoree, paced the Seahawks and ranked third in the CAA with 11 goals and 26 points. A two-time selection as the league's Offensive Player of the Week, the Charlotte, N.C., native scored five goals in the final three games of the regular season, including back-to-back two-goal efforts against Drexel and VCU.
One of four players in school history to register back-to-back 10-goal seasons, Goodrum started all 76 games in his college career and ranks fourth on the program's all-time list with 32 goals and 79 points.
"What a fantastic opportunity for Phillip to join such a tremendous organization as Atlanta United," Heaney said. "He also has family there, so he is delighted to be selected. Anyone who knows Phillip knows how hard he works and how committed he is to the sport. I am so happy to see him achieve one of his goals and I know he will continue to thrive at the next level."
Goodrum became the third Colonial Athletic Association player to be selected in the draft, joining Reynolds and James Madison senior midfielder Manuel Ferriol. Reynolds, the CAA's Defensive Player of the Year, was taken with the 35th overall pick while Ferriol, the CAA Player of the Year, followed with the No. 40 selection by FC Dallas.


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>> 29-year old sheriff's deputy, Zackari Parrish, was killed by a gunman when authorities responded to a disturbance call in a suburb outside of Denver, Sunday morning. The suspect, who was known to authorities, was also fatally shot. Parrish and three other deputies responded to a call of shots fired around 5 AM local time at Copper Canyon Apartments in Highlands Ranch, about 20 miles south of Denver.
>> I do know that all of them were shot very, very quickly, and they all went down almost within seconds of each other. So it was more of an ambush type of attack on our officers. He knew we were coming.>> Another officer later came to the scene and was shot, leaving four officers and two civilians injured by the gunman.
A code red was issued by the Sheriff's Office, instructing residents to shelter and place and avoid windows. Officials say the suspect was carrying a rifle. Parrish, who is survived by his wife and two children, had worked at the Sheriff's Office for only seven months.


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Time To Try An Economic Carrot Approach With North Korea
Harry G. Broadman Contributor
Kim Jong-un Looking Skyward
On the very eve of the (presumed) Summit between U.S. President Donald Trump and North Korean leader Kim Jong-un, it's high time to rethink the West's sole reliance on the stick approach of economic sanctions on the North. Kim's individual consumption habits are largely above the fray of international commerce, so sanctions by themselves have not directly affected him. And, to think that exacting economic pressure on North Korean society will incite meaningful revolt and an overthrow of the regime is fanciful. After all, Kim is a man who apparently didn't think twice about murdering his half-brother and uncle. It's imperative to get creative and consider adding in some positive reinforcement to the mix.
We economists love thinking in terms of a set of disincentives and incentives that might best bring about certain outcomes. To this end, what affirmative ingredients are missing?
Logically, the most effective ones should stem from the vision of the endgame we are hoping for with Pyongyang. However, it's rare to find a robust articulation of that goal these days — other than vague expressions of the desire to move towards a demilitarized Korean peninsula and, ultimately, an even more nebulously defined reunified North and South Korea. Does the absence of well-defined tangible strategic outcomes from U.S. overseas policy forays sound familiar?
At the risk of being cast as a dreamer, for the sake of argument I'll posit an objective that has concrete, recognizable features for which in the long run one might dare to aim: a U.S.-North-South Korea Free Trade Agreement (USNSK-FTA). Since the current U.S. administration has called the existing U.S.-South Korea Free Trade Agreement (KORUS-FTA) a 'bad deal' and has already signaled to Seoul that Washington will re-negotiate the agreement, maybe President Trump should completely catch Kim Jong-un off-guard and invite him to the FTA negotiating table?
This will sound very far-fetched to most. But compared to the current dangerous stalemate with the threatened use of nuclear weapons, maybe it's not so nutty.
If nothing else, the idea of such a trilateral agenda would certainly add juice to U.S.-China relations vis a vis the dynamics of their negotiations over North Korea. Why? Because China has a heavy stake in the game. On the commercial side, China currently accounts for 91% of North Korea's import purchases and 87% of its export sales. And, it also might induce a more constructive foreign policy dialog between Beijing and Washington vis a vis Pyongyang. That is sorely needed. One thing that Beijing does get that Washington does not, is the fear of a power vacuum in Pyongyang. It's amazing how potent having a crazy neighbor is in altering your thinking. In a world where the transboundary effects of bad neighbors increasingly know no bounds, we'd be wise to heed China's warnings.
