text:
\section*{Introduction}

One of the first surprising findings which one makes when
encountering diffeology for the first time is this inequality:
$L^{\infty}(V,W)<L(V,W)$, which says that the space of all
(diffeologically) smooth linear maps between two diffeological
vector spaces, $V$ and $W$, can be strictly smaller than that of all
maps that are simply linear.\footnote{Say what? It happens in
functional analysis already? Well, whatever. I prefer to keep it
simple (jokes apart, in diffeology it happens in finite dimension
already).} Stemming from that, an even more surprising situation
presents itself: a (finite-dimensional) diffeological vector space
that does not admit a smooth scalar product. This is a known fact
(see \cite{iglesiasBook}, p. 74, Ex. 70), and easily established at
that, yet, it is still surprising, in and of itself, and also for
how easily this can be illustrated, using the presence of just one
non-differentiable (in the usual sense) plot.

Thus, as mentioned in the above-given reference, any
finite-dimensional diffeological vector space that admits a smooth
scalar product is necessarily the usual $\matR^n$, with its usual
smooth structure (the diffeology that consists of all usual smooth
maps). The choices at this point are, to abandon the whole affair
(meaning to concentrate on infinite-dimensional spaces, where the
similar situation does not seem to occur), to consider a kind of
pseudo-metrics (meaning the sort of least degenerate symmetric
bilinear form that exists on a given space), or, finally, to
re-define scalar product as ones taking values in $\matR$ endowed,
not with the standard diffeology, but with the piecewise-smooth
diffeology. We concentrate on the second of the above approaches,
and it does lead to a few interesting conclusions, mainly regarding
the diffeological dual and the fact that a pseudo-metric induces an
isomorphism of it with a specific subspace, which is a smooth
summand. Furthermore, it gives rise to a smooth scalar product on
the dual, reflecting as much as possible of the usual duality for
the standard vector spaces.

Finally, a disclaimer: a lot of what is written here might be of a
kind of implicit knowledge for people working in the area. Part of
the aim of this paper is to collect these facts in one place, and to
make explicit what is implicit elsewhere.\footnote{Putting an
interesting statement as an exercise in a book is just an invitation
for somebody to re-discover it in good faith. Just sayin'.}

\paragraph{Acknowledgments} Some of the best things in life happen
by chance: a chance encounter, a chance phrase, a chance look...
This sounds trivial, but what is surprising is that I found this
true of professional life as well; some of my most productive
moments came about by chance. The very existence of this work is due
to chance, and for it, I would like to thank the people who,
incidentally, are also responsible for some of the best moments in
the last couple of years (all of them, of course, due to chance).
These people are: Prof. Riccardo Zucchi, Prof. Patrick
Iglesias-Zemmour, Prof. Danilo De Rossi. I also would like to thank
all the participants of the workshop ``On Diffeology etc.'' (Aix en
Provence, June 24-26, 2015) where I had my first opportunity to
discuss the reasoning that is at the origin of this paper.

\section{Diffeology and diffeological vector spaces}

We briefly recall here the main definitions regarding diffeological
spaces and diffeological vector spaces.

\paragraph{Diffeological spaces and smooth maps} We first recalling the notion of a
diffeological space and that of a smooth map between such spaces.

\begin{defn} \emph{(\cite{So2})} A \textbf{diffeological space} is a pair
$(X,\calD_X)$ where $X$ is a set and $\calD_X$ is a specified
collection of maps $U\to X$ (called \textbf{plots}) for each open
set $U$ in $\matR^n$ and for each $n\in\matN$, such that for all
open subsets $U\subseteq\matR^n$ and $V\subseteq\matR^m$ the
following three conditions are satisfied:
\begin{enumerate}
  \item (The covering condition) Every constant map $U\to X$ is a
  plot;
  \item (The smooth compatibility condition) If $U\to X$ is a plot
  and $V\to U$ is a smooth map (in the usual sense) then the
  composition $V\to U\to X$ is also a plot;
  \item (The sheaf condition) If $U=\cup_iU_i$ is an open cover and
  $U\to X$ is a set map such that each restriction $U_i\to X$ is a
  plot then the entire map $U\to X$ is a plot as well.
\end{enumerate}
\end{defn}

Usually, instead of $(X,\calD_X)$ one writes simply $X$ to denote a
diffeological space.

\begin{defn} \emph{(\cite{So2})} Let $X$ and $Y$ be two
diffeological spaces, and let $f:X\to Y$ be a set map. The map $f$
is said to be \textbf{smooth} if for every plot $p:U\to X$ of $X$
the composition $f\circ p$ is a plot of $Y$.
\end{defn}

The standard examples of diffeological spaces are smooth manifolds,
with diffeology consisting of all usual smooth maps $\matR^k\supset
U\to M$, for all $k\in\matN$ and for all domains $U$ in $\matR^k$. A
less standard example is, for instance, any Euclidean space
$\matR^n$ with the so-called \emph{wire diffeology}, namely, the
diffeology \emph{generated} (see the next paragraph) by the set
$C^{\infty}(\matR,\matR^n)$.

\paragraph{Generated diffeology} Let $X$ be a set, and let
$A=\{U_i\to X\}_{i\in I}$ be a set of maps with values in $X$. The
\textbf{diffeology generated by $A$} is the smallest, with respect
to inclusion, diffeology on $X$ that contains $A$. It consists of
all maps $f:V\to X$ such that there exists an open cover $\{V_j\}$
of $V$ such that $f$ restricted to each $V_j$ factors through some
element $U_i\to X$ in $A$ via a smooth map $V_j\to U_i$. Note that
this construction illustrates the abundance of diffeologies on a
given set: we can build one starting from \emph{any} set of maps; in
particular, given any map $p:U\to X$, there is (usually more than
one) diffeology containing it.

\paragraph{Fine diffeology and coarse diffeology} Given a set $X$, the set of all
possible diffeologies on $X$ is partially ordered by inclusion,
namely, a diffeology $\calD$ on $X$ is said to be \textbf{finer}
than another diffeology $\calD'$ if $\calD\subset\calD'$ (whereas
$\calD'$ is said to be \textbf{coarser} than $\calD$). Among all
diffeologies, there is the finest one, which turns out to be the
natural \textbf{discrete diffeology} and which consists of all
locally constant maps $U\to X$; and there is also the coarsest one,
which consists of \emph{all} possible maps $U\to X$, for all
$U\subseteq\matR^n$ and for all $n\in\matN$. It is called \emph{the}
\textbf{coarse diffeology}.

\paragraph{Subset diffeology} Let $X$ be a diffeological space, and
let $Y\subset X$ be a subset of it. The \textbf{subset diffeology}
on $Y$ is the \emph{coarsest} diffeology on $Y$ such that the
inclusion map $Y\hookrightarrow X$ is smooth. What this means
specifically, is that $p:\matR^k\supset U\to Y$ is a plot for the
subset diffeology on $Y$ if and only if its composition with the
inclusion map is a plot of $X$.

\paragraph{Products and sums of diffeological spaces} Let $\{X_i\}_{i\in I}$ be a
collection of diffeological spaces, where $I$ is a set of indices.
The \textbf{product diffeology} $\calD$ on the product $\prod_{i\in
I}X_i$ is the \emph{coarsest} diffeology such that for each index
$i\in I$ the natural projection $\pi_i:\prod_{i\in I}X_i\to X_i$ is
smooth. Note, in particular, that for the product of two spaces
$X_1\times X_2$ every plot is locally of form $(p_1,p_2)$, where
$p_i$ is a plot of $X_i$ for $i=1,2$.

If we now consider the disjoint union $\coprod_{i\in I}X_i$ of the
spaces $X_i$, this can be endowed with the \textbf{sum diffeology},
which by definition is the \emph{finest} diffeology such that all
the natural injections $X_i\hookrightarrow\coprod_{i\in I}X_i$ are
smooth. The plots for this diffeology are those maps that are
locally plots of one of the components of the sum.

\paragraph{Diffeological vector space} Let $V$ be a
vector space over $\matR$. The \textbf{vector space diffeology} on
$V$ is any diffeology of $V$ such that the addition and the scalar
multiplication are smooth, that is,
$$[(u,v)\mapsto u+v]\in C^{\infty}(V\times V,V)\mbox{ and }[(\lambda,v)\mapsto\lambda v]\in C^{\infty}(\matR\times V,V),$$
where $V\times V$ and $\matR\times V$ are equipped with the product
diffeology. A \textbf{diffeological vector space} over $\matR$ is
any vector space $V$ over $\matR$ equipped with a vector space
diffeology.

\paragraph{Fine diffeology on vector spaces} The \textbf{fine
diffeology} on a vector space $\matR$ is the \emph{finest} vector
space diffeology on it; endowed with such, $V$ is called a
\textbf{fine vector space}. As an example, $\matR^n$ with the
standard diffeology is a fine vector space.

The fine diffeology admits a more or less explicit description of
the following form: its plots are maps $f:U\to V$ such that for all
$x_0\in U$ there exist an open neighbourhood $U_0$ of $x_0$, a
family of smooth maps $\lambda_{\alpha}:U_0\to\matR$, and a family
of vectors $v_{\alpha}\in U_0$, both indexed by the same finite set
of indices $A$, such that $f|_{U_0}$ sends each $x\in U_0$ into
$\sum_{\alpha\in A}\lambda_{\alpha}(x)v_{\alpha}$:
$$f(x)=\sum_{\alpha\in A}\lambda_{\alpha}(x)v_{\alpha}\mbox{ for }x\in U_0.$$
A finite family $(\lambda_{\alpha},v_{\alpha})_{\alpha\in A}$, with
$\lambda\in C^{\infty}(U_0,\matR)$ and $v_{\alpha}\in V$, defined on
some domain $U_0$ and satisfying the condition just stated, is
called a \emph{local family} for the plot $f$.

Fine vector spaces possess the following property
(\cite{iglesiasBook}, 3.9), which in general not true: if $V$ is a
fine diffeological vector space, and $W$ is any other diffeological
vector space, then every linear map $V\to W$ is smooth, \emph{i.e.},
$L^{\infty}(V,W)=L(V,W)$.

\paragraph{Smooth scalar products} The existing notion of a
\textbf{Euclidean diffeological vector space} does not differ much
from the usual notion of the Euclidean vector space. A diffeological
space $V$ is Euclidean if it is endowed with a scalar product that
is smooth with respect to the diffeology of $V$ and the standard
diffeology of $\matR$; that is, if there is a fixed map $\langle ,
\rangle:V\times V\to\matR$ that has the usual properties of
bilinearity, symmetricity, and definite-positiveness and that is
smooth with respect to the diffeological product structure on
$V\times V$ and the standard diffeology on $\matR$. As has already
been mentioned in the Introduction, for finite-dimensional spaces
this implies that the space in question is just the usual $\matR^n$,
for appropriate $n$ (see \cite{iglesiasBook}). For reasons of
completeness we recall the precise statement and give a detailed
proof of this fact.\footnote{Special thanks go to Patrick
Iglesias-Zemmour and Yael Karshon who made first think of it; I
would have missed it otherwise.}

\begin{prop}\label{smooth:prod-smooth:plot}
Let $V$ be $\matR^n$ endowed with a vector space diffeology $\calD$
such that there exists a smooth scalar product. Then every plot $p$
of $\calD$ is a smooth map in the usual sense.
\end{prop}

\begin{proof}
Let $A$ an $n\times n$ non-degenerate symmetric matrix such that the
associated bilinear form on $V$ is smooth with respect to the
diffeology $\calD$, and let $\{v_1,\ldots,v_n\}$ be its eigenvector
basis. Let $\lambda_i$ be the eigenvalue relative to the eigenvector
$v_i$.

