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The term Stoozing came from a way of making money by treating 0% introductory rates on credit cards as interest free loans. The money that the stoozer borrows from the credit card issuer is invested in high interest savings accounts and the interest gained becomes the stoozer's profit. Stoozing contains comparison tables and calculators to help with credit cards and information on how to reduce interest costs or benefit from owning a card such as getting cashback.
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Usa News Post
Emma Raducanu explains why her hand blistered during Australian Open defeat
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Emma Raducanu's hand injury caused her problems in her second-round defeat (Picture: Andy Cheung/Getty Images)Emma Raducanu has explained a combination of factors led to her hand suffering a blister as she crashed out of the Australian Open. The British No.1 suffered a three-set defeat to world No.98 Danka Kovinic in the second round but was badly hindered by a blister to on her hand. Raducanu had to have medical treatment in the opening set to try and keep the pain at bay but then spent the rest of the game trying to play backhand shots and sliced forehands. As well as the heat and humidity, the 19-year-old feels her period in isolation after testing positive for coronavirus last month meant her hands suffered. 'It's very rare for me to be that long away from tennis so I thi...
More cities should follow Chicago's lead by sharing water billing data
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A kayaker paddles in August in Lake Oroville as water levels remain low due to continuing drought conditions in Oroville, Calif. According to three different reports released on Jan. 10, the United States staggered through a steady onslaught of deadly billion-dollar climate disasters in an extra hot 2021, while the nation's greenhouse gas emissions last year jumped 6% because of surges in coal and long-haul trucking, putting America further behind its 2030 climate change cutting goal. | Ethan Swope/APThe Earth is in what scientists say is the last decade in which there is an opportunity to ward off the worst effects of rising temperatures. Averting climate disaster should be a bipartisan issue. At his press conference last Wednesday, President Joe Biden asked: "What are Republicans for? Na
Florida Gov. Ron DeSantis is pushing legislation that takes aim at how history and racism are taught in public schools and even by private businesses. | APRon DeSantis and his GOP allies are pushing a bill in the Florida legislature that would prohibit public schools and private businesses from teaching or training about racism in ways that make white people squirm. Those poor, wittle babies. Their feelings are hurting. Some poor, wittle white people are uncomfortable about the hundreds of years of racism and hate that built this nation. Ron DeSantis, the famously infamous Republican governor of Florida, wants to relieve white Americans from any "discomfort" they may feel about our nation's racist past — and present. DeSantis and his GOP allies are pushing a bill in the Florida legislatur
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Woman who repaid £27k debt and now owns multiple properties shares her money secrets
Julie Haneline says a 'no spend month' each year helps her save (Picture: Jam Press/@juliehaneline)A woman who was in major debt due to student loans, credit cards, and frivolous shopping has revealed how she got herself out of it – and why she does a no-spend-month every year, saving £1,200 every time. Julie Haneline, 36, from Greenville, South Carolina, US, first got into saving after doing a finance course at university. Five years ago, she decided to give herself a challenge by not spending any money whatsoever – beyond necessary bills such as rent – and was blown away by just how much money she was wasting on frivolous purchases. The experiment was such as success that Julie has done it every year since, usually saving between $1,300-$1,700 USD (£950-£1,200) each time – racki
Christine Brown Kicks Out Kody: I Need Intimacy in My Marriage!
The beginning of the end. That's what Sister Wives fans will witness on this Sunday's new episode of the long-running TLC hit. Back on November 2, as you very likely know at this point, Christine Brown announced she was leaving spiritual husband Kody after over two decades as a couple. The split came as a surprise at the time. But now that we've been watching Sister Wives Season 16, it's clear the writing wasn't just on the wall for this couple a very long time ago. It was written on this wall in permanent marker. At various points over the past few installments, Kody has clashed with his spouses over his extremely strict COVID-19 guidelines. Aside from Robyn, each of Meri, Janelle and Christine has taken exception at different times to Kody's apparent lack of trust in...
A man was fatally shot while driving Jan. 22, 2022 on the Southwest Side. | Sun-times fileThe man, 33, was driving in the 4200 block of South Archer Avenue about 5:15 a.m. when an SUV pulled alongside him and someone inside opened fire, striking him in the head, police said. A man was shot to death early Saturday morning while driving in Brighton Park on the Southwest Side. The man, 33, was driving in the 4200 block of South Archer Avenue about 5:15 a.m. when an SUV pulled alongside him and someone inside opened fire, striking him in the head, Chicago police said. He was taken to Mount Sinai Hospital, where he was pronounced dead, police said. His name hasn't been released yet. No one was in custody. ...read more Source:: Chicago Sun Times...
Florent Malouda urges Chelsea to explore 'interesting' moves for Declan Rice and Ousmane Dembele
Declan Rice has been heavily linked with a return to Chelsea (Picture: Getty)Florent Malouda has urged Chelsea to add 'quality' to Thomas Tuchel's ranks and believes his former club should explore moves for Declan Rice and Ousmane Dembele. Chelsea enjoyed a strong start to the Premier League season on the back of their Champions League success last term, but results and performances dropped off a cliff over a difficult winter period. The Blues now find themselves 12 points adrift of league leaders Manchester City down in third place and Tuchel has been desperate to strengthen this month to make up ground on Pep Guardiola's champions. However, Chelsea are yet to any signings with just a week remaining of the January transfer window and there remains uncertainty over the future of s
Huge 15km lorry queues at Dover blamed on Brexit
Exasperated lorry drivers have reported queueing for up to eight hours to get through controls (Picture: Rex/PA)Huge 15km queues piling up at the Port of Dover have been blamed 'entirely' on extra controls brought in after Brexit. Exasperated lorry drivers have reported queueing for up to eight hours to get through controls as staff processing drivers struggle with complex new customs declarations and rules-of-origin forms. The queues of lorries waiting to be processed have become so long that the tailbacks can even been seen on Google Maps satellite images. Images from space show hundreds of lorries lining the roads around the port. It is not clear when the pictures were taken, with Google Earth only saying it is from March 2021 or newer. However they came to light as l
What could the Covid endemic look like in the UK? An expert weighs in
The UK is preparing to enter the endemic stage (Picture: Getty)Earlier this week, Prime Minister Boris Johnson announced that Plan B rules will soon be scrapped – triggering the start of the UK's transition from a Covid-19 pandemic to an endemic. The easing of restrictions comes as the country's coronavirus cases continue to plummet. As England prepares to remove their face masks and return to the office, many are starting to wonder what 'living with Covid-19' will entail. Here, Metro.co.uk spoke to an expert about what a post-pandemic world could look like. What is an endemic? You may have heard the term 'endemic' being thrown around by politicians, scientists, and the press over the past few weeks – but what does the term actually mean? Essentially, an 'endemic' is when
Dan Evans admits Andy Murray may regret his Australian Open gesture after 'panicking' in defeat
Dan Evans suffered a straight-sets defeat in the third round (Picture: Clive Brunskill/Getty Images)Dan Evans says he's expecting a message from his British compatriots after 'wasting' their time watching his defeat at the Australian Open. The British No.2 suffered a 4-6, 1-6, 1-6 defeat to world No.9 Felix Auger-Aliassime on Saturday, meaning all of Britain's male and female players are now out of the singles competitions. Despite making a solid start, Evans was blown away by the 21-year-old's talent in the final two sets and had few answers to the Canadian's serve. Fellow Brits Murray and Joe Salisbury were in the John Cain Arena to cheer on Evans but the latter didn't put on the performance his supporters deserved. 'It's a nice gesture that they come out. They're probably re
GREEN BAY, Wisc. — When Lambeau Field welcomes in the 49ers on Saturday night, it will mark the Packers' first playoff game since … the field goal. Remember last year, when Packers coach Matt LaFleur settled for a field goal rather than let Aaron Rodgers go for it on fourth-and-goal from the 8-yard line, down by eight points? That call got second-guessed as Tom Brady and the Tampa Bay Buccaneers ate the final two minutes to win the NFC Championship Game, 31-26, en route to their Super Bowl championship. It extended the Packers' Lombardi Trophy drought to 10 years and spelled the latest heartbreak for Aaron Rodgers. All of which is to say: the Packers won't dare take the ball — and their fate — out of Rodgers' hands again, not if they want to retain the services of the league's MVP going fo
Dear Abby: Boyfriend flirts with other women while we're out together
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redpajama_set_name: RedPajamaCommonCrawl


With New York City still recovering from a blizzard that immobilized the five boroughs, the Bloomberg administration said it was ready for another blast of snow Friday that could dump up to five inches on the city and its suburbs.
How was the Mayor's office supposed to know the snow was coming? Are we mystics, simply because we sit in an open space and are all very rich? It's not as if there's some all-knowing person with a crystal ball who can somehow forecast the weather. . . .
The snow is a mystery to me. One day, no snow. The next, so much snow. Where does it come from? Will it stay forever? If only we could fashion a large wall or blade and attach it to a big truck and somehow push the snow away. Perhaps in the future. Perhaps Apple is working on it. The iPlow . . .
Who knew there were four other boroughs? Brooklyn, Queens, Staten Island, and the other one . . .
All is snow. All around me. Snow. White. Not the Disney movie. And I don't care for the word "dwarf." Wandering around my home, avoiding the news, what book do I pick up this evening? Go ahead, guess. The Stieg Larsson one? No. "White Noise." Bizarre, no? Snow is white. And dampens noise. I leaf through. The book speaks of "the airborne toxic event." What is a snowstorm but an airborne snowing event? The American mystery deepens. Like a snowdrift. It's late and we are out of wine and my pajamas are itchy. . . .
The criticism mounts. Someplace called the Bronks (sp?) remains snowbound. Am I missing something? Yes, there is a lot of snow. Yes, we haven't plowed it. Yes, the subways and buses aren't really running so well. So why don't people simply use their helicopters? . . .
More stories. More criticism. Tension. Stress. Can't sleep more than eight hours at a stretch. I watch the local news on mute and listen only to the commercials. "They fit like my real teeth," said one commercial. What does it mean? And what is Rice-A-Roni? How is it different from normal rice? "Snow" sounds like "no," which sounds like "joe," which is another word for coffee. Coffee is nice. The critics do not bother me. I am covered in a metaphorical nonstick Teflon, which I have heard might be linked to cancer, so maybe that's not a good metaphor, and in fact I had the staff remove all Teflon cookware from the kitchen. Where is the kitchen?
Listening to "Kings of Convenience" and sipping a '61 Cheval Blanc (only three hundred bottles left), singing one of the two man parts, wishing New Yorkers in all of the six boroughs would sing the harmony of the other man part with me and like me more. A billionaire needs love, too, you know. I'm likable. Likable Mike, they called me in B-school. That's a lie. I like Mike. Mike is aw'right. Mikey B. Mike Berg. Mike the Man. Do you know who's a good dancer? This guy. Two hundred and ninety-nine bottles left. Snow. Weird thing? Take the "s" off and it's "now." Anyone else ever think of that? No.
Rode in a "plow." It was big and fun, but the man smelled like the inside of a ski hat, if that ski hat had also fallen in a public toilet in France and then somehow found its way into a cheese toilet factory for a year. His name was Ramón, and we got to talking. "Ramón," I said, "where do you stay when you're in the Hamptons?" He gave me a look I couldn't quite figure out, until I realized that he wanted me to ask him about St. Barts. Which I did. But again the look. Gulfstream V? Look. Maidstone Club? Look. Mustique? Look. Which is when I realized he was pulling my leg. "You're pulling my leg, aren't you, Ramón?" I chuckled when I said it, which, if you knew me, you'd know that when I chuckle I can be very charming. Ramón chuckled, too, and said something in Spanish and then stopped the plow (which was on my street) and told me to get out. . . .
Imagining is fun. Late at night, alone with an Etch-A-Sketch and a box of Yodels, I want the poorly dressed reporters to ask me the questions again, and this time I will give them the real answers.
THE HORDE: Mr. Mayor! they shout (except for the Post, who calls me Lady Mayor Tiny Pants, which I don't find amusing in the least). Why haven't the streets been plowed?
THE HORDE: Mr. Mayor! Huge parts of Brooklyn remain unplowed.
ME: I don't go to Brooklyn. I'm allergic to hip. Maybe they should make an organic hemp plow. Then go have a coffee and go to Trader Joe's with your fancy strollers.
THE HORDE: Mr. Mayor! Do you think it snows in outer space? On Mars? Are they all superheroes there with a maximum height of three feet? Are you a kind of superhero who could rule Mars because of your height?
ME: Now, that's an interesting question. And, yes, I believe I am.
But the aides won't let me talk like that. Follow the script, they say. The Yodels are gone. . . .
The only thing I've ever found harder than moving all that snow is making Business Week an interesting magazine. . . .


text:
Qeshlaq-e Malek Qaleh (, also Romanized as Qeshlāq-e Malek Qal'eh) is a village in Qomrud Rural District, in the Central District of Qom County, Qom Province, Iran. At the 2006 census, its population was 240, in 51 families.