Of course, there are abundant grounds to not even consider a U.S.-North-South Korea Free Trade Agreement in the realm of the possible. In part, this is because sadly it's unimaginable for this White House to take an economic 'high road' policy (it's hard to make the case it has done so to date). And, in part because the defense and foreign policy communities will dismiss the notion outright.
Still, such a proposal does provide for a very useful thought experiment for dispassionate self-examination of what exactly is our long-run objective with North Korea — not just on the part of U.S. policy-makers but also by the South Koreans.
History provides a guide to the latter.
Since the partition of the peninsula in 1945, there have been several attempts by the North and South towards various forms of economic integration: most notably South Korean President Roh Tae-woo's "Nordpolitik" initiative in the late 1980s that sought to open trade and facilitate exchange of visits with the North, and even included a slogan that called for a "Korean Community"; President Kim Dae-jung's "Sunshine Policy", which began in the late 1990s and led to the first Inter-Korean summit, between Kim Dae-jung and Kim Jong-il (Kim Jong-un's grandfather); and most recently the establishment in 2002 of the Kaesong Industrial Zone (KIZ) within North Korea, which, by 2013, was home to about 120 South Korean companies employing over 50,000 North Koreans, whose total annual wages amounted to approximately $90 million that the South Korean firms paid directly to the North Korean government (obviously a critical source of foreign exchange for Pyongyang). Shortly thereafter, however, as political tensions grew severe, the Kaesong Industrial Zone — which in reality accounted for the vast bulk of bona fide trade (that is, apart from smuggling) between North and South Korea — was shuttered up.
All this points to the need to ramp up the focus now on the economic dimensions of North Korea's future. The fact is that in large part, for most countries in transition away from socialism, transformation to more market-oriented systems rather than 'command and control' regimes — think Eastern, Central and Southern Europe; Russia and the CIS; and China and Vietnam; among others — the introduction of commercial incentives has played a key role in enhancing growth rates, reducing poverty, improving the state of health care and education, and advancing the standing of countries in the global community. While not pretending to know what's in Kim Jong-un's head (or how it operates), it's this last dimension that probably would resonate most forcefully with him.
In no way am I suggesting market-driven incentives in and of themselves have been a panacea (nor would any other economist worth his or her own salt who focuses on international development). Moreover, the degree of success of such an approach varies greatly across countries. In a nutshell, many other factors, including political and electoral reform, freedom of movement and speech, and establishment and respect for rules-based institutions, to name a few, are all equally and, in some cases, if not more, important. But unshackling the economy is certainly a bedrock. Arguably it's as good a place to start as any.
It will surely shock most readers to know that North Korea actually does already have a private sector, albeit of limited scale and scope — as best as those of us who assess such things can tell from the paucity of opportunities for first-hand observation in light of Pyongyang's political regime. And, these businesses have begun to grow materially over the last decade. Arguably, they may well be ripe for nurturing.
Just where did they come from and how do they operate?
Ironically it was the collapse of the Soviet Union's trade regime and the opening up of China in the early 1990s that provided the initial impetus for North Korea's private sector. Both countries had been the substantial supporters of, and engaged in trade with, North Korea. But such trade was carried out on terms that explicitly and implicitly provided subsidies to Pyongyang.
As Moscow's and Beijing's reforms inched towards market principles in those years, including the advent of entering into bona fide international trade transactions with capitalist countries within the world trading system, their appetite toward, and the capacity to provide, these subsidies abated, especially that of Russia, which for years prior began to perceive North Korea as an untrustworthy ally.
As the purveyors of fertilizer, chemicals and farm machinery to North Korea, Russia and China's pullback set the productivity and hence output of North Korea's agricultural sector into a rapid tailspin. In turn this engendered unrelenting strain on Pyongyang's archaic centralized system for physically allocating existing foodstuffs to the population through the use of ration cards to stringently regulate access to state-owned distribution centers and stores (where extraordinarily low nominal prices were charged).
The ensuing famine throughout the mid to late 1990s was horrific by any standard, with the death toll estimated to be as high as 13% of the nation's total population of 20-plus million. The state blamed the famine on floods (the existence of which to this day is questioned) and refused to acknowledge its own inability to provide for its citizenry. Indeed, Pyongyang's illusion of economic 'self-reliance' was perpetuated for years before, and largely remains so today.
