text:
\section{Introduction}

Let $Q_1(\b{x}),Q_2(\b{x}),Q_3(\b{x})\in K[\b{x}]$ be three quadratic forms in $n$
variables $\b{x}=(x_1,\ldots,x_n)$, defined over a number field $K$. This paper
will be concerned with the Hasse principle and weak approximation for points
over $K$ on the intersection
\[\cl{R}:\; Q_1=Q_2=Q_3=0.\]

We begin by reviewing the situation for individual quadrics and pairs
of quadrics.
In the case of a single quadratic form the Hasse principle is always valid,
while weak approximation holds for all nonsingular
quadratic forms in $n\ge 3$
variables. When one has a pair of forms the Hasse principle may fail, even
when the variety $Q_1=Q_2=0$ is nonsingular, as is shown by the example
\[Q_1=X_1X_2-(X_3^2-5X_4^2),\;\;\;
Q_2=(X_1+X_2)(X_1+2X_2)-(X_3^2-5X_5^2)\]
(with $K=\mathbb{Q}$) due to Birch and Swinnerton-Dyer \cite{BSD}.
However it is known that for a nonsingular intersection defined by a pair
of forms in 8 or more variables both the Hasse principle and weak
approximation hold, see Heath-Brown \cite[Theorem 1]{HB8}. One cannot
dispense with the smoothness condition here.  The example
\[6X_1^2-X_2^2-X_3^2=X_4^2+\ldots+X_n^2=0 \]
over the field
$\mathbb{Q}$, has points in every completion, as soon as $n\ge 6$, but has no
rational points.

The difficulty with the above example arises from the real completions, and
it was shown by
Colliot-Th\'{e}l\`{e}ne, Sansuc and Swinnerton-Dyer \cite[Theorem C]{CTSSD}
that any pair of quadratic forms defined over a totally complex number field
$K$ has a common zero over $K$ (and hence satisfies the Hasse principle) as
soon as the number of variables $n$ is at least $9$.  This enables us
to handle an intersection $\cl{R}$ of three quadrics over a totally
complex number field $K$, using the method of Leep \cite{leep}. If
$n\geq 21$, then the projective variety
$Q_3=0$ automatically contains a linear space of dimension 8 defined
over $K$, since it must split off 9 hyperbolic planes. However the pair
of forms $Q_1,Q_2$ will have a common zero over $K$ in this linear
space, by the above mentioned
result of Colliot-Th\'{e}l\`{e}ne, Sansuc and Swinnerton-Dyer, and
we deduce that any triple of quadratic forms in at least 21 variables,
defined over a totally complex number field $K$, will have a common
zero in $K$.

However we wish to handle general number fields, and so we will assume that
the variety $\cl{R}$ is nonsingular.
To be more precise, we shall require that the matrix
\[\left(\begin{array}{cc}\nabla Q_1(\b{x}) \\ \nabla Q_2(\b{x}) \\ \nabla Q_3(\b{x})
\end{array}\right)\]
has rank 3 for every point $[\b{x}]$ in $\cl{R}(\overline{K})$.  When this condition
holds we will say that $Q_1,Q_2,Q_3$ is a ``nonsingular system'' of quadratic
forms.  According to Lemma 3.2 of Browning and Heath-Brown
\cite{many}, this will ensure that $\cl{R}$ is an absolutely
irreducible variety of codimension 3 and degree 8. For nonsingular
systems over $\mathbb{Q}$ one can apply the well-known result of Birch
\cite{birch}, which was proved using the Hardy--Littlewood circle
method.  Indeed Birch's work was generalized to arbitrary number
fields by Skinner \cite{skin}, who explicitly considers the question of
weak approximation. The outcome is that, for a nonsingular system,
$\cl{R}$ will satisfy the Hasse principle and weak approximation provided
that $n>d+24$, where $d$ is the dimension of the ``Birch singular
locus''. In fact, for a nonsingular system of 3 quadratic forms the
Birch singular locus will have dimension at most 2, so that it
suffices to have $n\ge 27$.

Having described the relevant background we are ready to state our
principal result.
\begin{theorem}
Let $Q_1,Q_2,Q_3$ be a nonsingular system of quadratic forms in $n$
variables, defined over a number field $K$.  Then the variety
\[\cl{R}:\; \cl{Q}_1\cap\cl{Q}_2\cap\cl{Q}_3,\]
where $\cl{Q}_i$ is the quadric $Q_i=0$,
satisfies both the Hasse principle and weak approximation, as soon as
$n\ge 19$.
\end{theorem}
Thus we get both an improvement on the range $n\ge 21$ mentioned above
for the case of totally imaginary fields, and on the range $n\ge 27$
coming from the methods of Birch and Skinner. We should also observe
at this point that $\cl{R}$ will have points over any completion $K_v$
at a finite place $v$, as soon as $n\ge 17$.  This is established by
Heath-Brown \cite[page 138]{qbirs} when $K=\mathbb{Q}$, and the proof for
general number fields $K$ is completely analogous.  Thus as far as the
Hasse principle is concerned our theorem only requires solvability in
the real completions of $K$. In the same connection we mention that
$\cl{R}$ has local points over $K_v$ for finite places over primes $p\ge
37$ as soon as $n\ge 13$, see Heath-Brown \cite[Corollary 1]{HBQs}.

Our basic strategy for proving the theorem will be to try and find
a linear space $L$ of dimension 7, defined over $K$ and lying in the
quadric hypersurface $\cl{Q}_3$. If we can ensure that the variety
$\cl{Q}_1\cap \cl{Q}_2\cap L$ is nonsingular, we can then apply Theorem 1 of Heath-Brown
\cite{HB8}, as mentioned above. This approach will require us firstly
to establish the smoothness condition, and secondly to ensure that
$\cl{Q}_1\cap \cl{Q}_2\cap L$ has points everywhere locally.

A comment is required about ``smoothness'' requirements, where we
shall adopt a significant abuse of terminology. When we say that the
variety $\cl{Q}_1\cap \cl{Q}_2\cap \cl{Q}_3$ is nonsingular, for example, we mean that
the corresponding system of 3 quadratic forms is a nonsingular system,
as described above. It is of course quite possible for the variety
$\cl{Q}_1\cap \cl{Q}_2\cap \cl{Q}_3$ to be nonsingular in the usual sense, without
the corresponding system of forms being nonsingular. For example, if
all the forms $Q_i$ vanish identically then $\cl{Q}_1\cap \cl{Q}_2\cap
\cl{Q}_3=\mathbb{P}^{n-1}$, which is nonsingular. When we say that $\cl{Q}_1\cap
\cl{Q}_2\cap L$ is nonsingular we will have in mind a system consisting of 2
quadratic forms and a (minimal) set of linear forms defining $L$.  A
variety can be defined in many ways via a system of forms, and the
reader will have to decide from the context what constitutes an
appropriate system to use.  Since all the relevant varieties will be
complete intersections there should be little difficulty with this.