References

Populated places in Qom Province

redpajama_set_name: RedPajamaWikipedia


Mountain Equipment's take on what can be done with Gore-Tex Pro comes to life in the Lhotse Jacket. This best selling shell is made for mountaineers that require absolute protection. It is no surprise this piece has won numerous awards for function and craftsmanship few others can parallel. Do you want to climb an alpine north face in the winter or do you needing complete protection in the best blizzard Mother Nature can throw? This shell is all you will need.


Need Specialist (SQL, Delphi), for writing programs.
It is necessary to write a program to check the site for sql injections,POST,GET - and, or, error based.
Hello sir, I have vast experience of SQL Hope you look forward to work with me. Thanks!


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

In a previous work \cite{TV}, and related to the
so called inverse Galois problem, we raised the question of whether or not a given
effective group action on a manifold is determined, or ``described", by non-classic
tensors in general, or more specifically, by vector fields. More
precisely: Consider an effective action of a Lie group $G$ on an $m$-manifold $M$,
thus we can think of $G$ as a subgroup of the group
$\Diff(M)$ of diffeomorphisms of $M$. Given a vector field $X$ on $M$, we say $X$ is a {\it describing vector field} for the $G$-action if the following
hold:
\begin{enumerate}[label={\rm (\arabic{*})}]
\item $X$ is complete and its flow $\zF_t$ commutes with the action
of $G$; so $G\leq \Aut(X)$.
\item\label{(2)} The group homomorphism
$$\begin{array}{ccc}
G\zpor{\mathbb R}&\rightarrow& \Aut(X)\\
(g,t)&\mapsto&g\zci\zF_{t}
\end{array}$$
is an isomorphism.
\end{enumerate}

Notice that we compare $\Aut(X)$ with $G\zpor{\mathbb R}$ instead of $G$ since we always have to take into account the flow of $X$ (see Remark \ref{remA}).

Within this setting, the main result in \cite{TV} shows that
any finite group action on a connected
manifold admits a describing vector field.
Here we extend this result to toric actions.

\begin{theorem*}
Consider an effective action of the torus $\mathbb T^n$ on a
 connected $m$-manifold $M$. Assume that $n\zmei m-1$.
Then there exist a describing vector field for this action.
\end{theorem*}

The proof of the Theorem A involves three steps. First, the result is
established for free actions (Section \ref{secB}) and then extended to
effective ones (Section \ref{secC}), in both cases assuming $m-n\zmai 2$.
Finally the case $m-n=1$ is considered in Section  \ref{secD}.

\begin{remark}\label{remA}
{\rm
Notice that according to Proposition \ref{proAA},
if the $\mathbb T^n$-action on $M^m$ is effective then $n\zmei m$.

On one hand, when $n=m$, we can make the identification $M=\mathbb T^n$ endowed with
the natural  $\mathbb T^n$-action. In this case if $X$ is the fundamental vector field associated
to a dense affine vector field on $\mathbb T^n$ (see Section \ref{secA}), then
$\Aut(X)=\mathbb T^n$.

On the other hand, when $n<m$ no complete vector field $X$ on $M$ verifies
$\Aut(X)=\mathbb T^n$. Indeed, assume $\Aut(X)=\mathbb T^n$, and let
$X_1 ,\dots,X_n$ be a basis of the Lie algebra of fundamental vector fields
and $f_1 ,\dots,f_n$ any $\mathbb T^n$-invariant functions. Then
$[X_r ,\zsu_{j=1}^n f_j X_j ]=0$, $r=1,\dots, n$; so the flow of
$\zsu_{j=1}^n f_j X_j$ commutes with the action of $\mathbb T^n$ and,
as every element of the flow of $X$ belongs to $\Aut(X)$, with this flow too.
That is to say the  flow of $\zsu_{j=1}^n f_j X_j$ is included in $\Aut(X)$
and, necessarily, $\Aut(X)\znoi \mathbb T^n$ {\it contradiction}.

Thus our result is ``minimal" because the flow of $X$ is always included in $\Aut(X)$.}
\end{remark}

\begin{remark}\label{remB}
{\rm
Finally, notice that Theorem A cannot be extended to a general compact Lie group. In Section \ref{secE} we construct effective actions of
$SO(3)$ (Example \ref{ejeB}), and  of a non-connected
compact group of dimension two (Example \ref{ejeC}), for which there is no describing vector field.
}
\end{remark}

\noindent \textbf{Terminology:} The reader is supposed to be familiarized with our previous
paper  \cite{TV}. All structures and objects considered are real
$C^{\zinf}$ and manifolds are without boundary, unless another thing
is stated. For the general questions on Differential Geometry the reader is
referred to \cite{KN} and for those on Differential Topology to
\cite{HI}.
\medskip

\noindent \textbf{Acknowledgements:} The authors would like to thank
Prof.\ Arthur Wasserman for suggesting that our original result on finite
group actions could be extended to $S^1$-actions,
and for his helpful comments on the development of this work.

\section{Preliminary results on vector fields}\label{secA}
In this section we collect some results on vector fields that are needed in the following sections.
On $\mathbb R^k$ we set coordinates $x=(x_1 ,\dots,x_k )$, and define
 $\zx=\zsu_{j=1}^{k}x_{j}\zpar/\zpar x_{j}$.

\begin{lemma}\label{lemA}
For any function $g\colon \mathbb R^k \zfl\mathbb R$ with $g(0)=0$
there is a function $f\colon \mathbb R^k \zfl\mathbb R$ such that
 $\zx\zpu f=g$.
\end{lemma}
\begin{proof}
Although the proof of this result is routine we outline it here. Let
$\mathcal T$ denote the Taylor's series of $g$ at $0\zpe\mathbb R^k$.
Then there exists a series $\mathcal S$ such that formally
$\zx\zpu\mathcal S=\mathcal T$. According to Borel's Theorem \cite[Theorem 1.5.4]{NA}
there is a function $\zf$ whose Taylor's series at origin
equals $\mathcal S$, and therefore making $g-\zx\zpu\zf$ we may
suppose $\mathcal T=0$.

Since $\zx$ is hyperbolic at origin, following \cite[Theorem 10, page 38]{RRO}
there exist a function
$\tilde f\colon\mathbb R^k \zfl\mathbb R$ such that $g-\zx\zpu\tilde f$
vanishes on a open neighborhood $A$ of $0\zpe\mathbb R^k$.
Therefore it suffices to show the result when $g_{\zbv A}=0$.

But $\mathbb R^k -\{0\}$ can be identified to $S^{k-1}\zpor\mathbb R$
in such a way that $\zx=\zpar/\zpar t$ where $t$ is the variable in
$\mathbb R$ and $S^{k-1}\zpor(-\zinf,1)$ corresponds to
$B_\ze (0)-\{0\}\zco A$ for some radius $\ze>0$.

Finally set $f=\zil_{0}^{t}g{\operatorname d}s$ on $S^{k-1}\zpor\mathbb R$
and $f(0)=0$.
\end{proof}

Recall that a vector field $T$ on $\mathbb T^n$, endowed with coordinates
$\zh=(\zh_1 ,\dots,\zh_n )$, is named {\it affine} if
$T=\zsu_{r=1}^{n}a_{r}\zpar/\zpar \zh_{r}$ where
$a_1 ,\dots,a_n \zpe \mathbb R$. Then, the trajectories of an affine
vector field are dense if and only if $a_1 ,\dots,a_n$ are rationally
independent; {\it in this case we say that $T$ is dense}.

\begin{lemma}\label{lemB}
On $\mathbb R^k \zpor \mathbb T^n$ with coordinates
$(x,\zh)=(x_1 ,\dots,x_k ,\zh_1 ,\dots,\zh_n )$ consider the vector field
$X=\zx+T$, where $T$ is dense. Then $\L_X$, the set of all vector fields on $\mathbb R^k \zpor \mathbb T^n$
which commute with $X$, is a Lie algebra of dimension $k^2 +n$ with basis
$$\left\{ x_j {\frac {\zpar} {\zpar x_{\zlma}}},{\frac {\zpar}
{\zpar \zh_r}}\right\},\,\, j,\zlma=1,\dots,k;\, r=1,\dots,n.$$
\end{lemma}

\begin{proof}
Let $Y\in\L_X$ be the vector field defined as $Y=\zsu_{j=1}^k f_j (x,\zh)\zpar/\zpar x_j
+\zsu_{r=1}^n g_r (x,\zh)\zpar/\zpar \zh_r$, and let $\zF_t$
denote its flow. Since $\{0\}\zpor\mathbb T^n$ is compact,
$\zF_t$ is defined on $(\{0\}\zpor\mathbb T^n )\zpor(-\ze,\ze)$ for some
$\ze>0$.

Observe that $\{0\}\zpor\mathbb T^n$ is the set of all points whose
$X$-trajectory has compact adherence. Therefore
$\zF_t (\{0\}\zpor\mathbb T^n )\zco\{0\}\zpor\mathbb T^n$ for any
$t\zpe(-\ze,\ze)$, what implies that $Y$ is tangent to
$\{0\}\zpor\mathbb T^n$, and therefore $f_j (0,\zh)=0$, for $j=1,\dots,k$.

A computation, taking into account that every $\zpar/\zpar\zh_r$ commutes
with $X$, yields $$[X,Y]=\widetilde Y
+\zsu_{r=1}^n (X\zpu g_r )\zpar/\zpar \zh_r$$ where $\widetilde Y$ is a
functional combination of $\zpar/\zpar x_i$'s.
Therefore $X\zpu g_r =0$, i.e.\ $g_r$ is constant
along the trajectories of $X$, for $r=1,\dots,n$. But the $\zw$-limit of these trajectories is
$\{0\}\zpor\mathbb T^n$ and $g_r$ is constant on this set because
$\{0\}\zpor\mathbb T^n$ is the adherence of the trajectory of any of its
points, so each  $g_r$ is constant on
$\mathbb R^k \zpor \mathbb T^n$.