Let $p:U\to V$ be a plot of $\calD$; we wish to show that it is
smooth as a map $U\to\matR^n$. Recall that
$\langle\cdot|\cdot\rangle_A$ being smooth implies that for any two
plots $p_1,p_2:U\to V$ the composition
$\langle\cdot|\cdot\rangle_A\circ(p_1,p_2)$ is smooth as a map
$U\to\matR$. Let $c_i:U\to V$ be the constant map $c_i(x)=v_i$; this
is of course a plot of $\calD$. Set $p_1=p$ and $p_2=c_i$; then the
above composition map writes as $\lambda_i\langle p(x)|v_i\rangle$,
where $\langle\cdot|\cdot\rangle$ is the canonical scalar product on
$\matR^n$.

Since $A$ is non-degenerate, all $\lambda_i$ are non-zero; this
implies that each function $\langle p(x)|v_i\rangle$ is a smooth
map. And since $v_1,\ldots,v_n$ form a basis of $V$, this implies
that for any $v\in V$ the function $\langle p(x)|v\rangle$ is a
smooth one. In particular, this is true for any $e_j$ in the
canonical basis of $\matR^n$; and in the case $v=e_j$ the scalar
product $\langle p(x)|e_j\rangle$ is just the $j$-th component of
$p(x)$. Thus, we obtain that all the components of $p$ are smooth
functions, therefore $p$ is a smooth map.
\end{proof}

\section{The degeneracy of smooth forms on non-standard spaces}

As mentioned in the introduction, pretty much all finite-dimensional
diffeological vector spaces do not have smooth non-degenerate
bilinear forms; there is only one for each dimension that does, and
that is the standard one. For exposition purposes, we start with a
detailed illustration of how this happens, via an example (this part
is however easily deduced from the proof of Proposition
\ref{smooth:prod-smooth:plot} above). We then consider the degree of
degeneracy of smooth bilinear forms on a given vector space (what
does it mean, for a given $V$, to be the least-degenerate bilinear
form on it?), going on to the question of subspaces which do, or do
not, split off as smooth direct summands, and finally considering
the diffeological dual.

\subsection{Pseudo-metrics}

A \emph{pseudo-metric} is, roughly speaking, the best possible
substitute for the notion of a smooth scalar product in the case of
a finite-dimensional vector space whose diffeology is not the
standard one.

\paragraph{The first example} The following example, already
considered in \cite{multilinear}, is presented for illustrative
purposes.

\begin{example}
Let $V=\matR^n$, and let $v_0\in V$ be any non-zero vector. Let
$p:\matR\to V$ be defined as $p(x)=|x|v_0$; let $\calD$ be any
vector space diffeology on $V$ that contains $p$ as a
plot.\footnote{Such diffeology does certainly exist; for instance,
the coarse diffeology would do.} Suppose that $A$ is a symmetric
$n\times n$ matrix, and assume that the bilinear form $\langle
v|w\rangle_A=v^tAw$ associated to $A$ is smooth with respect to
$\calD$ and the standard diffeology on $\matR$. We claim that $A$ is
degenerate.

Indeed, $\langle v|w\rangle_A$ being smooth implies, in particular,
that for any two plots $p_1,p_2:\matR\to V$ of $V$ the composition
map $\langle \cdot|\cdot\rangle_A\circ (p_1,p_2):\matR\to\matR$ is
smooth in the usual sense; this map acts as $\matR\ni x\mapsto
(p_1(x))^tAp_2(x)$. Let $w\in V$ be an arbitrary vector; denote by
$c_w:\matR\to V$ the constant map that sends everything to $w$,
$c_w(x)=w$ for all $x\in\matR$. Such a map is a plot for any
diffeology on $V$. But then $(\langle \cdot|\cdot\rangle_A\circ
(p,c_w))(x)=|x|v_0^tAw$; the only way for this to be smooth is to
have $v_0^tAw=0$, and since there was no assumption on $w$, this
implies that $\langle v_0|\cdot\rangle_A$ is identically zero on the
whole of $V$, \emph{i.e.}, that $A$ is degenerate. In other words,
$V$ does not admit a smooth scalar product.
\end{example}

Note that in the above example we could have taken $p(x)=f(x)v_0$
with $f(x)$ \emph{any} function $\matR\to\matR$ that is not
differentiable in at least one point; this suggests that there are
numerous diffeological vector spaces that do not admit
diffeologically smooth scalar products. In fact, Proposition
\ref{smooth:prod-smooth:plot} shows that the phenomenon is much more
general: in order to have a smooth scalar product, we must ensure
that all plots are smooth maps.

\paragraph{The signature of a smooth bilinear form} Suppose that we
have a finite-dimensional diffeological vector space $V$; let $A$ be
a symmetric $n\times n$ matrix (with $n=\dim V$) such that the
associated bilinear form $\langle\cdot|\cdot\rangle_A$ on $V$ is
smooth. Let $(\lambda_+,\lambda_-,\lambda_0)$ be the signature of
this form; recall that $V^*$ stands for the \emph{diffeological
dual} of $V$ (see \cite{vincent} and \cite{wu}), \emph{i.e.} the set
of all smooth linear functionals on $V$.

\begin{lemma}
Let $V^*$ be the diffeological dual of $V$. Then
$$\lambda_0\geqslant n-\dim(V^*).$$
\end{lemma}

\begin{proof}
Choose a basis $\{v_1,\ldots,v_n\}$ of eigenvectors of $A$ such that
the last $\lambda_0$ vectors are those relative to the eigenvalue
$0$. For $i=1,\ldots,n-\lambda_0$ let $v^i\in V^*$ be the dual
function, that is, $v^i(w)=\langle v_i|w \rangle_A$ (it is obviously
smooth since $\langle\cdot|\cdot\rangle_A$ is smooth, so is an
element of $V^*$). It remains to notice that by standard reasoning
the elements $v^1,\ldots,v^{n-\lambda_0}$ are linearly independent
(they belong to non-zero eigenvalues), so
$n-\lambda_0\leqslant\dim(V^*)$, hence the conclusion.
\end{proof}

Notice that we can always find a smooth bilinear (symmetric) form
for which we have $\lambda_0=n-\dim(V^*)$. It suffices to take any
basis $\{f_1,\ldots,f_k\}$ of $V^*$ (with $k=\dim(V^*)$) and
consider $\sum_i f_i\otimes f_i$; this is obviously a symmetric
bilinear form on $V$, and it being smooth follows from Theorem 2.3.5
of \cite{vincent}. By an obvious argument, we can see that all the
eigenvalues are non-negative; indeed, $(\sum_i f_i\otimes
f_i)(v\otimes v)=\sum_i(f_i(v))^2\geqslant 0$. Finally, the
multiplicity of the eigenvalue $0$ is precisely $n-k$, since
$f_1,\ldots,f_k$ form a basis of $V^*$.

Thus, the following definition makes sense.

\begin{defn}
Let $V$ be a diffeological vector space of finite dimension $n$. A
\textbf{diffeological pseudo-metric} on $V$ is a smooth bilinear
symmetric form on $V$ such that the eigenvalue $0$ has multiplicity
$n-\dim(V^*)$, and all the other eigenvalues are positive.
\end{defn}

A pseudo-metric is the best notion of a smooth\footnote{Which means
a bilinear form that takes values in $\matR$ with standard
diffeology and is smooth with respect to the latter.} metric that
can exist for an arbitrary finite-dimensional diffeological vector
space; note that if a given space is standard, \emph{i.e.} is of the
only type which admits a smooth metric, then obviously a
pseudo-metric is a usual Euclidean metric. Of course, one can look
for true metrics which are not smooth but are at least
piecewise-differentiable, or simply continuous, something that we
will do in the section that follows; meanwhile, we explore whatever
applications pseudo-metrics might have.

\begin{rem}
While it is more usual to use scalar products for various
constructions of (multi)linear algebra, we do note that a number of
them hold in part for symmetric bilinear forms. A specific example
would be a Clifford algebra, which is defined for a vector space
endowed with a symmetric bilinear form that \emph{a priori} does not
have to be a scalar product (an extension of this notion into the
diffeological context was considered in \cite{clifford}). Obviously,
if it is not then one cannot speak of unitary action on the exterior
algebra, for example; yet, various constructions hold.
\end{rem}

\subsection{Smooth direct sums and pseudo-metrics}

One easily observed fact that distinguishes diffeological vector
spaces from their usual linear counterpart is that, in the
diffeological case, there are two types of direct sum: one that is
smooth (in the sense described just below) and the other that is
not.

\paragraph{Smooth splittings as direct sums} In general, if we have a
diffeological vector space $V$ (of finite dimension) that, as a
usual vector space, decomposes into a direct sum $V=V_1\oplus V_2$,
then \emph{a priori} the following two diffeologies are different:
\begin{itemize}
  \item the given diffeology of $V$, and
  \item the vector space sum diffeology on $V_1\otimes V_2$ obtained from the
  subset diffeologies on $V_1$ and $V_2$.\footnote{The vector space sum diffeology
  on a direct sum of diffeological vector spaces is the finest vector space
  diffeology that contains the sum diffeology; it is formed of all maps
  that locally are (formal) sums of plots of the summands.}
\end{itemize}

\begin{example}
Let us consider an example where the two might actually be
different. Take $V=\matR^3$ endowed with the vector space diffeology
that is generated by the map $p:\matR\to V$ acting by
$p(x)=|x|(e_2+e_3)$. Set $V_1=\mbox{Span}(e_1,e_2)$ and
$V_2=\mbox{Span}(e_3)$. Observe that the subset diffeology on each
of these subspaces is just the standard diffeology of $\matR^2$ and
$\matR$ respectively.\footnote{Let us prove this claim. Let $q:U\to
V_1$ be a plot for the subset diffeology on $V_1$; let
$q_i:U\to\matR$ be the projection on $\mbox{Span}(e_i)\leqslant V_1$
for $i=1,2$. It suffices to show that $q_i$ must necessarily be a
smooth map (in the usual sense). Note that by definition of the
subset diffeology $Q(u)=(q_1(u),q_2(u),0)$ must be a plot for the
diffeology of $V$. By definition of the latter diffeology in the
neighbourhood of zero it writes as $Q(u)=\sum_{i=1}^3
f_i(u)e_i+(p\circ F)(u)$, where $f_1,f_2,f_3,F$ are the usual smooth
maps $U\to\matR$; therefore up to smooth summands $Q(u)$ writes as
$(0,|F(u)|,|F(u)|)$, from which we conclude that $|F(u)|\equiv 0$ in
the neighbourhood of $u=0$. This implies that $q_1,q_2$ are the
usual smooth maps, and the claim is proved for $V_1$; the reasoning
is analogous in the case of $V_2$.} This means that the vector space
sum diffeology on their direct sum is again the standard one, which
the diffeology on $V$ is not; indeed, the projection on the second
coordinate is smooth (in the usual sense) for the standard
diffeology, which does not happen for the diffeology of $V$.
\end{example}

