text:
224. Indian director Rahul Jain's debut, long-documentary film "Machines" (2016): Hard-hitting and real perspective of modern India
India produces some of the world's most attractive textiles that contribute to making lives in India and elsewhere colourful and comfortable-- whether it be the clothes one wears or the cloth-based furnishings in one's dwellings. Few realize the oppressive conditions in which textile printing workers in India toil to make the lives of billions of diverse people across the world happy and content. Machines and human beings together contribute to those lovely printed textiles. The contribution of human beings in the process is rarely in the limelight. Toiling within dingy factories, these human beings gradually become dehumanized and mechanical in their actions in their sheer desperation to earn a regular income to keep themselves afloat above the abject line of hunger and poverty. They become machines not out of choice but more from a lack of choice.
Film director Rahul Jain's honest perspective is not focussed on the machines that manufacture and print the textiles but more on the faceless tens of thousands of workers, exploited and dehumanized to work like machines for extended work hours, deprived of basic rights of hygiene, medical safety, statutory limitations of working hours and legal age and, of course, fair compensation for their time and toil. The film Machines underscores the no-win situation of migrant workers within India caught between poverty and survival, in the clutches of heartless contractors and factory owners, who spin profits for themselves sitting in contrasting distant cosy comfort.
Cinematography ( Rodrigo Villanueva) picked up two
important international awards
Machines has won several accolades worldwide. Apart from winning the Golden Eye award at the Zurich Film Festival in 2017 for the Best International Documentary Film, it picked up the cinematography award at the Sundance film festival, the Silver Gateway award at the Mumbai film festival, the best cinematography award of the International Documentary Association and three awards/prizes at the Thessaloniki documentary festival. What is it that makes Machines tick?
Diegetic sound recorded and mixed by the Indo-German crew
is laudable
Machines could have been made in diverse ways. Mr Jain could have opted to make a film contrasting machines and human beings with music matching the visuals on the lines of the Dutch maestro Bert Haanstra's 11-minute Oscar-winning wordless sublime film Glass (1958) on the Dutch glass factories. Jain's film consciously does not use music—his attempt was not to capture the beauty, but the sweat and grime of the workers much in contrast with the workers in Europe. Machines could have been made without words to mirror the French director Louis Malle's Humain, Trop Humain (Human, all too human) (1974), which is roughly the same length as Machines. That French film looked at the Citroen automobile factories in France and compared the human workers with the machines on the assembly line without words spoken except for brief pitches of the sales staff selling the cars. There are commonalities between Jain's and Malle's film: same length, human workers who appear and work like machines, and no music. The big difference in Machines compared to the two European filmmakers is that the punch of the Indian film comes from the honest spoken lines of the workers captured by the camera replacing the silence of the European works. Malle probably thought that he conveyed a lot by choosing as the title of his film to be same as Nietzche's last book which appeared to revise all his earlier written works. But little did Malle realize that all filmgoers need not be as well read as he was to make the bigger connection beyond what was obvious within the film's visuals and sounds of the factory.
Spoken words matter in the film
Words when spoken in Machines sock you on the jaw. The workers have fled their villages because incomes from crops are undependable compared to grimy, sleep-deprived, and low-paid work that in sharp contrast can be depended on as steady income. It is steady as long as you don't upset the apple cart by protesting the raw deal meted out by the contractors and the factory owners.
The few spoken words are stronger than the visuals. The workers state they have never seen the factory's owners—but the owners watch them on closed circuit TV in comfortable offices. The workers can't afford to buy cigarettes and instead ingest the cheaper semi-dry mix of raw tobacco and slaked lime locally called khaini while the factory owner ironically justifies the low wages as being more than double of what it was 10 years before, especially when workers were comparatively more committed to their work than today, casting a blind eye to the rising costs of living. (Khaini is proven to be injurious to health as much as it is to work with chemicals and dyes without adequate physical protection.) Equally disturbing is the logic of a teenage boy (it is illegal for children to work in factories in India) who claims that working at his age would develop him into a superior and sharper worker when he grows up compared to others who didn't have his experience. Or of another boy who reaches the gates of the factory each day and wishes soon after entering it that he could run out of the factory from another gate but chooses not to. More disturbing are the statements of a worker that any potential unionist seeking better compensation and hygienic conditions would be knocked off, while fearfully looking over his shoulder if someone heard him make that statement.
A factory worker reminiscent of a
Thomas Hardy novel
The critical decision that goes in the favour of Machines is that the spoken words are not preceded by questions. Questions don't matter. Those have to be imagined. When the workers do ask inconvenient questions of the filmmakers, the answers too are not heard. The film as the finished product is the answer. The brief silence before the end credits is loud and punchy.
After hard labour, a brief nap in the factory
The crucial bit beyond making of the film was revealed by the young talented director—the film having won all the global awards is yet to be widely seen within India because it is awaiting a Censor Certificate from the Government. Few can deal with truth, fewer with injustice. Economic growth for those who matter is the mantra of the day. If the film is indeed seen widely, the question asked by the workers at the end of the film would be answered. Nietzsche could be smiling in his grave.
P.S. Director Rahul Jain, who grew up near a family owned small textile mill in India and studied in the US will soon be teaching at a prominent US University and hopefully continue to make hard hitting films. Terrence Malick used to teach at MIT. Both wear similar hats.
Posted by Jugu Abraham at 12:33 AM
Labels: Finland , Germany , India , Mumbai International Film Festival winner , Sundance winner , Thessaloniki winner , Zurich winner
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224. Indian director Rahul Jain's debut, long-docu...
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\section{Introduction}

CAT(0) spaces are generalizations of nonpositively curved Riemannian manifolds. Nonsmooth spaces, which are also called singular spaces, may occur in CAT(0) spaces. A classical example of a singular CAT(0) space is a tree, which is an one-dimensional graph without loops. The vertices are the singular set. A CAT(0) space could be no proper, that is, not locally compact. The properness of a tree depends on whether it is locally finite. Actually a locally infinite tree is the simplest example for CAT(0) spaces which are not locally compact. The first part of this paper will focus on parabolic isometries of complete CAT(0) spaces which may be no proper.

Let $M$ be a complete CAT(0) space. An isometry $\gamma$ of $M$ is a map $\gamma:M \to M$ which satisfies $dist(\gamma \cdot p, \gamma \cdot q)=dist(p,q)$, for all $p,q\in M$. The set of isometries on a metric space is a group. An isometry can be classified as elliptic, hyperbolic, or parabolic. An isometry is called \textsl{elliptic} if it has a fixed point in $M$. Any finite order element in the isometry group of a complete CAT(0) space is always elliptic (see lemma \ref{fpt} in section \ref{np}). An isometry $\gamma$ is called \textsl{hyperbolic} if there exists a geodesic line $c: (-\infty,+\infty)\rightarrow M$ such that $\gamma$ acts on $c(\mathbb{R})$ by a non-trivial translation. The fundamental group of a closed nonpositively curved Riemannian manifold consists of hyperbolic isometries except the unit. If an isometry is neither elliptic nor hyperbolic, then we call it to be \textsl{parabolic}. Parabolic isometries may occur in the fundamental group of an open nonpositively curved Riemannian manifold.

Let $\gamma$ be an isometry of a complete CAT(0) space $M$. We define the \textsl{translation length} $|\gamma|$ of $\gamma$  as
\[|\gamma|:=\inf_{p\in M} dist(\gamma \cdot p,p).\]
From the definition of translation length, $|\gamma|$ may be not achieved. If $|\gamma|$ is achieved in $M$, $|\gamma|=0$ corresponds to the elliptic case, and $|\gamma|>0$ corresponds to the hyperbolic case. If $|\gamma|$ can not be achieved in $M$, $\gamma$ is parabolic. One can see more details in \cite{BH,BGS}. Let us look at the following two examples. Let $\mathbb{H}^2$ be the upper half plane endowed with the hyperbolic metric and define $\gamma :\mathbb{H}^2\rightarrow \mathbb{H}^2$ to be $\gamma\cdot(x,y)=(x+1,y)$. It is easy to see that $|\gamma|=0$, which can not be achieved in $\mathbb{H}$. Thus, $\gamma$ is parabolic. Similarly consider $\mathbb{R}\times \mathbb{H}^2$ and define $\gamma :\mathbb{R}\times \mathbb{H}^2\rightarrow \mathbb{R}\times \mathbb{H}^2$ to be $\gamma\cdot(z,(x,y))=(z+1,(x+1,y))$. It is easy to see that $\gamma$ is parabolic and $|\gamma|=1$. So parabolic isometry with positive translation length may occur in CAT(0) spaces.

A visible CAT(0) space, introduced by Eberlein and O'Neill in \cite{EO}, needs the space to be more curved. In some sense it means that for any two different points at ``infinity", they can be viewed from each other along the space, which can not happen in $\mathbb{R}^{n}$. We call a complete CAT(0) space $M$ is \textsl{visible} if for any $x\neq y \in M(\infty)$ there exists a geodesic line $c:(-\infty,+\infty)\to M$ such that $c(-\infty)=x$ and $c(+\infty)=y$, where $M(\infty)$ is the visual boundary of $M$ consisting of asymptotic geodesic rays (see \cite{BH} for details). In particular, a complete CAT(0) space, whose visual boundary is empty, is a visible space. When the visual boundary is not empty, classical examples for visible spaces include trees, which are singular, and complete simply connected Riemannian manifolds whose sectional curvatures are bounded above by negative numbers, which are smooth.

For the example $\mathbb{R}\times \mathbb{H}^2$ above which contains parabolic isometries with positive translation lengths, there exists embedding totally geodesical Euclidean planes in it. In particular, $\mathbb{R}\times \mathbb{H}^2$ is not visible. It was shown in \cite{Wu12} that any parabolic isometry of a complete proper visible CAT(0) space has zero translation length, which was also served as a bridge to find a counterexample for the Eberlein conjecture which said that \textsl{a complete open manifold $M$ with sectional curvature $-1\leq K_{M}\leq 0$ and finite volume is visible if the universal covering space $\tilde{M}$ of $M$ contains no imbedded flat half planes}. And the proof in \cite{Wu12} requires that the space is proper. So it is natural to ask
\begin{question}\label{gq-1}
Does every parabolic isometry of a complete visible CAT(0) space have zero translation length?
\end{question}

In general, the answer to the question above is not true, one can see the example in remark \ref{tushar}. If we just consider complete visible CAT(0) spaces, which may be no proper, we still would like to know what kind of spaces such that parabolic isometries have zero translation length. A semi-uniformly visible CAT(0) space, roughly speaking, means a visible CAT(0) space which can not contain flat strips with arbitrary large widths, one can see definition \ref{dosuvs} for details. The motivation that we introduce semi-uniformly visible CAT(0) spaces is that the counterexample in remark \ref{tushar} for question \ref{gq-1} contains flat strips with arbitrary large widths. Proposition \ref{sosuv} in section 2 tells that both complete proper visible CAT(0) spaces and complete Gromov-hyperbolic CAT(0) spaces are semi-uniformly visible CAT(0) spaces. The following theorem gives an affirmative answer to question \ref{gq-1} for semi-uniformly visible CAT(0) spaces.

\begin{theorem}\label{mt-1}
Let $M$ be a complete semi-uniformly visible CAT(0) space. Then any parabolic isometry has zero translation length. i.e., for any parabolic isometry $\gamma$ on $M$ we have $|\gamma|=0$.
\end{theorem}

Buyalo in \cite{BU} proved theorem \ref{mt-1} when $M$ is a complete Gromov-hyperbolic CAT(0) space. Since a complete Gromov-hyperbolic CAT(0) space is a complete semi-uniformly visible CAT(0) space (see proposition \ref{gmhiv} in section \ref{np}), theorem \ref{mt-1} gives a generalization. And the method here is different as what Buyalo did in \cite{BU}.  For the manifold case, Bishop and O'Neill in \cite{BO} proved that any parabolic isometry of a complete simply connected manifold M with sectional curvature $K_{M}\leq -1$ has zero translation length. In \cite{HH} Heintze and Hof used the geometry on horospheres to give a new proof for the manifold case.
\newline

As a consequence of theorem \ref{mt-1}, we have the following classification on isometries of a complete semi-uniformly visible CAT(0) space in terms of translation lengths.

\begin{theorem}\label{class}
Let $M$ be a complete semi-uniformly visible CAT(0) space, and $\gamma$ be an isometry of $M$. Then,

(1): $\gamma$ is elliptic if and only if $\gamma$ has a fixed point in $M$.