By considering $Y-\zsu_{r=1}^n g_r \zpar/\zpar \zh_r$ one may suppose
$Y=\zsu_{j=1}^k f_j (x,\zh)\zpar/\zpar x_j$ where each
$f_j (\{0\}\zpor\mathbb T^n)=0$. Now from $[X,Y]=0$ follows
$X\zpu f_j =f_j$, $j=1,\dots,k$.

On the other hand given $f\colon\mathbb R^k \zpor\mathbb T^n
\zfl\mathbb R$ such that $f(\{0\}\zpor\mathbb T^n)=0$ and $X\zpu f=f$,
then $f$ does not depend on $\zh$ and it is linear on $x$. Indeed if $k=0$
it is obvious; assume $k=1$ by the moment. Then $f=xg$ for some function
$g$ since $f$ vanishes on $\{0\}\zpor\mathbb T^n$, and $X\zpu f=f$
becomes $X\zpu g=0$. Therefore $g$ is constant along the trajectories of $X$
and, by the same reason as before, constant on
$\mathbb R\zpor\mathbb T^n$.

Now suppose $k\zmai 2$. Let $E$ be any vector line in $\mathbb R^k$. As
$X$ is tangent to $E\zpor\mathbb T^n$ and the restriction of $\zx$ to $E$
is still the radial vector field, $f\colon E\zpor\mathbb T^n \zfl\mathbb R$ is
independent of $\zh$ and linear on $E$ (it is just the case $k=1$).
Since the union of all the vector lines
$E$ equals $\mathbb R^k$, it follows that $f$ does not depend on $\zh$ and
$f\colon\mathbb R^k \zfl \mathbb R$ is linear on each $E$. But, as it is well
known, this last property implies that $f\colon\mathbb R^k \zfl \mathbb R$
is linear.

In short every $f_j$ is linear on $x$ and independent of $\zh$.
\end{proof}

\begin{corollary}\label{cor_de_2.2}
If $F\colon \mathbb R^k \zpor \mathbb T^n \zfl
 \mathbb R^k \zpor \mathbb T^n$ is an automorphism of
$X=\zx+T$ and $T$ is dense, then there exist an isomorphism $\zf\colon\mathbb R^k
\zfl\mathbb R^k$ , and an element $\zl\zpe \mathbb T^n$ such that
$F(x,\zh)=(\zf(x),\zh+\zl)$.
\end{corollary}

\begin{proof}
The diffeomorphism $F$ induces an isomorphism on $\L_X$, the Lie algebra described in Lemma~\ref{lemB}. Now observe that:
\begin{enumerate}
\item The
only elements in $\L_X$ with singularities are the elements $\zsu_{j,\zlma=1}^k a_{j\zlma}x_j\zpar/\zpar x_{\zlma}$, where
$(a_{j\zlma})\zpe\GL(n,\mathbb
R)$. Besides they give rise to the foliation
$d\zh_1 =\dots=d\zh_n =0$ (extend it to $\{0\}\zpor\mathbb T^n$ by
continuity); so this foliation is an invariant of $F$.

\item The center of $\L_X$
is spanned by $\zx, \zpar/\zpar\zh_1 ,
\dots,\zpar/\zpar\zh_r$. But the adherences of the trajectories of a vector
field  $b\zx+b_1 \zpar/\zpar\zh_1 +\dots+b_r \zpar/\zpar\zh_r$
are always tori if
and only if $b=0$, so $F$ sends the Lie subalgebra
spanned by $ \zpar/\zpar\zh_1 ,\dots,\zpar/\zpar\zh_r$ into itself.
\end{enumerate}
These two facts imply that $F(x,\zh)=(\zf(x),\zq(\zh))$ where
$\zq\colon\mathbb T^n \zfl\mathbb T^n$ is an affine transformation
of $\mathbb T^n$; that is
$$\zq(\zh)=\left(\zsu_{j=1}^{n}c_{1j}\zh_{j},\dots,
\zsu_{j=1}^{n}c_{nj}\zh_{j})\right)+\zl$$
where $(c_{\zlma j})\zpe\GL(n,\mathbb Z)$ and $\zl\zpe\mathbb T^n$.
As $X=T$ on $\{0\}\zpor\mathbb T^n$, which is an invariant of $F$,
it follows  that $(a_1 ,\dots,a_n )$ is an eigenvector
of $(c_{\zlma j})$ whose eigenvalue
equals 1. But in this case $a_1 ,\dots,a_n$ are rationally dependent unless
$(c_{\zlma j})=Id$. In short $\zq(\zh)=\zh+\zl$.

On the other hand $\zf\colon\mathbb R^k \zfl\mathbb R^k$ has to be an automorphism of $\zx$, which implies that $\zf$ is an isomorphism
\cite[Lemma 3.4]{TV}.
\end{proof}

\begin{lemma}\label{lemC}
On $\mathbb R^k \zpor \mathbb T^n$ one considers the vector field
$\widetilde X =\tilde\zx +\widetilde T$ where
$\tilde\zx=\zsu_{j=1}^{k}\tilde f_{j}(x)\zpar/\zpar x_{j}$ and
$\widetilde T=\zsu_{i=1}^{n}\tilde g_{r}(x)\zpar/\zpar \zh_{r}$.
Assume that on $\mathbb R^k$ the following hold:
\begin{enumerate}[label={\rm (\alph{*})}]
\item $\tilde\zx$ is complete,
\item $\tilde\zx(0)=0$ and its linear part at the origin is a positive
multiple of identity,
\item the outset of the origin equals $\mathbb R^k$.
\end{enumerate}

Then there exists a self-diffeomorphism of $\mathbb R^k \zpor \mathbb T^n$
that commutes with the natural $\mathbb T^n$-action and
transforms $\widetilde X$ into
$$b\zx+\zsu_{r=1}^{n}b_{r} {\frac {\zpar} {\zpar \zh_r}}$$
where $b\zpe\mathbb R^+$ and $b_1 ,\dots,b_n \zpe\mathbb R$.
\end{lemma}

\begin{proof}
By Sternberg's linearization theorem \cite{SST},
see \cite[Proposition 2.1]{TV}, there exists a diffeomorphism $f\colon\mathbb R^k \zfl\mathbb R^k$
transforming $\tilde\zx$ into $b\zsu_{j=1}^k x_j \zpar/\zpar x_j$ with $b>0$.  Dividing by $b$ we may suppose
$\tilde\zx=\zx$.

By Lemma \ref{lemA} there are functions $\zf_1 ,\dots,\zf_n \colon\mathbb
R^k \zfl\mathbb R$ such that $\zx\zpu\zf_r =g_r -g_r (0)$, $r=0,\dots,n$.
Now, if $\zf=(\zf_1 ,\dots,\zf_n )$ and
$\tilde\zp\colon\mathbb R^n \zfl\mathbb T^n$ is the canonical covering, then the diffeomorphism $F\colon \mathbb R^k \zpor \mathbb T^n \zfl
 \mathbb R^k \zpor \mathbb T^n$ given by $F(x,\zh)
=(x,\zh-\tilde\zp\zci\zf)$,
transforms $\widetilde X$ into $\zx+\zsu_{r=1}^n g_r (0)\zpar/\zpar\zh_r$.
\end{proof}

\begin{remark}\label{remB2}
{\rm If $\widetilde X$ matches the hypotheses of Lemma \ref{lemC} and
$h\colon \mathbb R^k \zfl \mathbb R$ is a positive bounded function, then
$h\widetilde X$ satisfies these hypotheses too (when $h$ is regarded  as
 a function on  $\mathbb R^k \zpor \mathbb T^n$ in the obvious way).
Therefore $h\widetilde X$ can be written as in Lemma \ref{lemC} for a
suitable choice of coordinates, and thus the control of its automorphisms
becomes simple.}
\end{remark}

\section{Free actions}\label{secB}
In this section we assume there is a free $\mathbb T^n$-action on the
connected $m$-manifold $M$, where $m-n\zmai 2$. This gives rise to a principal fibre
bundle $\zp\colon M\zfl B$ whose structure group is $\mathbb T^n$ and
$B$ is a connected manifold of dimension $k=m-n\zmai 2$. Then, we construct
a suitable vector field on $B$ that is later on lifted
to $M$ by means of a connection.

In order to construct the vector field on $B$, we closely follow along the lines in
 \cite[Section 3]{TV} applied to the case of the trivial group action on $B$.

Consider a Morse function $\zm\colon B\zfl{\mathbb R}$ that is
proper and non-negative. Denote by $C$ the set of its
critical points, which is closed and discrete, and therefore countable. As $B$ is
paracompact, there exists a locally finite family $\{A_{p}\}_{p\zpe C}$
of disjoint open set such that $p\zpe A_{p}$, $p\zpe C$.

Following along the lines in \cite[Section 3]{TV}, there exist a
Riemannian metric ${\tilde g}$ on $B$ such that the gradient vector
field $Y$ of $\zm$ is complete and, besides, around each $p\zpe C$
there are coordinates $(x_{1},\ldots,x_{k})$ with $p\zeq 0$ and
$Y=\zsu_{j=1}^{k}\zl_{j}x_{j}\zpar/\zpar x_{j}$,
$\zl_1 ,\dots,\zl_k \zpe\mathbb R -\{0\}$, where

\begin{enumerate}[label={\rm (\arabic{*})}]
\item $\zl_1 =\dots=\zl_k >0$  if $p$ is a source of $Y$, that is a
 minimum of $\zm$,
\item $\zl_1 =\dots=\zl_k <0$  if $p$ is a sink of $Y$ (a maximum
of $\zm$),
\item some $\zl_j$ are positive and the remainder negative if $p$ is
a saddle.
\end{enumerate}

Note that these properties still hold when $Y$ is multiplied by a positive
bounded function, since they only depend on the
Sternberg's Theorem.

Let $I$ be the set of local minima of $\zm$, that is the set of
sources of $Y$, and $S_{i}$, $i\zpe I$, the outset of $i$ relative
to $Y$. Now Lemma 3.3 of \cite{TV} becomes:

\begin{lemma}\label{lemD}
The family $\{ S_{i}\}_{i\zpe I}$ is locally finite and the
set $\zung_{i\zpe I}S_{i}$ is dense in $B$.
\end{lemma}

In what follows, and by technical reasons, one makes use of the notion of
\emph{order of nullity} instead of \emph{chain}. More exactly, for every $i\zpe I$ we
choose a subset $P_i$ of $A_i$ with $k+1$ points close enough to $i$
but different from it, in such a way that the linear $\za$-limits of their
trajectories are in general position (see \cite[pags.\ 319 and 320]{TV}
for definitions).

Set $P=\zung_{i\zpe I}P_i$. Consider an injective map
$p\zpe P\mapsto n_p \zpe\mathbb N -\{0\}$. Let $\mathbb N'$ be the
image of $P$.

By definition a differentiable object has {\it order $r$ at point $q$} if its
$(r-1)$-jet at this point vanishes but its $r$-jet does not; for instance
$Y$ has order one at sources, sinks and saddles.

Since $\{A_i \}_{i\zpe I}$ is still locally finite, one may construct a bounded
function $\zt\colon B\zfl\mathbb R$ such that $\zt$ is positive on $B-P$
and has order $2n_p$ at every $p\zpe P$.

Put $Z=\zt Y$. Then $Z^{-1}(0)=Y^{-1}(0)\zun P$; that is the singularities
of $Z$ are the sources, sinks and saddles of $Y$ plus the points of $P$,
that we call {\it artificial singularities} and whose order is even $\zmai 2$.
Note that two different artificial singularities have different orders.