\paragraph{Standard subspaces that split off as direct summands}
The example given in the previous paragraph easily leads to some
more general observations, such as the following.

\begin{lemma}\label{proj:smooth:lem}
Let $V$ be a finite-dimensional diffeological vector space, let
$V_1\leqslant V$ be its subspace such that its respective subset
diffeologies is standard, and let $V=V_1\oplus V_2$ be its
decomposition into a direct sum for some subspace $V_2\leqslant V$.
This decomposition is smooth if and only if there exists a basis
$\{v_1,\ldots,v_k\}$ of $V_1$ such that the projection\footnote{By
this we mean the map $V\to\matR$ acting by
$\pi_i(\sum_i\alpha_iv_i+w)=\alpha_i$, where $w\in V_2$.} on each
vector $v_i$ is a smooth linear functional on $V$.
\end{lemma}

\begin{proof}
The part ``only if'' of the statement follows from the definition of
the (vector space) direct sum diffeology. Indeed, suppose that the
diffeology on $V$ coincides with the direct sum diffeology formed
from the subset diffeologies on $V_1$ and $V_2$; recall that the
plots of standard diffeology are of form $\sum_if_i(x)v_i$, for some
smooth functions $f_i$ and a basis $\{v_1,\ldots,v_k\}$.

Let $\pi_i:V\to\matR$ be the projection on the vector $v_i$. Recall
that each plot $p:U\to V$ of $V$ locally writes as $p=p_1\oplus
p_2$, where (in particular) $p_1$ is a plot of $V_1$, hence
$(\pi_i\circ p)(u)=\pi_i(p_1(u))$. The latter is a smooth map
$U\to\matR$, since the diffeology of $V_1$ is standard; therefore
$\pi_i$ is smooth as a map $V\to\matR$.

Let us now prove the ``if'' part. Let, again, $\pi_i:V\to\matR$ be
the projection on the vector $v_i$; by assumption, this is a smooth
map, \emph{i.e.}, for any plot $p:U\to V$ of $V$ the composition
$\pi_i\circ p:U\to\matR$ is smooth in the usual sense. Now, in the
usual linear algebra sense $p$ does write as $p=p_1\oplus p_2$,
where $p_1:U\to V_1$ and $p_2:U\to V_2$; we need to show that each
of these $p_i$ is a plot of $V_i$.

Let us first prove this for $p_1$. Note that we have
$p_1(u)=\pi_1(u)v_1+\ldots+\pi_k(u)v_k$ by definitions of the maps
involved; since all $\pi_i$'s are usual smooth maps by assumption,
this is of course a plot of $V_1$. Note also that the formula just
given allows to regard $p_1$ also as a plot of $V$. Then $p_2=p-p_1$
(where $p_2$ is also considered as a map in $V$ in the obvious
sense), and since the diffeology of $V$ is a vector space
diffeology, $p_2$ is a plot of $V$. By definition of the subset
diffeology, as a map in $V_2$ it is a plot of it. This implies that
the diffeology of $V$ is indeed the (vector space) sum diffeology.
\end{proof}

We have already given an example for which this claim produces a
non-trivial result. It also can be generalized as follows (the
statement is rather complicated and probably not very useful in
itself; it does underline what truly matters in the above proof).

\begin{oss}
Let $V$ be a finite-dimensional diffeological vector space, and let
$V_1,V_2\leqslant V$ be its usual vector space subspaces such that
$V=V_1\oplus V_2$. This decomposition is smooth if and only if there
exist bases $\{v_1,\ldots,v_k\}$ of $V_1$ and
$\{v_{k+1},\ldots,v_{k+l}\}$ of $V_2$ satisfying the following
condition: if $\calD_i$ for $i=1,2$ is the finest diffeology on
$\matR$ that contains all projections $V_i\to\matR$ on the vectors
of the corresponding basis, then either $\calD_1\subset\calD_2$ or
$\calD_2\subset\calD_1$.
\end{oss}

\paragraph{The maximal standard subspace relative to a pseudo-metric}
As follows from the above discussion, a pseudo-metric exists on any
diffeological vector space (and we cannot do any better). On the
other hand, given a pseudo-metric $\langle\cdot|\cdot\rangle_A$ on a
fixed diffeological vector space $(V,\calD)$ of finite dimension,
the subspace $V_0$ of $V$ spanned by all the eigenvectors belonging
to the positive eigenvalues of $\langle\cdot|\cdot\rangle_A$ and
considered with the subset diffeology $\calD_0$ induced by $\calD$
is endowed, via the restriction of $\langle\cdot|\cdot\rangle_A$,
with a true metric; it follows then from \cite{iglesiasBook}, Ex.
70, p.74 that $(V_0,\calD_0)$ is the standard Euclidean space of
dimension $n-\lambda_0=n-\dim(V^*)$.

\begin{lemma}\label{V_0:smooth:summand:lem}
The subspace $V_0$ splits off as a smooth direct summand.
\end{lemma}

\begin{proof}
Denote by $V_1$ the subspace generated by all the eigenvectors
relative to the eigenvalue $0$. Obviously, $V$ splits as the direct
sum $V=V_0\oplus V_1$ in the usual sense; we need to check that this
decomposition is smooth, that is, that the diffeology on $V$
coincides with the direct sum diffeology on $V_0\oplus V_1$. Now, by
the lemma in the preceding paragraph, it suffices to check that the
projection on each eigenvector in some basis of $V_0$ is a smooth
linear functional on the whole of $V$.

The matrix $A$ being symmetric, there exists an orthonormal (with
respect to the usual scalar product on $\matR^n$ underlying $V$)
basis $\{v_1,\ldots,v_n\}$ of $V$ composed of eigenvectors of $A$,
where we can obviously assume that the eigenvectors relative to all
the positive eigenvalues (and so forming a basis of $V_0$) are the
first $k$ vectors, $v_1,\ldots,v_k$. Let $v_i$ be one of these,
relative to the eigenvalue $\lambda_i$, let $\pi:V\to\matR$ be the
corresponding projection, and let $p:U\to V$ be any plot of $V$; we
need to show that $\pi\circ p:U\to\matR$ is smooth in the usual
sense. Write $p(x)=\sum_{j=1}^n p_j(x)v_j$; recall that
$p_i(x)=\pi\circ p$.

Since $A$ defines a smooth bilinear form on $V$, and all the
constant maps are plots, the assignment $x\mapsto\langle v_i|p(x)
\rangle_A$ defines a map $U\to\matR$ which is smooth in the usual
sense. However, we have $\langle v|p(x)
\rangle_A=\sum_{j=1}^n\lambda_jp_j(x)\langle v_i|v_j
\rangle=\lambda_i||v_i||^2(\pi\circ p)(x)$, where $||v_i||$ is the
usual Euclidean norm of $v_i$. Since neither it nor $\lambda_i$ are
zero, we conclude that $\pi\circ p$ is smooth, as wanted.
\end{proof}

Furthermore, the following is true.

\begin{prop}\label{V_0:maximal:prop}
The subspace $V_0$ is a maximal subspace of $V$ with the subset
diffeology that is standard and which splits off as a smooth direct
summand.
\end{prop}

\begin{proof}
Suppose that $V_1>V_0$ is a bigger subspace of $V$ such that its
subset diffeology is standard. Let $v_1,\ldots,v_k$ be all the
(linearly independent) eigenvectors of $A$ relative to the positive
eigenvalues; by definition of $V_0$, they form a basis of it. Let
$v\in V_1\setminus V_0$ be a vector linearly independent from
$v_1,\ldots,v_k$.

Set $\varphi(w)=\langle w|v\rangle$ (the canonical scalar product on
$V$); let us show that $\varphi$ is smooth as a linear functional on
$V$. Indeed, if $p:U\to V$ is a plot of $V$ then $\varphi(p(x))v$ is
the usual orthogonal projection of $p(x)$ on the line generated by
$v$. Since $V_0$ splits off as a smooth direct summand, by Lemma ??
the projection on $v$ is smooth, therefore $\varphi$ belongs to the
diffeological dual $V^*$ of $V$. Recall now that $\dim(V^*)=k$, and
we can obtain its basis $f_1,\ldots,f_k$ by setting $f_i(w)=\langle
w|v_i\rangle_A$ for each $i=1,\ldots,k$. Then $\varphi$ writes as
$\varphi=\alpha_1f_1+\ldots+\alpha_kf_k$ for some
$\alpha_1,\ldots,\alpha_k$ not all of which are zero. Since $v_i$'s
are eigenvectors, we then get $\langle
w|v\rangle=\alpha_1\lambda_1\langle
w|v_1\rangle+\ldots+\alpha_k\lambda_k\langle w|v_k\rangle$; since
this is true for an arbitrary $w$ and the scalar product is the
canonical one, this implies that
$v=\alpha_1\lambda_1v_1+\ldots+\alpha_k\lambda_kv_k$. Since by
assumption all $\lambda_1,\ldots,\lambda_k$ are positive, and $v$ is
linearly independent from $v_1,\ldots,v_k$, we get a
contradiction.
\end{proof}

In other words, a pseudo-metric on $V$ allows to extract, in a
sense, from $V$ its biggest standard part, on which the diffeology
includes essentially the usual smooth maps. Let us be more precise.

\begin{lemma}
Let $V=W_0\oplus W_1$ be a smooth decomposition such that
$\dim(W_0)\geqslant\dim(V^*)$, and the subset diffeology on $W_0$ is
standard. Then there exists a pseudo-metric
$\langle\cdot|\cdot\rangle_A$ on $V$ such that $W_0$ is the space
generated by the eigenvectors of $A$ relative to the positive
eigenvalues, and $W_1$ is the space generated by the eigenvectors
relative to the eigenvalue $0$.
\end{lemma}

In particular, $\dim(W_0)=\dim(V^*)$.

\begin{proof}
Since the decomposition is smooth, each plot $q$ of $V$ is locally
of form $q_0+q_1$ for some plots $q_0,q_1$ of $W_0,W_1$
respectively. Now, the subset diffeology on $W_0$ being standard,
there exists a basis $\{w_1,\ldots,w_k\}$ of $W_0$ such that each
plot $q_0$ is (locally) of form $q_0(u)=\sum_{i=1}^kq_0^i(u)w_i$.
Take some basis $\{w_{k+1},\ldots,w_n\}$ of $W_1$; then
$\{w_1,\ldots,w_n\}$ is of course a basis of $V$, and it suffices to
define $B$ as $(B)_{ii}=1$ for $i=1,\ldots,k$ and $(B)_{ij}=0$
otherwise.
\end{proof}

To summarize, all subspaces of $V$ with standard sub-diffeology that
in addition split off as smooth direct summands are associated to
some pseudo-metric, provided they have a sufficiently large
dimension. We can actually show more: there is only one of them.

\begin{prop}
The subspace $V_0$ is an invariant of the diffeological space $V$,
\emph{i.e.}, it does not depend on the choice of a pseudo-metric
$\langle\cdot|\cdot\rangle_A$.
\end{prop}

\begin{proof}
Let $W_0\neq V_0$ be another subspace of $V$ of dimension
$k=\dim(V^*)$ such that the subset diffeology on $W_0$ is standard
and $W_0$ splits off as a smooth summand, $V=W_0\oplus W_1$. Let
$w_0\in W_0\setminus W_1$; it writes uniquely as $w_0=v_0+v_1$. Note
that by Lemma \ref{proj:smooth:lem} the projections on both $w_0$
and $v_0$ are smooth linear functionals; therefore so is the
projection on $v_1$. That would imply that $\mbox{Span}(V_0,v_1)$
has standard diffeology and splits off as a smooth summand; since
its dimension is strictly greater than $\dim(V_0)=\dim(V^*)$, this
is a contradiction.
\end{proof}

\subsection{The metric on $V^*$ induced by a pseudo-metric on $V$}

Here we consider a finite-dimensional diffeological vector space $V$
(we assume that its underlying vector space is identified with
$\matR^n$), endowed with a pseudo-metric
$\langle\cdot|\cdot\rangle_A$ given by a symmetric matrix $A$. As
for usual vector spaces, the pseudo-metric induces a symmetric
bilinear form on the diffeological dual $V^*$ of $V$; in this
section we deal with this induced form.