(2): $\gamma$ is hyperbolic if and only if $|\gamma|>0$.

(3): $\gamma$ is parabolic if and only if $|\gamma|=0$ and $\gamma$ does not have any fixed point in $M$.
\end{theorem}

Let $S_{g,n}$ be a surface of $g$ genus with $n$ punctures, and $\mathbb{M}(S_{g,n})$ be the moduli space of $S_{g,n}$. There are a lot of canonical metrics on $\mathbb{M}(S_{g,n})$ like the Teichm\"uller metric, the Weil-Petersson metric, and so on (see \cite{IT92}). Kravetz in \cite{Kravetz} proved the Teichm\"uller metric has negative curvature. However, a mistake in the proof of Kravetz's theorem was found by Linch in \cite{Linch}. Masur in \cite{Masur75} proved that the Teichm\"uller metric is not nonpositively curved except several cases. Although the Weil-Petersson metric is negatively curved (see \cite{Wolpert86}), but not complete (see \cite{Wolpert75}). The McMullen metric, constructed by McMullen in \cite{McMullen00}, is a complete K\"ahler-hyperbolic metric in the sense of Gromov. Liu, Sun and Yau in \cite{LSY05} perturbed the Weil-Petersson metric to construct the so-called perturbed Ricci metric, which is complete and whose Ricci curvatures are pinched by two negative numbers. There is a question in Brock-Farb's paper \cite{BF} which states
\begin{question}[Brock-Farb-McMullen]
Does $\mathbb{M}(S_{g,n})$ admit a complete, nonpositively curved Riemannian metric?
\end{question}

Since $\mathbb{M}(S_{g,n})$ is an orbifold, here a Riemannian metric on $\mathbb{M}(S_{g,n})$ means a Riemannian metric on the Teichm\"uller space $\mathbb{T}(S_{g,n})$, the universal covering space of $\mathbb{M}(S_{g,n})$, on which the natural action of the mapping class group $\Mod_{S_{g,n}}$ on $\mathbb{T}(S_{g,n})$ is an isometric action.

We call a Riemannian metric on $\mathbb{M}(S_{g,n})$ is called \textsl{visible CAT(0) Riemannian} if the sectional curvature of $\mathbb{T}(S_{g,n})$ is nonpositive and $\mathbb{T}(S_{g,n})$ is visible. If $3g+n\leq 4$ and $\mathbb{M}(S_{g,n})$ has positive dimension, then $(g,n)$ must be one of $\{(1,0),(1,1),(0,4)\}$. For these three cases, it is well known that the Teichm\"uller metric on $\mathbb{M}(S_{g,n})$ is a complete hyperbolic metric (see \cite{FW10}). In particular, the Teichm\"uller metric on $\mathbb{M}(S_{g,n})$ is complete visible CAT(0) Riemannian. Hence, we can always assume that $3g+n\geq 5$ for Brock-Farb-McMullen's question. The following theorem answers Brock-Farb-McMullen's question in the visible case.

\begin{theorem}\label{mt-2}
If $3g+n\geq 5$, then $\mathbb{M}(S_{g,n})$ admits no complete visible CAT(0) Riemannian metric.
\end{theorem}

Since a complete Riemannian manifold with sectional curvatures bounded above by a negative number is a visible CAT(0) manifold, the following theorem follows immediately from theorem \ref{mt-2}.
\begin{theorem}\label{ccnba}
If $3g+n\geq 5$, $\mathbb{M}(S_{g,n})$ admits no complete Riemannian metric such that the sectional curvature $K(\mathbb{M}(S_{g,n}))\leq -1$.
\end{theorem}

Ivanov in \cite{Iv88} showed that $\mathbb{M}(S_{g,n})$ $(3g+n\geq 5)$ admits no complete, finite volume Riemannian metric whose sectional curvature is pinched by two negative numbers. McMullen in \cite{McMullen00} stated that $\mathbb{M}(S_{g,n})$ $(3g+n\geq 5)$ admits no complete Riemannian metric whose sectional curvature is pinched by two negative numbers, which was proved by Brock and Farb in \cite{BF}. Moreover, the authors in \cite{BF} showed that  $\mathbb{M}(S_{g,n})$ $(3g+n\geq 5)$ admits no complete, finite volume Riemannian metric such that the universal covering space is Gromov-hyperbolic. For the visible and finite volume case, one can refer to \cite{Wu12}. What is new for theorem \ref{mt-2} and \ref{ccnba} is that there is no finite volume restriction. For related topics, one can also see \cite{Ji02, KM99, KN04, LSY04, McP99, MW95}.
\newline

Throughout this paper, we always assume that the geodesics use arc-length parameters.
\newline

\textbf{Plan of the paper.} In section \ref{np} we set out necessary backgrounds, and prove some basic properties on CAT(0) spaces and mapping class groups, which will be applied in subsequent sections. Section \ref{mt12} establishes theorem \ref{mt-1} and theorem \ref{class}. Theorem \ref{mt-2} is proved in section \ref{somp}.

\ \
\newline
\textbf{Acknowledgments.} The author is indebted to Andy Putman for the discussions on the proof of theorem \ref{mt-2}, in particular for his suggestion on writing this article and correction on English for the original manuscript. Thank also to Benson Farb and Mike Wolf for the discussions on the proof of theorem \ref{mt-2}.  The author also would like to thank Tushar Das for the discussions on theorem \ref{mt-1} and sharing his idea in remark \ref{tushar}.

\section{Notations and Preliminaries}\label{np}

\subsection{CAT(0) spaces} A CAT(0) space is a geodesic metric space in which each geodesic triangle is no fatter than a triangle in the Euclidean plane with the same edge lengths.
\begin{definition}
let $M$ be a geodesic metric space. For any $a,b,c \in M$, three geodesics $[a,b],[b,c],[c,a]$ form a geodesic triangle $\Delta.$ Let $\overline{\Delta}(\overline{a},\overline{b},\overline{c})\subset \mathbb{R}^2 $ be a triangle in the Euclidean plane with the same edge lengths as $\Delta$. Let $p,q$ be points on $[a,b]$ and $[a,c]$ respectively, and let $\overline{p},\overline{q}$ be points on $[\overline{a},\overline{b}]$ and $[\overline{a},\overline{c}]$ respectively, such that $dist_{M}(a,p)=dist_{\mathbb{R}^2}(\overline{a},\overline{p}), dist_{M}(a,q)=dist_{\mathbb{R}^2}(\overline{a},\overline{q})$. We call $M$ a \textbf{CAT(0) space} if for all $\Delta$ the inequality $dist_{M}(p,q)\leq dist_{\mathbb{R}^2}(\overline{p},\overline{q})$ holds.
\end{definition}

Let $M$ be a complete CAT(0) space. The ideal boundary, denoted by $M(\infty)$, consists of asymptotic rays. For each point $p \in M$ and $x \in M(\infty)$, since the distance function between geodesics is convex, it is not hard to see that there exists a unique geodesic ray $c$ which represents $x$ and starts from $p$ (see \cite{BH}). We write $c(+\infty)=x$. Although a complete CAT(0) space may be singular, the definition of CAT(0) spaces can guarantee that the notation of the angle, like the smooth case, still make sense (see \cite{BH}).  Given two points $x,y$ in $M(\infty)$. Let $\angle_{p}(x,y)$ denote the angle at $p$ between the unique geodesics rays which issue from $p$ and lie in the classes $x$ and $y$ respectively. The angular metric is defined to be $\angle(x,y):=\sup_{p\in M}{\angle_{p}(x,y)}$. It is easy to see that $\angle(x,y)=0$ if and only if $x=y$. On a complete visible CAT(0) space $M$, for any $x\neq y \in M(\infty)$, $\angle(x,y)=\pi$. So the angular metric gives a discrete topology on the ideal boundary of a complete visible CAT(0) space.

The following lemma will be used in next section, which gives us a way to compute the angular metric.
\begin{lemma}\label{pro:angular}
Let $M$ be a complete CAT(0) space with a basepoint $p$. Let $x,y \in M(\infty)$ and $c,c'$ be two geodesic rays with $c(0)=c'(0)=p,$ $c(+\infty)=x$ and $c'(+\infty)=y$. Then,
\begin{eqnarray*}
2\sin(\frac{\angle(x,y)}{2})=\lim_{t\rightarrow +\infty}\frac{dist(c(t),c'(t))}{t}.
\end{eqnarray*}
\end{lemma}
\begin{proof}
See proposition 9.8 on page 281 in chapter II.9 of \cite{BH}.
\end{proof}

\subsection{Product} Let $X_{1}$ and $X_{2}$ be two metric spaces. The product $X=X_{1}\times X_{2}$ has a natural metric which is called the product metric. Let $\gamma_{i}$ be an isometry of $X_{i}$ $(i=1,2)$. It is obvious that $\gamma=(\gamma_{1},\gamma_{2})$ is an isometry of $X$ under the natural action. The following lemma tells when the converse is true.
\begin{lemma}\label{product}
Let $X=X_{1}\times X_{2}$. Then an isometry $\gamma$ on $X$ decomposes as $(\gamma_{1},\gamma_{2})$, with $\gamma_{i}$ be an isometry of $X_{i}$ $(i=1,2)$, if and only if, for every $x_{1}\in X_{1}$, there exists a point denoted $\gamma_{1}\cdot x_{1} \in X_{1}$ such that $\gamma \cdot (\{x_{1}\}\times X_{2})=\{\gamma_{1}\cdot x_{1}\}\times X_{2}$.
\end{lemma}
\begin{proof}
See proposition 5.3 on page 56 in chapter I.5 of \cite{BH}.
\end{proof}

The following product decomposition theorem will be applied for several times in this article.
\begin{proposition}\label{pdt}
Assume that $M$ is a complete CAT(0) space. Let $c:\mathbb{R} \to M$ be a geodesic line and $P_{c}$  be the set of geodesic lines which are parallel to $c$. Then, $P_{c}$ is isometric to the product $P'_{c}\times \mathbb{R}$ where $P'_{c}$ is a closed convex subset in $M$.
\end{proposition}
\begin{proof}
See theorem 2.14 on page 183 in chapter II.2 of \cite{BH}.
\end{proof}

\subsection{Semi-uniformly visible CAT(0) spaces}

Recall a metric space $M$ is called \textsl{Gromov-hyperbolic} if there exists a $\delta>0$ such that every geodesic triangle is $\delta$-thin. Where a \textsl{$\delta$-thin geodesic triangle} means that each of its sides is contained in the $\delta$-neighborhood of the union of the other two sides. A $\mathbb{R}$-tree is a Gromov-hyperbolic space which holds for any $\delta>0$. For more details one can see \cite{BH, Gromov87}. The following proposition tells that Gromov-hyperbolic CAT(0) spaces are stronger than visible CAT(0) spaces.
\begin{proposition}\label{gmhiv}
Every complete Gromov-hyperbolic CAT(0) space is a visible CAT(0) space.
\end{proposition}

\begin{proof}
See proposition 10.1 in \cite{BU}. For the proper case, one can also see proposition 1.4 in chapter III.H of \cite{BH}.
\end{proof}

Before we define semi-uniformly visible CAT(0) spaces, let us consider the following two examples.

\begin{example}\label{exmp1}
For every positive integer $n$, let $I_{n}$ be the vertical segment in $\mathbb{R}^2$ as follows
\[I_{n}:=\{(x,y)\in \mathbb{R}^2; \ \ x=\frac{1}{n}, 0\leq y \leq n\}\]
and $I$ be the horizontal segment with unit length as follows
\[I:=\{(x,y)\in \mathbb{R}^2; \ \ 0\leq x \leq 1, y\equiv0\}.\]
Consider the space
\[M:=((\bigcup_{n=1}^{+\infty} I_{n})\bigcup I,ds^2)\]
where $ds^2$ is the induced metric from $\mathbb{R}^2$.