Let $R_i$, $i\zpe I$, be the $Z$-outset of $i$. As $S_i -R_i$ is the union of
$k+1$ half-trajectories of $Y$ one has:

\begin{lemma}\label{lemE}
The family $\{ R_{i}\}_{i\zpe I}$ is locally finite and the
set $\zung_{i\zpe I}R_{i}$ is dense in $B$.
\end{lemma}

On the principal fibre bundle $\zp\colon M\zfl B$ consider a connection
$\mathcal C$ which is a product around every fibre $\zp^{-1}(p)$,
$p\zpe C$ (that is there exist an open set $p\zpe A\zco B$ and a fiber bundle
isomorphism between  $\zp\colon \zp^{-1}(A)\zfl A$
and $\zp_1 \colon A\zpor\mathbb T^n \zfl A$ in such a way that
$\mathcal C$, regarded on $\zp_1 \colon A\zpor\mathbb T^n \zfl A$,
is given by $\mathcal C (q,\zh)=T_q A\zpor \{0\}\zco
T_{(q,\zh)}(A\zpor\mathbb T^n )$).
This kind of connection always exists because $\{ A_{p}\}_{p\zpe C}$ is
 locally finite.

Let $Y'$ denote the lift of $Y$ to $M$ by means of $\mathcal C$; that
is $Y'(u)\zpe\mathcal C (u)$ and $\zp_{*}(Y'(u))=Y(\zp(u))$ for every
$u\zpe M$. By construction $Y'$ is $\mathbb T^n$-invariant and
 $Y'(u)=0$ if and only if $Y(\zp(u))=0$.

Let $T$ be a dense affine vector field on $\mathbb T^n$ and $T'$ the
fundamental vector field, on $M$, associated to $T$ through the action. As
describing vector field we take $X'=(\zt\zci\zp)(Y'+T')$, which clearly is
$\mathbb T^n$-invariant and complete.

The remainder of this section is devoted to show that $X'$ is a
describing vector field. First we study the behavior of $X'$ near some
fibres. If $p$ is a source of $Y$ there exist coordinates $(x_1 ,\dots,
x_k )$, about $p\zpe B$, with $p\zeq 0$ and
$Y=a\zsu_{j=1}^{k}x_{j}\zpar/\zpar x_{j}$, $a>0$. As around
$\zp^{-1}(p)$ the connection is a product, these coordinates can be
prolonged to a system of coordinates $(x,\zh)$ on a product open set
$A\zpor\mathbb T^n$, with the obvious identifications, while
$\mathcal C$ is given by the first factor. In this case
$$Y'+T'=a\zsu_{j=1}^{k}x_{j}{\frac {\zpar} {\zpar x_j}}+T$$
since $T'$ is just $T$ regarded as a vector field on
$A\zpor\mathbb T^n$.

The same happens when $p$ is a sink but $a<0$. If $p$ is a saddle then
the model of the first part is
$\zsu_{j=1}^{k}\zl_{j}x_{j}\zpar/\zpar x_{j}$ with some $\zl_j$ positive
and the others negative.

Thus, when $p\zpe C$, the torus $\zp^{-1}(p)$  is the adherence of a
trajectory of $X'$, this vector field never vanishes on $\zp^{-1}(p)$ and,
besides:
\begin{enumerate}[label={\rm (\alph{*})}]
\item If $p\zpe I$, then $\zp^{-1}(p)$ is the $\za$-limit of some external
trajectories but never the $\zw$-limit.
\item If $p$ is a sink, then $\zp^{-1}(p)$ is the $\zw$-limit of some external
trajectories but never the $\za$-limit.
\item If $p$ is a saddle, then $\zp^{-1}(p)$ is the $\za$-limit of some external
trajectories and the $\zw$-limit of other ones.
\end{enumerate}

On the other hand $(X')^{-1}(0)=\zp^{-1}(P)$. Moreover if $p\zpe P$
then $X'$ has order $2n_p$ at each point of $\zp^ {-1}(p)$.

If $X'$ is multiplied by a positive, bounded and $\mathbb T^n$-invariant
function $\zr\colon M\zfl\mathbb R$, then $\zr=\tilde\zr\zci\zp$ for some
positive and bounded function $\tilde\zr\colon B\zfl\mathbb R$. Thus
$\zr X'= ((\tilde\zr \zt)\zci\zp)(Y'+T')$. As $\zt$ and $\tilde\zr \zt$ have
 the same  essential properties, the foregoing description still holds for
$\zr X'$. In other words, this description is geometric and independent
of how trajectories are parameterized.

Consider $i\zpe I$ and identify its $Y$-outset $S_i$ to $\mathbb R^k$ in
such a way that $i\zeq 0$ and
$Y=a\zsu_{\zlma=1}^{k}x_{\zlma}\zpar/\zpar x_{\zlma}$, $a>0$. As $S_i$
is contractible the fibre bundle $\zp\colon\zp^{-1}(S_i )\zfl S_i$ is trivial;
so it can be regarded like
$\zp_1 \colon\mathbb R^k \zpor\mathbb T^n\zfl\mathbb R^k$ while
$$Y'+T'=a\zsu_{\zlma=1}^{k}x_{\zlma}{\frac {\zpar} {\zpar x_{\zlma}}}
+\zsu_{r=1}^{n}g_{r}(x){\frac {\zpar} {\zpar \zh_r}}.$$

Finally Lemma \ref{lemC} allows us to suppose
$$Y'+T'=a\zsu_{\zlma=1}^{k}x_{\zlma}{\frac {\zpar} {\zpar x_{\zlma}}}
+\zsu_{r=1}^{n}a_{r}{\frac {\zpar} {\zpar \zh_r}}$$
with $a>0$ and $a_1 ,\dots,a_n \zpe\mathbb R$. Moreover since
$T'=\zsu_{r=1}^{n}a_{r}\zpar/\zpar \zh_{r}$ at any point of
$\{0\}\zpor\mathbb T^n$, scalars $a_1 ,\dots,a_n$ are rationally
independent (recall that $Y'(u)=0$ whenever $Y(\zp(u))=0$).

Now is clear that, for every $p\zpe P_i$ and $u\zpe \zp^{-1}(p)$, there
exists a trajectory of $X'=(\zt\zci\zp)(Y'+T')$ whose $\za$-limit and $\zw$-limit are
$\zp^{-1}(i)$ and  $u$ respectively. As $n_p \znoi n_{p'}$ when
$p\znoi p'$, the existence of this kind of trajectories shows that any
automorphism of $X'$ has to send $\zp^{-1}(i)$ in itself and, consequently,
the $X'$-outset of $\zp^{-1}(i)$ in itself too (here outset means the set
of points whose trajectory has its $\za$-limit included in $\zp^{-1}(i)$).

The next question is to determine this outset. First we identify the outset
$R_i$, of $i$ with respect to $Z=\zt Y$, to $\mathbb R^k$ in such a way
that $i\zeq 0$ and $Z=b\zsu_{\zlma=1}^{k}x_{\zlma}\zpar/\zpar
x_{\zlma}$, $b>0$. Again  $\zp\colon\zp^{-1}(R_i )\zfl R_i$ is trivial and
reasoning as before, taking into a account that $X'$ is complete on
$\zp^{-1}(R_i )$ which allows to apply Lemma \ref{lemC}, leads us to the
case $\zp^{-1}(R_i )\zeq\mathbb R^k \zpor \mathbb T^n
\zfl\mathbb R^k \zeq R_i$ and
$$X'=b\zsu_{\zlma=1}^{k}x_{\zlma}{\frac {\zpar} {\zpar x_{\zlma}}}
+\zsu_{r=1}^{n}b_{r}{\frac {\zpar} {\zpar \zh_r}}$$
where $b>0$ and $b_1 ,\dots,b_n \zpe \mathbb R$ are rationally
independent. Therefore the $X'$-outset of $\zp^{-1}(i)$ equals
$\zp^{-1}(R_ i)$.

Let $F\colon M\zfl M$ be an automorphism of $X'$. Then
$F\colon\zp^{-1}(R_ i)\zfl \zp^{-1}(R_ i)$ is a diffeomorphism. As the
trajectories of $\zsu_{r=1}^{n}b_{r}\zpar/\zpar \zh_{r}$ are dense, from
Corollary \ref{cor_de_2.2} it follows that
$F(x,\zh)=(\zf(x),\zh+\zl)$, where $\zl\zpe \mathbb T^n$ and
$\zf\colon\mathbb R^k\zfl\mathbb R^k$ is an isomorphism.

For any $p\zpe P_i$ some trajectories of $X'$ with $\za$-limit
$\zp^{-1}( i)$ have, as $\zw$-limit, a point of $\zp^{-1}(p)$, that is a
singularity of order $2n_p$. Since $n_p \znoi n_{p'}$ when $p\znoi p'$,
the set of these trajectories has to be an invariant of $F$.
Regarded on $R_i \zco B$, and taking into account that $X'$ projects in
$Z$, this fact implies that $\zf$ has to map the trajectory of
$Z=b\zsu_{\zlma=1}^{k}x_{\zlma}\zpar/\zpar x_{\zlma}$
of $\za$-limit $i$ and $\zw$-limit $p$ into itself. Thus the direction vector of this
curve is an eigenvector of $\zf$ with positive eigenvalue. But there are
$k+1$ eigenvector (as many as points in $P_i$) and they are in general
position, so $\zf$ is a positive multiple of identity.

Therefore there exists $t_i$ such that
$\zF_{t_i}(x,\zh)=(\zf(x),\zh+\tilde\zl_i)$ where $\zF_{t}$ denotes the
flow of $X'$ and $\tilde\zl_i \zpe\mathbb T^n$. Since $\zF_{t}$ and the
action of $\mathbb T^n$ commute, that shows the existence of
$t_i \zpe\mathbb R$ and $\zl_i \zpe\mathbb T^n$ such that
$F=\zl_i \zci\zF_{t_i}$ on $\zp^{-1}(R_i )$.

It easily seen that the family $\{\zp^{-1}( R_{i})\}_{i\zpe I}$ is locally
finite and the set $\zung_{i\zpe I}\zp^{-1}( R_{i})$ is dense in $M$.

On each $\zp^{-1}( R_{i})$ the action of $\mathbb T^n$ and $F$ commute,
so they do on $M$. Thus $F$ induces a diffeomorphism $f\colon B\zfl B$
such that $f\zci\zp=\zp\zci F$. Besides, since $Z$ is the projection of $X'$,
our $f$ is an automorphism of $Z$. Now from the expression of
$F\colon\zp^{-1}( R_{i})\zfl\zp^{-1}( R_{i})$ it follows that $f=\zf_{t_i}$ on
$R_i$, $i\zpe I$, where $\zf_t$ is the flow of $Z$.

\begin{lemma}\label{lemF}
All $t_i$'s are equal and $f=\zf_t$ for some $t\zpe\mathbb R$.
\end{lemma}

\begin{proof} Notice that $X$ has no regular periodic trajectories
(here dimension of $B$ $\zmai 2$ is needed). Then, the lemma follows from
the proof of Lemma 3.7 in \cite{TV}, when $G$ is the trivial group and $X$ becomes $Z$.
\end{proof}

Therefore composing $F$ with $\zF_{-t}$ allows to suppose that $f$ is
the identity and $F=\zl_i$ on each $\zp^{-1}( R_{i})$, $i\zpe I$.