\section{Geometric Considerations}

We begin by replacing the forms $Q_1,Q_2,Q_3$ by more convenient ones.
We will write $\b{Q}=(Q_1,Q_2,Q_3)$ for our triple of
quadratic forms, and proceed to consider linear combinations $\b{t}.\b{Q}
=t_1Q_1+t_2Q_2+t_3Q_3$.  The determinant $d_1(\b{t}):=\det(\b{t}.\b{Q})$
is a form in $t_1,t_2,t_3$ of degree $n$. We also define the
determinant
\[\delta(X,Y;\b{t}^{(1)},\b{t}^{(2)}):=\det(X\b{t}^{(1)}.\b{Q}+Y\b{t}^{(2)}.\b{Q})\]
and the discriminant
\[d_2(\b{t}^{(1)},\b{t}^{(2)}):=
{\rm Disc}(\delta(X,Y;\b{t}^{(1)},\b{t}^{(2)})).\]
Thus $\delta(X,Y;\b{t}^{(1)},\b{t}^{(2)})$ is a form of degree $n$ in
$X$ and $Y$, and
$d_2(\b{t}^{(1)},\b{t}^{(2)})$ is bihomogeneous in the entries of
$\b{t}^{(1)}$ and $\b{t}^{(2)}$.

It will be convenient to record some properties of nonsingular systems
of two quadratic forms.  The following lemma follows from Heath-Brown
and Pierce \cite[Proposition 2.1]{HBP}, for example.
\begin{lemma}\label{hbp}
Let $R_1(x_1,\ldots,x_m)$ and $R_2(x_1,\ldots,x_m)$ be quadratic forms
over an algebraically closed field $k$ of characteristic zero, and
suppose that they constitute a nonsingular system, so that $\nabla R_1(\b{x})$
and $\nabla R_2(\b{x})$ are linearly independent for any non-zero $\b{x}$
satisfying $R_1(\b{x})=R_2(\b{x})=0$.  Then every non-trivial linear
combination $aR_1+bR_2$ has rank at least $m-1$.
Moreover $\det(XR_1+YR_2)$ is not
identically zero, and has
distinct linear factors over $k$. Conversely, this last condition is
equivalent to the statement that $R_1$ and $R_2$ form a nonsingular
system.
\end{lemma}

We now use the following result of Aznar \cite[\S 2]{aznar}.
\begin{lemma}\label{lem:aznar}
Let  $V\subset \mathbb{P}^{n-1}$ be a nonsingular complete intersection  of
codimension $r$, which is defined over a field $k$ of characteristic
zero.  Then there is a system of generators $F_1,\ldots,F_r\in k[\b{x}]$ of the
ideal of $V$, with
\[\deg F_1\geq \cdots \geq \deg F_r,\]
such that the varieties
\[W_i: \quad F_1=\cdots=F_i=0, \quad (i\leq r),\]
are all non-singular. Moreover if $r<n-1$ then $V$ will be irreducible
with degree equal to the product of the degrees of the $F_i$.
\end{lemma}
The last statement in the lemma follows from Heath-Brown and Browning
\cite[Lemma 3.2]{many}.
When $V$ is originally defined by $r$ forms $G_j$ of the same
degree $d$, the proof of Aznar's result
shows that one can take the $F_i$ to be linear
combinations of the $G_i$.

In our case Lemma \ref{lem:aznar} implies in particular
that there are linear combinations
$\b{m}^{(1)}.\b{Q}$ and $\b{m}^{(2)}.\b{Q}$, with
$\b{m}^{(1)},\b{m}^{(2)}\in \overline{K}^3$, such that both the hypersurface
$\b{m}^{(1)}.\b{Q}=0$ and the intersection
$\b{m}^{(1)}.\b{Q}=\b{m}^{(2)}.\b{Q}=0$ are nonsingular.  In particular we
will have $d_1(\b{m}^{(1)})\not=0$, so that the form $d_1(\b{t})$ does
not vanish identically. Moreover it follows from Lemma \ref{hbp}
that the intersection
$\b{m}^{(1)}.\b{Q}=\b{m}^{(2)}.\b{Q}=0$ is nonsingular if and only if
$d_2(\b{m}^{(1)},\b{m}^{(2)})$ is non-zero, and so we deduce that the form
$d_2(\b{t}^{(1)},\b{t}^{(2)})$ does not vanish identically.

We therefore see that if we choose any three vectors $\b{m}^{(1)},
\b{m}^{(2)}, \b{m}^{(3)}\in K^3$ such that none of
\[d_1(\b{m}^{(3)}),\;\; d_2(\b{m}^{(1)},\b{m}^{(2)}),\;\;
d_2(\b{m}^{(1)},\b{m}^{(3)}),\;\;\mbox{or}\;\;
\det\big(\b{m}^{(1)}|\b{m}^{(2)}|\b{m}^{(3)}\big),\]
vanishes,
then the three forms $Q_i'(\b{x})=\b{m}^{(i)}.\b{Q}(\b{x})$, for $i=1,2,3$ are
defined over $K$ and
generate the same linear system as do $Q_1,Q_2,Q_3$.  Moreover $Q_3'=0$
is nonsingular, as are the varieties $Q_1'=Q_2'=0$ and $Q_1'=Q_3'=0$.
Thus without loss of generality we will assume that our
original forms satisfy these conditions.

We will also require $\cl{Q}_3$ to contain suitable linear spaces defined
over the real completions of $K$.  Suppose as above
that $Q_1$ and $Q_3$ form a
nonsingular system, and that $K_v$ is real.  Then the argument of
Heath-Brown \cite[Lemma 12.1]{HB8} shows that there is a real $\theta_v$ for
which $(\cos\theta_v)Q_1+(\sin\theta_v)Q_3=0$ is nonsingular and contains
a linear space of dimension at least $(n-4)/2$ over $K_v$.  (The
argument of \cite{HB8} does not explicitly produce a nonsingular
quadratic form, but since the functions $n_+$ and $n_-$ are everywhere
locally minimal one can change $\theta_v$ slightly, if necessary.)
By weak
approximation in $\mathbb{P}^1(K)$ we deduce that there exists $c\in K$ for
which $cQ_1+Q_3=0$ is also nonsingular and contains
linear spaces of dimension at least $(n-4)/2$ over each real
completion $K_v$.  We now replace $Q_3$ by $cQ_1+Q_3$ so that $\cl{Q}_3$ is
nonsingular and has linear spaces over
each real completion, of dimension at least $(n-4)/2$.
For finite places $v$ it is automatic that $\cl{Q}_3$
contains linear spaces over $K_v$, of dimension at least $(n-5)/2$.  We can
therefore conclude that $\cl{Q}_3$ has a linear space over $K$, with
dimension at least $(n-5)/2$. The existence of a single such linear space
is enough to ensure that there is one through every $K$-point of
$\cl{Q}_3$. An alternative way to express the above facts is to say that
$Q_3$ splits off at least $(n-5)/2$ hyperbolic planes over $K$.