It is easy to see that $M$ is a complete tree. So $M$ is a complete Gromov-hyperbolic space. By proposition \ref{gmhiv}, $M$ is also a complete CAT(0) visible space. It is not hard to see that $M$ is not locally compact (around the point $(x,y)=(0,0))$, unbounded, and the visual boundary of $M$ is empty.

\end{example}

Let $M$ be a complete CAT(0) space. We call $M$ has \textsl{an infinite-flat-strip} if there exists a totally geodesical convex subset $U\times \mathbb{R} \subseteq M$ where $U$ is unbounded. From proposition \ref{pdt} we can always assume that $U$ is closed convex. The following example tells that the convex hull of an infinite-flat-strip may not contain a half flat plane.

\begin{example}\label{exmp2}
Let $M$ be the metric space in example \ref{exmp1}. Consider the product space

\[N:=M\times \mathbb{R}\]
which is endowed with the product metric.

It is obvious that $N$ is a complete CAT(0) space, but not locally compact. Actually the visual boundary $N(\infty)$ of $N$ consists of two points, which are the positive and negative directions of the $\mathbb{R}$ component. In particular they can be joined by a geodesic line in $N$. Thus $N$ is a complete visible CAT(0) space. It is easy to see that $N$ contains an infinite-flat-strip. And $N$ is not a Gromov-hyperbolic CAT(0) space.

\end{example}
Example \ref{exmp2} tells us that a complete visible CAT(0) space may contain an infinite-flat-strip. The following definition will exclude complete CAT(0) spaces with infinite-flat-strip.

\begin{definition}\label{dosuvs}
Let $M$ be a complete CAT(0) space. $M$ is called \textsl{semi-uniformly visible} if $M$ is visible and $M$ can not contain an infinite-flat-strip.
\end{definition}

It is obvious that any complete simply connected Riemannian manifold whose sectional curvatures are bounded by a negative number is a semi-uniformly visible CAT(0) space. The following proposition tells that semi-uniformly visible CAT(0) spaces contain a lot of standard visible spaces.
\begin{proposition}\label{sosuv}
(1): Every complete proper visible CAT(0) space is a semi-uniformly visible CAT(0) space.\\
(2): Every complete Gromov-hyperbolic CAT(0) space is a semi-uniformly visible CAT(0) space.
\end{proposition}

\begin{proof}
\textit{Proof of part (1):} If not. Then we can assume that $M$ is a complete proper visible CAT(0) space which contains an infinite-flat-strip $U\times \mathbb{R}$ where $U$ is unbounded. Let $c:(-\infty,+\infty)\to M$ be the geodesic line $x_{0}\times \mathbb{R}$ where $x_{0}\in U$. And let $P_{c}$ be the set of geodesic lines which are parallel to $c$. By proposition \ref{pdt}, $P_{c}$ is isometric to the product $P'_{c}\times \mathbb{R}$ where $P'_{c}$ is a closed convex subset in $M$. Hence $U\subseteq P'_{c}$. Since $U$ is unbounded, $P'_{c}$ is also unbounded. Thus $P'_{c}$ a complete, unbounded, proper space.
The Arzel\`a Ascoli theorem would guarantee that there exists a geodesic ray $d:[0,+\infty)\to P'_{c}$. Hence $M$ contains a flat half plane $[0,+\infty)\times \mathbb{R}$ in $M$ which is impossible because $M$ is visible.
\newline
\\
\textit{Proof of part (2):} By proposition \ref{gmhiv} it suffices to show that every complete Gromov-hyperbolic CAT(0) space is semi-uniform. Assume not. Then we can assume that $M$ is a complete Gromov-hyperbolic CAT(0) space which contains an infinite-flat-strip $U\times \mathbb{R}$ where $U$ is unbounded. Let $\delta>0$ be the number such that every geodesic triangle in $M$ is $\delta$-thin. Let $c:(-\infty,+\infty)\to M$ be the geodesic line $x_{0}\times \mathbb{R}$ where $x_{0}\in U$. And let $P_{c}$ be the set of geodesic lines which are parallel to $c$. By proposition \ref{pdt}, $P_{c}$ is isometric to the product $P'_{c}\times \mathbb{R}$ where $P'_{c}$ is a closed convex subset in $M$. Hence $U\subseteq P'_{c}$.  Since $U$ is unbounded, we can find a flat strip $[0,k]\times \mathbb{R}$ with width $k$ where $k$ is an arbitrary positive number. If we choose $k$ to be greater enough than $\delta$, it is not hard to find a geodesic triangle $\Delta$ in $[0,k]\times \mathbb{R}$ such that $\Delta$ is not $\delta$-thin, which is a contradiction.

\end{proof}

\begin{remark}\label{suv-1}
From part (1) of the proposition above, if the space $M$ is complete, locally compact, CAT(0) space, then semi-uniform visibility is equvilent to visibility. Hence, the interesting aspect for semi-uniform visibility is for complete CAT(0) spaces which are not locally compact.
\end{remark}

\subsection{Horospheres in CAT(0) spaces} Let $M$ be a complete CAT(0) space and $x \in M(\infty)$. $c:[0,+\infty)\to M$ is a geodesic ray with $c(+\infty)=x$, which determines a \textsl{Busemann function} $f=f_{c}$ at $x$ given by $f(p)=\lim_{t\to +\infty}(dist(p, c(t))-t)$. $f$ is a convex function. For each $t\in \mathbb{R}$, the set $B(t):=\{q \in M, f(q)\leq t\}$ is a \textsl{horoball} at $x$, whose boundary,  a level set of $f$, is a \textsl{horosphere} at $x$. Every horoball is convex because of the convexity of $f$.  If $x$ is fixed by some parabolic isometry $\gamma$ of $M$, then each horoball at $x$ is $\gamma$-invariant. In particular each horosphere at $x$, the boundary of some horoball, is also $\gamma$-invariant. One can see more details in \cite{BH}. The following lemma will be applied to prove theorem \ref{mt-2}.

\begin{proposition}\label{hsie}
Let $M$ be a complete, simply connected, nonpositively curved Riemannian manifold and $c:[0,+\infty)\to M$ be a geodesic ray. Then every horosphere at $c(+\infty)$ is diffeomorphic to $\mathbb{R}^{k}$ where $k=dim(M)-1$.
\end{proposition}

\begin{proof}
Let $H$ be a horosphere at $c(+\infty)$ and $B$ be the horoball whose boundary is $H$. Choose a point $q$ in the complement of $B$ and consider the distance function $f:H \to (0,+\infty)$, given by $f(p)=dist(q,p)$. The idea is prove that $f$ only has one critical point on $H$, which is the intersection point of $H$ and the geodesic ray, starting at $q$, which goes to $c(+\infty)$. Then the conclusion follows from standard Morse theory. One can see lemma 3-4 on page 511 in \cite{Fukaya84} for details.
\end{proof}

\subsection{Isometries on CAT(0) spaces}
Let $M$ be a complete CAT(0) space and $\gamma$ be an isometry of $M$. If the translation length $|\gamma|$ is achieved in $M$, $\gamma$ is either elliptic or hyperbolic. The following lemma tells us any isometry on $\mathbb{R}^n$ can achieve its translation length, which will be applied in subsequent sections.
\begin{lemma}\label{R1}
Let $\gamma$ be an isometry of the Euclidean space $\mathbb{R}^n$ where $n$ is a positive integer. Then $\gamma$ is either elliptic or hyperbolic.
\end{lemma}
\begin{proof}
One possible way is to argue it by contradiction. Assume that $\gamma$ is parabolic. Then there exists a horosphere $H$ such that $\gamma$ acts on $H$ as an isometry. Since $H$ is a horosphere of $\mathbb{R}^n$, $H$ is isometric to $\mathbb{R}^{n-1}$ which is totally geodesic in $\mathbb{R}^n$. The conclusion follows by induction on the dimensions. For the details we leave it as an exercise to the reader.
\end{proof}

From definition an elliptic isometry $\gamma$ of $M$ has at least one fixed point. The following is one of the basic fixed point theorems in CAT(0) geometry.
\begin{lemma}\label{fpt}
An isometry $\gamma$ of a complete CAT(0) space $M$ has a fixed point if and only if there exists a $\gamma$-invariant bounded subset in $M$.
\end{lemma}
\begin{proof}
The idea is to prove that, for any bounded subset in $M$ there exists a unique centre point for this bounded subset, and then prove this point is fixed by $\gamma$. One can see the proof of proposition 6.7 on page 231 in chapter II.6 of \cite{BH}.
\end{proof}

Recall that the translation length $\gamma$ of an isometry of $M$ is defined as the infimum of translation of $\gamma$ over $M$. The following lemma gives us another viewpoint for the translation length of a single isometry.

\begin{lemma}\label{tl}
Let $M$ be complete CAT(0) space and $\gamma$ be an isometry of $M$. Then, for all $p\in M$, we have
\[|\gamma|=\lim_{n\rightarrow+\infty}{\frac{dist(\gamma^{n} \cdot p,p)}{n}}.\]
\end{lemma}
\begin{proof}
See lemma 6.6 on page 83 in \cite{BGS} or lemma 3.4 in \cite{Wu12}.
\end{proof}

As a direct corollary,
\begin{lemma}\label{22}
Let $M$ be complete CAT(0) space and $\gamma$ be an isometry of $M$. Then
\[|\gamma|^2=2|\gamma|.\]
\end{lemma}
\begin{proof}
Let $p \in M$. By lemma \ref{tl},
\[|\gamma^2|=\lim_{n\rightarrow+\infty}{\frac{dist(\gamma^{2n} \cdot p,p)}{n}}=2 \lim_{n\rightarrow+\infty}{\frac{dist(\gamma^{2n} \cdot p,p)}{2n}}=2|\gamma|.\]
\end{proof}