\paragraph{The functional diffeology on $V^*$} Recall \cite{wu} that in
general, the diffeological dual $V^*$ of $V$ is \emph{not} even
isomorphic to $V$, so not diffeomorphic to it; this occurs already
in the finite-dimensional case (in addition to \cite{wu}, see
Example 3.1 in \cite{multilinear}). Thus, it might well occur that
while the diffeology on $V$ is not standard, the corresponding
functional diffeology on its dual is. The following statement makes
it precise.

\begin{thm}\label{dual:diff-gy:standard:thm}
Let $V$ be a finite-dimensional diffeological vector space, and let
$V^*$ be its diffeological dual. Then the functional diffeology on
$V^*$ is standard (\emph{i.e.}, $V^*$ is diffeomorphic to some
$\matR^k$ with standard diffeology, for the appropriate $k$).
\end{thm}

\begin{proof}
Let us choose any pseudo-metric $\langle\cdot|\cdot\rangle_A$ on $V$
(as we have already noted, we can always find one). Let
$\{v_1,\ldots,v_n\}$ be an orthonormal basis of $V$ such that the
eigenvectors relative to the positive eigenvalues are the first $k$
vectors; by an above lemma, this means that the dimension of $V^*$
is equal to $k$. Denote by $\lambda_i$ the eigenvalue relative to
the eigenvector $v_i$; note that for each $i=1,\ldots,k$ the map
$f_i:V\to\matR$ given by $f_i(w)=\frac{1}{\lambda_i}\langle
w|v_i\rangle_A$ is a smooth linear functional on $V$, and so an
element of $V^*$. Finally, observe that each $f_i$ coincides with
$v^i$, the usual dual of the vector $v_i$, from which we conclude
that $\{f_1,\ldots,f_k\}$ is a basis of the diffeological dual
$V^*$.

To show that the diffeology of $V^*$ is the standard one, it is
sufficient to show that any plot $p:U\to V^*$ of it locally writes
as $p(u)=p_1(u)f_1+\ldots+p_k(u)f_k$ for some smooth (in the usual
sense) functions $p_i:U\to\matR$. That it does write so, for
\emph{some} functions $p_i$, obviously follows from the fact that
$\{f_1,\ldots,f_k\}$ is a basis, so we only need to show that each
$p_i$ is a smooth function.

Recall (\cite{iglesiasBook}, Section 1.57) that, the diffeology of
$V^*$ being functional, $p$ is a plot of it if and only if the
function $\varphi_p:U\times V\to\matR$ given by
$\varphi_p(u,v)=p(u)(v)$ is smooth (for the product diffeology). By
definition of the latter, in particular, the following map is
ordinarily smooth: $\varphi_p\circ(\mbox{Id}_U,c_{v_i}):U\to\matR$,
where $\mbox{Id}_U$ is the identity map on $U$ and $c_{v_i}$ is the
constant map that sends each $u\in U$ in $v_i$. But
$(\varphi_p\circ(\mbox{Id}_U,c_{v_i}))(u)=p(u)(v_i)=p_i(u)$, since
the vectors $v_i$ form an orthonormal basis. This allows us to
conclude that each $p_i$ is indeed an ordinary smooth map
$U\to\matR$, which, as already mentioned, implies that the
diffeology of $V^*$ is the standard one.
\end{proof}

\paragraph{The subspace $V_0$ and the diffeological dual} The proofs
of Lemma \ref{V_0:smooth:summand:lem} and of Theorem
\ref{dual:diff-gy:standard:thm} suggest that each pseudo-metric on
$V$ gives a natural identification between the corresponding $V_0$
and $V^*$ (that the two spaces are \emph{a priori} diffeomorphic is
now obvious, since they are both standard spaces of the same
dimension). More precisely, the following is true.

\begin{thm}\label{diffeo:V_0:dual:thm}
Let $V$ be a finite-dimensional diffeological vector space, and let
$\langle\cdot|\cdot\rangle_A$ be a pseudo-metric on $V$. The
restriction on the subspace $V_0$ of the induced map $\Psi:V\to V^*$
(\emph{i.e.}, the map given by $v\mapsto[w\mapsto\langle
w|v\rangle_A]$) is a diffeomorphism.
\end{thm}

\begin{proof}
As has been already observed in the proof of Theorem
\ref{dual:diff-gy:standard:thm}, the subspace $V_0$ admits an
orthonormal (with respect to the canonical scalar product on
$\matR^n$ underlying $V$) basis $\{v_1,\ldots,v_k\}$ composed of
eigenvectors of $A$; furthermore, if $f_i=\Psi(v_i)$ for
$i=1,\ldots,k$ then $f_i$ form a basis of $V^*$. In particular, the
restriction of $\Psi$ to $V_0$ is a bijection with $V^*$, hence an
isomorphism.

Now, applying the reasoning from the proof of Lemma
\ref{V_0:smooth:summand:lem} we find that each plot $p$ of $V_0$ is
of form $p(u)=p_1(u)v_1+\ldots+p_k(u)v_k$ for some smooth
real-valued maps $p_1,\ldots,p_k$. Hence the composition $\Psi\circ
p$ writes as $(\Psi\circ p)(u)=p_1(u)f_1+\ldots+p_k(u)f_k$, which is
obviously a plot of $V^*$ (the maps $p_1,\ldots,p_k$ being smooth,
this is a plot for any vector space diffeology on $V^*$), therefore
$\Psi$ is smooth. \emph{Vice versa}, arguing as in the proof of
Theorem \ref{dual:diff-gy:standard:thm}, every plot $q$ of $V^*$
writes as $q(u)=q_1(u)f_1+\ldots+q_k(u)f_k$ for some smooth
real-valued maps $q_1,\ldots,q_k$. Thus, the composition
$\Psi^{-1}\circ q$ is of form $(\Psi^{-1}\circ
q)(u)=q_1(u)v_1+\ldots+q_k(u)v_k$, which is a plot of $V_0$. We
conclude that $\Psi$ is smooth with smooth inverse, hence the
conclusion.
\end{proof}

\paragraph{The induced metric on $V^*$} It is now easy to see that
any pseudo-metric on $V$ induces a true metric on $V^*$, via the
smooth surjection $\Psi$ described in Theorem
\ref{diffeo:V_0:dual:thm}. Let us describe this precisely; notice
first of all that if $f\in V^*$ then there is a single element
$v_0\in\Psi^{-1}(f)$ such that $v_0\in V_0$. Furthermore, every
other element in $\Psi^{-1}(f)$ is of form $v_0+v_1$, where $v_1\in
V^{\perp}$, the orthogonal of the whole $V$ with respect to the
pseudo-metric $\langle\cdot|\cdot\rangle_A$ (equivalently, the
subspace $V_1$ spanned by all the eigenvectors relative to the
eigenvalue $0$). Let now $g\in V^*$ be another element of the dual
$V^*$, and let $w_0+w_1\in\Psi^{-1}(g)$. Then obviously, $\langle
v_0+v_1|w_0+w_1\rangle_A=\langle
v_0|w_0\rangle_A=\langle(\Psi|_{V_0})^{-1}(f)|(\Psi|_{V_0})^{-1}(g)\rangle_A$.
The first equality implies that the pushforward of the pseudo-metric
$\langle\cdot|\cdot\rangle_A$ to $V^*$ is well-defined (it could
have been replaced by a reference to \cite{wu}, Proposition 3.12
(1), of which it is a partial instance); the second equality means
that this pushforward is a true metric. We summarize this discussion
in the following statement.

\begin{cor}
Any pseudo-metric $\langle\cdot|\cdot\rangle_A$ on $V$ induces a
true metric on the diffeological dual $V^*$ of $V$, via the natural
pairing that assigns to each $v\in V$ the smooth linear functional
$\langle\cdot|v\rangle_A$.
\end{cor}

Using the reasoning employed so far, we can also establish the
inverse of this statement, namely, that \emph{every smooth metric
(scalar product) on $V^*$ is induced by a pseudo-metric on $V$.}
Indeed, let $\langle\cdot|\cdot\rangle_B$ be smooth metric on $V^*$,
given by a $k\times k$ matrix $B$. Let $\{f_1,\ldots,f_k\}$ be an
orthonormal basis of $V^*$ composed of eigenvectors of
$B$.\footnote{Note that the basis $\{f_1,\ldots,f_k\}$ generates the
standard diffeology on $V^*$, in the sense that each plot $q:U\to
V^*$ is locally of form $q(u)=q_1(u)f_1+\ldots+q_k(u)f_k$ for some
smooth functions $q_1,\ldots,q_k:U\to\matR$. Indeed, each plot does
write in this manner, simply by virtue of $\{f_1,\ldots,f_k\}$ being
a basis; furthermore, $\langle\cdot|\cdot\rangle_B$ being smooth,
and each constant map being a plot, implies that $\langle
q(u)|f_i\rangle_B=q_i(u)$ is smooth in the usual sense as a map
$U\to\matR$.} As has already been noted, the form
$\sum_{i=1}^kf_i\otimes f_i$ gives a certain pseudo-metric
$\langle\cdot|\cdot\rangle_A$ on $V$; it remains to check that this
pseudo-metric does induce $\langle\cdot|\cdot\rangle_B$ on $V^*$,
and this follows from the fact that $\{f_1,\ldots,f_k\}$ is an
orthonormal basis.