Out of desperation, small-holding individual and family growers began to appear — using fields often, though not always, out of public view — with any surplus sold to others. In parallel, barter exchanges and small open-air markets arose, where anything of value was sold, including parts of machinery and the like stolen from factories in which they "worked". And, of course smuggling along the Chinese border blossomed.
Local government officials turned a blind eye to virtually all of such non-state, i.e., private, activity in return for bribes. In fact, the state refused to acknowledge the existence of private enterprise — the official stance of Pyongyang.
In time, especially in the early 2000s, these micro "enterprises", barter exchanges and shuttle traders earned surplus revenues, which prompted the search for investment opportunities across the country. The result has been the rapid emergence of small to medium-sized workshops producing clothes and household products; larger factories manufacturing fabric, building supplies, tools and machines; cafes, restaurants, hotels, and private apartments; fisheries; provision of transport services, including the ownership of trucks, buses and cars; and foreign trade enterprises. Indeed, Kim Jong-un's father, Kim Jong-il, had little choice but to recognize that the "shadow" private sector had become the growth engine of North Korea's economy, and up until 2004 actually instigated limited reforms to explicitly encourage it.
However, as the surpluses generated by the non-state sector ballooned, evidenced tangibly by 'conspicuous consumption' but little in the way of formal tax payments to Pyongyang, Kim Jong-il reversed course and not only applied administrative and legal restrictions curbing the private sector, but in 2009 also instituted currency reforms.
These actions actually had little effect on the moneyed-class, but depleted savings of much of the rest of the population, fomenting broad hostility. By the time of his father's death and the start of Kim Jong-un's rule in 2011, these constraints were largely lifted. Yet Pyongyang refused to acknowledge officially the presence of the private sector. That policy largely remains in effect today.
With an official stance of non-admission of the existence of the private sector, it is truly impossible to know with any semblance of accuracy its contribution to the country's economy today; indeed, Pyongyang itself may have only the roughest of ideas. The best educated guesses based on select case studies by outsiders point to a wide range of the role of the private sector, approximately 30% to 60% of North Korea's GDP.
The one area of the economy, however, Kim Jong-un simply cannot pretend the private sector does not play a critical role, is the country's need to increase its integration into global trade. For two mutually-reinforcing reasons. First, he knows private sector-led trade, especially its ability to earn hard currency, lies at the heart of his rule, inasmuch as North Korea chronically runs a significant trade deficit (although in recent years it is far smaller than it was one and a half decades). But second and doubtlessly far more important to Kim personally, is that robust international trade — which in the case of the North Korean economy can only be propelled by its private sector — feeds his psychological craving for global stature.
Here's from where the opportunity for an FTA stems.
Critically, over the last four years, North Korea's total trade with the rest of the world has precipitously declined. From its peak in 2013, when combined exports and imports amounted to $8 billion, in 2016 it stood at $6 billion, a 25% fall. A significant culprit is the lack of product diversification of the country's exports, which are mainly minerals and metals, metal products (including weapons), and coal. In contrast, its imports are mostly comprised of high value-added machinery (both electrical and non-electrical) and motor vehicles.
And, with China accounting for the overwhelming bulk of both the country's exports and imports, the lack of diversification in trading partners is equally a ball and chain for Kim. Among advanced countries, while France is North Korea's 6th largest export market, the volume of French imports from North Korea amounted to only $11 million. Germany is the 8th largest exporter to North Korea, but North Korea was able to purchase only $6 million of German goods. In both cases these are miniscule amounts.
It may well be too much of leap to even contemplate a U.S.-North-South Korea Free Trade Agreement in our lifetimes. But at the very least, the concept can be an eye-opening discussion point, inducing little if any harm. And even if it were to become a reality, by no means would it imply any weakening of our export controls, allowing Kim access to purchase high risk products from the U.S. In fact, arguably Kim could end up with a vested interest in the success of the U.S. economy so as to not kill his 'golden goose'.
And who knows, if it were to succeed, Donald Trump could walk away with both the Nobel Peace and Economic Prizes.
Harry G. Broadman
I advise, testify as an expert witness, speak, and write on multinational business growth, innovation and risk-mitigation strategies; international trade and investment ...