A group $G$ acting on a metric space $X$ is called \textsl{properly} if for each compact subset $K\subset X$, the set $K \cap gK$ is nonempty for only finitely many $g$ in $G$.
\begin{lemma}\label{z2ip}
Let $M$ be a complete visible CAT(0) Riemannian manifold and $G=\mathbb{Z}\oplus \mathbb{Z}$, a free abelian group of rank 2, act properly on $M$ by isometries. Then for any nontrivial $g\in G$, $g$ is parabolic.
\end{lemma}

\begin{proof}
Since $G$ acts properly on $M$, for any nontrivial $g\in G$, $g$ is either parabolic or hyperbolic. Assume that $g$ is hyperbolic. Let $Min(g):=\{p\in M, dist(g\cdot p,p)=|g|\}$. From theorem 6.8 on page 231 in chapter II.6 of \cite{BH} we know that $Min(g)$ is isometric to the product $Y\times \mathbb{R}$ on which $g$ acts trivially on the $Y$ component, where $Y$ is a closed convex subset in $M$. Since $M$ is a visible CAT(0) manifold, $Y$ is bounded, otherwise there exists a flat half plane $[0,+\infty)\times \mathbb{R}$ in $X$ which is impossible in visible CAT(0) spaces (this step needs that the space is proper). Let $h \in G$ such that the group $<g,h>$, generated by $g$ and $h$, is a free abelian group of rank 2. Since $gh=hg$, $Min(g)$ is $h$-invariant. Since $h$ is an isometry, it must send a geodesic line to another geodesic line. Thus, by lemma \ref{product}, $h$ splits as $(h_{1},h_{2})$ where $h_{1}$ is an isometry on $Y$ and $h_{2}$ is an isometry on $\mathbb{R}$. Firstly $Y$ is a complete CAT(0) space because $Y$ is closed convex in $M$. Since $Y$ is bounded, by lemma \ref{fpt}, there exists $x_{1} \in Y$ such that $h_{1} \cdot x_{1}=x_{1}$. Hence $x_{1}\times \mathbb{R}$ is $<g,h>$-invariant. Since $G$ acts properly on $M$ and $<g,h>$ is a subgroup of $G$, $<g,h>$ also acts properly on $x_{1}\times \mathbb{R}$, which is impossible because the rank of $<g,h>$ is 2.
\end{proof}

\subsection{Mapping class groups} Let $S_{g,n}$ be a Riemann surface of genus $g$ with $n$ punctures, and $\Mod_{S_{g,n}}$ be the mapping class group of $S_{g,n}$, i.e., the group of isotopy classes of self-homeomorphisms of $S_{g,n}$ which preserve the orientation and the punctures. The following proposition lists the basic properties of $\Mod_{S_{g,n}}$, which will be used.
\begin{proposition}\label{mcgp}
Let $\Mod_{S_{g,n}}$ be the mapping class group of $S_{g,n}$. Then

(1): $\Mod_{S_{g,n}}$ acts properly  on the Teichm\"uller space.

(2): $\Mod_{S_{g,n}}$ is finitely generated by Dehn-twists along simple closed curves.

(3): There exist torsion-free subgroups of finite index in $\Mod_{S_{g,n}}$.
\end{proposition}
\begin{proof}
See the details in \cite{FM}.
\end{proof}

Bestvina, Kapovich and Kleiner in \cite{BKK} defined the \textsl{action dimension} of a group $G$, denoted by $actdim(G)$, to be the minimum dimension of a contractible manifold on which $G$ properly acts. For examples, the action dimension of $\mathbb{Z}$ is $1$. The action dimension of the fundamental group of a closed hyperbolic surface is $2$.  In \cite{BKK} there is a notation, which is called the \textsl{obstructor dimension} of a group $G$, denoted by $obdim(G)$ (see \cite{BKK}).
\begin{proposition}[BKK] \label{obdim}
(1): For any group $G$, $actdim(G)\geq obdim(G)$.

(2): The obstructor dimension is a quasi-isometric invariance.

\end{proposition}
\begin{proof}
See theorem 1 in \cite{BKK} for part (1) and remark 11 in \cite{BKK} for part (2).
\end{proof}

In \cite{Despto} Z. Despotovic proved that the action dimension of the mapping class group $actdim(\Mod_{S_{g,n}})=6g-6+2n$ if $3g+n\geq 5$ (see page 6 of \cite{FW10}). In fact, she proved the obstructor dimension $obdim(\Mod_{S_{g,n}})=6g-6+2n$ and then used part (1) of proposition \ref{obdim} to conclude $actdim(\Mod_{S_{g,n}})=6g-6+2n$. The following result tells that the action dimension is preserved by subgroups of mapping class groups, up to finite index.
\begin{proposition}[Despotovic] \label{adom}
If $3g+n\geq 5$, then for any finite index subgroup $G$ of $\Mod_{S_{g,n}}$, $actdim(G)=6g-6+2n$.
\end{proposition}

\begin{proof}
Let $G$ be a subgroup of $\Mod_{S_{g,n}}$ with finite index. So $G$ endowed with a word metric is quasi-isometric to $\Mod_{S_{g,n}}$. Since the obstructor dimension of $\Mod_{S_{g,n}}$ $obdim(\Mod_{S_{g,n}})=6g-6+2n$ (see \cite{Despto}), by part (2) of proposition \ref{obdim} we have $obdim(G)=6g-6+2n$. Thus from part (1) of proposition \ref{obdim} we know that $actdim(G)\geq 6g-6+2n$.

On the other hand it is obvious that $actdim(G)\leq 6g-6+2n$ because $G$ acts properly on the Teichm\"uller space which is contractible. Hence $actdim(G)=6g-6+2n$.

\end{proof}

\section{Proofs of theorem \ref{mt-1} and theorem \ref{class}}\label{mt12}
Before we go to prove theorem \ref{mt-1}, let us control the size of the fixed points of parabolic isometries.

\begin{proposition}\label{lt1}
Let $M$ be a complete semi-uniformly visible CAT(0) space and $\gamma$ be a parabolic isometry on $M$. Then, the number of the fixed points of $\gamma$ satisfy
\[\#\{x \in M(\infty): \gamma \cdot x=x\}\leq 1.\]
\end{proposition}

\begin{proof}
Assume not. That is $|Fix(\gamma)|\geq 2$, where $Fix(\gamma)$ denotes the set of the fixed points of $\gamma$ in $M(\infty)$. Let $x\neq y \in Fix(\gamma)$. Since $M$ is visible, there exists a geodesic line $c: \mathbb{R}\to M$ such that $c(-\infty)=x$ and $c(+\infty)=y$. Let $P_{c}$ be the set of geodesic lines which are parallel to $c$. By proposition \ref{pdt}, $P_{c}$ is isometric to the product $P'_{c}\times \mathbb{R}$ where $P'_{c}$ is a convex subset in $M$. Since $M$ is semi-uniformly visible, $P'_{c}$ is bounded, otherwise there exists an infinite-flat-strips in $M$, which contradicts the definition of semi-uniformly CAT(0) spaces.

Since $c(-\infty),c(+\infty) \in Fix(\gamma)$, $\gamma \cdot (c(\mathbb{R}))$ is also geodesic line which is parallel to $c$. In particular, $P_{c}=P'_{c}\times \mathbb{R}$ is a $\gamma$-invariant subset in $M$. From lemma \ref{product}, $\gamma$ splits as $(\gamma_{1},\gamma_{2})$ where $\gamma_{1}$ is an isometry on $P'_{c}$ and $\gamma_{2}$ is an isometry on $\mathbb{R}$. Firstly since $P'_{c}$ is convex in $M$, $P'_{c}$ is also a CAT(0) space. Since $P'_{c}$ is bounded, by lemma \ref{fpt}, there exists $x_{1} \in P'_{c}$ such that $\gamma_{1} \cdot x_{1}=x_{1}$. By lemma \ref{R1}, $\gamma_{2}$ is either elliptic or hyperbolic.

\textsl{Case 1: $\gamma_{2}$ is elliptic.}

There exists $x_{2}\in \mathbb{R}$ such that $\gamma_{2}\cdot x_{2}=x_{2}$. In particular $\gamma=(\gamma_{1},\gamma_{2})$ fixes the point $(x_{1},x_{2})$, which means that $\gamma$ is elliptic, which contradicts the assumption that $\gamma$ is parabolic.

\textsl{Case 2: $\gamma_{2}$ is hyperbolic.}

Since $\gamma_{1}\cdot x_{1}=x_{1}$, $\gamma=(\gamma_{1},\gamma_{2})$ acts on the line $x_{1}\times \mathbb{R}$ as a translation. From the definition of hyperbolic isometries, $x_{1}\times \mathbb{R}$ is an axis for $\gamma$. In particular, $\gamma$ is hyperbolic, which also contradicts our assumption that $\gamma$ is parabolic.

\end{proof}

\begin{remark}\label{2}
If $M$ is proper, i.e., locally compact,  it is not hard to prove the existence of fixed points for parabolic isometry (see proposition 8.25 on page 275 in chapter II.8 of \cite{BH}). By part (1) of proposition \ref{sosuv}, proposition \ref{lt1} tells that that the fixed point of any parabolic isometry of a complete proper visible CAT(0) space is unique. One can also see \cite{FNS}.
\end{remark}

\begin{remark}\label{efgh}
If $M$ is a complete Gromov-hyperbolic CAT(0) space (may be no proper!), the existence of fixed points of any parabolic isometry is also guaranteed (see theorem 0.3 in \cite{BU} or one can also see \cite{Ghys}). I am grateful to Tushar Das for this point.
\end{remark}

\begin{proposition}\label{gt2}
Let $M$ be a complete CAT(0) space and $\gamma$ be an isometry on $M$. If $|\gamma|>0$, then
\[\#\{x \in M(\infty): \gamma \cdot x=x\}\geq 2.\]
\end{proposition}
\begin{proof}
Since $|\gamma|>0$, $\gamma$ is either hyperbolic or parabolic.

If $\gamma$ is hyperbolic, let $c:(-\infty,+\infty) \to M$ be an axis for $\gamma$. The conclusion follows from the fact that $\{c(-\infty),c(+\infty)\}$ belongs to $\{x \in M(\infty): \gamma \cdot x=x\}$.

If $\gamma$ is parabolic. Since $|\gamma|>0$, a special case of a result of Karlsson and Margulis \cite{KM} shows that $\gamma$ has a unique fixed point $x\in M(\infty)$ such that for every $p \in M$ and every geodesic ray $c:[0,+\infty)\to M$ with $c(0)=p$ and $c(+\infty)=x$ we have
\[\lim_{n\rightarrow +\infty}{\frac{dist(\gamma^n \cdot p, c(n |\gamma|))}{n}}=0.\]

Similarly $|\gamma^{-1}|=|\gamma|>0$, $\gamma^{-1}$ has a unique fixed point $y\in M(\infty)$ such that for every $p \in M$ and every geodesic ray $c':[0,+\infty)\to M$ with $c'(0)=p$ and $c'(+\infty)=y$ we have
\[\lim_{n\rightarrow +\infty}{\frac{dist(\gamma^{-n} \cdot p, c'(n |\gamma|))}{n}}=0.\]

By triangle inequality, we have
\[\lim_{n\rightarrow+\infty}{\frac{dist(c(n |\gamma|), c'(n |\gamma|))}{n}}=\lim_{n\rightarrow+\infty}\frac{dist(\gamma^{n} \cdot p,\gamma^{-n}\cdot p)}{n}.\]

Since $\gamma$ is an isometry, by lemma \ref{pro:angular} and lemma \ref{tl}, we have
\[2 |\gamma| \sin(\frac{\angle(c(+\infty),c'(+\infty))}{2})=|\gamma^2|.\]

From lemma \ref{22}, we know that
\[2 |\gamma| \sin(\frac{\angle(c(+\infty),c'(+\infty))}{2})=2|\gamma|.\]

Since $|\gamma|\neq 0$, $\angle(c(+\infty),c'(+\infty))=\pi \neq 0$. That is $\angle(x,y)\neq 0$. Since $x, y \in Fix(\gamma)$, we have
\[\#\{x \in M(\infty): \gamma \cdot x=x\}\geq 2.\]

\end{proof}

\begin{remark}
The result of Karlsson and Margulis above was also used by M. Bridson in \cite{Bridson10} to show that any Dehn-twist has zero translation length when a mapping class group of a closed surface with genus $g\geq 3$ acts on a complete CAT(0) space by isometries.
\end{remark}

Now we are ready to prove theorem \ref{mt-1} and theorem \ref{class}.

\begin{proof}[Proof of theorem \ref{mt-1}]
Suppose not, we assume that $|\gamma|>0$. From proposition \ref{gt2} we know that
\[\#\{x \in M(\infty): \gamma \cdot x=x\}\geq 2.\]
On the other hand, since $M$ is complete semi-uniformly visible and $\gamma$ is parabolic, by proposition \ref{lt1}, we have
\[\#\{x \in M(\infty): \gamma \cdot x=x\}\leq 1\]
which is a contradiction.