\subsection{A simple example of a pseudo-metric}

Let us return to an above example, considering $V=\matR^3$ endowed
with the vector space diffeology generated by the plot $p:\matR\to
V$ given by $p(x)=|x|(e_2+e_3)$. Observe first that the
diffeological dual of $V$ is generated by maps $e^1$ and $e^2-e^3$
(with $\{e^1,e^2,e^3\}$ being the canonical basis of the usual dual
of $\matR^3$). In particular, $\dim(V^*)=2$.

It is also easy to see that any smooth symmetric bilinear form on
$V$ is given by a matrix of form
${\small{\left(\begin{array}{ccc} c & a & -a \\
                                  a & b & -b \\
                                  -a & -b & b \end{array}\right)}}$
for some $a,b,c\in\matR$ (the vector $(0,|x|,|x|)$ must be
orthogonal to any other vector of $\matR^3$ with respect to the
associated bilinear form). For this matrix to define a
pseudo-metric, we must have $b,c>0$ and $a^2<bc$.

To give a specific example, we can take $a=1$, $b=c=2$, obtaining
$A={\small{\left(\begin{array}{ccc} 2 & 1 & -1 \\
                                  1 & 2 & -2 \\
                                  -1 & -2 & 2 \end{array}\right)}}$.
The two positive eigenvalues of $A$ are $\lambda_1=3+\sqrt3$ and
$\lambda_2=3-\sqrt3$; the corresponding (non unitary) eigenvectors
are $v_1'={\small{\left(\begin{array}{c} 3+\sqrt3 \\
                            3+2\sqrt3 \\
                        -3-2\sqrt3 \end{array}\right)}}$ and
$v_2'={\small{\left(\begin{array}{c} 3-\sqrt3 \\
                            3-2\sqrt3 \\
                        2\sqrt3-3 \end{array}\right)}}$. We can
however easily find that the subspace $V_0$ can be (better)
described as $V_0=\mbox{Span}(e_1,e_2-e_3)$; the restriction of
$\langle\cdot|\cdot\rangle_A$ to $V_0$, with respect to the basis
$\{e_1,e_2-e_3\}$ has the matrix ${\small{\left(\begin{array}{cc} 2 & 2 \\
                           2 & 8 \end{array}\right)}}$
(and yields indeed a true matrix). We also note that the subspace
$V_1$ (generated by eigenvectors relative to $0$) is $\mbox{Span}(
{\small{\left(\begin{array}{c} 0 \\
                               1 \\
                               1 \end{array}\right)}})$.

Finally, let us calculate the matrix of the induced metric on $V^*$
with respect to its basis $\{e^1,e^2-e^3\}$. For this, we need to
find their pre-images with respect to $\Psi|_{V_0}$ (we'll just
write $\Psi$ for brevity). By easy calculation
$\Psi^{-1}(e^1)=\frac23 e_1-\frac16(e_2-e_3)$ and
$\Psi^{-1}(e^2-e^3)=-\frac13 e_1+\frac13(e_2-e_3)$. Calculating
their pairwise products with respect to the pseudo-metric
$\langle\cdot|\cdot\rangle_A$, we find the matrix
$\frac19{\small{\left(\begin{array}{cc} 6 & 5 \\
                           5 & 6 \end{array}\right)}}$.


text:
Jordan Haworth Peele (born February 21, 1979) is an American actor, comedian, director, writer, and producer known for his film and television work in the comedy and horror genres.
Peele's breakout role came in 2003, when he was hired as a cast member on the FOX sketch comedy series Mad TV, where he spent five seasons, leaving the show in 2008. In the following years, he and his frequent Mad TV collaborator, Keegan-Michael Key, created and starred in their own Comedy Central sketch comedy series Key & Peele (2012–2015). In 2014, they recurred together as FBI agents in season one of FX's anthology series Fargo.[3] He co-created the TBS comedy series The Last O.G. (2018–present) and the YouTube Premium comedy series Weird City (2019–present). He will also host, narrate and produce the CBS All Access anthology series The Twilight Zone (2019–present).
Born Name: Jordan Haworth Peele
Born: Feb 21, 1979 in New York City, NY, United States
Read More On: Wikipedia
The Sidekick
Horror Noire: A History of Black Horror
Abruptio
GenreAdventure, Animation, Comedy, Family, Fantasy
RatingG
GenreHorror, Mystery, Thriller
GenreDocumentary
GenreHorror
GenreComedy, Romance

redpajama_set_name: RedPajamaCommonCrawl


* What's changed for Orca 0.2.6?
* Addition of verbalized punctuation settings.
* Addition of spoken indentation information for lines.
* Adjustments to brief verbosity settings for speech.
* Additional work on Evolution to better support configuration wizards.
* Very preliminary support for the Java platform.
* Moved a number of utilities from various places to util.py.
* Fixes for reading table row and column headers.
settings to be set properly from the configuration GUI.
* More work on bug 319652 to reduce even more hangs.
* Updates to the style guide.


For a woman, her wedding day is perhaps the most important day of her life. She will be meeting a host of people, including relations of her husband. They would expect her to look best and perfect in the wedding ensemble. She will also be photographed, and videoed a number of times. These images and videos will stay with her for the rest of her life. With so many important aspects of the wedding and high expectations of people, a woman may feel confused regarding her wedding wear.
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With Christmas just around the corner, there are so many exciting things to see and do in Liverpool and the city region over Christmas and the New Year.
With handy benefits like mobile tickets, free Wi-Fi and charging points, hopping on the bus is the best way of making the most of Christmas in Liverpool.
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To find out more about why it's Better By Bus, visit www.betterbybus.org.
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text:
Palestinian Prime Minister: Restructuring national economy will create jobs for ...
Palestinian Prime Minister: Restructuring national economy will create jobs for Palestinians
The Prime Minister of the Palestinian Authority Mohamad Shtayyeh was amongst a large number of world leaders who expressed strong support for the ILO as it marks its 100th anniversary at the Centenary International Labour Conference.
News | 12 June 2019
High-level address by H.E. Mr Mohammed SHTAYYEH, Prime Minister of the Palestinian Authority, to the 108th (Centenary) Session of the International Labour Conference. Geneva, 12 June 2019. © Crozet / Pouteau
The Prime Minister of the Palestinian Authority Mohammad Shtayyeh was among world leaders who gathered at the Centenary International Labour Conference (ILC) to reaffirm their commitment to the principle of social justice upon which the ILO was founded 100 years ago.
A total of over 40 heads of government and state will address the two-week long Conference, which is the 108th session of the ILC, running from June 10 to 21.
Addressing the Conference plenary, Prime Minister Shtayyeh said Palestinian workers, employers and government officials have long been strongly represented in the ILO. He thanked the ILO for its technical and financial support, including through ongoing work towards developing a comprehensive social protection system in the Palestinian Territory.
Shtayyeh recounted the many economic, labour market, political and humanitarian challenges facing Palestinians in the Occupied Palestinian Territory.
He recounted a number of initiatives and reforms his government was taking in order to counter these challlenges, including restructuring the Palestinian economy so that it gradually moves towards self sufficiency, largely through import substitution industrialization.
"Such a strategy will enable us to concentrate on the productive sectors of agriculture and industry in order to create jobs, generate income and increase the buying power of citizens, thus helping to address the imbalances in the labour market," Shtayyeh told delegates.
"Investing in people's capabilities and in institutions is important for the Palestinians, who are striving to build an independent Palestinian State that is capable of serving its citizens and meeting their needs," the Prime Minister added.
During the Conference nearly 6,000 delegates – representing governments, workers and employers – will discuss transformative changes in the world of work, and will consider the adoption of a landmark ILO Centenary Declaration on the future of work. They will also discuss violence and harassment at the workplace with a view to adopting a new international instrument.
The ILO and the Occupied Palestinian Territory (OPT) last year embarked on the second Palestinian Decent Work Programme, for the period 2018-2022. The ILO has continued to support the OPT in strengthening labour market analysis capacity, sustainable employment creation and labour market development in the OPT. The ILO is also working with partners in the OPT to strengthen the legislative framework on labour matters, including the re-initiation of the Labour Law reform process, as well as to support the establishment of the Palestinian Social Security Corporation through both direct financial and technical support.
Prime Minister Shtayyeh's full address to the ILC can be viewed here.
The ILO Director-General's 2019 report on the situation of workers of the occupied Arab territories can be accessed here.
Tags: employment, armed conflict, International Labour Conference
Regions and countries covered: Palestinian Territory, Occupied


At Beat IT 2014 event, MSI announced its new X99 chipset motherboard, which will support the Intel Haswell-E processors and eight DDR4 memory slots. If you like the previous release from MSI which is X99S Gaming 9 AC, then I'm damn sure you will like this Bad boy more than its older sibling. The new MSI X99S Gaming 9 ACK in terms of specifications is almost identical to its sibling, but it's equipped with improved Internet connectivity.
MSI X99S Gaming 9 ACK comes with built-in "Streaming Engine" AverMedia capture card, and it is styled to match most readily-available gaming hardware. Apart from other specifications the most eye catching feature is that, it's equipped with an Intel adapter for a Killer 1525 from Qualcomm. Unlike previous Gaming Series motherboards from MSI which came with a Killer E2200 NIC and an Intel Wireless AC adapter.
The Gigabit Ethernet connectivity (Killer E2205) with WLAN (WiFi 802.11ac Killer 1525) you can achieve a bandwidth of up to 1.866 Gbps with low latency, very useful for streaming as fashionable as the Gigabit Ethernet network can have the focus when playing DOTA /LOL. While on WiFi you can upload content to Twitch real time, thus ensuring the lowest latency possible.
MSI has not disclosed the price of MSI X99S Gaming 9, but we can expect that It would be more than $399.
P.S: The Wireless Killer interface is not more than $50.


Mostly sunny. Freezing level near 5000 feet.
Mostly cloudy. A chance of showers. Snow level near 3500 feet.
Mostly cloudy. A chance of showers. Snow level near 2000 feet.
Mostly cloudy with a slight chance of showers in the evening, then mostly clear after midnight. Snow level near 2500 feet.
Sunny. Freezing level near 3500 feet increasing to 4500 feet in the afternoon.
Mostly clear. Freezing level near 5500 feet.
Mostly sunny with a slight chance of showers. Snow level near 3500 feet.
Partly cloudy with a slight chance of showers. Snow level near 4000 feet.
Mostly sunny. Freezing level near 6000 feet.
Partly cloudy. Freezing level near 6500 feet.
Partly sunny with a slight chance of showers. Snow level near 4000 feet.
Partly sunny with a chance of showers. Snow level near 4000 feet.


(1) Comforter 90"x92" (2) Pillow Shams 20"x26"+2"
We are dedicated to providing our customers with different styles of great comforters for all the seasons.Enjoy these group of special offers to making your home great in a minute.Save a lot but get a lot.
The naturally relaxed feel of the SARAGOTA Comforter Set will accentuate the decor of your bedroom. Ruffled petals have been freely dispersed on a ruffled field which wonderfully complements the design. The whole effect is that the comforter is incredibly soft and inviting. Revel in the softness that this comforter will lovingly offer you.
Please Note: Due to different computer/screen display, the color of the actual item may vary slightly from the displayed images.