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Gambling.com » Sports Betting » March Madness 2023 Betting - How To Bet On The NCAA Tournament Updated: Feb 8th, 2023
March Madness 2023 Betting - How To Bet On The NCAA Tournament
The 2023 NCAA men's basketball tournament should be an exciting time for sports bettors. More states are on board this year and it is one of the top betting events in the country. The 2023 national champion will be crowed at NRG Stadium in Houston.
Make sure to check out the legal online US sportsbooks to find out the best betting odds and get in on the action!
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March Madness Betting Tips
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How Does the NCAA Basketball Tournament Work?
There are 68 available spots in the NCAA Tournament: 32 teams get in by way of winning their respective conference tournaments while the remaining 36 "at-large" teams are chosen by the NCAA selection committee, a 10-person group made up of athletic directors and conference commissioners.
The committee takes into consideration a number of different criteria to determine the best 36 teams that did not win their conference tournament. The committee then seeds the teams 1-68, which makes up the four separate 1-16 regions (plus the four play-in games).
This year, Selection Sunday takes place on March 12 at 6 p.m. ET. The selection show will be aired on CBS with the bracket being revealed to show the 68 teams who qualify for this year's tournament. Games begin with the First Four on March 14-15. First-round games start on March 16-17, with second-round contests taking place on March 18-19. The Sweet 16 will be held on March 23-24 before the Elite Eight on March 25-26. Houston will host the Final Four on April 1 with the national title game tipping off on April 3.
Check Out: NJ Sports Betting | PA Sports Betting sites for legal betting options.
2023 NCAA Basketball Tournament Winners Odds
Duke and Kentucky are the early favorites at +1000, Arkansas, Gonzaga and Kansas at +1200 and North Carolina and UCLA at +1400.
Duke +1000
Kentucky +1000
Arkansas +1200
Gonzaga +1200
Kansas +1200
North Carolina +1400
UCLA +1400
Arizona +1600
Baylor +1800
Villanova +1800
Houston +1800
Michigan +2500
Alabama +2500
Texas Tech +5000
Odds are current as of publication and subject to change.
Betting on the March Madness 2023 Tournament
March Madness is one of the most wagered on events of the year. Given the number of games and the unpredictable nature of the tournament, it always draws betting interest. This will be the first March Madness tournament residents in markets such as Ohio and Maryland can place bets online through their mobile devices now that Maryland and Ohio online sportsbook apps will be live by then.
Bettors can expect to wager on point spreads, totals, futures, parlays and even in-game wagering. Expect some point spreads to reach the 30s with favored No. 1 seeds taking on relatively unknown underdogs from smaller conferences. This can also lead to inflated moneyline odds with heavily-favored teams.
Depending on your state, individual player props may also be a wagering option. These can include the Over-Under on an individual's player points or rebounds totals. Some states also have individual props on who will be named the Most Outstanding Player of the tournament.
Futures odds continue to update throughout the tournament, too. Bettors can wager on a team to make the Sweet 16, Final Four, or win the national title. Sportsbooks also give players the option to bet on which conference the national champion will come from.
Enhance your experience with some of the most popular sports betting features:
✅ Moneyline betting: A staple wager for casual bettors and seasoned pros alike, moneylines simply require you to pick the winning team in a one-off sporting contest.
✅ Futures betting: Essentially a moneyline bet for the long haul, the futures market enables you to speculate on the outcome of distant events such as the next Stanley Cup winners.
✅ Spread betting: A slightly more complex market, spread betting enables you to bet on the margin of victory between two teams. The narrower the spread, the better the odds will be.
✅ Betting on the over/under: A simple and easy-to-understand betting option that allows you to predict if the final collective score will be over or under the oddsmakers set points line.
✅ Prop betting: Got a hunch about a player's form going into game? Back them to do something special with prop betting – a market that allows you to bet on specific actions in a game at increased odds.
Best Sportsbooks for Betting on March Madness 2023
Sports betting has continued to expand across the country as more states have legalized it in recent years. Some states have opted for an online-only market, such as Wyoming, while others like South Dakota only permit retail sports betting in the city of Deadwood.
When online sports betting in New York officially launched in January 2022, it became the biggest state to have sports wagering. The state has produced higher handle totals in a week than most states do in an entire month.
In addition to the ease and speed that online sports betting provides, it also allows players to shop the best line across multiple sportsbooks, which is critical for profiting long-term in sports betting. For instance, the UCLA Bruins could be favored by -7 at FanDuel, but Caesars Sportsbook could have the Bruins as -6-point favorites. Searching for the best line across all the games is critical for every sports bettor, and a state with multiple online books benefits the bettors.