From now on we will fix the forms $Q_1$, $Q_2$ and $Q_3$.  We remind
the reader that the varieties $\cl{R}$, $\cl{Q}_1\cap\cl{Q}_2$ and $\cl{Q}_3$ are
all nonsingular, and that we have arranged that $\cl{Q}_3$ contains a linear
space, defined over $K$, of dimension at least $(n-5)/2$.

For the remainder of this section we work over an arbitrary
algebraically closed field $k$ of characteristic zero.
After the above preliminary manoeuvres we are ready to prove our first
key result.

It will be convenient to define
\[F_t=F_t(\cl{Q}_3):=\{L\in\mathbb{G}(t,n-1):\, L\subset \cl{Q}_3\}.\]
\begin{lemma}\label{existL}
Suppose that the forms $Q_1,Q_2,Q_3$ are such that the varieties
$\cl{R}=\cl{Q}_1\cap\cl{Q}_2\cap\cl{Q}_3$, $\cl{Q}_1\cap\cl{Q}_2$, and $\cl{Q}_3$ are all
nonsingular.
Then for every integer $t$ in the range $3\le
t\le(n-5)/2$ there is an $L\in F_t$,
such that $L\cap\cl{Q}_1\cap\cl{Q}_2$ is nonsingular.
\end{lemma}

We should perhaps be more specific as to this last condition.  If $L$
has codimension $c$ say, and is given by linear equations
$\ell_1(\b{x})=\ldots=\ell_c(\b{x})=0$, then according to our convention,
$L\cap\cl{Q}_1\cap\cl{Q}_2$ is nonsingular if the partial derivatives of
$Q_1,Q_2$ and $\ell_1,\ldots,\ell_c$ are linearly independent at any
non-zero vector $\b{x}\in k^n$ such that
\[Q_1(\b{x})=Q_2(\b{x})=\ell_1(\b{x})=\ldots=\ell_c(\b{x})=0.\]
Thus for example, if $L\subset \cl{Q}_3$ is a line disjoint from
$\cl{Q}_1\cap\cl{Q}_2$, then $L$ fulfils the condition vacuously.

For the proof we will need some information about $t$-planes lying in
quadric hypersurfaces. We begin by introducing some notation.
Let $Q(x_1,\ldots,x_n)$ be a quadratic form of rank $r$, over
$k$, and let $\cl{Q}$
be the quadric $Q=0$, with dimension $n-2$.  Let $F(n,r,t)$ be the
Fano variety of $t$-planes in $\cl{Q}$, and let $F(n,r,t;P)$ be the
subvariety of such planes passing through a given point $P\in \cl{Q}$.
The variety $F(n,r,t)$ will be non-empty when $t\le n-r/2-1$.
Write $d_0(n,t,r)=\dim F(n,t,r)$, which will be independent of the
particular quadratic form $Q$.  Similarly let
$\dim F(n,t,r;P)=d_1(n,t,r)$ for $P$ a smooth point of $\cl{Q}$.
These dimensions are given by the following lemma.

\begin{lemma}\label{fano}
We have the following statements.
\begin{enumerate}
\item[(i)] $d_1(n,t,r)=d_0(n-2,t-1,r-2)$ if $t\ge 1$ and $r\ge 2$.
\item[(ii)] $d_0(n,t,r)=n-2-t+d_0(n-2,t-1,r-2)$ if $t\ge 1$ and
$2\le r\le 2n-2t-2$.
\item[(iii)] $d_0(n,t,r)=(t+1)(n-2-3t/2)$ if $2t+2\le r\le 2n-2t-2$.
\end{enumerate}
\end{lemma}

We take $P=[\b{e}_1]$ and extend to a basis
$\b{e}_1,\ldots,\b{e}_n$ of $K^n$. Then $Q$ takes the shape
$x_1L(x_2,\ldots,x_n)+Q'(x_2,\ldots,x_n)$ with respect to this basis,
with $L\not=0$. A further change of basis simplifies this to $x_1x_2+
Q''(x_3,\ldots,x_n)$. Here $Q''$ will have rank $r-2$.  One then sees
that $t$-planes in $\cl{Q}$ containing $\b{e}_1$ correspond
to $(t-1)$-planes in $Q''=0$, and the result (i) follows.

For part (ii) we consider the incidence correspondence
\[I=\{(P,L):P\in L\in F(n,t,r)\}.\]
The projection $\pi_2$ onto the second factor takes $I$ onto
$F(n,t,r)$, and each fibre has dimension $t$, so that
$d_0(n,t,r)=\dim(I)-t$. The condition $r\le 2n-2t-2$ ensures that
$d_0(n-2,t-1,r-2)\ge 0$, whence $d_1(n,t,r)\ge 0$ by part (i).  Thus,
for the projection $\pi_1$ onto the first factor, we see that
$\pi_1(I)$ contains every smooth point of $\cl{Q}$.  Hence
$\pi_1(I)=\cl{Q}$, and $\pi_1^{-1}(P)$ will have dimension $d_1(n,t,r)$ for
smooth points $P$.  Thus $\dim(I)=\dim(\cl{Q})+d_1(n,t,r)$, and
the result follows from part (i).

Finally, part (iii) follows from part (ii) by induction on $t$, the
result being clearly true for $t=0$.

We can now move to the proof of Lemma \ref{existL}
By Lemma \ref{fano} we have
\[\dim(F_t)=d_0(n,t,n)=(t+1)(n-2-3t/2)\]
if $2t+2\le n$.
We proceed to consider
the variety $F^{\dagger}$ defined to be the set of $t$-planes $L\in F$ for
which $L\cap\cl{Q}_1\cap\cl{Q}_2$ is singular.  Let
\[I:=\{(L,[\b{x}],[\b{t}])\in F_t\times\cl{R}\times\mathbb{P}^1:\,
[\b{x}]\in L,\,\b{y}^T(t_1Q_1+t_2Q_2)\b{x}=0\,
\forall [\b{y}]\in L\}.\]
If $L\cap\cl{Q}_1\cap\cl{Q}_2$ has a singular singular point $\b{x}$, then
$\b{x}\in L\cap\cl{R}$ and there is some $[\b{t}]\in\mathbb{P}^1$ such that
$\b{y}^T(t_1Q_1+t_2Q_2)\b{x}=0$ for every $[\b{y}]\in L$.
Thus if $\pi_1$ is the projection from $I$ onto its first factor,
one has $\pi_1(I)=F^{\dagger}$. It follows that $\dim(F^{\dagger})\le\dim(I)$.