Categories: All, Gyokuro, Genmaicha, Fukamushi, Kukicha, Hojicha, and more, Hojicha, Iced Tea, Leaf, Low or no caffeine Tags: hojicha benefits, how to pronounce hojicha	SKU: HJ-40R.
The slightly sweet round flavor goes well with any meal or sweets. Enjoy chilled with any style of meal. The comparison between Hojicha loose leaf and Hojicha powder: If you like a clear, fresh note, Hojicha loose leaf is for you. If you like a savory, richer flavor, you will enjoy Hojicha powder.
Preparation: Brew 1 heaping teaspoonful to 1 Tbsp. with 8 oz hot water (185º F) for 2 to 5 minutes.


Question How can I stop my students speaking their mother tongue during task-based lessons? Answer There is plenty of advice on how to cope with L1 use during pair-work in our book Doing Task-based Teaching, Chapter 10, pages 220 to 221.
Question Why is music especially suited to CLIL? Answer Music helps learners to develop acute listening skills, respond to the ranges in tone, pitch and expression in the voice and concentrate hard on details of pronunciation, stress and rhythm.


Following the publication of CP18/29, which outlined the key features of the proposed financial services contracts regime (FSCR), the FCA has published a further consultation setting out additional details of the regime and proposals for the rules that will apply to firms within the FSCR.
The aim of the FSCR is to allow firms which have not entered the temporary permissions regime (TPR) to run-off their existing contracts with UK customers and exit the UK market in an orderly fashion. The actions covered by the FSCR will be limited to regulated activities which are necessary for the execution of a contract entered into prior to the exit day, to manage financial risk or to transfer property, rights or liabilities under a pre-existing contract.
A PI or RAISP which, at the end of the TPR, does not have the necessary authorisation to continue operating within the UK.
Firms do not need to notify the FCA in advance for the SRO period to apply. Firms under the SRO will be subject to all existing and future product intervention rules made by the FCA, however, the protection of the Financial Services Compensation Scheme will not apply.
The FCA has published a copy of a letter sent to The Rt Hon Nicky Morgan MP, Chair of the Treasury Committee, by Chief Executive Andrew Bailey which outlines the work the regulator is doing to improve mortgage switching options for customers who are currently unable to do so.
120,000 are customers of unregulated firms.
As part of its Mortgage Market Study Interim Report, the regulator has proposed measures to help customers of active firms switch to a better deal and welcomes the agreement between members of UK Finance, the Building Societies' Association and the Intermediary Mortgage Lenders' Association to enable these customers to switch, subject to certain criteria.
Now it is looking to take immediate action to help customers of inactive and unregulated firms. One of the key issues for this customer segment is that internal switches are typically unavailable because the lender either does not, or cannot, offer new loans. In the regulator's view, the solution is to enable them to switch to an active lender. However, the rules state that an affordability assessment must be carried out for each new customer, even if they are not borrowing a large amount. The regulator plans to consult on changes to its responsible lending rules by changing the affordability test from an absolute test to a relative test. This would change the focus to whether the new mortgage costs are more affordable than the current arrangement.
Once these regulatory barriers are removed, the industry needs to be willing to offer remortgaging options to consumers. A recent industry round-table indicates that there is a willingness to consider remortgage options, if it is commercially viable to do so.
The FCA has launched a new online portal designed to make it easier for mutual societies to manage their information, submissions and applications, as well as providing a gateway to the regulator's Mutuals Team.
Store and retrieve recent society documentation.
Prior to their annual return submission deadline, all mutual societies will receive details of the portal via post.
The regulator has published consolidated data on the fines issued so far this year. Two fines have been imposed on individuals in January 2019, totalling £79,240,787, compared to 15 fines imposed against both firms and individuals, totalling £60,467,212, during the whole of 2018.


text:
Q: Show number of factories behind categories in sidebar. codeigniter I have a sidebar with categories. when i click on one categorie it shows me the factories who are located in that category.
I did this using joins in codeigniter.
Now i want to show how much factories are in a category. so for example:
Categories.

Cars (2)
Books (7)
Animals (45)

So it simple has to show how much factories have that specific category.
i tried to do a simple count_all_results but then i get the total count of factories. but i want them to count by the specific id of categories.
my model function:
function categorieen_getall($segment3)
{
    $this->db->where('categorieen.idcategorieen', $segment3);
    $this->db->select('*');
    $this->db->from('bedrijfcategorieen');
    $this->db->join('categorieen', 'categorieen.idcategorieen = bedrijfcategorieen.idcategorieen');
    $this->db->join('bedrijven', 'bedrijven.idbedrijven = bedrijfcategorieen.idbedrijven');
    $query = $this->db->get();
    $result = $query->result();
    /*
    echo '
';
    print_r($result);
    */
    return $result;
}

My controller function:
    function get_All()
    {
        $data['cat'] = $this->categorieen_model->categorieen_getall($segment3);
        $this->load->view('views/sidebar', $data);
    }

My view:

    <?php
    echo '
Categorieën
';
    echo ($this->db->count_all_results('categorieen'));

    ?>

        <?php
        echo '
';
        if(isset($cat) && is_array($cat)){
            foreach($links as $k => $value){

                echo '
';
                echo '
';
                echo 'idcategorieen.' ">' .$value->Categorie. '';
                echo '
';

            }
        }
        echo '
';
        /*
        if(isset($cat ) && is_array($cat )){
            foreach($cat as $key => $row){
                echo "Categorie:"; echo $row->Categorie;
                echo "
";
                echo $row->idcategorieen;
                echo "
";
                echo $row->Bedrijfsnaam;
            }
        }
        */
        ?>

My database scheme:
Factories
--------
idfactories
factoryname
adress
email
...

Categories
----------
idcategories
Category

Factorycategories
-----------------
idfactorycategories
idcategories
idfactories

A: This is not quite suitable because I know there is a better way to do this, but this will do the work for you:
// Get categories
$categories = array();

$query = $this->db->query("select * from `Categories` order by `idcategories`");

if($query->num_rows() > 0)
{
  foreach($query->result_array() as $row)
  {
    $query_count = $this->db->query("select `idfactorycategories` from `Factorycategories` where `idcategories` = {$row['idcategories']}");
    $factory_count = $query_count->num_rows();

    $categories[] = array(
      'count' => $factory_count
    );
  }
}

For ease, you can first get categories in a foreach loop and then for each of them submit a query for the factories table to get relevant count of factories with that related category id.

A: I asked another question. with the same problem and i solved it.
Look at my own answer with the solution i got!
Count results in joined table and show between () in sidebar
Thanks for the people who helped me :)

redpajama_set_name: RedPajamaStackExchange


text:
Q: Disable selecting other cells after didSelectRowAtIndexPath I need to disable selecting other cells after selecting right cell in my app.
So, if user selects wrong answer, he can select other cell, but if select right answer, he can't do it.
Help me to resolve my problem.
My code is:
- (void) tableView: (UITableView *) tableView didSelectRowAtIndexPath: (NSIndexPath *) indexPath {

UITableViewCell *cell = [tableView cellForRowAtIndexPath:indexPath];
NSInteger rowNumber = [indexPath row];
NSArray *array = [self getAnswers];
if (rowNumber == 0)
{
    cell.contentView.backgroundColor = [UIColor whiteColor];
    cell.selectionStyle = UITableViewCellSelectionStyleNone;
}
if (rowNumber == [array[array.count - 2] intValue])
{
    cell.contentView.backgroundColor = [UIColor colorWithRed:0.0 green:1.0 blue:0.0 alpha:1.0];
    [_rightAnswersArray addObject:[NSNumber numberWithInt:_index+1]];
    NSLog(@"rightAnswersArray = %@",_rightAnswersArray);
    for (rowNumber = 1; rowNumber < array.count - 1; rowNumber++)
    {
        cell.userInteractionEnabled = NO;
    NSLog(@"%hhd",cell.userInteractionEnabled);
    }
}
else
{
    cell.contentView.backgroundColor = [UIColor colorWithRed:1.0 green:0.0 blue:0.0 alpha:1.0];
    [_wrongAnswersArray addObject:[NSNumber numberWithInt:_index+1]];
    NSLog(@"wrongAnswersArray = %@",_wrongAnswersArray);
    [self showComment];
}
}

UPD: Solved with this code.
if (rowNumber == [array[array.count - 2] intValue])
{
cell.contentView.backgroundColor = [UIColor colorWithRed:0.0 green:1.0 blue:0.0 alpha:1.0];
[_rightAnswersArray addObject:[NSNumber numberWithInt:_index+1]];
NSLog(@"rightAnswersArray = %@",_rightAnswersArray);
for (rowNumber = 1; rowNumber < array.count - 1; rowNumber++)
self.tableView.allowsSelection = NO;
}

A: You should implement UITableViewDelegate method
// Called before the user changes the selection. Return a new indexPath, or nil, to change the proposed selection.
- (NSIndexPath *)tableView:(UITableView *)tableView willSelectRowAtIndexPath:(NSIndexPath *)indexPath;

It should look like this
- (NSIndexPath *)tableView:(UITableView *)tableView willSelectRowAtIndexPath:(NSIndexPath *)indexPath
{
    return _isRightAnswerAlreadySelected ? nil : indexPath;
}

_isRightAnswerAlreadySelected should be set up in (void) tableView: (UITableView *) tableView didSelectRowAtIndexPath: (NSIndexPath *) indexPath


text:
Adrianus (Adriaan) van Veldhuizen (Groningen, 4 februari 1932 – Winsum (Gr.), 6 april 2013) was een Nederlands politicus namens de PvdA.

Van Veldhuizen doorliep het Marnix Gymnasium in Rotterdam en studeerde theologie aan de Rijksuniversiteit Leiden en deed daar ook de predikantenopleiding voor de Nederlandse Hervormde Kerk. Hij was adjunct-directeur van een Volkshogeschool in Uithuizen en was actief in de lokale en provinciale politiek voor hij in 1977 in de Eerste Kamer kwam. Hij was woordvoerder op het gebied van volkshuisvesting, justitie en binnenlandse zaken.

In 1990 werd Van Veldhuizen onderscheiden als Ridder in de Orde van de Nederlandse Leeuw. Hij is een zoon van schrijver en predikant Gijsbert van Veldhuizen en de kleinzoon van hoogleraar theologie Adrianus van Veldhuizen. Zijn broer is voormalig burgemeester Eppo van Veldhuizen.

PvdA-politicus (Nederland)
Eerste Kamerlid
Nederlands gemeenteraadslid
Lid van de Provinciale Staten van Groningen

redpajama_set_name: RedPajamaWikipedia


text:
Q: Store Polygon points to sqlite database In my app I create fields on the map by using polygons.The polygons are created by the user by clicking on the map and creating points/markers and the clicking a button to create the polygon.
Now I want these fields/polygons to be stored on my SQLite database but the points are given in lat/lng data type.
One way I thought of doing this is by storing the string of the polygon.getPoints() function:
Polygon p;
String latlng = p.getPoints().toString();

The output of latlng is:
[lat/lng: (-3.987413925092502,-18.76223847270012), lat/lng: (-4.878691546616136,-2.84572284668684), lat/lng: (-7.610723424687183,-11.666696257889273), lat/lng: (-3.987413925092502,-18.76223847270012)]

And then when retrieving it from the database convert all the points by first removing all the unnecessary text from the string using replaceAll() and then putting every single point on the list and then again converting them to lat/lng.
It seems a little bit like overkill is there a better way to handle this?