\end{proof}

Since a complete proper visible CAT(0) space is semi-uniformly visible (see part (1) of proposition \ref{sosuv}), theorem \ref{mt-1} implies
\begin{theorem}(\cite{Wu12})
Let $M$ be a complete proper visible CAT(0) space. Then any parabolic isometry of $M$ has zero translation length. i.e., for any parabolic isometry $\gamma$ of $M$ we have $|\gamma|=0$.
\end{theorem}

Since a complete Gromov-hyperbolic CAT(0) space is semi-uniformly visible (see part (2) of proposition \ref{sosuv}), by theorem \ref{mt-1} we immediately obtain
\begin{theorem}[Buyalo]
Let $M$ be a complete Gromov-hyperbolic CAT(0) space. Then any parabolic isometry of $M$ has zero translation length. i.e., for any parabolic isometry $\gamma$ of $M$ we have $|\gamma|=0$.
\end{theorem}

\begin{remark}
We call a manifold $M$ is tame if $M$ is the interior of some compact manifold $\overline{M}$ with boundary. Phan conjectures in \cite{Phan} that let $M$ be a tame, finite volume, negatively curved manifold, then $M$ is not visible if the fundamental group of $M$ contains a parabolic isometry of $\tilde{M}$ with positive translation length. This conjecture is confirmed in \cite{Wu12}. Since the proof in this paper is different as the one in \cite{Wu12}, theorem \ref{mt-1} gives a new proof to confirm this conjecture.
\end{remark}

\begin{remark}\label{tushar}
 A complete visible CAT(0) space may contain parabolic isometries with positive translation lengths. One possible example is the following: \textsl{Firstly find a complete CAT(0) space $M$ such that there exists a parabolic isometry $\gamma_{0}$ on it and the visual boundary is empty (one may construct $M$ by taking the convex hull of the orbits of zero under $\alpha$ on page 6 in section 1.3.3 in \cite{Vale}, which is modified by Edelstein's example in \cite{Edel64}). Then take the product of $M$ with $\mathbb{R}$ we get a space $N:=M\times \mathbb{R}$. Since the visual boundary of $N$ consists of two points which can be joined by a geodesic line, $N$ is a complete visible CAT(0) space. Consider the isometry $\gamma:N \to N$ which is defined as $\gamma \cdot (m,t)=(\gamma_{0}\cdot m,t+1)$. It is easy to see that $\gamma$ is a parabolic isometry on $N$ whose translation length $|\gamma|>0$.} I am greatly indebted to Tushar Das to share his idea on this example.

\end{remark}

Theorem \ref{class} is an easy application of theorem \ref{mt-1}.

\begin{proof}[Proof of theorem \ref{class}]  \texttt{Proof of (1)}: By definition.

\texttt{Proof of (2)}: If $\gamma$ is hyperbolic, by the definition we know that $|\gamma|>0$.

If $|\gamma|>0$, assume that $\gamma$ was not hyperbolic, so $\gamma$ is parabolic. From theorem \ref{mt-1} we have $|\gamma|=0$ which contradicts with our assumption.

\texttt{Proof of (3)}: If $\gamma$ is parabolic, it is obvious that $\gamma$ does not have fixed points. $|\gamma|=0$ follows from theorem \ref{mt-1}.

If $\gamma$ does not have fixed points and $|\gamma|=0$, the conclusion that $\gamma$ is parabolic follows from part (2).
\end{proof}

\section{Proof of theorem \ref{mt-2}}\label{somp}
Before we go to prove theorem \ref{mt-2}, let us make some preparations. The following proposition has been proven in different literature (see \cite{BF, KN04, McP99}). For completeness we still give the proof here.

\begin{proposition}\label{fpfm}
Let $M$ be a complete, visible CAT(0) Riemannian manifold. Assume that the mapping class group $\Mod_{S_{g,n}}$ acts properly on $M$. If $3g+n\geq 5$, then there exists a point $x\in M(\infty)$ such that $\gamma \cdot x=x$ for all $\gamma \in \Mod_{S_{g,n}}$.
\end{proposition}

\begin{proof}

Let $\alpha$ be a simple closed curve on $S_{g,n}$ and $\tau_{\alpha}$ be the Dehn twist along $\alpha$ (see the definition of Dehn twist in \cite{FM}). Since $3g+n\geq 5$, there exists a simple closed curve $\beta$ which is disjoint with $\alpha$. Let $\tau_{\beta}$ be the Dehn twist along $\beta$. Since $<\tau_{\alpha},\tau_{\beta}>$ is a free abelian group of rank 2, by lemma \ref{z2ip}, $\tau_{\alpha}$ is parabolic. Since $M$ is proper visible, by remark \ref{2} there exists a unique $x \in M(\infty)$ such that $Fix(\tau_{\alpha})=\{x\}$.

\textsl{Claim 1: If $\tau_{\alpha}\cdot \tau_{\beta}=\tau_{\beta}\cdot \tau_{\alpha}$, then $Fix(\tau_{\beta})=Fix(\tau_{\alpha})=\{x\}$.}

\textsl{Proof claim 1:} Since $\tau_{\alpha}\cdot \tau_{\beta}=\tau_{\beta}\cdot \tau_{\alpha}$ and $\tau_{\alpha}\cdot x=x$, $\tau_{\alpha}\cdot (\tau_{\beta}\cdot x)=\tau_{\beta}\cdot x$. $Fix(\tau_{\beta})=\{x\}$ follows from the fact that $Fix(\tau_{\alpha})=\{x\}$.

\textsl{Claim 2: For any simple closed curve $\beta$ on $S_{g,n}$, $Fix(\tau_{\beta})=\{x\}$.}

\textsl{Proof claim 2:} Let $\beta$ be a simple closed curve. Since $3g+n\geq 5$, the curve complex is connected (see theorem 4.3 in \cite{FM}). In particular, there exists a sequence of simple closed curves $\{\alpha_{i}\}_{i=1}^{k}$ such that $\alpha_{1}=\alpha, \alpha_{k}=\beta$ and $\alpha_{i}\bigcap \alpha_{i+1}=\emptyset$. Hence, by claim 1,
\[Fix(\tau_{\beta})=Fix(\tau_{\alpha_{k-1}})=Fix(\tau_{\alpha_{k-2}})=\cdots =Fix(\tau_{\alpha_{1}})=\{x\}.\]

The conclusion follows from part (2) of proposition \ref{mcgp} and claim 2.
\end{proof}

\begin{proposition}\label{mcape}
Let $M$ be a complete, visible CAT(0) Riemannian manifold and the mapping class group $\Mod_{S_{g,n}}$ acts properly on $M$. If $3g+n\geq 5$, then any infinite ordered element $\phi \in \Mod_{S_{g,n}}$ acts as a parabolic isometry.
\end{proposition}
\begin{proof}
Suppose that there exists an element $\phi \in \Mod_{S_{g,n}}$ with infinite order which acts on $M$ as a hyperbolic isometry. Then there exists a geodesic line $\gamma:\mathbb{R}\rightarrow M$, an axis for $\phi$, such that $\phi \cdot \gamma(t)=\gamma(|\phi|+t)$ for all $t \in \mathbb{R}$. Since $M$ is a visible CAT(0) space, it is not hard to see that $Fix(\phi)=\{\gamma(+\infty),\gamma(-\infty)\}$. From proposition \ref{fpfm} we assume that $\gamma(+\infty)$ is fixed by $\Mod_{S_{g,n}}$.  Let $\sigma \in \Mod_{S_{g,n}}$, since $\sigma$ fixes $\gamma(+\infty)$ there exists some a $C>0$ such that $dist(\sigma \cdot \gamma(n\cdot|\phi|),\gamma(n\cdot|\phi|))\leq C$ for all $n>0$. Hence $dist((\phi^{-n}\cdot \sigma\cdot \phi^n)\cdot \gamma(0),\gamma(0))\leq C$. Since the action is proper, there exists a subsequence $\{n_{i}\}$ such that $\phi^{-n_{i}}\cdot \sigma \cdot \phi^{n_{i}}\equiv \phi^{-n_{1}}\cdot \sigma\cdot \phi^{n_{1}}$, hence $\phi^{n_{1}-n_{i}}\cdot \sigma= \sigma\cdot \phi^{n_{1}-n_{i}}$. Since $\sigma$ is arbitrary and $\phi$ has infinite order in $\Mod_{S_{g,n}}$, we can choose $\sigma$ to be pseudo-Anosov such that $\{\sigma,\phi\}$ generates a free group of rank 2 (see \cite{Iv92}). Since $\phi^{n_{1}-n_{i}}\cdot \sigma= \sigma\cdot \phi^{n_{1}-n_{i}}$, the group $<\sigma,\phi>$ contains a free abelian subgroup of rank $2$, which is a contradiction since $<\sigma,\phi>$ is a free group.
\end{proof}

\begin{proposition}\label{mcghs}
Let $M$ be a complete, visible CAT(0) Riemannian manifold and the mapping class group $\Mod_{S_{g,n}}$ acts properly on $M$. If $3g+n\geq 5$, then there exists a horosphere $H$ such that every torsion free subgroup of $\Mod_{S_{g,n}}$ acts properly on $H$.
\end{proposition}

\begin{proof}
From proposition \ref{fpfm}, there exists a point $x \in M(\infty)$ such that $\Mod_{S_{g,n}}$ fixes $x$. Let $H$ be a horosphere at $x$ and $G$ be a torsion free subgroup of $\Mod_{S_{g,n}}$. By proposition \ref{mcape} we know that $G$ consists of parabolic isometries except the unit. Hence, by proposition 8.25 on page 275 in chapter II.8 of \cite{BH}, $H$ is $G$-invariant because $G$ fixes $x$. Let $d$ be the metric of $M$ and  $d_{H}$ be the induced metric on $H$. It is obvious that $d_{H}(p,q)\geq d(p,q)$ for all $p,q \in H$. The conclusion that $G$ acts properly on $H$ follows easily from the facts that $G$ acts properly on $M$ and $d_{H}(p,q)\geq d(p,q)$.
\end{proof}

Now we are ready to prove theorem \ref{mt-2}.

\begin{proof}[Proof of theorem \ref{mt-2}]
We argue it by contradiction. Assume that $\mathbb{M}(S_{g,n})$ admits a complete, visible CAT(0) Riemannian metric $ds^2$. Let $\mathbb{T}(S_{g,n})$ be the universal covering space of $(\mathbb{M}(S_{g,n}),ds^2)$, which is the Teichm\"uller space endowed with the pull back metric $ds^2$. By part (3) of proposition \ref{mcgp} we can assume that $G$ is a torsion free subgroup of $\Mod_{S_{g,n}}$ with finite index. By proposition \ref{mcghs}, there exists a horosphere $H$ such that $G$ acts properly on $H$. Proposition \ref{hsie} tells us that $H$ is homeomorphic to $\mathbb{R}^{6g-7+2n}$, in particular $H$ is a contractible manifold. By the definition of action dimension we know that
\[actdim(G)\leq 6g-7+2n.\]
On the other hand, since $3g+n\geq 5$ and $G$ is a finite index subgroup of $\Mod_{S_{g,n}}$, by proposition \ref{adom}, we have
\[actdim(G)=6g-6+2n\]
which is a contradiction.
\end{proof}

\begin{remark} If we carefully check the proof of theorem \ref{mt-2}, we can conclude as follows: \textsl{The Teichm\"uller space $\mathbb{T}(S_{g,n})$ $(3g+n\geq 5)$ admits no complete $\Mod_{S_{g,n}}$-invariant CAT(0) Riemannian metric such that every Dehn twist has only one fixed point in the visual boundary of $\mathbb{T}(S_{g,n})$.} I am very grateful to Benson Farb for this point.
\end{remark}

\begin{remark}
If we consider the case that $n=0$ and $g\geq 3$. From lemma 4.5 in \cite{Wu12}, the proof of theorem \ref{mt-2} can also conclude as follows: \textsl{The Teichm\"uller space $\mathbb{T}(S_{g})$ $(g\geq 3)$ admits no complete $\Mod_{S_{g}}$-invariant CAT(0)  Riemannian metric such that the Dehn twist on some non-separating simple closed curve has only one fixed point in the visual boundary of $\mathbb{T}(S_{g})$.}
\end{remark}

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Still cant believe you are gone daddy, we watched you suffer in pains and couldnt do anything to help you..
I have always heard about cancer but i never knew it would come to you. you were my hero..a good man, my brother,my friend..