Now consider the projection $\pi_{2,3}$ onto the second and third
factors. The fibre above the pair $([\b{x}],[\b{t}])$ will be
\beql{fib}
\{L\in F_t:\,[\b{x}]\in L,\,\b{y}^T(t_1Q_1+t_2Q_2)\b{x}=0\,\forall [\b{y}]\in L\}.
\end{equation}
We write $Q=t_1Q_1+t_2Q_2$ and
\[H(\b{x},\b{t})=\{[\b{y}]:\,\b{y}^TQ\b{x}=0\}\]
for convenience. Thus the fibre (\ref{fib}) may be written as
\[\{L\in F_t:\, L\subseteq H(\b{x},\b{t}),\,[\b{x}]\in L\}.\]
Since $[\b{x}]\in\cl{R}$, and $\cl{R}$ is nonsingular, we must have
$Q\b{x}\not=\b{0}$, so that $H(\b{x},\b{t})$ is a hyperplane. Hence
$H(\b{x},\b{t})$ intersects $\cl{Q}_3$
to produce a quadric hypersurface $\cl{Q}'$ in
$\mathbb{P}^{n-2}$, with rank, $r$ say, at least $n-2$. The point $[\b{x}]$
must be a smooth point of $\cl{Q}'$, since otherwise $Q\b{x}$ and $Q_3\b{x}$
would be proportional, contradicting the fact that $[\b{x}]$ is a smooth
point of $\cl{R}$.

It follows that
the dimension of the fibre (\ref{fib}) will be $d_1(n-1,t,r)$.
According to Lemma \ref{fano} we have
\[d_1(n-1,t,r)= d_0(n-3,t-1,r-2)=t\big(n-5-3(t-1)/2\big)\]
if $2(t-1)+2\le r-2\le 2(n-3)-2(t-1)-2$.
Since we are assuming that $t\le (n-5)/2$ the
required condition on $r$ certainly holds. We therefore deduce that
\begin{eqnarray*}
\dim(F^{\dagger})\leq\dim(I)&=&\dim(\cl{R}\times\mathbb{P}^1)
+\dim\pi_{2,3}^{-1}([\b{x}],[\b{t}])\\
&=& \big((n-4)+1\big)+t\big(n-5-3(t-1)/2\big)\\
&=& (t+1)\big(n-2-3t/2)-1\\
&=& \dim(F_t)-1
\end{eqnarray*}
so that $F^{\dagger}$ must be a proper subvariety of $F_t$.
This completes the proof of Lemma \ref{existL}.
\bigskip

We turn now to our second key result.
\begin{lemma}\label{existL2}
Suppose that the forms $Q_1,Q_2,Q_3$ are such that the varieties
$\cl{R}=\cl{Q}_1\cap\cl{Q}_2\cap\cl{Q}_3$, $\cl{Q}_1\cap\cl{Q}_2$, and $\cl{Q}_3$ are all
nonsingular.
Then for every non-negative integer $t\le(n-5)/2$ there is an
$L\in F_t$ such that $(Q_3L)^{\perp}\cap\cl{R}$ is nonsingular.
\end{lemma}
Here we define
\[(Q_3L)^{\perp}:=\{[\b{y}]:\,\b{y}^TQ_3\b{x}=0,\,\forall [\b{x}]\in L\}.\]

We prove this by induction on $t$.
For the base case $t=0$ the space $L$ is a single point $P$, say. It then
suffices that $P\in \cl{Q}_3$ and $P\not\in
Q_3^{-1}(\cl{R}^*)$, where $\cl{R}^*$ is the dual variety to $\cl{R}$, and
$Q_3^{-1}(\cl{R}^*)$ is its image under the linear map $Q_3^{-1}$.
There will always
be a suitable point $P$ if $\cl{Q}_3\not\subseteq Q_3^{-1}(\cl{R}^*)$. However $\cl{R}^*$
is a proper subvariety of $\mathbb{P}^{n-1}$.  We claim that it cannot be a
nonsingular quadric, whence we cannot have $\cl{Q}_3=Q_3^{-1}(\cl{R}^*)$. To prove the
claim we merely observe that if $\cl{R}^*=\cl{Q}$, say, then
$\cl{R}=\cl{R}^{**}=\cl{Q}^*$. However $\cl{Q}^*$ is itself a nonsingular quadric,
whereas $\cl{R}$ has degree 8, by Lemma \ref{lem:aznar}. The lemma then
follows in the case of dimension zero.

To establish the induction step we suppose we have a suitable space
$L$ of dimension $t$, and look for an appropriate $L'$ of dimension
$t+1$. Indeed we shall restrict our attention to linear spaces satisfying
\[L\subset L'\subseteq\cl{Q}_3\cap (Q_3L)^{\perp}.\]
This requirement on $L'$ allows us to restrict all our varieties to the
subspace $(Q_3L)^{\perp}$. We write $\cl{Q}_i'=\cl{Q}_i\cap(Q_3L)^{\perp}$
for $i=1,2,3$, and $\cl{R}'=\cl{R}\cap(Q_3L)^{\perp}$. Thus $\cl{R}'$ is nonsingular.
More concretely, we choose
$[\b{e}_0],\ldots,[\b{e}_t]$ spanning $L$, and extend these to a set of points
$[\b{e}_0],\ldots,[\b{e}_{n-t-2}]$ spanning $(Q_3L)^{\perp}$, and then to a set
$[\b{e}_0],\ldots,[\b{e}_{n-1}]$ spanning $\mathbb{P}^{n-1}$. We proceed to
write
\[Q_i'(x_1,\ldots,x_{n-t-1})=Q_i(\sum_{j=1}^{n-t-1}x_j\b{e}_{j-1})
\;\;\;(i=1,2,3)\]
so that these quadratic forms correspond to the quadrics $\cl{Q}_i'$,
seen as varieties in $\mathbb{P}^{n-t-2}$. We note in particular that
$Q_3'$ is singular, with rank $n-2t-2$. Indeed, as a subvariety of
$\mathbb{P}^{n-t-2}$ the quadric $\cl{Q}_3'$ hypersurface will be a cone with
vertex set $L\subset\mathbb{P}^{n-t-2}$.