A: Try this.
Pattern LAT_LNG = Pattern.compile("([-\\d.]+),([-\\d.]+)");
String latlng = "[lat/lng: (-3.987413925092502,-18.76223847270012),"
    + "lat/lng: (-4.878691546616136,-2.84572284668684),"
    + "lat/lng: (-7.610723424687183,-11.666696257889273),"
    + "lat/lng: (-3.987413925092502,-18.76223847270012)]";
Matcher matcher = LAT_LNG.matcher(latlng);
while (matcher.find()) {
    double lat = Double.parseDouble(matcher.group(1));
    double lng = Double.parseDouble(matcher.group(2));
    System.out.println("lat=" + lat + " lng=" + lng);
}

output:
lat=-3.987413925092502 lng=-18.76223847270012
lat=-4.878691546616136 lng=-2.84572284668684
lat=-7.610723424687183 lng=-11.666696257889273
lat=-3.987413925092502 lng=-18.76223847270012


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powers and faculties slumber undeveloped;?
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always.continue. The fruit is bittei while green?
upon the Globe compose their infancy ; n long ca?
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the principles of a true System of Astronomy.
follow in the moral order of tilings front a true, com?
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passional (that is, moral and intellectual) Beauti?
idea of the Glon and Wisdom of the Divinity.
those engaged in them unceasingly to useful In?
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call t!ie attenbon of Country Mercliants to his stock ot th?
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el, r-. it can be folded and pnt in an ??rdinary \'.il...-.
Newark, Match. 1342. (m22 lm*] RICHARD WARD.
aii my busine?.- at my old Oar E-tal.li-ument. 4"2 Waier?t..
street,5 doors ibove Catherine Market.
Laces, Button-,"plumes, MusJietS,Rules, .'^c. i.e. belowco-t.
nersoaS t'urchx-in^ to sell a^am will uo weB to end.
jt^ads. kc kc of the most modem and improved patterns.
m inner, bv anpivmg at the >*~ ? t-d S.
? ,. S ! : ? L. TRI'SriiEL.
?ICK NO. :?? A NX-STREET.
-. ss on given immediately-?Kent low. Inqu re of J. Locke.
ortor kiltie manufacturing purposes. Apply to George s.
112 GK1NNELL, MIN I IHN k CO.,78 South-*.
in sti part of ?Vestt eet?would make two good Store?.
**' - i Dwelling between K-ntrtti a-id'Emu-""?"
a!4.'u- w\|. ||. Pl.N'KXKV. 179 Bowery, 2 story.
ami Lot nf Ground. No. If> Hester-street, T ?
isreplete with every convenience. Title indispuin?
Uiiard ii?j house', having two finished Basements. Inq 1 ?
.-;;A TO LET?The elegant House No.
J?jgj?, Lafayette Place, with the st?hle in the rear.?
.il''fiS ed three story Duelling House, No. 7j Diianc-st.
near Broadwny,..?uitalilo lor the residence of a privr-i.- fa?
May, ihrasdourtlis of the fourth story ol the Frank?
litres,"Stove. &c. Possession can !)?? ha<i immediately. In?
ii'' 'it iin the premises.
a< if s.\.M'L. KOOME, 121 Walkers?.
one acre set out witli a handsome growth fit peach tr*1''-.
MTO LET;?To a small family with?
and cloilies proses; also rear basement with cellar under?
i;OL?.MS Tp LET at.iUl Brwadway.
MTO LET?The Store H) liroadu ay.
D vvnlitig^di" the House aisri; consisting of a good large Par?
Ferry. The corner has a basement -mre. whieh would be !
nrick ilweilint! Houses, now finisli?hg; they will be !
the North River, and well soiled for anv business wber?
light mach uery is required, po-^ssine a d?"*ek and i w?arr? !
antlv -it.wted in a small mid pleasant vill;ur-'?/"?? |M.;'" f?
Pub ic noose, and a very ,???iya?.l. Rent onlv S100.
fn e and clear of al incunjbrance, ?* ith ne w House and ' ?ut- ?
'l b*- alcove i- o?en d a*. pp\;**e sale uutil the 18th day ol ?
ag-.-e-l family; y ttti 2 var eis i ; eh ncr fnnunthes^<on.
yestmew, or a part ia view of improving t?e a.ijoinii,.
twtt'i water privri-L'.-?, in thenurth Qu^rtero! St.:teu !?!??. .
?An to Let ?>? Lea*. ,orie ami a half mi es from Hacken.
FOR SAl.K.? '-i.-i.-ai.i^ l>-w>;;iiti<' ?
Eor sals <+r exehangr. *a tract* - i baud-onie Lan-1.
is ?!? ??? oak, luckih-> , bcaea, ?-a!:r:t. maple aad'^ryca'nwre.
nl tf SMITH i. will ?,-J;"?:K y--. Jo?n sl n;> ?airs.
-.i.^.. Itockawsy. L. i. ? .; ..- ,1 * ?> t ?Iis ti ? e 1 . the >larin?
??s-titte gnotl gr?sV'or iheadow land, old sbnn wo<x3land.
f? r. It l>(- divided into farms of different quantites.?
-Vj-.-I OE.VSKEO. : iiioc licest species and *arie.
contninuig the results ol hLs22 yeai>' practice tn New-York.
\- tin. work ha. been honored by the award of a Gold Me?
len Products, i,? ;!'>-r couiuienl i- deemed unneces-ary.
. -,j.r,~ acres in ttown ol Siuithtown, L I. mi the nun?
'calf titulier, and nU.in good fence, in a healthy neighbor?
hood,- w*itli schools* near. The l<and is'g?v>d anifsoisiiuatr.-!
houses for their families near this land, and is very desirable.
It w Ii be -old at -ii*> per acre.
Also, a Parin of 2-10 acre- near the Sound and a cood har?
1 ms lor his si?dents; and as regnrtls his mode ol' instruc?
struct. TIKI.MAS JUNKS, 183 Broadway.
PEALE'S ? Old Curiosity Shcp,' No.
oi specimens ol NatHral History ever oi'Vr?sd tor sale 111 thi?
received and for sale,.ih< larg? collection of the late Dr.
? 11 1 specin ' nn of* very other branch of natuial science.
Pet Birds and \uilllllls stutled a* natural as hie.
lisingiiiC Pi r-iii - taught and furnished withapparatus, Lc.
N B.?Pet Birds and Animals ?tu(Ted as natural as life.
in rcbants suppli< d at llie shoi lest notice.
name or lirm under which ?uith copartnership 1- to he con?
uarv, in 1I1? year one thousand r-i^'.t hundred and lorty-five.
? D^ted New-York. April II. !->g.
fa<uil!es, w ho aie now usint' nie an.icie.
Dr*m!*i'id bira-seJI;; before 'h?- public in the advertising col?
QJLVEli F1 IkTEVVru. Thornson, Pio.
VYarieiof the best description and of-the latest pa item*.?
ricultural; Honjcultural and Agricultural Societi??.
Onters by letter punctually aUended to.
lie beauty os the workmanship.
x) TO THE QI'EEN.?CAUTION.?The hi-rh ehanu.
Ill,,- ?rve, the e^nti'.ne Pens are ar?* all marked in furl?
tin?thai eacl gn-;- ?? i,-- a fnc simile of his signature.


Sparkloft Media's client, The Beaches of Fort Myers & Sanibel, was honored with a Silver Adrian Award in Social Media for its outstanding Pinterest "bucket list" sweepstake.
"Earning this award in this international travel marketing competition recognizes our work as some of the best in the world and acknowledges the outstanding results we are achieving for our destination," Tamara Pigott, VCB Executive Director, said in response to the award from the Hospitality Sales & Marketing Association International (HSMAI).
The "Pin Your Way to The Beaches of Fort Myers & Sanibel" campaign harnessed the power of photography by asking users to build custom Pinterest "bucket lists" highlighting their Southwest Florida dream vacations.
The Adrian Awards, which honors outstanding achievement in advertising, public relations and digital marketing in the travel industry, recognized this campaign for its innovative design that successfully raised awareness of the destination and cultivated a community of passionate destination ambassadors.
During the sweepstake, the client gained 300 new Pinterest followers and received 4,363 repins of their destination images. Promotion of the sweepstake on Facebook and Twitter reached more than 45,000 fans and 269,000 followers, respectively.
Sparkloft Media provides a wide range of services, including social media strategy development, social media campaign management and implementation services, as well as social software development. The agency's clients include some of the most innovative organizations in travel and tourism around the world in addition to consumer / lifestyle brands, government, non-profit and economic development organizations.
Gold Crest Care Center continues support for National Heart Month.


text:
Il Torneo dei quattro trampolini 2022-23 si è svolto nelle quattro sedi tradizionali di Oberstdorf, Garmisch-Partenkirchen, Innsbruck e Bischofshofen, situate in Germania e Austria, tra il 28 dicembre 2022 e il 6 gennaio 2023. È la 71ª edizione della manifestazione , valevole per la classifica della Coppa del mondo di salto con gli sci 2023.

Programma e risultati

Oberstdorf
 HS137 Schattenbergschanze, Germania

29 dicembre 2022

Garmisch-Partenkirchen
 HS142 Große Olympiaschanze, Germania

1 Gennaio 2023

Innsbruck
 HS128 Bergiselschanze, Austria

4 Gennaio 2023

Bischofshofen
 HS142 Paul-Ausserleitner-Schanze, Austria

6 Gennaio 2023

Classifica generale

Note

Voci correlate

 Torneo dei quattro trampolini
 Coppa del Mondo di salto con gli sci 2023
 Schattenberg (trampolino)
 Große Olympiaschanze
 Bergisel (trampolino)
 Trampolino Paul Ausserleitner

Collegamenti esterni

 Sito ufficiale
 Sito della FIS

Torneo dei quattro trampolini
Sci nel 2023


"Transitions" is a four-part international cultural exchange between the cities of Chicago, Bogotá, Goma, and Sarajevo. This multimedia exhibition highlights several national journeys of transitional justice in post-conflict environments. Through collaboration, community conversations, and workshops, the four teams of curators and photographers have explored and reflected upon the unique and shared experiences of a country's evolution from war to peace, alongside the lessons learned by their colleagues and audiences.
After its openings in Chicago and Bogotá, "Transitions" premiered in Sarajevo on the iconic Skenderija Bridge during WARM Festival 2017. In summer 2018, "Transitions" opened in Goma, DRC.