Zara is an independent art consultant, lecturer and exhibition curator with specialist knowledge of Buddhist art; focusing on the art and culture of Tibet and the Himalayas. She also leads treks and tours to Tibet, Bhutan, Ladakh, Zanskar, Nepal and other areas of the Himalayas and led the first British tour to Tibet in 1981.
She is the Vice President of the Tibet Society and the Tibet Relief Fund of the UK.


If you are alive on Earth, there's a good chance you're microscopic.
"From a diversity perspective, 95%" of life on this planet is smaller than the eye can see, Laura Hug, an environmental microbiologist at the University of Waterloo in Canada says. And that collection of microorganisms just got a lot more complex.
More than 1,000 previously undocumented microbes are reported in a paper published Monday in the journal Nature Microbiology, authored by Dr. Hug, as a postdoctoral fellow at the University of California, Berkeley, along with colleagues there. These additions greatly expand the tree of life, a model used by biologists to describe the phylogenetic relationships between all known species, living and extinct.
"The tree itself is a depiction of the relationship of all of the organisms that are on the planet, all living things," Hug tells The Christian Science Monitor in an interview. "That includes humans and plants and the things we can see, and it also includes all of the microorganisms that are too small to see with the naked eye."
The new additions are largely bacteria and archaea, two microscopic ancient lineages of life that were once thought to be the same. Together, they make up the vast majority, 95%, of life forms on Earth.
"Keep in mind that they've had 2 billion years longer to develop and to diversify than the eukaryotes [beings, like animals, with more complex cellular structure]. The eukaryotes emerged later, so they're running on a disadvantage," Hug says.
This new study was made possible by a scientific breakthrough. Previously, scientists looking to identify new organisms by their genomes would isolate and cultivate the cells in a lab. But not all organisms can survive and grow in that environment. Many of these microorganisms depend on other organisms. But now, using genome-based approaches developed in the past two decades, scientists are able to dig into the DNA of microbes in their environment.
The organisms in this study were found in a wide range of places. Some were found in mundane places like sediments, while others were in extreme environments, like a hot spring in Yellowstone National Park, or salt flats in the Atacama Desert. The researchers even found new diversity inside a dolphin's mouth.
"The tree of life is an underlying structure of diversity of life on Earth," Hug says. And adding these lineages to the tree highlights where researchers can hunt for further lineages and map out the relationships between different organisms.
But Gary Olsen, a microbiologist at the University of Illinois who was not part of this study, is skeptical of how these new, diverse lineages fit into the tree, and what those relationships actually are. "I have no good sense of how profound is this biological diversity" based on the new data, he tells the Monitor.
Although Hug's research ultimately suggests a revision of the tree of life, she says the team's goal was to investigate the diversity of life on Earth rather than use this structure to answer questions about the evolution of life over time.
One of the big open questions, says Hug, is about the relationship between the three domains of life: bacteria, archaea, and eukaryotes. But "this dataset and this study does not shed any light on that question whatsoever, and it wasn't meant to."
Hug expects scientists to further expand on her work, both by identifying more previously unknown organisms and by focusing in on certain lineages to investigate evolutionary relationships.
"This is a dataset that people can build on," she says. In fact, "it might already be out of date" if other research teams are already adding genome sequences.


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The Mission of the Western Magic Valley REALTORS® is to support it's members in their business pursuits by promoting the highest professional standards ethics and being the primary provider for quality services, education and real estate services in our market area. WMVR has over 500 REALTOR® members and an increasing number of Affiliates.
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Religious Liberty Threatened
Last Sunday Catholics at the Kings Bay Naval Submarine Base in Georgia were banned from attending Mass because of the partial shutdown of the Federal Government. Father Ray Leonard, a Catholic priest, was prohibited from volunteering to celebrate Mass without pay and was told if he violated that order, he could be subject to arrest.
However, Protestant services there continued as usual.
"This is an astonishing attack on religious freedom by the Federal Government," asserted the Thomas More Law Center, which filed a federal lawsuit on behalf of Fr. Leonard. The Pentagon later apologized, but the Center's suit will continue.
This is one example of hundreds of new cases of attacks on religious liberty by our government. In fact, the 2,200 attendees at the Family Research Council's Value Voters Summit last weekend – declared that multiple threats to religious liberty was their deepest concern.
Todd Starnes, who served in the U.S. Army for 25 years before becoming a Fox News commentator, told attendees, "The military profession is different because people are risking their lives for others. The tradition of religious liberty in the military is being transformed into a culture of intimidation and fear.
"If they express support for traditional marriage, they fear consequences for their careers. The Army has one value now – tolerance."
In a column for Fox News this week, Starnes assailed the U.S. Army for including the American Family Association as a "hate group" like the Ku Klux Klan or Black Panthers. Starnes accurately called the group a "well-respected Christian ministry" that simply supports "traditional family values."
Again, the Army later apologized, but Tim Wildmon, President of AFA, asserted, "We are probably going to be taking legal action. The Army has smeared us."
Master Sgt. Philip Monk was asked by his superior officer, Major Elisa Valenzuela, a lesbian, what he thought about same-sex marriage. Fearing to express his opposition due to his Christian convictions, Monk replied that because of his religious beliefs, he could not answer her question. He was summarily dismissed from his unit, leaving the impression on his record that "he might have done something wrong," says Michael Berry, an attorney with the Liberty Institute.
The greater threat to religious liberty is with how Obamacare is being implemented. The Liberty Institute has identified 39 for profit companies and 37 non-profit groups such as Wheaton College and Catholic Dioceses who are objecting to being required to offer their employees free birth control and abortion drugs, such as "morning after pills."
Hobby Lobby, a company with 561 stores is owned by a Catholic who objects to the Health and Human Services (HHS) mandate to provide pills, sterilization and abortion drugs. It has won a preliminary injunction from the Tenth Federal Circuit of Appeals. However, other companies have lost an HHS case in the Third Circuit. That conflict likely to lead to the Supreme Court.
The U.S. Conference of Catholic Bishops has not been so fortunate. When they protested the HHS mandate, the Administration created an "accommodation" for religious employers, like Catholic University or the Diocese of Evanville - to have the objectionable products provided indirectly to their employees by a third party, their insurers.
New York Cardinal Timothy Dolan, President of the USCCB, said the accommodation provided "only minor changes. Instead of spending our time, energy and treasure on increasing access to health care, as we have done for many decades, we're now forced to spend those resources on determining how to respond to recently enacted government regulations that restrict and burden our religious freedom. We are united in our resolve to continue to defend our right to live by our faith, and our duty to serve the poor, heal the sick…and insure our people."
Religious liberty is not something Americans have worried about.
Ever since 1791 when the Bill of Rights was added to the Constitution, the First Amendment guaranteed that government "shall make no law respecting an establishment of religion, or prohibiting the free exercise thereof."
However, the Obama Administration seems to have forgotten that freedom of conscience is built into the American DNA. We expect countries like Egypt or Russia to deny freedom of religion to its citizens.
But our own government has been forcing Catholic hospitals, which provide 40 percent of the nation's inpatient care - to give its employees birth control, sterilization and abortion-inducing drugs in violation of Catholic conscience – and for free.
This is an outrage which people of all political persuasions should all condemn.


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Home News Apple may be soon launching a lite version of AirPods Pro
NewsApple
Apple may be soon launching a lite version of AirPods Pro
by Nick Randall February 23, 2020
written by Nick Randall February 23, 2020
It's not too long Apple launches AirPods Pro, now rumors have suggested that Apple will be appearing soon with their next TWS wireless earphone, which may be called Apple AirPods Lite. However, its official name and features are currently unavailable.
Latest News: Mobvoi TicWatch Pro 2020 features Wear OS, 1GB RAM, and dual display
A new information coming from Digitimes speculates that Apple is gearing up to release a "lite" version of its AirPods Pro.
According to the website, Apple will launch, within the next six months, something they strangely call "AirPods Pro-Lite" – that is, a cheaper version of AirPods Pro. Yes, at first this doesn't make much sense: well, cheaper version of AirPods Pro already exists and is called – AirPods.
image credit: Apple
According to another report, Taiwan -based company Universal Scientific Industrial (USI) is said to be the main supplier of AirPods Pro Lite and the product will meet the user in the middle of the year.
Apple AirPods Lite expected price
The new Lite version is expected to be positioned at a price between standard AirPods (2nd generation) and AirPods Pro. These ear-tips would be a cheaper version of the AirPods Pro model. As a reminder, this equipment in the United States was valued at $ 249. For AirPods 2 you have to pay $ 159. Thus, the new Apple headphones, most likely named AirPods Pro Lite, should be valued at around $ 200.
AirPods Lite expected release date?
At present, there's no information available about the specs and features of the AirPods Pro Lite. The report further claims that the production for the same is likely to begin by the end of the 2nd Quarter 2020 or later. If the report is true, we will find more information soon.
Source: 9to5mac
Next Read: Samsung Introduces Galaxy Buds Plus earbuds with massive battery life
Mobvoi TicWatch Pro 2020 features Wear OS, 1GB RAM, and dual display
Wyze is planning to release a fitness tracker with Alexa support
Xiaomi Mi Band 5 is expected to...
Apple Watch 4 Review | Buy For Best...
Samsung Introduces Galaxy Buds Plus earbuds with...