We construct
$L'$ as $<L,[\b{x}]>$ with $[\b{x}]\in\cl{Q}_3'$ but $[\b{x}]\not\in L$.
This will ensure that $L'\subseteq\cl{Q}_3'$ and that $\dim(L')=t+1$.
Moreover we will have
$(Q_3L')^{\perp}=(Q_3L)^{\perp}\cap(Q_3\b{x})^{\perp}$, so that
$(Q_3L')^{\perp}\cap\cl{R}'$ will be nonsingular provided that $[Q_3\b{x}]$
is not in $(\cl{R}')^*$. Since $\dim(L)=t<\dim(\cl{Q}_3')=n-t-3$, the generic
$[\b{x}]\in\cl{Q}'_3$ will satisfy $[\b{x}]\not\in L$.  Moreover it will also
satisfy $[Q_3\b{x}]\not \in (\cl{R}')^*$ by a similar reasoning to that given
in the dimension zero case above.  Specifically, $(\cl{R}')^*$ is a
proper subvariety of $\mathbb{P}^{n-t-2}$, so the only situation to rule
out is that in which it is equal to $\cl{Q}_3'$. However if $Q_3'$ has
rank $n-2t-2$ then the dual $(\cl{Q}_3')^*$ will be a quadric of dimension
$n-2t-4$, which cannot possibly be the variety $\cl{R}'$, since the
latter will have degree 8.

It follows that we can take $L'=<L,[\b{x}]>$ for a generic
$[\b{x}]\in\cl{Q}_3'$. This completes the induction step and so establishes
the lemma.
\bigskip

Following on from Lemmas \ref{existL} and \ref{existL2} we make
the following definitions.
\begin{definition}
Let $Q_1,Q_2,Q_3$ be a nonsingular system of quadratic forms in $n$
variables as above, and let
$L\subseteq\mathbb{P}^{n-1}$ be a linear space contained in
the quadric $\cl{Q}_3$.  Then we say that $L$ is ``admissible'' if and
only if both $L\cap\cl{Q}_1\cap\cl{Q}_2$ and $(Q_3L)^{\perp}\cap\cl{R}$ are nonsingular.
If $L$ has dimension $t$ in the range $3\le t\le 7$
we say that $L$ is ``chain-admissible'' if
there exist admissible linear spaces $L=L_t\subset
L_{t+1}\subset\ldots\subset L_7$ with $\dim(L_i)=i$ for $t\le i\le 7$.
\end{definition}

We then have the following result.
\begin{lemma}\label{existL3}
For each $t\in[3,7]$ there is a Zariski-closed proper subset $Z_t\subset
F_t$, such that $L\in F_t$ is chain-admissible if and only if
$L\not\in Z_t$.
\end{lemma}

It is clear that the $L\in F_t$ for which $L\cap\cl{Q}_1\cap\cl{Q}_2$ is
singular form a closed subset $A_t$ say, of $F_t$.  Similarly those
for which $(Q_3L)^{\perp}\cap\cl{R}$ is singular form a closed subset
$B_t$ say. By Lemmas \ref{existL} and \ref{existL2} these are
proper closed subsets of $F_t$.  Moreover, since $F_t$ is absolutely
irreducible, the union $A_t\cup B_t=C_t$, say, is also a closed proper
subset of $F_t$.  A linear subspace $L\in F_t$ is admissible if and
only if $L\not\in C_t$. We proceed to prove Lemma \ref{existL3}
by downwards induction, and it is clear that we can take $Z_7=C_7$.

For $t<7$ if $L\in F_t$ fails to be chain-admissible then either $L\in
C_t$ (because $L$ is not itself admissible) or $L'$ fails to be
chain-admissible for every $L'\in F_{t+1}$ containing $L$.  This
latter case holds precisely when $L'\in Z_{t+1}$ for every such $L'$. Write
$D_t$ for the set of $L\in F_t$ with the property that $L'\in Z_{t+1}$
for every $L'\in F_{t+1}$ containing $L$.  We claim that $D_t$ is a
proper closed subset of $F_t$.  Once this is established the induction
step of the proof is completed by taking $Z_t=C_t\cup D_t$.

To handle $D_t$ we consider
\[I:=\{(L,L')\in F_t\times Z_{t+1}:\, L\subset L'\}.\]
For the projection onto the first factor we have
\[\dim(\pi_1^{-1}(L))=\dim\{L'\in Z_{t+1}:\,L\subset L'\}\leq n-2t-4,\]
with equality exactly when $L'\in Z_{t+1}$ for every linear space $L'\in
F_{t+1}$ containing $L$. Thus $D_t$ is the set of $L$ for which
$\dim(\pi_1^{-1}(L))$ is maximal, whence $D_t$ is
Zariski-closed. Moreover we have
\[\dim(D_t)\le \dim(I)-(n-2t-4).\]
On the other hand, for the projection onto the second factor we have
\[\dim(\pi_2^{-1}(L'))= t+1,\]
whence
\[\dim(I)=\dim(Z_{t+1})+t+1.\]
Since $Z_{t+1}$ is a proper subset of $F_{t+1}$ by the downward
induction hypothesis we deduce that
\begin{eqnarray*}
\dim(D_t)&\le& \dim(I)-(n-2t-4)\\
&=& \dim(Z_{t+1})+t+1-(n-2t-4)\\
&<& \dim(F_{t+1})-n+3t+5\\
&=& (t+2)(n-2-3(t+1)/2)-n+3t+5\\
&=& (t+1)(n-2-3t/2) \\
&=& \dim(F_t).
\end{eqnarray*}
Here we have used Lemma \ref{fano} to compute
\[\dim(F_{t+1})=d_0(n,t+1,n)=(t+2)(n-2-3(t+1)/2)\]
and
\[\dim(F_t)=d_0(n,t,n)=(t+1)(n-2-3t/2).\]
The above calculation shows that $\dim(D_t)<\dim(F_t)$, so that $D_t$
is a proper subset of $F_t$, as required.

We conclude this section with two easy results in a similar vein.
\begin{lemma}\label{R3chain}
There is a Zariski-closed proper subset $\cl{R}_0\subset\cl{R}$
such that every $P\in\cl{R}-\cl{R}_0$ is contained in a chain-admissible
linear space $L\in F_3-Z_3$.
\end{lemma}

Let $\cl{R}_0$ be the set of points $P\in\cl{R}$ such that $L\in Z_3$ for every
$L\in F_3$ which contains $P$.  We need to show that $\cl{R}_0$ is a
proper closed subset of $\cl{R}$.  It follows from Lemma \ref{existL3}
that there is at least one $L\in F_3-Z_3$.  Since
$\dim(L)+\dim(\cl{R})=3+(n-4)=n-1$ it follows that $L\cap\cl{R}$ is
non-empty, containing $P$ say.  Then $P\in \cl{R}-\cl{R}_0$, so that $\cl{R}_0$
is a proper subset of $\cl{R}$.