Consider a $\mathbb T^n$-invariant Riemannian metric on $M$. Then
$F$ is an isometry on every $\zp^{-1}( R_{i})$ so, by continuity, on $M$.
Take $i_0 \zpe I$; then the isometries $F$ and $\zl_{i_0}$ agree on
$\zp^{-1}( R_{i_0})$. But on connected manifolds, isometries are
determined by their $1$-jet at any point. Therefore $F=\zl_{i_0}$
on $M$. In other words
$(\zl,t)\zpe \mathbb T^n\zpor{\mathbb R}\zfl \zl\zci\zF_t \zpe\Aut(X)$
is an epimorphism.

We now prove injectivity. Assume $\zl\zci\zF_t =Id$, that is
$\zF_t =(-\zl)$. Then $\zf_t\colon B\zfl B$ equals the identity because
$(-\zl)$ induces this map on $B$.
Since $Z$ has no periodic regular trajectories this implies $t=0$ and,
finally, $\zl=0$ because the action of $\mathbb T^n$ is effective.
{\it In short $X'$ is a describing vector field for free actions.}

\begin{remark}\label{remC}
{\rm
Note that if $\zr\colon M\zfl\mathbb R$ is a $\mathbb T^n$-invariant,
positive and bounded function, then $\zr X'$ is a describing vector field too.
Indeed, there exists a function $\tilde\zr\colon B\zfl\mathbb R$ such that
$\zr=\tilde\zr\zci\zp$ and it suffices reasoning with $\tilde\zr\zt$
instead of $\zt$.}
\end{remark}

\section{Effective actions}\label{secC}
Throughout this section we assume that $m-n\zmai 2$ and the
 $\mathbb T^n$-action is effective. Our next goal is to construct a describing vector
field on $M$. Let $S$ be the set of those
 points of $M$ whose isotropy (stabilizer)
is non trivial. By Proposition \ref{proAA}
 the set $M-S$ is dense, open, connected and $\mathbb T^n$-invariant. Moreover the
$\mathbb T^n$-action on $M-S$ is free, so we can consider on this set a
describing vector field $X'$ as in Section \ref{secB}. Let $\zf$ be a function
like in Proposition~\ref{proAB} for $X'$ and $M-S$, and
$\widehat X'$ be the vector field on $M$ given by $\zf X'$ on $M-S$ and
zero on $S$.

Set $\zq=h\zci\zf$ where $h\colon\mathbb R\zfl\mathbb R$ is defined
by $h(t)=0$ if $t\zmei 0$ and $h(t)=exp(-1/t)$ if $t>0$. The function
$\zq$, which is $\mathbb T^n$-invariant, vanishes at order infinity at
every point of $S$, i.e.\ all its $r$-jets vanish. Besides, it is bounded on
$M$ and positive on $M-S$.

Now put $\widetilde X=\zq\widehat X'$. Clearly $\widetilde X$
vanishes at order
infinity at  $u$ if and only if $u\zpe S$; therefore $S$ is an invariant of
$\widetilde X$. Moreover $\widetilde X$ is complete on $M$ and $M-S$
respectively. By Remark~\ref{remC} our $\widetilde X$ is a describing
vector field on $M-S$ since $\widetilde X=(\zq\zf)X'$ on this set.

If $F\colon M\zfl M$ is an automorphism of $\widetilde X$ then $F(S)=S$,
and  $F\colon M-S\zfl M-S$ is an automorphism of $\widetilde X$; so on
$M-S$ one has $F=\zl\zci\widetilde\zF_t$, where $\widetilde\zF_t$ is
the flow of $\widetilde X$. By continuity $F=\zl\zci\widetilde\zF_t$
everywhere, which implies that $\widetilde X$ is a describing vector field
on $M$ (the homomorphism injectivity is inherited from $M-S$).

\section{The case $m-n=1$}\label{secD}
First assume the action is free, which gives rise to a principal fibre bundle
$\zp\colon M\zfl B$ with $B$ connected and of dimension one. Therefore $B$
is $\mathbb R$ or $S^1$ and $\zp\colon M\zfl B$ can be identified to
$\zp_1 \colon B\zpor\mathbb T^n \zfl B$. One will need the following result
whose proof is routine.

\begin{lemma}\label{lemG}
On a open set $0\zpe A\zco\mathbb R$ consider a vector field $X$ such that
its $(r-1)$-jet at origin vanishes but its $r$-jet does not, $r\zmai 1$.
Let $\zf_t$ be the flow of $X$. Given $f\colon A\zfl\mathbb R$ and
$t_1 ,t_2 \zpe\mathbb R$, if $f=\zf_{t_1}$ on $A\zin (0,\zinf)$ and
$f=\zf_{t_2}$ on $A\zin (-\zinf,0)$ then $t_1 =t_2$.
\end{lemma}

Set $Y=q(q^2 +1)^{-1}\zpar/\zpar x$, where
$q=x(x-1)(x-2)(x-3)(x-4)$ when $B=\mathbb R$, and
$Y=\sin(3\za)\zpar/\zpar\za$ if $B=S^1$ endowed with the angular
coordinate $\za$. Clearly $Y$ is complete. In the first case the sources are
$0,2,4$ and the sinks $1,3$, and in the second one $0,2\zp/3,4\zp/3$
and $\zp/3,\zp,5\zp/3$ respectively.

When $\dim B\zmai 2$ we have created new singularities called artificial. Now
instead of that one will increase the order of sinks (otherwise the non-singular
set has too many components). Let $\zt\colon B\zfl \mathbb R$ be  a bounded
function which is positive outside sinks and has order two at $1$ and
$\zp/3$, order four at $3$ and $\zp$ and, finally, order six at
$5\zp/3$. Set $Z=\zt Y$, which is a complete vector field.

This time the $Z$-outsets $\{R_i \}_{i\zpe I}$, where $I=\{0,2,4\}$ if
$B=\mathbb R$ or $I=\{0,2\zp/3,4\zp/3\}$ if $B=S^1$, equal those
of $Y$, and any of them is an invariant of $Z$ because the $\zw$-limits
of its trajectories have different orders or are empty; even more,
every side of the outset has to be preserved.

On $M=B\zpor\mathbb T^n$ with coordinates $(x,\zh)$ or $(\za,\zh)$ set
$X'=\zt(Y+T)$, where $T$ is a dense affine vector field and $\zt, Y, T$
are regarded on $M=B\zpor\mathbb T^n$ in the natural way.
Now $(X')^{-1}(0)=(B-\zung_{i\zpe I}R_i )\zpor\mathbb T^n$. Therefore
$(X')^{-1}(0)$ is the union of two ($B=\mathbb R$) or three ($B=S^1$)
fibres; moreover points of different fibres have different orders.

Let $F\colon M\zfl M$ be an automorphism of $X'$; reasoning as in the case
$\dim B\zmai 2$ shows that $F=\zl_i \zci \zF_{t_i}$ on
$R_i \zpor\mathbb T^n$ for each $i\zpe I$, where $\zF_t$ is the flow of
$X'$. Thus the induced automorphism $f\colon B\zfl B$ of $Z$ equals
$\zf_{t_i}$ on $R_i$, where $\zf_t$ is the flow of $Z$. Now Lemma
\ref{lemG} implies that $t_i =t_j$ if $R_i$ and $R_j$ are contiguous. In short
$f=\zf_t$ for some $t\zpe\mathbb R$. The remainder of the proof is
similar to that of $\dim B\zmai 2$.

Finally notice that $\zr X'$ is a describing vector field too if
$\zr\colon M\zfl\mathbb R$ is $\mathbb T^n$-invariant, positive and
bounded; therefore the result can be extended to the case of an effective ç
action following the lines in Section \ref{secC}.

\section{Examples}\label{secE}

One starts this section by giving two examples of effective toric actions. The
first one is a general construction on the Lie groups. In the second example
a describing vector field is constructed for the usual action of $\mathbb T^3$
on $S^3$.

On the other hand two examples more show that the main theorem fails for
general compact Lie groups. More exactly for effective actions of
$SO(3)$ (Example \ref{ejeB}) and for effective actions of a non-connected
compact group, of dimension two, with abelian Lie algebra (Example \ref{ejeC}).

\begin{example}\label{ejeAcero}
{\rm Let $G$ be a connected Lie group, with center $ZG$, and
two (non necessary maximal) tori $H,\widetilde H\leq G$. Then there exists a
$(H\zpor\widetilde H)$-action on $G$ given by
$(h,\tilde h )\zpu g= hg{\tilde h}^{-1}$, whose kernel $K$ equals
$\{(h,h)\zbv h\zpe H\zin\widetilde H\zin ZG\}$. Thus an effective action
 of the torus $(H\zpor\widetilde H)/K$ on $G$ is induced.

Now suppose that $G$ is compact with rank $r$ and $ZG$ finite,
that is the center of the Lie algebra of $G$ is zero. Let $H$ be a
maximal torus of $G$; set  $\widetilde H=H$. Then one obtains
an effective action of  $\mathbb T^{2r}\zeq (H\zpor H)/K$ on $G$.
 Moreover the isotropy group of any
$g\zpe G$ has two or more elements if and only if
$(gHg^{-1})\zin H$ is not included in $ZG$; by Proposition \ref{proAA}
 this happens for almost no $g\zpe G$.}
\end{example}

\begin{example}\label{ejeA}
{\rm On $S^5 =\{y\zpe \mathbb R^6 \zbv y_{1}^2 +\dots+y_{6}^2 =1\}$
consider the action of $\mathbb T^3$ given by the fundamental vector fields
$U_j =-y_{2j}\zpar/\zpar y_{2j-1}+y_{2j-1}\zpar/\zpar y_{2j}$,
$j=1,2,3$. In order to construct a describing vector field for this action we follow along the lines of
Sections \ref{secB} and \ref{secC} up to some minor changes.
First observe that the singular set for this action is
$S=\{y\zpe S^5 \zbv (y_{1}^2 +y_{2}^2)(y_{3}^2 +y_{4}^2)
(y_{5}^2 +y_{6}^2)=0\}$, so the action of $\mathbb T^3$ on $S^5 -S$
is free.

Let $\zp\colon S^5 \zfl\mathbb R^2$ be the map given by
$\zp(y)=(y_{1}^2 +y_{2}^2 ,y_{3}^2 +y_{4}^2 )$, and $B$ be the interior of the
triangle of vertices $(0,0),(1,0),(0,1)$. Then
$\zp\colon S^5 -S\zfl\mathbb B$ is the $\mathbb T^3$-principal bundle associated
to the action of $\mathbb T^3$. A connection $\mathcal C$ for this principal bundle
is defined by $Ker(\za_1 \zex\za_2 \zex\za_3 )$ where each
$\za_j =(y_{2j-1}^2 +y_{2j}^2 )^{-1}(-y_{2j}dy_{2j-1}+y_{2j-1}dy_{2j})$,
which is flat since $d\za_1 =d\za_2 =d\za_3 =0$.

The vector fields
$$V_r =(y_{5}^2 +y_{6}^2 )\left(  y_{2r-1}{\frac {\zpar} {\zpar y_{2r-1}}}
+y_{2r}{\frac {\zpar} {\zpar y_{2r}}}\right)
-(y_{2r-1}^2 +y_{2r}^2 )\left(  y_{5}{\frac {\zpar} {\zpar y_{5}}}
+y_{6}{\frac {\zpar} {\zpar y_{6}}}\right)\, ,$$
$r=1,2$, are tangent to $\mathcal C$ and project in
$$2(1-x_{1}-x_{2}){\frac {\zpar} {\zpar x_{r}}}\, ,$$
$r=1,2$, on $B$ endowed with coordinates $x=(x_1 ,x_2)$.