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Full fibre broadband rollout will offer residents and businesses ultrafast internet speeds
Published: Monday, December 21, 2020
Tens of thousands of households in Wandsworth can look forward to ultrafast internet speeds with the unveiling of new plans to install full fibre broadband across the borough.
Infrastructure provider G.Network has announced it will be investing more than £105m in upgrading Wandsworth's broadband connections - many of which still rely on copper based wiring.
In the New Year the company will begin installing brand new, state-of-the-art, contiguous full fibre broadband infrastructure throughout the borough.
It is planning to install cable stretching nearly 250 miles which will guarantee ultrafast download and upload speeds to more than 100,000 homes and businesses in Wandsworth.
And as part of that rollout the company plans to create more than 1,250 jobs with the hope that many are filled by Wandsworth residents.
The announcement has been warmly welcomed by council leader Ravi Govindia who said: "This is tremendous news for local residents and businesses, who will soon be enjoying some of the country's fastest and most reliable broadband services. This is just one way the council is working with its partners to improve people's connectivity.
"I know just how important this is for our residents and businesses. It will give people from all walks of life new opportunities to get on in life and is a hugely important milestone in efforts to eliminate digital poverty."
Sasho Veselinski, G.Network CEO, also commented on the announcement. "We're delighted to be bringing upgraded broadband connections to businesses and residents of Wandsworth. The pandemic has demonstrated that there is a strong need for full fibre connectivity across many London boroughs including Wandsworth. We look forward to working closely with the Council in 2021 and beyond."
G.Network's infrastructure will be made available on an 'Open Access' principle. This means the company will work with other broadband providers so they can serve their customers via G.Network infrastructure in future.
More than 30,000 households on Wandsworth housing estates already enjoy superfast internet speeds as a result of a joint council project with Community Fibre. Tenants and leaseholders in around 70 per cent of Wandsworth's social housing stock are able to access this service.


UK-based sales and financing house IAC Films and Will Clarke's UK distributor Optimum Releasing are partnering on a UK release of IAC's To Walk With Lions.
The two companies will give the picture a nationwide release in May with support from several promotional partners and charities. The film, the continuation of Born Free's story, stars Richard Harris.
The deal is anticipated as the first in an ongoing partnership between IAC and Optimum.
IAC has also struck a pact with video operation Mosaic Movies for rental, sell-through and DVD rights in the UK on the film.
Since launching last Cannes, Optimum has quickly made a name for itself with canny marketing campaigns such as a re-release of Carol Reed's The Third Man, the only film to open against Star Wars: The Phantom Menace.


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Category Archives: 245
160(1), 245, 249.1, GAAR
Difference Between Astute and Abusive Tax Avoidance
January 3, 2017 Sas Ansari Leave a comment
594710 B.C. Ltd. v The Queen, 2016 TCC 288
The Court was asked to draw a line between permissible and impressible tax avoidance strategies. The correctness of the application of the General Anti-Avoidance Rule (GAAR) was at issue when applied (i) at the partnership level and (ii) at the holding corporation level.
The TCC held that the GAAR was incorrectly applied in both instances, allowing the taxpayer's appeal. In the first instance, upon which the second relied, the failure occured at the abuse and misuse level. The MNR had failed to demonstrate a particular provisions that was abused or misused, relying on a general policy against loss tranding between unrelated parties (the court also held that the transaciton was profit trading not loss trading).
The taxpayer was a holding company, wholly owned by a member of a family involved in the real estate business, that wholly owned a corporation that was a limited partner in a limited partnership. The other limited partners were wholly owned holding corporations of other family members' wholly owned holding corporations. The sole business of each partnership corporation was the participation in the limited partnership.
Nuinsco Resources Limited was an arm's length Canadian public corporation with large tax pools (losses and resource-related deductions) that eventually purchased al of the shares of the partner-Cos.
The plan permitted the partnership income to be paid to the corporate partner tax-free while, for tax purposes, the income was allocated almost entirely to an arm's length corporation with accumulated losses sufficient to reduce taxable income.
Rather than allocated the taxable income of the partnership to each of the partner-Cos, a series of transactions were undertaken:
HoldCo subscribed for additional shares of PartnerCo, paid for by set-off against debt owed by PartnerCo to HoldCo, increasing ACB of the shares held in PartnerCo by the debt amount (GAAR was applied to this);
The Limited Partnership lent cash in an amount almost equal to what would have been each limited partner's after-tax income for that fiscal period to each PartnerCo;
Each PartnerCo declared and paid a series of sequential stock dividends to each HoldCo, consisting of preferred shares equal in value to the loan amounts (each with a PUC and ACB of $1);
absent GAAR, The issuance of the stock dividends would have resulted in a dividend to the HoldCo that was taxable to the HoldCo and also deductible to HoldCo (s 112), but there would have been no change in the ACB of the shares:
PartnerCo used the proceeds from the loan to redeem the preferred shares issued in the stock dividends;
absent GAAR, there would be no deemed dividend received by the HoldCo, and the preferred shares would have been redeemed at ACB, resulting in no gain or loss;
A service contract and a put-option were entered into by the partnership with non-arm's length parties;
A second stock dividend (preferred shares) was declared by PartnerCo to HoldCo;
Each PartnerCo sold all of its preferred shares in each HoldCo, at a price equal to the second stock dividend, and all common shares at a price equal to the ACB (stepped up by the loan-fueled share subscription) by the Loss Public-Corp (which also acquire all the shares of the GP for $1);
The acquisition of shares by the Loss Public-Corp was funded by a loan from the partnership to it;
but for GAAR, the acquisition of shares, occurring at ACB, would result in no gain or loss, an acquisition of control would occur resulting in a deemed year-end for the PartnerCo, and no income was allocable to PartnerCo for its new taxation year;
The Loss Public-Corp wound up each PartnerCo, becoming the sole limited partner; and
the partnership allocated its net income to the Loss Public-Corp and the GP, each PartnerCo was dissolved resulting in distributions, the remaining units were sold pursuant to the option (back to the family), plus some other transactions that ended in the dissolution of the partnership.
The MNR assumed that these transactions were pre-ordained as a series and that all were avoidance transactions. The agreed statement of facts agreed that most but not all steps were part of a series fur GAAR purposes.
Initial Issues
One of the initial issues was whether the appellant had complied with the large corporation rules, specifically, whether the argument that the assessments were issued outside of the normal reassessment period should be considered given that it did not form part of the issues raised in the Notice of Objection as required by subsection 165(1.11) and therefore precluded from raising it on appeal to the TCC pursuant to subsection 169(2.1).
The TCC held that an out of time assessment is void (Blackburn Radio Inc. v The Queen, 2012 TCC 255 and in The Queen v Canadian Marconi Co., [1992] 1 FC 655 (FCA)) and section 169.1 is aimed at precluding the TCC from doing anything but vacating or varying an assessment (para 37). If an assessment is out of time, the court held that it is void ab initio, and t=does not exist and cannot be given effect. Subsection 152(8) does not apply to an out of time assessment (Lornport Investments Ltd. et al v The Queen, [1992] 1 CTC 351 (FCA)).
The pertinent period here was the tax year ending June 1, 2006, and, in this case, the reassessment of the PartnerCo the tax year is not altered by "the fact that the Minister has reassessed Partnerco to include the tax consequences of transactions that would have otherwise fallen outside of the taxation year" (para 45). Therefore, the court would have held that the assessment was timely and valid.
The Appellant argued that the Minister cannot invoke GAAR as a result of an amount of tax owing as a result of a separate GAAR assessment of another taxpayer (para 49). The Court disagreed. The Minister assesses each taxpayer separately and, although the Minister cannot stack GAAR assessments with respect to the same taxpayer, this does not extend to third parties with derivative liability assessments (para 51). The TCC referred to the SCC decision in Copthorne Holdings Ltd. v Canada, 2011 SCC 63, where at paragraph 72 it was said:
The analysis will then lead to a finding of abusive tax avoidance: (1) where the transaction achieves an outcome the statutory provision was intended to prevent; (2) where the transaction defeats the underlying rationale of the provision; or (3) where the transaction circumvents the provision in a manner that frustrates or defeats its object, spirit or purpose (Trustco, at para. 45; Lipson, at para. 40). These considerations are not independent of one another and may overlap. At this stage, the Minister must clearly demonstrate that the transaction is an abuse of the Act, and the benefit of the doubt is given to the taxpayer.
The GAAR cannot be interpreted to allow abusive tax avoidance to go uncaught where it is subsidiary to and arising out of another act of abusive tax avoidance transaction (para 53). The consequences of the application of GAAR, pursuant to s 245(7), can include an assessment under subsection 160(2) (para 55).
Application of GAAR
The court repeated the principles in applying GAAR set ou in Inter-Leasing Inc. v Ontario (Minister of Revenue), 2014 ONCA 575 (paras 49-52), and reminded itself that the GAAR is a provision of last resort (Canada Trustco Mortgage Co. v Canada, 2005 SCC 54, para 21). The Court must, inter alia:
weigh the entirety of the evidence to identify, objectively, the purposes that were primary in the transactions, assessing the reasonableness of the evidence of the purposes (para 57);
determine if there was a series, what transactions make up the series, what tax benefit resulted from the series, and if each transaction in the series was primarily for a bona fide non-tax purpose (Copthorne Holdings Ltd. v Canada, 2011 SCC 63, para 40);
put the burden on the taxpayer to establish that a transaction or series of transactions was for a primary bona fide purpose other than to obtain a tax benefit ( 1207192 Ontario Ltd. v The Queen, 2012 FCA 259, para 6); and
remember that a tax benefit need not occur in the year in question (para 64) or to the taxpayer to whom the tax consequences are re-determined by application of the GAAR (para 70).
PartnerCo Reassessment
Tax Benefit – The Appellant argued that the comparator should be if Nuinsco has acquired PartnerCo without being able to utilize the tax attributes – therefore the tax benefit was to the Public Co. The Respondent argued that the comparison should be if the Public Co had never acquired PartnerCo, as the only benefit to Public Co were the deal fees paid to it by the Appellant. The Court held that the Appellant had failed to demonstrate the reasonableness of its alternative on the facts, and a tax benefit arose (paras 72-73).
Misuse or Abuse – The Minister alleges that there is a general policy of preventing loss sharing between unrelated taxpayers. But, the Court said that it is fatal to the Minister position that the policy is broad and not attached to the provisions at issue (Copthorne Holdings Ltd. v Canada, 2011 SCC 63, para 118). Specific provisions must be associated with the tax benefit and be abused (para 81). The court noted that the distinction is a fine line between transactions that offend a provision in a manner that circumvent it and transactions that fall outside of the scope of the provision (para 90). However, the provisions of the ITA referred to prevent only certain types of loss trading, not profit or gain trading (para 90).
HoldCo Reassessment
The TCC did not need to deal with this reassessment but made the following observations in Obiter:
A stock dividend is not a transfer of property as required by s 160 (para 132);
A share redemption is a transfer of property (para 133);
A transfer of property can be indirect in the context of the steps of a particular series (paras 136-147);
Misuse or abuse – The purpose of s 160 is to enable the Minister to collect a tax debt of one person from another to the extent that the value of property transferred is for less than its FMV (therefore not preserving the value of the tax debtor's property) (paras 151-52). The Court held that the reliance on the deemed year-end ruled in s249.1, in the context of the series, was abusive of s 160 of the ITA as it prevented the operation of the s 160 by divorcing the partnership income from the allocation of it for tax purposes.
Sas Ansari, BSc BEd PC JD LLM PhD (exp) CPA In-Depth Tax 1, 2 &3
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161(1), 245, GAAR, Tax Court of Canada
Timing of Arrears Interest in GAAR Assessments – Sas Ansari
October 3, 2016 Sas Ansari
Timing of Arrears Interest in GAAR Assessments
Quinco Financial Inc v The Queen, 2016 TCC 190
This was a Rule 58 Application under the Tax Court of Canada Rules (General Procedure). The question to be determined was whether, where the Minister relies on a GAAR assessment to deny capital losses, does arrears interest payable under subsection 161(1) of the Income Tax Act accrue in respect to any period after the taxpayer's balance due date and before the issuance of the reassessment?
In other words, the Court was asked to determine when arrears interest would start to accrue where GAAR is used to recharacterize the transaction.
Although GAAR is different than the provisions in the Income Tax Act, it is still a part of the Act and part of the other provisions of the Act. A court must "to the extent possible, contemporaneously give effect to both the GAAR and the other provisions of the ... [Act] ... relevant to a particular transaction" (para 30) - Canada Trustco. The ITA must be interpreted as a functional, coherent whole with respect to all statutory schemes engaged by the transaction.
The TCC held that "an assessment under GAAR [...] is not an assessment divorced from the other provisions of the Act [, rather,] GAAR is entirely dependent upon a textual, contextual and purposive analysis of the object, spirit or purpose of the very provisions which allegedly confer the tax benefit" (para 31). The Minister nullifies the tax benefit through employing other non-GAAR provisions of the ITA, giving rise to tax consequences under those other provisions at the time of the transaction (para 33).
When engaging in tax planning, the taxpayer's advisors must weight the consequences of a GAAR reassessment like any other assessment (para 39). A taxpayer possibly subject to GAAR could have filed by deducting the future-impugned capital loss, but calculated tax payable by applying GAAR (para 41). If this was done, then upon a GAAR assessment interest would not accrue. All taxpayers who are directly subject to GAAR assessments are required to consider and apply GAAR (para 42). The taxpayer does not need to ask the Minister's permission to apply GAAR - STB Holdings Ltd.
Subsection 161(1) is the default provision governing interest accrual for all tax liability arising under an assessment or reassessment under the ITA. The Court concluded that on a plain reading of the arrears interest provisions, a taxpayer is liable to arrears interest beginning on the day after the balance due date on a GAAR assessment (para 45). This is because "tax payable" is fixed by an assessment or reassessment, subject to an appeal - ITA s 248(1) "tax payable". Where taxes payable exceed the taxed paid before the balance due date, on the day after the balance due date, arrears interest applies. A GAAR assessment fixes or adds to the tax payable of a taxpayer, causing (where there has not been a payment in anticipation of the denial of the tax benefit) the tax payable to exceed the taxes paid.
The Court found support for the imposition of arrears interest from the balance due date in the GAAR itself. GAAR defines "tax benefit" to include any reduction of other amounts that would be payable under the ITA. Were interest not charged until a GAAR assessment was issued, tax deferral is created during the period between the balance due date and the date of the GAAR assessment (para 53). GAAR is meant to deny tax benefits not allow them.