To show that $\cl{R}_0$ is Zariski-closed we consider
\[I=\{(P,L)\in \cl{R}\times Z_3:\, P\in L\}.\]
For the projection onto the first factor we have
\[\dim(\pi_1^{-1}(P))\leq d_1(n,3,n),\]
in the notation of Lemma \ref{fano}, with equality exactly when $L\in
Z_3$ for every $L\in F_3$ which contains $P$.  Thus $\cl{R}_0$ is the set
of $P$ for which $\dim(\pi_1^{-1}(P))$ is maximal, and it follows that
$\cl{R}_0$ is Zariski-closed, as claimed.

\begin{lemma}\label{good4}
Given $L\in F_3-Z_3$, there is a Zariski-closed proper subset $\cl{R}_1$
of $\cl{R}\cap(Q_3L)^{\perp}$, such that $<L,P>\in F_4-Z_4$ for every
$P\in\cl{R}\cap(Q_3L)^{\perp}-\cl{R}_1$. Similarly, given $L\in F_t-Z_t$, for some
$t\in[3,6]$, there is a Zariski-closed proper subset $E_t$ of
$\cl{Q}_3\cap(Q_3L)^{\perp}$, such that $<L,P>\in F_{t+1}-Z_{t+1}$ for every
$P\in\cl{Q}_3\cap(Q_3L)^{\perp}-E_t$.
\end{lemma}

To prove the first part of the lemma we let
\[I:=\{(P,L')\in(\cl{R}\cap(Q_3L)^{\perp})\times Z_4:\, P\in L',\,L\subset L'\},\]
and take $\cl{R}_1=\pi_1(I)$, so that $\cl{R}_1\subseteq\cl{R}\cap(Q_3L)^{\perp}$
is clearly Zariski-closed.  Since $L\in F_3-Z_3$ is chain-admissible
there is at least one linear space $L_0\in F_4-Z_4$ containing
$L$. Then $L_0\cap\cl{R}=L_0\cap\cl{Q}_1\cap\cl{Q}_2$ has dimension $2$, since
$L_0$ is admissible, and similarly $L\cap\cl{R}=L\cap\cl{Q}_1\cap\cl{Q}_2$
has dimension $1$. We may therefore find a point $P\in
L_0\cap\cl{R}-L\cap\cl{R}$. We claim that $P\in\cl{R}\cap(Q_3L)^{\perp}-\cl{R}_1$, which
shows that $\cl{R}_1$ is a proper subset of $\cl{R}\cap(Q_3L)^{\perp}$, and so proves the
first part of the lemma.  Since $L\subset L_0\subset\cl{Q}_3$ it follows that
$L_0\subseteq(Q_3L)^{\perp}$, whence $P\in L_0\cap\cl{R}\subseteq (Q_3L)^{\perp}\cap\cl{R}$.  On the
other hand if we had $P\in\cl{R}_1$ there would be a linear space $L'$
such that $(P,L')\in I$.  Then we would have $P\in L'$ and $L\subset
L'$, and therefore $L'=<L,P>$, since $P\not\in L$ by our choice of
$P$. However the same reasoning shows that $L_0$ is also equal to
$<L,P>$, so that $L'=L_0$.  This gives us a contradiction, since
$L'\in Z_4$ while $L_0\in F_4-Z_4$.

Turning to the second part of the lemma, we consider
\[I:=\{(P,L')\in(\cl{Q}_3\cap(Q_3L)^{\perp})\times Z_{t+1}:\, P\in L',\,L\subset L'\},\]
and take $E_t=\pi_1(I)$, so that $E_t\subseteq\cl{Q}_3\cap(Q_3L)^{\perp}$
is clearly Zariski-closed.  Since $L\in F_t-Z_t$ is chain-admissible
there is at least one linear space $L_0\in F_{t+1}-Z_{t+1}$ containing
$L$. We claim that $P\in\cl{Q}_3\cap(Q_3L)^{\perp}-E_t$ for any point $P\in L_0-L$,
which will show that $E_t$ is a proper subset of $\cl{Q}_3\cap(Q_3L)^{\perp}$.
Since $L\subset L_0\subset\cl{Q}_3$ it follows that
$L_0\subseteq(Q_3L)^{\perp}$, whence $P\in\cl{Q}_3\cap(Q_3L)^{\perp}$.  On the
other hand if we had $P\in E_t$ there would be a linear space $L'$
such that $(P,L')\in I$.  Then we would have $P\in L'$ and $L\subset
L'$, and therefore $L'=<L,P>$, since $P\not\in L$ by our choice of
$P$. However the same reasoning shows that $L_0$ is also equal to
$<L,P>$, so that $L'=L_0$.  This gives us a contradiction, since
$L'\in Z_{t+1}$ while $L_0\in F_{t+1}-Z_{t+1}$.

\section{Global 3-planes in $\cl{Q}_3$}

In this section we shall make repeated use of three key principles.
The first of these is the fact that we have weak approximation on
quadrics.  The second is that if $V$ is an absolutely irreducible
projective variety defined over the number field $K$, with a smooth
point $P_v$ in some completion $K_v$, then the $K_v$-points of $V$ are
Zariski-dense in any given neighbourhood of $P_v$. (See Browning,
Dietmann and Heath-Brown
\cite[Lemma 3.4]{BDHB}, for example. The proof is an application of
the implicit function theorem.)

The third general principle is embodied in the following lemma.
\begin{lemma}\label{approxV}
Let $V$ be a projective
algebraic variety defined over a completion $K_v$ of $K$
by equations
\[f_1(x_0,\ldots,x_m)=\ldots=f_r(x_0,\ldots,x_m)=0,\]
and suppose that $P$ is $K_v$-point on $V$ at which the vectors
$\nabla f_1,\ldots,\nabla f_r$ are linearly independent. Suppose we
are given varieties $V^{(j)}$ defined over $K_v$ by equations
\[f_1^{(j)}(x_0,\ldots,x_m)=\ldots=f_r^{(j)}(x_0,\ldots,x_m)=0,\]
of bounded degree,
in which $f_i^{(j)}\rightarrow f_i$ (under the metric induced from
$K_v$) as $j\rightarrow\infty$.  Then for sufficiently large $j$ there
are $K_v$-points $P^{(j)}$ on $V^{(j)}$ such that $P^{(j)}\rightarrow
P$ as $j\rightarrow\infty$. Moreover $P^{(j)}$ will be a nonsingular
point of $V^{(j)}$, in the sense above.
\end{lemma}

For the proof we suppose firstly that $v$ is a finite place. We let
$P=[\b{t}]$ say, and we rescale $\b{t}$ and the polynomials $f_j$ so
as to have $v$-adic integer coefficients. Since $f_i^{(j)}\rightarrow f_i$ it
follows that $f_i^{(j)}$ also has integral coefficients if $j$ is large
enough. By hypothesis, the matrix
formed from the rows $\nabla f_1(\b{t}),\ldots,\nabla f_r(\b{t})$ has
rank $r$. We may therefore suppose without loss of generality that the
determinant, $\Delta$ say, of the first $r$ columns is non-zero. Let
$\Delta_j$ be the corresponding determinant formed from
$\nabla f_1^{(j)}(\b{t}),\ldots,\nabla f_r^{(j)}(\b{t})$. Thus
$\Delta_j\rightarrow\Delta$, so that $|\Delta_j|_v=|\Delta|_v\not=0$ if $j$ is
large enough.