Set
$$Y=2(1-x_{1}-x_{2})x_1 x_2 \left[ \left( x_1 -{\frac {1}  {4}}\right)
{\frac {\zpar} {\zpar x_{1}}}+ \left( x_2 -{\frac {1}  {4}}\right)
{\frac {\zpar} {\zpar x_{2}}}\right]\, ,$$
whose lifted vector field through $\mathcal C$ is

$$Y'= (y_{1}^2 +y_{2}^2)(y_{3}^2 +y_{4}^2)
\left(( y_{1}^2 +y_{2}^2 -1/4)V_1 +
( y_{3}^2 +y_{4}^2 -1/4)V_2\right).$$

Note that $Y'$ extends in a natural way to $S^5$. Moreover
$Y'$ vanishes on $S$.

Clearly $Y$ on $B$ and $Y'$ on $S^5$ and $S^5 -S$ are complete.
Besides $(1/4,1/4)$ is the only source of $Y$.

Let $\zt\colon\mathbb R^2 \zfl\mathbb R$ be the function defined by
$$\zt(x)=\zr(x)\left((x_1 -1/8)^{2}+(x_2 -1/8)^{2}\right)
\left((x_1 -1/8)^{2}+(x_2 -1/4)^{2}\right)^2
\left((x_1 -1/4)^{2}+(x_2 -1/8)^{2}\right)^3$$
where $\zr(x)=x_{1}^{10}x_{2}^{10}(1-x_{1}-x_{2})^{10}$, whose
zeros on in $B$ are $(1/8,1/8)$ with order two,  $(1/8,1/4)$ with order four
and  $(1/4,1/8)$ with order six.

Now $X'=(\zt\zci\zp)(Y'+U_1 +e U_2 +e^2 U_3)$, defined on $S^5$, is
a describing vector field for the action of $\mathbb T^3$.

Indeed, the only difference with respect to the construction of Sections
\ref{secB} and \ref{secC} is that every point of $S$ is a singularity of
$X'$ with order $\zmai 10$ instead of infinity, but it is not important because
the order of the remainder singularities of $X'$ is always $\zmei 6$.}
\end{example}

\begin{example}\label{ejeB}
{\rm Let $H$ be a closed subgroup of a connected Lie group $G$ and
$G/H$ be the (quotient) homogeneous space associated to the equivalence
relation $g_1 \mathcal R g_2$ if and only of $g_2 =g_1 h$ for some
$h\zpe H$. As it is well known $G$ acts on $G/H$ by setting
$g\zpu\overline{ g'}=\overline{gg'}$.

Now assume $H$ discrete; then the canonical projection
$\zp\colon G\zfl G/H$ is a covering. Moreover a vector field $V$ on $G/H$
commutes with the action of $G$, that is to say with every fundamental
vector field, if and only if its lifted vector field $V'$, on $G$, is at the same
time left $G$-invariant and right $H$-invariant.

If one suppose that $V'$ is left $G$-invariant this property is equivalent
to say that $V'(e)$, where $e$ is the identity of $G$, is invariant by
(the adjoint action of) $H$. Therefore if no element of $T_e G-\{0\}$
is invariant by $H$, then any vector field on $G/H$ which commutes with
the action of $G$ identically vanishes.

Set $G=SO(3)$ and let $H$ be the subgroup of order four consisting of the
identity plus the three rotations of angle $\zp$ around any of the coordinates
axes (i.e.\ $H$ is the Klein four-group). Then no element of  $T_e SO(3)-\{0\}$ is invariant by $H$.

Consider on $M=\mathbb R^k \zpor(SO(3)/H)$, $k\zmai 1$, the action of
$SO(3)$ given by $g\zpu(x,\overline{g'})=(x,\overline{gg'})$. {\it This action is
effective but it does not have any describing vector field.} Indeed, take a vector
field $U$ on $M$ which commutes with the action of $SO(3)$. Then $U$ has
to respects the foliation defined by the orbits of $SO(3)$, so
$U=\zsu_{j=1}^k f_j (x)\zpar/\zpar x_j +V$ where $V$ is a vector field
tangent to the second factor and $x=(x_1 ,\dots,x_k )$ the canonical
coordinates in $\mathbb R^k$.

Since $U$ and $\zsu_{j=1}^k f_j (x)\zpar/\zpar x_j$ commute with the
action, $V$ has to commute with the fundamental vector
fields on each orbit, hence $V=0$. In other words $U=\zsu_{j=1}^k f_j (x)\zpar/\zpar x_j$.

On the other hand if $\zf\colon SO(3)/H\zfl SO(3)/H$ is a diffeomorphism, then
$$F\colon\mathbb R^{k}\zpor(SO(3)/H)\zfl\mathbb R^k \zpor(SO(3)/H)$$ given
by $F(x,\overline{g})=(x,\zf(\overline{g}))$ is an automorphism of $U$, so
$\Aut(U)$ is strictly greater than $SO(3)\zpor\mathbb R$.

Another possibility is to consider the action of $SO(3)$ on the sphere $S^2$.
Then for each $p\zpe S^2$ there exists a fundamental vector field $X$ such
 that $p$ is an isolated singularity of $X$. Therefore if $V$ commutes with $X$
then $V(p)=0$; consequently if $V$ commutes with the action of $SO(3)$ on
$S^2$ necessarily $V=0$.

By the same reason as before, the action of $SO(3)$ on
$\mathbb R^k \zpor S^2$, $k\zmai 1$, defined by $g\zpu(x,p)=(x,g\zpu p)$
has no  describing vector field.}
\end{example}

\begin{example}\label{ejeC}
{\rm Let $G$ be the group of affine transformations $\zf\colon\mathbb T^2
\zfl\mathbb T^2$ defined by $\zf(\zh)=a\zh+\zl$ where
$a=\zmm 1$ and $\zl\zpe\mathbb T^2$.
The fundamental vector fields of the natural action
of $G$ on $\mathbb T^2$ are $b_1 \zpar/\zpar\zh_1
+b_2 \zpar/\zpar\zh_2$, $b_1 ,b_2 \zpe\mathbb R$.
If $V$ is a vector field on $\mathbb T^2$ which commutes with the action of $G$
it has to commute with the fundamental vector fields, therefore
$V=c_1 \zpar/\zpar\zh_1  +c_2 \zpar/\zpar\zh_2$, $c_1 ,c_2 \zpe\mathbb R$.

But, at the same time, $V$ commutes with $\tilde\zf(\zh)=-\zh$
since $\tilde\zf\zpe G$, which implies $V=0$.
Now reasoning as in Example \ref{ejeB} shows that the effective action of
$G$ on $\mathbb R^k \zpor\mathbb T^2$, $k\zmai 1$, defined by
$\zf\zpu(x,\zh)=(x,\zf(\zh))$ has no describing vector field.

Observe that $G$ is a compact Lie group with abelian Lie algebra, but
it is not connected.}
\end{example}

\section{Two auxiliary results}\label{secAA}

Here we include two complementary results which were
needed before. The first one is a straightforward consequence of the Principal Orbit Theorem
on the structure of the orbits of the action of a compact Lie group (see
\cite[Theorem IV.3.1]{bredon}).

\begin{proposition}\label{proAA}
Consider an effective action of $\mathbb T^n$ on a connected
$m$-manifold $M$. Let $S$ be the set of those points of $M$ whose
isotropy group has two or more elements, i.e.\ the isotropy group is non trivial. Then the set $M-S$ is connected,
dense, open and $\mathbb T^n$-invariant. Moreover $n\zmei m$.
\end{proposition}

The second result shows how, for connected compact Lie group actions,
locally defined invariant vector fields give rise to invariant
vector fields defined on the whole manifold.

\begin{proposition}\label{proAB}
Consider an action of a connected compact Lie group $G$ on a
$m$-manifold $M$. Given a vector field $X$ on an open set $A$ of $M$,
both of them $G$-invariant, then there exists a $G$-invariant
bounded function $\zf:M\zfl{\mathbb R}$,  which is positive on $A$ and
vanishes on $M-A$, such that the vector field ${\hat X}$ on $M$ defined
 by ${\hat X}=\zf X$ on $A$ and ${\hat X}=0$ on $M-A$ is differentiable
and $G$-invariant.
\end{proposition}

First let us recall some elementary facts on actions. Let $Z$ be a vector
field on $M$ and $\widetilde Z$ the vector field depending on a parameter
$g\zpe G$ given by $\widetilde Z(g,p)=(g_*)^{-1}(Z(g\zpu p))$. On $G$
consider a bi-invariant volume form whose integral equals $1$ and
 the measure associated to it, that is the Haar measure. Then since
$\widetilde Z(\{p\}\zpor G)\zco T_p M$ the formula
$$Z'(p)=\zil_G \widetilde Z(g,p)$$
defines a $G$-invariant vector field $Z'$ on $M$. Moreover if $Z=\zr U$
where $U$ is a $G$-invariant vector field, then $Z'=\zq_\zr U$ where
$\zq_\zr$ is the $G$-invariant function constructed from $\zr$ in the usual way,
that is $$\zq_\zr(p)=\zil_G \zr(g\zpu p)\, .$$

In order to prove Proposition \ref{proAB}, we start considering, for $X$ and $A$,
a function $\zf\colon M\zfl\mathbb R$ like in
\cite[Proposition 5.5, pag.\ 329]{TV}, and the vector field $\hat X$ defined by $\hat X=\zf X$ on $A$
and $\hat X=0$ on M-$A$.

Now observe that $\hat X'=\zq_\zf X$ on $A$ because on this open set
$X$ is $G$-invariant and $\hat X=\zf X$. Thus $\zq_\zf$ is the required
function  (up to the name) since $\zq_\zf $ and $\hat X'$ vanish on
$M-A$; so {\it Proposition \ref{proAB} is proved}.

redpajama_set_name: RedPajamaArXiv


text:
Antonio Jorge Gauna, apodado Tata (nacido el  en San Lorenzo, provincia de Santa Fe),  es un exfutbolista argentino. Se desempeñaba como delantero y su primer club fue Rosario Central.

Carrera
Su debut se produjo en el cotejo de la 26.° fecha del Campeonato de Primera División 1948, disputándose con futbolistas juveniles, ya que se encontraba en desarrollo una huelga; en dicho partido, celebrado el 14 de noviembre, Rosario Central venció a Racing Club 6-2, con dos tantos de Gauna. Jugó los restantes cuatro encuentros del torneo, sin volver a marcar.

En las dos temporadas siguientes tuvo poca participación, logrando titularidad en 1951, cuando Central debió jugar el Campeonato de Segunda División, tras haber perdido la categoría el año anterior. La recuperación canalla fue pronta, ya que se consagró en el certamen, consiguiendo también el retorno a la máxima categoría.

En el Campeonato de Primera División 1952 marcó su primera presencia en la red en un clásico rosarino, al convertir el primer gol centralista en la victoria 2-0 ante Newell's Old Boys como visitante, partido disputado el 12 de octubre.

Tuvo su más destacada actuación en el derbi rosarino en 1953, cuando anotó los tres goles de su equipo (uno de ellos de chilena) en la victoria 3-1 en el Parque Independencia.