While we are on the subject of demon, who is down for a nerf or adjustment of some kind?
One suggestion i heard was to make it lose health like oxymora when it fires.
Other suggestions include do less damage, have less health, etc.
Demon is definitely too powerful right now.
The main reason is because it's so expensive to kill and if you don't kill it, it stays there forever.
My favorite option is #5. Even before Oxymora existed, @CharmedImSure floated the idea of making Demon lose health every time it shoots a projectile. I think this would work very well for several reasons, and it should be a fairly easy tweak to implement. Demon losing 5 HP for each projectile would probably work pretty well, without having to change its starting health from 50.
Alternatively, they could just make Stone prison or whatever it is called actually CC it permanently or for a significant duration so that the cost of summoning a demon vs keeping it at bay is more balanced. Not that I don't think making it lose health is a good idea too, but then it becomes much less useful as well.
It would be fun to be able to curse the Demon, which would then turn it against the player who cast it.
"You forgot 7 Missiles for 84 mana. Oh, and 7 Blood Bolt for 28 HP, which seems pretty risky. You'd have a massive mana advantage though."
"In practice, it's usually not quite as expensive to kill as you're claiming though, because you can combine multiple spells. Notice that Lightning does 40 damage, and then one reflection with Shield is enough to finish it. That's maybe 70-75 instead of 120. That's not as cheap as just reflecting, but it's a lot faster. Most viable loadouts can at least approximately break even vs a Demon, if you plan ahead."
That all depends on your loadout and where your opponent puts their Demon. It's often more expensive to kill in practice if you don't have the right spells and if you don't want to take damage from Demon (since you have to spend mana teleporting or distracting it). Demon is so powerful that I specifically account for it in all of my competitive loadouts. Shield/Fist is the best combo. Twins/Lightning is pretty effective too.
I think demon still costs more than the "84 for missle" etc because you arent counting the fact that you have to teleport or shield while youre killing it unless you want to take damage, meanwhile your opponent can also attack.
Also i had another thought for a nerf which is to reduce the fire rate. That would help cut down on teleports or shields needed while killing a demon.


PenDance requires that address lists and guest lists in be provided to us in spreadsheet format. We'll merge the spreadsheet into our templates for either hand calligraphy or computer printing, then send back your list in label format so that you can review it for errors before we start addressing.
You can sign up on our Home page to receive our Free Guide for Brides-to-Be which will explain the etiquette rules for addressing envelopes.
You will be able to catch errors or omissions more easily when you see it in label format.
It's easier for the calligrapher to visualize how the addresses should look on the envelopes.
Proofreading is more efficient when the addresses appear exactly the same on the list as on the envelopes.
If you would like to use PenDance's Excel template to prepare your address list, you can download it here.
If you prefer that your out-of-town guests' envelopes be addressed first, sort them to the top of the list.
If you have a "B" list, sort them to appear at the end of your list.
Alphabetize the list by last name, if possible.
If you need to add additional names or make changes to your list after sending us the original list, please email ONLY the changes, NOT the entire list over again. If we have to process additional spreadsheets an administrative fee will apply each time.
Spell out Number or Apartment. "Apartment Number" is redundant.
Include punctuation for titles such as Mr. and Mrs. There is no period after "Miss".
Spell out postal and state abbreviations (Post Office Box, Boulevard, Avenue, Lane, etc.).
Zip codes can be 5-digits for USA addresses.
Married couples with different last names are noted with "and" between their names and their names appear on separate lines. Unmarried couples' names are not joined with "and" and their names appear on separate lines as well.
Inner envelope names should be in separate columns on your spreadsheet.
Indicate whether or not the "G" in "Guest" should be capitalized (or not), and whether or not the "F" in "Family" should be capitalized.
You can spell out "Doctor" for medical doctors.
If you have any questions about formatting your list, please contact us!


Corporate media outlets are reporting that thousands of members of the Boston Teachers Union walked out Friday to protest school budget cuts and contract concessions demanded by the school department. The protest follows two student walkouts, also in protest of budget cuts.
May 25, 2016, West Roxbury, Boston, MA -- Out of concern over global climate change and public safety, 16 local religious leaders were arrested for blocking the construction of the West Roxbury Lateral pipeline with an interfaith worship service. The peaceful civil disobedience occurred on the morning of Wednesday May 25th in West Roxbury.
Cambridge, MA (May 18, 2016) - Last night 75 community members gathered to launch a human rights campaign calling for local and state governments to end contracts with Hewlett-Packard (HP). Speakers introduced a newly-released report entitled Apartheid Technology, and detailed the company's role in maintaining the Israeli occupation of Palestine, U.S. deportation of immigrants, mass incarceration, and global militarization.
As of this writing, there is no obvious path to the White House for Donald Trump. As Harry Enten recently pointed out, Trump's strong unfavorability rating (not his unfavorability rating, his strong unfavorability rating) stands at 53%, a full 16 points ahead of Hillary Clinton's heroic-in-its-own-right 37%. While some combination of electoral fraud and liberal third party candidate could conceivably send Trump to the Oval Office, if the election were to be held today he would lose. Trump has proven to the world that with enough hard work and dedication there is such a thing as bad publicity. What a lot of people don't seem to realize though, is that the election isn't the whole fight, it's only the first round.
Thousands turned out today for a rally/march for $15 an hour wages for exploited low wage workers.Part of global actions today.


In this eye-opening series, mums down tools and escape to a health spa leaving dads in charge of the house... and the kids.
In this eye-opening series, mums down tools and escape to a health spa leaving dads in charge of the house - and the kids.
But will dad cope if mum goes on strike? Will dad rise to the challenge - change the nappies, cook the dinner and do the laundry without putting red socks in with the whites? Or will it be utter chaos?
Two mums head off to the health spa and leave the dads to see if they can survive on their own... without the TV remote, the computer games or the golf clubs. Anna Raeburn offers parenting advice.


Extremely low levels of I-131 can travel great distances in molecular or organic form and are trapped on high efficiency absorbent thermoplastic cartridges which contain an impregnated activated carbon and detected using a 1 1/4" x 1" NaI (TI) scintillation detector. These iodine detection systems can be configured on either mobile or fixed platforms.
The IM 201L monitor continuously measures the gamma volumetric activity of radioactive iodine samples, in both molecular and organic forms contained in air drawn from stacks, ventilation ducts or working areas.
The IM 201S monitor continuously measures the gamma volumetric activity of radioactive iodine sample, in both molecular and organic forms contained in air drawn from stacks, ventilation ducts or working areas. Can withstand seismic conditions.


If there is one thing that seven years at Camp Unirondack has taught me, it is that camp is not just for kids. I know, I know, camp director rhetoric if you've ever heard it right? But let me tell you why it's true. It stems from a struggle that I think many people share. If there is one thing almost all people on this planet have one thing in common it is that feeling of doubt inside. Am I going to pass this test? Will I graduate? Am I good enough? I have yet to meet someone who hasn't felt it. Part of that feeling is an ever-present wonder about what others think of us. Some night in our lives we will lay in bed and ponder what other think of us. It doesn't make us weak, it makes us human.
Interestingly, I discovered this reality through finding its solution. Before I ever knew it was such a universal experience I started working for a place which I have since found to be a sanctuary for those who need to answer that lingering question — are there people out there who care about me? At Unirondack, the answer was yes. I started in 2009 and in that time I have had many conversations with people who go to camps about the reasons they go back year after year. The reality is, camps create a place where the self-doubt is left behind. There simply isn't room for it.
I've helped women in their 50s and 60s experience rock climbing for the first time in their lives. I've seen a nine year-old boy teach a college student the lesson of patience. I've seen young adults leave camp in tears on their third day. How does it happen? It's the power of experiencing the creation of an intentional community and actively engaging in it. I think what makes a Unitarian Universalist congregation inspiring is when congregants come together to create. At camp, that creation is ongoing and ever-present.
The result of this experience is equal for both campers and staff – and how could it not be? Working at a camp requires a surprising amount of selflessness. Finding a group of people willing to work 16-hour days to create something generates two things. First, an incredible amount of pride in the construction of such an experience. Second, a community with universal appreciation of its creation. The dog-eat-dog world we live in practically force feeds competition and a need to promote the self over others. At camps, it's for the good of all. Bottom line — there is no better place to learn about the self than a place created to care for others.
Find someone you care about and get them involved in a camp. I strongly believe in their ability to heal a part of the human condition that is sadly far too commonplace. There are many excellent camps and conference centers throughout the country, find one to call home. Visit the Unitarian Universalist Alliance of Camps & Conference Centers (UUACCC) at www.uuaccc.org for more info.
Kristofer Fiore is the camp director at Camp Unirondack in the Adirondack Mountains of New York. Unirondack is a member of the Unitarian Universalist Alliance of Camps and Conference Centers, a group whose aim is to promote the success and well-being of Unitarian Universalist centers across the county.
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Marissa, 100 feet in the air (where robots fear to go), and her human smile showing love for Camp Unirondack.
Humans connecting in the Camp Unirondack Art Shop as Blue Rabbit looks for robots. None found.
Inner humans bursting forth from outer robots at Camp Unirondack.


Healing the Human Animal™, is a framework, built around our primal heritage as a wild homosapien on planet earth and the implications it presents to us in terms of our personal evolution. It seeks to put the process of spiritual growth, personal transformation and physical healing within this naturalistic context. This system is about tending to the roots and recovering a relationship to the elemental self.