We serve Waccabuc residents from our Danbury office.
After you complete your consultation you can schedule your first appointment with your orthodontist, complete any necessary paperwork, and schedule a second appointment for your x-rays. During your time with us one of our friendly Waccabuc orthodontic assistants will teach you all about your new orthodontic appliance and show you how to place it. Your second appointment will last around 2 hours and our entire staff is here to help calm any nerves that you may have.
In order to ensure that your orthodontic piece is working properly we suggest that our patients visit our Waccabuc office every four to eight weeks. We do everything we can to work around your hectic schedule and create an appointment that is convenient. In order to ensure that your orthodontic piece is working properly we suggest that our patients visit our Waccabuc office every four to eight weeks. We do everything we can to work around your hectic schedule and create an appointment that is convenient.
For those who are unable to make their appointment give our Waccabuc office a call to reschedule. If you cannot plan ahead just yet feel free to give us a call when you have more availability, we are here to help and accommodate each of our patients.
Common orthodontic issues can be crowding, widely spaced teeth, jaw growth problems, protruding teeth, or bad bites which in children are often inherited or a result of injury. They can also be formed by thumb-sucking or early or late tooth loss. Children's orthodontic issues may correct themselves, though their jaws will harden by adulthood. Adult orthodontic problems won't get better with age which may result in surgery.
In order to help prevent any orthodontic issues in adulthood it is best to correct them at an early age. If your child is 7 years or older we recommend making an initial consultation to go over any problems they may be experiencing. The earlier your child is seen the better their outcome will be. Everyone at our Waccabuc office is here to put your child at ease and get them smiling confidently.
We recommend our Two-Phase treatment plan, which consists of three steps, to help get your child's smile corrected.
The first phase starts off with our Waccabuc Orthodontists recommending treatment for your child's jaw growth, overcrowding or spacing issues. By doing so your child's adult teeth will be aligned correctly.
Your child's permanent teeth will then naturally grow into position during a resting period, without the need of a retainer.
For those looking for further information please call our qualified team members at our Waccabuc office and let us assist you.
Most patients who receive orthodontic care are over the age of 21, and here at CT Braces in Waccabuc we are ready to give you a smile that you can be proud to show off. Adult orthodontics differ from childrens in the fact that their jaws are harder, which may cause some issues. There is a higher risk of gum or bone loss, worn or missing teeth, or incomplete orthodontic treatment as a child or teen.
CT Braces of Waccabuc understands that adults and children have a variety of different issues when it comes to orthodontic needs. We we assure you that our staff is here to make the process as easy as possible. Call us today to schedule a consultation or to learn more about your specific needs.
Here at CT Braces in Waccabuc we will first use orthodontic surgery to align jaws when necessary, we will then provide you with orthodontic braces to help move teeth into their correct position. Our orthodontics will work with an oral and maxillofacial surgeon to perform your surgery in a hospital.
Your orthodontic surgery will take several hour to complete, and afterwards we recommend you rest for two to four weeks. Some patients are fitted for braces after surgery that should be worn for six to twelve months, then they should use a retainer to help maintain the placement of their teeth. Orthodontic surgery is considered a major medical procedure that we are extremely experienced with, but we understand that you may have questions and we are here to help to calm any nerves or anxiety you may have. Call our Waccabuc office today to discuss your particular needs.
Our Waccabuc office is happy to provide Invisalign® to our patients who are looking for a straighter smile without the need of braces. Invisalign® is a great alternative to braces and allows your teeth to align slowly using clear and comfortable trays which are updated every two weeks. While wearing your Invisalign® retainers you will be able to eat, drink, floss, and brush just as you always would.
Your Invisalign® treatment will likely be complete within nine to fifteen months. Our orthodontists recommend visiting our Waccabuc office every six weeks to review your progress. To hear more about Invisalign® and its benefits call our Waccabuc office today for more information.
CT Braces has provided expert orthodontic care in Waccabuc and all throughout New York to our beloved patients for years. We continue to provide our patients with top doctors and advanced technologies, which has allowed us to grow our business throughout the state.


We help individuals and families meet the challenges of funding lifetime health care.
If you have it, you can have financial goals.
If you lose it, you have only one goal – get it back.
Understand Health, Retirement and Care Expenses.
Health & Care Contingencies for Family Focus on immediate family.
Form an Elder Care Funding Strategy.
Define Specific Health Insurance Choices, Funding Sources, & Implementation Tactics.
Monitor financial plan and performance vs. current and anticipated health conditions.
Confirm any expected roles of family members.
Monitor Financial Plan vs. Anticipated Health Conditions (of both spouses).
Assure needs & desires are met.


Gardermoen Raceway, NORWAY 26-29 July 2018: Sverre Dahl raced to victory in Top Fuel Motorcycle at the Drag Challenge at Gardermoen Raceway, round three of the EDRS Pro Nordic Motorcycle Championship. The Norwegian rider stormed to a stunning 5.89 seconds, a new personal best and track record, to defeat Rikard Gustafsson in an exciting final.
In Super Twin Motorcycle Marcus Christiansen scored his second consecutive win of the season, while Kalle Lyrén was in a class of his own in Pro Stock Motorcycle with two 6.99-second runs and a win. In Super Street Bike Mogens Lund took the win in a great 7.03 seconds to defeat the so far this season unbeaten Mathias Bohlin in the final.
The Drag Challenge at Gardermoen Raceway in Norway was the third round of the EDRS Pro Nordic Motorcycle Championship. After heavy rain in the night, the track crew did an excellent job again in preparing a record setting track. Several personal bests and even track records were reset.
In Top Fuel Motorcycle Sverre Dahl gave the home crowd and the track crew something to cheer about. The Norwegian rider stormed to a great 5.94 seconds in the semifinals to face championship leader Rikard Gustafsson in the final. In a great race Dahl raced to a stunning 5.896 seconds, a new personal best and new track record. Gustafsson had a good run as well, but the Swede's 6.01 seconds couldn't match Dahl's fabulous elapsed time.
In Super Twin Motorcycle the numbers one and two qualifiers, Svein Olaf Rolfstad and Marcus Christiansen, didn't make any mistakes and advanced to the final. Rolfstad couldn't copy his stunning 6.23 seconds from qualifying, but the Norwegian rider was too quick for Lasse Leponiemi and Anders Hörnström.
Christiansen won his first round on a holeshot, to improve on his qualifying performance in the semifinal to beat Samu Kemppainen in 6.40 seconds. In the final Christiansen improved once again to 6.38 seconds and as Rolfstad suffered serious engine problems, it was an easy win for the championship leader from Denmark.
After his win at Alastaro it was Christiansen's second consecutive victory of the EDRS Pro Nordic MC Super Twin Motorcycle championship.
In Super Street Bike the numbers one and two qualifiers, Mathias Bohlin and Mogens Lund didn't make any mistakes either and had to face each other in the final. After severe engine damage in the first round, Bohlin had to use a second engine in the next round, to storm to a stunning 7.019 seconds, a new personal best for the Swedish rider.
Alex Hope couldn't match Bohlin's 7.06 seconds in the semifinal either and so the championship leader advanced to the final. Lund improved on his qualifying performance with one 7.12 and two 7.04-second runs to defeat Bianca Nilsson, Roberth Häggblom and Vesa Ruhanen.
In the final Bohlin pulled a red light while Lund stormed to an even better 7.03 seconds to take his first EDRS Pro Nordic MC win of the season.
In Pro Stock Motorcycle Kalle Lyrén stormed to a stunning 6.999 seconds in the semifinal to advance to the final. In the other semifinal reigning champion Thomas Lysebraate Olsen suffered problems when the Norwegian rider tried to fire up his engine, handing the win to Bo Thorselius.
But in the final it was Thorselius who had problems when firing up the engine. In a bye run Lyrén improved to 6.990 seconds to take a well deserved win as the former champion was in a class of his own at Gardermoen.
In Super Comp Bike the numbers two and three qualifiers, Peter Östlund and Tomas Jonsson advance to the final. After an almost identical reaction time, Jonsson had the better run and took the win.
In Super Gas Bike Jami Oksanen and Alastaro winner Jenni Pekkarinen advanced to the final. Oksanen had a better reaction time while Pekkarinen suffered some problems at the start line. The Finnish female rider had the better run, but it was not enough to beat Oksanen. After his Tierp victory, it was the Finnish rider's second win of the season.
The EDRS Pro Nordic Motorcycle Championship continues August 23-26 with the final race of the season, the Summit Racing Equipment Internationals at Tierp Arena.


The Pro Bowl has long been considered a laughable representation of the NFL game. It reached a new level of comedy as several players swapped positions during the annual all-star game.
Orlando, Fla. — The Pro Bowl has long been considered a laughable representation of the NFL game.
The last two were played in sloppy weather, with the latest one also coming amid temperatures in the mid-50s. It was far from ideal conditions, raising speculation about the game's future in Orlando, but fairly fitting considering the effort players provided. It was two-hand touch most of the day, with officials blowing plays dead at the slightest hint of contact.
Regardless of the elements, the AFC made the plays the NFC didn't.
Kansas City's Patrick Mahomes completed an 18-yard touchdown pass to Indianapolis' Eric Ebron on the opening possession, helping Mahomes earn the offensive Most Valuable Player award. Mahomes pleaded with voters to give it to Chiefs fullback Anthony Sherman, who caught three passes for 92 yards and ran for a score.
Ramsey got in on offense late, catching a 6-yard slant pass from Houston's Deshaun Watson with 19 seconds remaining. Los Angeles Chargers rookie safety Derwin James failed to haul in the 2-point conversion.
New York Giants running back Saquon Barkley, Dallas running back Ezekiel Elliott, Tampa Bay receiver Mike Evans and New Orleans' Alvin Kamara all got in on defense for the NFC. Evans notched an interception.
The AFC led 20-0 early in the fourth quarter, looking like it might record the first shutout in Pro Bowl history. But Dallas' Dak Prescott found Atlanta's Austin Hooper for a 20-yard score on fourth down with 9:09 remaining.
The NFC had plenty of chances before that. The conference failed to score on a fourth-and-goal run early. Chicago's Mitchell Trubisky, Minnesota receiver Adam Thielen and Prescott threw interceptions.
Trubisky was sacked by Adams on a flea flicker, and Dallas' Amari Cooper had a wide-open touchdown pass bounce off his face mask.
Seattle's Russell Wilson also was sacked four times.
The New England Patriots have the Super Bowl send-off routine down pat.
Break out the "Super Bowl Bound" banner. Queue up the highlight video. Take a few shots at those loser cities that don't have annual championship pep rallies. And round up the usual suspects: owner Robert Kraft, coach Bill Belichick, quarterback Tom Brady and the rest of the team captains.
The five-time Super Bowl champion then led the crowd in a chant of "We're still here!" before tossing the microphone aside to more cheers, and the Patriots headed off for buses that will start them on their way to Atlanta. They'll play the Los Angeles Rams on Sunday seeking their sixth NFL title since the 2001 season.
"This is the biggest send-off we've ever had," Belichick said.
Defensive Back Patrick Chung took a video with the crowd chanting "Beat LA!" in the background. Special teams captain Matthew Slater, whose father played 20 years for the Rams and is a member of the team's Hall of Fame, said he grew up rooting for the team he will face when he tries for his third Super Bowl victory.
"But on Sunday, I don't know nothing about no Rams," he said.