Continuó en la titularidad de la delantera canalla durante varias temporadas, compartiendo la línea ofensiva con futbolistas de la talla de Alberto Massei, Humberto Rosa, Oscar Mottura, Roberto Appicciafuocco, Juan Apolonio Vairo, entre otros. Marcó un quinto gol en clásicos, aunque en esta oportunidad su equipo cayó 3-2 como visitante el 15 de agosto de 1954.

Continuó en Rosario Central hasta el Campeonato de Primera División 1956, acumulando en su trayectoria por la institución de Barrio Arroyito 152 presencias y 29 goles convertidos.

Su siguiente destino fue en el exterior, jugando en O'Higgins de Chile entre 1957 y 1958. Prosiguió en Rampla Juniors de Uruguay, para luego retornar a Argentina al fichar por Colón, mientras que su último club fue Sport Club Cañadense.

Clubes

Palmarés

Referencias

Futbolistas sanlorencinos
Futbolistas del Club Atlético Rosario Central en los años 1940
Futbolistas del Club Atlético Rosario Central en los años 1950
Futbolistas del Rampla Juniors Fútbol Club
Futbolistas del Club Deportivo O'Higgins en los años 1950
Futbolistas del Club Atlético Colón en los años 1950


Asymmetric Y junctions allow the development of a new class of optical frequency mixers that utilize higher-order waveguide modes for signal processing. We measure high-contrast (>30 dB) mode sorting in asymmetric Y junctions by use of a novel technique: efficient TM00, TM10, and TM20 mode mixing in a periodically poled lithium niobate waveguide. We also demonstrate an odd-to-even mode wavelength converter capable of spectral inversion without offset or bidirectional wavelength conversion.


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Broken family situations are sometimes breeding grounds for sexual misconduct. Many rapes are carefully planned callously implemented and viciously carried out.
Young women need to protect themselves by only dating men who share their religious values. They need to be clear about what kind of relationship they want before it ever begins. And they need to be ready to end a relationship if they are pressured to commit fornication.


text:
<?php declare(strict_types=1);

namespace PharIo\Phive;

class ReleaseException extends \Exception implements Exception {
}

redpajama_set_name: RedPajamaGithub


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Here at Synergem we believe our size is our strength. Our small team is built up of experienced directors and senior consultants, each with an indepth knowledge of the industries we cover.
We treat every client and candidate as individual. Being a small business means we are not bound by the processes and demands of a larger firm, allowing us to offer a bespoke and flexible service, suited towards your business needs and objectives.


text:
Oxford Biomedica named as winner of The Oxford Science Park Innovation Award 2019
By rowandev18th June 2019May 28th, 2020No Comments
Leading gene therapy manufacturer recognised as part of Oxfordshire Business Awards
Oxford, UK, June 18 2019 – Its work on life-changing treatments for serious diseases has earned Oxford Biomedica, a leading gene and cell therapy group, The Oxford Science Park Innovation Award 2019. Given as part of the Oxfordshire Business Awards, it recognises the company as the sole manufacturer of the gene therapy vector used in the ground-breaking leukaemia treatment Kymriah®, which was launched in 2017. This is the fourth year that The Oxford Science Park, one of the UK's leading parks for science and technology companies, has sponsored the Innovation Award.
In an Award open to companies in all sectors across the County, the judges were looking for an innovative product or process which had made progress over the past three years, driving corporate development in a significant way. Oxford Biomedica is the only group in the world with a GMP1-approved facility for commercial scale lentiviral vector manufacturing. Its technology is being used in the development of its own products and that of its partners (such as Novartis' Kymriah®).
Established as a spin-out from the University of Oxford in 1995, the company now employs more than 430 people in several locations across the County.
Kyriacos Mitrophanous, PhD, Chief Scientific Officer of Oxford Biomedica, said, 'Personalised, gene and cell-based therapy is now a reality for patients and we're proud to be at the heart of this emerging area of medicine. Our capabilities have been honed over 20 years and we are honoured to be recognised for our achievements with the Oxfordshire Business Awards' Innovation Award. Delivering innovation is one of our core values at Oxford Biomedica and it is recognition such as this that inspires us to continue to push the boundaries and set new standards for the sector.'
Piers Scrimshaw-Wright, Managing Director of The Oxford Science Park, said, 'Oxford Biomedica's key role in the development and supply of Kymriah®, which is for rare blood cancers in adults and children, makes it a worthy winner of The Oxford Science Park Innovation Award 2019. We congratulate the company on its success, and the impact it is having on the Oxfordshire life sciences landscape.'
'We are delighted to have been involved for the fourth year with the Oxfordshire Business Awards, which celebrate excellence across the County. Several of The Oxford Science Park's occupiers, including Exscientia, Fuel3D, Oodle, Sensyne Health and Sesui, were shortlisted or winners in other categories, and we congratulate them.'
Award pictures available
1 Good Manufacturing Practice
Emma Palmer Foster, Strategic Communications Consultant
emmapf@oxfordsp.com
About The Oxford Science Park
The Oxford Science Park is owned and managed by Magdalen College, Oxford. Created in 1991, the Park upholds the College's heritage and provides one of the most influential science & technology environments in the UK. There is approaching 750,000 square feet of workspace accommodation across the Park, which is now home to more than 2,500 people and more than 90 businesses. These range from start-ups based in the Magdalen Centre innovation hub to major international companies and include Blue Earth Diagnostics, Circassia, OrganOx, OxSonics, Oxford Nanopore Technologies, Sharp Life Science, Oxford Genetics, ProImmune, Oxford Sciences Innovation, Evox Therapeutics, Vaccitech and Exscientia.
In addition to being a key property investment, the Park is at the heart of Magdalen College's strategy to support discovery, innovation and entrepreneurship. It will continue to develop The Oxford Science Park as a long-term strategic asset, with ambitious plans to create an additional 300,000+ sq ft of office and laboratory space on the remaining 10+ acres of land over the next 3-5 years. This additional capacity will support the growth of businesses already based on the Park, providing flexible workspace accommodation, and enabling new companies to enjoy the Park's exceptional environment and collegiate and collaborative ethos. The Oxford Science Park is located approximately four miles south-east of Oxford city centre, just off the City's southern ring road. It has easy access to the M40 and A34, as well as to Heathrow Airport and mainline train services. For further information, please visit: www.oxfordsp.com or follow us on twitter @OxfordSciencePK
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text:
describe('environment', function() {

	describe('document', function() {
		it('should be defined', function() {
			expect(document).toBeDefined();
		});
	});

	describe('window', function() {
		it('should be defined', function() {
			expect(window).toBeDefined();
		});

		it('should update globals', function() {
			window.globalTest = 0;
			// eslint-disable-next-line no-undef
			expect(globalTest).toEqual(0);
		});
	});
});

describe('app', function() {
	beforeEach(function() {
		$('<div/>')
			.attr('id', 'test-container')
			.appendTo(document.body);
	});

	afterEach(function() {
		$('#test-container').remove();
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	it('should be defined', function() {
		expect(app).toBeDefined();
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		expect($('#test-container')).toExist();

		app.main('#test-container');
		var el = $('#test-container > span');
		expect(el).toExist();
		expect(el.html()).toEqual('test');
	});
});


text:
Sheepstown Church
N 52° 29' 05.0" W 007° 14' 36.0"
Nearest town
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S 51455 37277
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The ruins of the church from a distance.
This parish church is 90 metres east of the road, on the grounds of a farm. To reach it I had to pass three electric fences and two hens.
It's dated from around 1200 and was dedicated to St. Muicín. Parish churches in Ireland were common in the 15th and 16th century, when the Irish Church was reorganised on a diocesan basis rather than on monastic basis. This church, though, is a very rare example of a parish church of the 13th century.
The building is small, it measures 15 metres in length and 7 metres in width.
The main doorway is on the west side, but there's a smaller doorway on the south side as well. There aren't decorations on the building except for four beaded mouldings on the four corners of the church. On the northwest corner of the building there's a corbel that the informative sign calls "The Clock Stone". One of the stones on the north wall carries a simple but beautiful decoration, but it's clear enough that it comes from a different site. This stone is broken and only a small part of the original carving can be seen here.
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A couple of developments have already taken place in the courts after Prime Minister Modi's surprise announcement of his decision to discontinue recognition of five hundred and thousand-rupee currency notes as legal tender.
While the Bombay lawyers Jamshed Mistry and Jabbar Shaikh argue that before the November 8 notification which was issued last night, there was a November 2 notification. This Nov 2 notification says that a few number of new 100 rupee notes were to be introduced as a pilot project within 15 days and that was to cover ten per cent of all ATM's. They have reasoned that within 6 days how could the banks have been prepared for such a drastic step. In 1946 an ordinance was passed, wherein currencies of 100, 1000 and 10,000 rupee denominations were discontinued after a separate Section 26(a) was added to the RBI Act. In 1978, there was an ordinance and a separate act called The High Denomination Bank Notes Act, 1978.
The petition in Supreme Court has called the Prime Minister's decision a "tuglaki farman" citing several inconveniences that the general public will face as a result of this decision.
While SC may hear this petition tomorrow, the Bombay HC has directed the lawyers to approach a regular bench once court reopens on a regular basis after the Diwali vacations end. Hence a writ petition will now be filed and mentioned on November 15.


text:
Kimmel Wins His First ARCA Race of the Season
Series Vet Holds Off Young Charger at Iowa
The win marked Kimmel's 69th career series victory in his 335th series start. The victory also marked the ninth different winner in as many events in 2007.
"It was a great race," said Kimmel. "We had some issues. We had a few mishaps on a pit stop. This is a very good racetrack. This series is so competitive anymore; it's unreal. Bill (Kimmel/crew chief) and the guys worked really hard on the car.
"I knew Michael (McDowell) was coming; he was really charging hard. But he raced us clean and we both came out with great finishes. I had a ball today. There was some serious, clean racing going on out there."
McDowell, after erasing Kimmel's 10-length lead inside of five laps remaining, caught Kimmel on the final lap and drew up alongside the eventual winner coming to the checkered flag. McDowell, in the Z Line-Make A Wish-Beck Dodge, fell a half-length short at the final stripe.
"It was a good race," said McDowell. "Unfortunately we came down for a green flag stop and ended up a victim of a caution. For a few laps there I was overdriving. Once I calmed down it was fine. It's hard because we were chasing down our first victory and we should have won a few times already. It makes you very anxious."
Rookie Michael Annett, after earning the SIM Factory Pole Award in his career-first attempt, finished third in the TMC Dodge.
"I've watched a lot of races on television where they talked about how important track position was, but I never really understood it until today," said Annett. "If you get stuck in the back
Frank Kimmel hoists another ARCA winner's trophy. (ARCA Photos)
ARCA RE/MAX Series veteran Frank Kimmel, in the TransAmerica-Fort Dodge Animal Health-Pork Ford, held off a late-race charge from rookie Michael McDowell to win the Prairie Meadows 250 Saturday night at Iowa Speedway.
sometimes it doesn't matter how good your car is, you will never be able to make it up to the front. We definitely had a car that could have won. I can't say enough about the guys at Country Joe Racing."
Phillip McGilton, in the SI Performance Ford, finished fourth with Bryan Clauson, in the Memorex Dodge, trailing in fifth.
In all, six drivers exchanged the lead 11 times throughout the 250-lap event while nine cautions for 49 laps, all for minor incidents, slowed the field nine times. There were no injuries.
Kimmel led the most laps with 94 over McDowell's 74 laps-led.
Bobby Gerhart finished sixth in the Lucas Oil Chevrolet.
Justin Allgaier was a major player most of the way, leading three times for 47 laps. However, Allgaier scraped the wall in turn two in an effort to avoid an accident and had to pit for tires and fender repair late in the race. Allgaier, in the AG Tech Chevrolet, battled back to finish seventh.
Justin South finished eighth in the HD Supply Dodge in front of Josh Krug, who finished a career-best ninth in the Bigford Enterprises Chevrolet, the last car on the lead lap. Krug, who started 23rd, also earned the Hoosier Tire Hard Charger Award for advancing the most positions.
Kimmel (#46) crosses the finish line just ahead of McDowell (#2).
Jason Hedlesky finished 10th in the Hardcore Motorsports Ford.
The ARCA RE/MAX Series travels next to Pocono Raceway for the 44th Pocono 200 Saturday, June 9th. The race will air same-day on SPEED at 8 pm ET.