Tips to Consider When Betting on March Madness
✅ Be sure to find the best line for whatever bet it is you like to make. This can benefit in the long run, and the last thing you want is to have a 6.5-point favorite only win by six when you could have possibly gotten a –5.5 on a different sportsbook. The same idea goes for betting the Over-Under and the moneyline. If one sportsbook has the Michigan State Spartans as +195 underdogs and another has them at +215, it's beneficial to take the higher number in case Michigan State does pull off the upset.
✅ Do your research. Sure, firing off bets on a whim can be fun but it likely won't result in any type of long-term profitability. Your best chance to finish ahead this March Madness tournament is to research all the teams and try to find an edge. Whether that's how a team's perimeter defense matches up or the size and length of a roster, these are all things worth considering when trying to bet on a game.
✅ Don't chase your losses. Games begin early in the afternoon and can go well past midnight. If you have a rough go at it early, don't unload your entire bankroll on the west coast game tipping off at 10 p.m. Eastern hoping to make all your money back. This is not responsible bankroll management, and the last thing you want is to blow all your money on the first day of the tournament.
✅ Stay disciplined with your bankroll. This goes with the last tip but remember: the tournament spans nearly a month. There will be emotional highs and lows, and it's important to remain level-headed and responsible with your bankroll.
Have Fun and Bet Responsibly
March Madness is one of the most thrilling events in sports and can be enjoyed further with responsible gaming. A standard bet should be 1% to 2% of your entire bankroll, meaning if you have $1,000 bankroll, a standard wager would be $10 or $20 depending on your unit size. Betting responsibly is key to enjoying March Madness when it comes to betting on all the games.
Guides Every Bettor Needs to Read
NCAAB Betting: How to Bet on College Basketball
Betting NCAA College Basketball Over/Under Odds and Markets
When is March Madness?
March Madness starts on March 14-15 with the First Four games and ends on April 3 with the national title contest.
When and Where are the First and Second Rounds?
The first round takes place on March 16-17, while the second round is from March 18-19. These games will take place across the country, including in Birmingham, Ala., Des Moines, Iowa, Orlando, Sacramento, Calif., Albany, N.Y., Columbus, Ohio, Denver and Greensboro, N.C.
When is the Sweet 16?
The Sweet 16 will be played on March 23-24.
When is the Elite Eight?
Elite Eight games will be held on March 25-26.
When and Where is the Final Four?
The Final Four is taking place at the NRG Stadium in Houston with games tipping off on April 1.
When and Where is the National Championship?
Houston will host the national title game on April 3.
Can I bet on March Madness?
Yes, if you live in one of the states with a form of legalized sports betting.
Who is the favorite to win the 2023 title?
Early odds have Duke and Kentucky at +1000, Arkansas, Gonzaga and Kansas at +1200 and North Carolina and UCLA at +1400.
Hi, I'm Bryce Derouin
"Bryce joined Gambling.com after spending the last nine years covering high school and college sports throughout Michigan. He's served as the primary Division II beat writer for Grand Valley State and Michigan Tech. Along with his newspaper background, he's worked in marketing and recently co-founded Upbeat — the Upper Peninsula of Michigan's first subscription-based prep sports website. His favorite sports to gamble on are football, basketball, baseball and Formula 1."


PRINCE RUPERT- Nathan Cullen (Skeena- Bulkley Valley) said this afternoon's announcement that the BC government will fund a temporary new bus service to provide limited long distance public transportation across the top of the province will bring much-needed peace of mind to communities and residents.
"This interim measure buys everyone time to work together to come up with a permanent solution for long haul passenger transportation service in northern BC," Cullen said.
The provincial government today announced the launch of a 12-month interim long-haul northern bus service to fill the gap Greyhound will leave behind with the elimination of its northern BC routes on June 1.
Dubbed BC Bus North, the short-term transport program starts Monday, June 4 and provides two round-trips per week between Prince Rupert and Prince George, as well as connector services north and east of Prince George.
In February, days after the BC Passenger Transportation Board approved Greyhound's bid to eliminate all its northern BC routes due to increased regional transit and ride sharing, Cullen called on First Nations, municipal and regional governments to join with him to press the federal government for a northern BC bus service.
Cullen said these efforts will continue as the BC government works with communities for a permanent solution to public passenger transportation options for in the North.