We also set
\[\delta_j=\max\{|f_1^{(j)}(\b{t})|_v,\ldots,|f_r^{(j)}(\b{t})|_v\},\]
and note that $\delta_j$ tends to
\[\max\{|f_1(\b{t})|_v,\ldots,|f_r(\b{t})|_v\}=0,\]
since $P\in V$. Thus if $j$ is large enough we will have
$\delta_j<|\Delta|_v^2$. This condition allows us to use Hensel's Lemma,
which provides a point $P^{(j)}=[\b{t}^{(j)}]$ on $V^{(j)}$, with
\beql{henb}
\max_{i}|t_i^{(j)}-t_i|_v\le \delta_j/|\Delta|_v.
\end{equation}
It follows that $P^{(j)}$ tends to $P$ as required. Since
$\Delta_j\not=0$ for large $j$ the nonsingularity condition also holds.

When $v$ is an infinite place we use a completely analogous argument,
replacing Hensel's Lemma by Newton Approximation. We begin by
normalizing so that the entries of $\b{t}$, and the coefficients of
the $f_i$, all have modulus at most 1.  The condition
$\delta_j<|\Delta|^2$ has to be replaced by $\delta_j<C|\Delta|^2$
with a constant $C$ depending on $m,r$ and the degrees of the
polynomials involved. Similarly, the bound in (\ref{henb}) becomes
$C'\delta_j/|\Delta|$ with a corresponding constant $C'$.  With these
changes the proof goes through as before.
\bigskip

For our theorem we will assume that we are given local points
$[\b{x}_v]\in \cl{R}(K_v)$ for every place $v$ of $K$.  We will also be
given a finite set of places $S$ and a (small) positive $\varepsilon$, and our
challenge will be to find a point $[\b{x}]\in\cl{R}(K)$ such that
\beql{appr}
\left|\b{x}-\b{x}_v\right|_v<\varepsilon\;\;\;\mbox{for all}\; v\in S.
\end{equation}
Without loss of generality we will include all infinite places in $S$,
as well as all finite places above rational primes up to 37.
In particular, $S$ will be non-empty.
From now on we will assume that the number $n$ of variables in our
quadratic forms satisfies $n\ge 19$.

The variety $\cl{R}$ is nonsingular, and for each
$v\in S$ the $K_v$-points of $\cl{R}$ are therefore Zariski-dense in
every neighbourhood of $[\b{x}_v]$, by the second principle above.
It follows that if $\cl{R}_0$ is as in Lemma \ref{R3chain} then
there is a point
$[\b{x}_v']\in \cl{R}-\cl{R}_0$, defined over $K_v$, in the neighbourhood
$\left|\b{x}_v-\b{x}_v'\right|_v<\varepsilon/2$. Thus it suffices to find a
$K$-point of $\cl{R}$ with
\[\left|\b{x}-\b{x}_v'\right|_v<\varepsilon/2\;\;\;\mbox{for all}\; v\in S,\]
where now there is a chain-admissible 3-plane $L_v\in F_3$ through
$[\b{x}_v']$.  We therefore change our
notation, replacing $\b{x}_v'$ by $\b{x}_v$ and $\varepsilon$ by $\varepsilon/2$ so that is
still suffices to work with the condition (\ref{appr}).
Note that $L_v$ may be
defined over $\overline{K_v}$ rather than $K_v$. None the less
the existence of a single chain-admissible $L_v$ shows that the generic
3-plane $L\subset\cl{Q}_3$ through $[\b{x}_v]$ is also chain-admissible.
Thus we can in fact assume that $L_v$ is defined over
$K_v$.

Our plan now is to produce a sequence of 3-planes $L^{(m)}$ defined
over $K$, which approximate $L_v$ for each $v\in S$, in the following
sense.
\begin{lemma}\label{Lappr}
For each $v\in S$ let $L_v\in F_3-Z_3$ be a chain-admissible
3-plane defined over $K_v$, and suppose that
$[\b{e}_{0,v}],\ldots,[\b{e}_{3,v}]$ is a basis of $L_v$.
Then there are sequences of $K$-points
$[\b{e}_{0}^{(m)}],\ldots,[\b{e}_{3}^{(m)}]$ spanning chain-admissible
3-planes
$L^{(m)}\subset\cl{Q}_3$, such that
\beql{lim!}
\lim_{m\rightarrow\infty}\b{e}_i^{(m)}=\b{e}_{i,v}\;\;\;\mbox{for
}0\le i\le 3,
\end{equation}
for every $v\in S$.
\end{lemma}

(We teach our students that a sequence can have at most one limit!
There is of course some abuse of notation above, and strictly speaking
we should have said that $\iota_v(\b{e}_i^{(m)})$ tends to
$\b{e}_{i,v}$, where $\iota_v$ is the embedding of $K^n$ into $K_v^n$.)
In effect the result is saying that the Fano variety of $3$-planes in
$\cl{Q}_3$ satisfies weak approximation.
We will produce the sequences $[\b{e}_t^{(m)}]$ by induction on $t$,
the case $t=0$ merely being an instance of weak approximation on $\cl{Q}_3$.

We therefore consider the induction step, and suppose we already have
suitable sequences of vectors
$\b{e}_{0}^{(m)},\ldots,\b{e}_{t-1}^{(m)}$. Thus the conditions
required for $\b{e}_t^{(m)}$ are that
\[[\b{e}_t^{(m)}]\in V^{(m)}:=\{[\b{x}]\in\cl{Q}_3: \b{x}^TQ_3\b{e}_i^{(m)}=0
\mbox{ for }0\le i\le t-1\},\]
and that
\beql{lim}
\lim_{m\rightarrow\infty}\b{e}_t^{(m)}=\b{e}_{t,v}
\end{equation}
for every $v\in S$.  Notice that (\ref{lim!}) (for $i\le t-1$)
and (\ref{lim}) automatically ensure
that the vectors $\b{e}_{0}^{(m)},\ldots,\b{e}_t^{(m)}$ are linearly
independent, if $m$ is large enough.
The variety $V^{(m)}$ is a quadric of rank at least
$n-2t\ge 5$, so that it must have smooth points over $K_v$ for every finite
place $v$. For every $v\in S$, and in particular for every infinite
place, the varieties $V^{(m)}$ are approximations to
\[V_v:=\{[\b{x}]\in\cl{Q}_3: \b{x}^TQ_3\b{e}_{i,v}=0
\mbox{ for }0\le i\le t-1\},\]
in the sense given by Lemma \ref{approxV}.  Moreover $[\b{e}_{t,v}]$
is a nonsingular point on $V_v$, in the sense of the lemma. It
follows that for large enough $m$ the variety $V^{(m)}$ has a smooth point
$[\b{f}_v^{(m)}]$ for every $v\in S$, such that
\[\lim_{m\rightarrow\infty}\b{f}_v^{(m)}=\b{e}_{t,v}.\]
In particular $V^{(m)}$ has points everywhere locally, and so has a
$K$-point, by the Hasse principle.