Print La vie est belle L'Eclat L'Eau de Parfum is a radiant composition made of the noblest natural ingredients. Its sparkling heart composed of the most luminous flower, the Orange Blossom, is enlightened by a smile of fruity delicacies.
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This first Tasty Orange Blossom is dedicated for women who seek a radiant sophistication in La vie est belle.
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The sunny facets of the Orange Blossom reflects the light through an elegant Iris. Jasmin, the most precious flower, spreads its sophistication and brings feminity to the fragrance.
Finally the accord of Patchouli and Vanilla gives its gourmand facet to the fragrance.
Print La vie est belle L'Eclat L'Eau de Parfum is a masterpiece. The multi facetted smile captures the light of happiness and shares it brillantly.
Print This first Tasty Orange Blossom is dedicated for women who seek a radiant sophistication in La vie est belle.
A burst of fresh and fruity sparkling notes.
Take a stroll in a garden surrounded by the summer scent of this flower of light.
L'Éclat has inspired our perfumers to create a luminous and modern fragrance around the orange blossom, a solar and radiant flower.
Our two brightest smiles associate to light up your life. Julia Robert is the perfect embodiment of grace and radiance. As vibrant as her smile, La Vie Est Belle Le'Eclat, Lancôme's new fragrance shines bright, let it lift your day.
Highlight your natural radiance with Teint Miracle 03 fluide. Apply gently, starting with the middle of your face, where you need more coverage.
Apply Hypnôse mascara in Black from the roots to the tips of your lashes to leave them looking longer.
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Add some colour to your cheeks by applying Blush Subtil in shade 01 along your cheekbones.
Spray La Vie Est Belle L'Éclat onto your pulse points to crown your look.
L'Eclat has inspired our perfumers to create a luminous and modern fragrance sheding the lights on the Orange Blossom, the most solar and radiant flower.


I long for war but the media just teases me with the idea of it like the boy who cried wolf. A real war is the only way the subhuman filth can be cleansed from this earth and natural selection can actually take place.
LMAO it´s true, genetic elite and spring of the nation actually die in the war and mostly cripples and women survive.
Imagine if Japan was destroyed. Dear god what would all the western weebs do without animu.


text:
from django.shortcuts import render_to_response
from models import UserProfile
from django.contrib.auth.decorators import login_required
from django.http import HttpResponseRedirect, HttpResponseNotAllowed
from django.views.generic import simple
from openflow.optin_manager.opts.models import UserFlowSpace, UserOpts, OptsFlowSpace, AdminFlowSpace
from openflow.optin_manager.admin_manager.models import *
from openflow.optin_manager.users.forms import *
from django.contrib.auth.forms import PasswordChangeForm

def index(request):
    if request.user.is_authenticated():
        return HttpResponseRedirect('/dashboard')
    else:
        return HttpResponseRedirect('/accounts/login')

@login_required
def change_profile(request):
    '''
    The view function to change profile information for a user or admin.
    '''
    profile = UserProfile.get_or_create_profile(request.user)
    if request.method == "GET":
        user_form = UserForm(instance=request.user)
        pass_change_form = PasswordChangeForm(request.user)
        if profile.is_net_admin:
            admin_form = AdminProfileForm(instance=request.user.get_profile())
    elif request.method == "POST":
        user_form = UserForm(request.POST, instance=request.user)
        valid1 = user_form.is_valid()

        if profile.is_net_admin:
            admin_form = AdminProfileForm(request.POST, instance=request.user.get_profile())
            valid3 = admin_form.is_valid()
        else:
            valid3 = True

        if (request.POST['old_password']!=""):
            pass_change_form = PasswordChangeForm(request.user, request.POST)
            valid2 = pass_change_form.is_valid()
        else:
            pass_change_form = PasswordChangeForm(request.user)
            valid2 = True

        if (valid1 and valid2 and valid3):
            user_form.save()
            if (request.POST['old_password']!=""):
                pass_change_form.save()
            if profile.is_net_admin:
                admin_form.save()

            return HttpResponseRedirect("/dashboard")

    else:
        return HttpResponseNotAllowed("GET", "POST")

    if (profile.is_net_admin):
        return simple.direct_to_template(request,
            template = 'openflow/optin_manager/users/change_profile_admin.html',
            extra_context = {
                'user_form':user_form,
                'pass_form':pass_change_form,
                'admin_form':admin_form
            }
        )
    else:
        return simple.direct_to_template(request,
            template = 'openflow/optin_manager/users/change_profile_user.html',
            extra_context = {
                'user_form':user_form,
                'pass_form':pass_change_form,
            }
        )

@login_required
def dashboard(request):
    '''
    The dashboard view function
    '''
    profile = UserProfile.get_or_create_profile(request.user)

    if (not profile.is_net_admin):
        next_steps = 0
        opts = []
        ufs = UserFlowSpace.objects.filter(user=request.user)
        if (len(ufs) == 0):
            rufs = RequestedUserFlowSpace.objects.filter(user=request.user)
            if (len(rufs) == 0):
                next_steps = 0
            else:
                next_steps = 1
        else:
            opts = UserOpts.objects.filter(user=request.user).order_by('-priority')
            if (len(opts) == 0):
                next_steps = 2
            else:
                next_steps = 3

        return simple.direct_to_template(request,
                            template = 'openflow/optin_manager/dashboard_user.html',
                            extra_context = {
                                    'user':request.user,
                                    'next_steps':next_steps,
                                    'opts': opts,
                            },
                        )
    else: #Admin
        adminfs = AdminFlowSpace.objects.filter(user=request.user)

        opts = UserOpts.objects.filter(user=request.user).order_by('-priority')
        optsfs = []
        for opt in opts:
            fses = opt.optsflowspace_set.all()
            for fs in fses:
                optsfs.append(fs)

        num_user_reqs = RequestedUserFlowSpace.objects.filter(admin=request.user).count()
        num_admin_reqs = RequestedAdminFlowSpace.objects.filter(admin=request.user).count()
        num_admin_promotes = AdminPromoteRequest.objects.filter(admin=request.user).count()
        return simple.direct_to_template(request,
                            template = 'openflow/optin_manager/dashboard_admin.html',
                            extra_context = {
                                    'user': request.user,
                                    'optsfs': optsfs,
                                    'adminfs': adminfs,
                                    'num_admin_reqs': num_admin_reqs,
                                    'num_user_reqs': num_user_reqs,
                                    'num_admin_promotes': num_admin_promotes,
                            },
                        )


Most expect school hallways to be empty during the summer, but camp is in session at some Grand Erie elementary schools this month.
In the third year of Camp SAIL (Summer Adventures in Learning), teachers are enhancing primary and junior school students' abilities in mathematics and technology, while also giving them a true summer camp experience.
During Camp SAIL, you'll find students (aged 5-10) engaging in experiential learning activities such as salsa dancing, cooking, yoga, painting rocks and even juggling.
The three-week summer program runs from August 13-31, and is intended to support students from Senior Kindergarten to Grade 5. The primary focus of the program is to support students with numeracy expectations, and to sustain or increase individual skill development, engage parents and develop student confidence.
When arriving at camp, each student is given a backpack, a water bottle, several picture books with a growth mindset or Indigenous education focus and math manipulatives they can use at home to practice camp learnings with their parents.
During camp, student abilities are assessed individually. They are able to complete learning activities in small groups of three to five students based on their skill level. In these smaller groups, teachers are able to target activities based on individual learning levels and enhance every student's learning.
Many activities at Camp SAIL make use of technology and hands-on manipulatives for math. While some students are coding robots to drive using an iPad, younger students are learning how to make estimates by guessing how many candies are in a jar.
Teachers aren't the only ones getting involved in camp. Many recently graduated Grade 8 students are fulfilling their volunteer hours for secondary school by helping out campers.
More than 200 students are registered for Camp SAIL across seven Grand Erie sites, including: Branlyn Community School, Caledonia Centennial Public School, North Ward School, Prince Charles Public School, Waterford Public School, West Lynn Public School and Woodman-Cainsville School. The Camp SAIL program is free for parents and is coordinated through the Council of Ontario Directors of Education.
The lessons taught in Camp SAIL link directly to the Achievement indicator of Grand Erie's Multi-Year Plan and Success for Every Student by increasing student understanding of effective learning strategies and how to use them.
Through the use of games and technology students are able to stay engaged and enhance their abilities at the same time. With Technology at the forefront of Grand Erie's Multi-Year Plan, providing up-to-date tools for staff is a priority.


text:
Low-Cost 3D-Printed Holographic Microscope Could Lead to Rapid COVID-19 Testing
from hackster.io
A team of scientists has developed a method for quickly finding COVID-19 in blood samples, using a deep learning system and a 3D-printed low-cost digital holographic microscope (DHM).
"We propose the use of a deep learning cell classification system using the reconstructed phase profiles output from a compact and field-portable shearing digital holographic microscope in the rapid screening of RBCs for COVID-19," the researchers explain of the work behind their letter.
"Features are extracted from the phase profile of segmented RBCs at each time frame of the reconstructed video data, then input into a bi-directional long short-term memory (Bi-LSTM) network to classify the cells based on their spatiotemporal behavior. The system is capable of successfully classifying between diseased and healthy samples without chemical processing and with fast turnaround times."
The 3D-printed prototype microscope is based around an off-the-shelf red laser diode, a one-dimensional translation stage, a 40x-magntification objective lens, a glass plate, and a CMOS image sensor. The laser illuminates the blood sample and a magnified image is projected onto the glass plate, forming an interference pattern captured by the image sensor as a video and sent to a computer for analysis.
The resulting video is fed into a neural network running on a workstation with an NVIDIA Quadro P4000 graphics processor, which proved capable of classifying a single patient's red blood cells in under 10 seconds — allowing it to quickly detect patients with COVID-19 infections with 80 per cent sensitivity and nearly 93 per cent specificity.
"The advantages of the presented system include cost, accessibility, and time to results," the researchers note, "but further investigations with larger patient pools including mild or asymptomatic infections are needed to validate the system performance more rigorously."
The team's work has been published in the journal Optics Letters under open-access terms, with more information available on IEEE Spectrum.
Filed in: covid


text:
Trichotosia malleimentum är en orkidéart som först beskrevs av Johannes Jacobus Smith, och fick sitt nu gällande namn av S.Thomas, Schuit. och De Vogel. Trichotosia malleimentum ingår i släktet Trichotosia och familjen orkidéer. Inga underarter finns listade i Catalogue of Life.

Källor

Orkidéer
malleimentum


text:
Hrušínský ist der Familienname folgender Personen:

 Jan Hrušínský (* 1955), tschechischer Schauspieler