We can now complete the induction step.  Given $\eta>0$ we choose
$m_0(\eta)$ so that
\[|\b{f}_v^{(m)}-\b{e}_{t,v}|_v<\eta/2\]
for all $v\in S$, and all $m\ge m_0(\eta)$. Moreover we can use weak
approximation on $V^{(m)}$ to find points $\b{e}_t^{(m)}$ on $V^{(m)}(K)$ such that
\[|\b{e}_t^{(m)}-\b{f}_v^{(m)}|_v<\eta/2\]
for all $v\in S$, and all $m\ge m_0(\eta)$. Then for all $v\in S$ and
all $m\ge m_0(\eta)$ we will have
\[|\b{e}_t^{(m)}-\b{e}_{t,v}|_v<\eta,\]
whence (\ref{lim}) holds, as required.
Finally, since $L_v\not\in Z_3$ we will have $L^{(m)}\not\in Z_3$ for
large enough $m$, by
continuity, so that $L^{(m)}$ is also chain-admissible.

\section{Completion of the Argument}

Our strategy now is to consider the intersection of $\cl{Q}_1\cap\cl{Q}_2$
with $L^{(m)}$. We begin with the following result.
\begin{lemma}\label{3p}
Let $\varepsilon>0$ be given.  If $m$ is large enough, for every
$v\in S$ there will be a $K_v$-point $[\b{y}_v^{(m)}]
\in\cl{Q}_1\cap\cl{Q}_2\cap L^{(m)}$ such that
\[|\b{y}_v^{(m)}-\b{x}_v|_v<\varepsilon/2.\]
\end{lemma}

This is a further application of Lemma \ref{approxV}.  The varieties
\[V^{(m)}:=\cl{Q}_1\cap\cl{Q}_2\cap L^{(m)}\]
are approximations to
$V:=\cl{Q}_1\cap\cl{Q}_2\cap L_v$, in the sense of the lemma, and $P=[\b{x}_v]$
lies on $V$, since we have both $P\in\cl{R}$ and $P\in L_v$. Moreover
$L_v$ is chain-admissible, and hence in particular is admissible, whence
$V$ is nonsingular. The lemma therefore produces appropriate points
$[\b{y}_v^{(m)}]$.
\bigskip

We now fix a suitable $m$ in Lemma \ref{3p}, and write
$L=L^{(m)}$ and $\b{y}_v=\b{y}_v^{(m)}$ accordingly.  Thus $L$ is a
chain-admissible 3-plane, defined over $K$. Moreover for each
$v\in S$ we have $[\b{y}_v]\in\cl{Q}_1\cap\cl{Q}_2\cap L$.  Finally, to
prove our theorem it will suffice to find a point $[\b{x}]\in\cl{R}(K)$ with
\[|\b{y}_v-\b{x}|_v<\varepsilon/2\]
for each $v\in S$.

For each $v\in S$ the variety $\cl{Q}_1\cap\cl{Q}_2\cap L$ has a local point, namely
$[\b{y}_v]$.  Moreover it is nonsingular, since $L$ is chain-admissible.
Thus there are local points at all but
finitely many places.  Let $T$ be the set of places for which there
are no local points.  Thus $S$ and $T$ are disjoint, so that $T$ is a
finite set of finite places $v$ each of which lies over a prime $p\ge
37$.
To handle this remaining set of places it
suffices to intersect $\cl{Q}_1\cap\cl{Q}_2$ with a suitable 4-plane, as our
next result shows.
\begin{lemma}\label{fin-}
If $T$ is nonempty there is a chain-admissible 4-plane $L'$,
defined over $K$, such that $L\subset L'$, for which
$\cl{Q}_1\cap\cl{Q}_2\cap L'$ has $K_v$-points for every $v\in T$.
\end{lemma}
It follows of course that $\cl{Q}_1\cap\cl{Q}_2\cap L'$ has points over
every completion of $K$. Naturally, if $T$ were empty the same would
already be true for $\cl{Q}_1\cap\cl{Q}_2\cap L$.

To prove Lemma \ref{fin-} we choose
$K$-points $[\b{e}_0],\ldots,[\b{e}_3]$ spanning $L$, and look for an
additional $K$-point $[\b{x}]=[\b{e}_4]$ such that
\[L':=<[\b{e}_0],\ldots,[\b{e}_4]>\]
fulfils the necessary conditions. In order to have $L'\subset\cl{Q}_3$ we
will require $[\b{x}]\in\cl{Q}_3\cap(Q_3L)^{\perp}$
We would like the variety $L'$ to be
chain-admissible, and so we will require that $L'\not\in Z_4$.
However $L$ itself is chain-admissible, so that there is at least one
point $P_0$, (which might be defined over $\overline{K}$) for which
\[<L,P_0>\in F_4-Z_4.\]
Such a point $P_0$ will be smooth point of
$\cl{Q}_3\cap(Q_3L)^{\perp}$. It follows that the set of points $P$ for
which $<L,P>$ is a chain-admissible 4-plane, is a nonempty
Zariski-open subset ($U$, say) of $\cl{Q}_3\cap(Q_3L)^{\perp}$.

To arrange that
$\cl{Q}_1\cap\cl{Q}_2\cap L'$ has a $K_v$-point it will be helpful if $[\b{x}]$
is ``near'' to a $K_v$-point of $\cl{Q}_1\cap\cl{Q}_2$.  We would
therefore like the variety
\[\left(\cl{Q}_3\cap(Q_3L)^{\perp}\right)\cap\left(\cl{Q}_1\cap\cl{Q}_2\right)
=(Q_3L)^{\perp}\cap\cl{R}\]
to contain a $K_v$-point $[\b{x}_v]$ for every $v\in T$. However
$(Q_3L)^{\perp}$ has dimension $n-5\ge 14$, so that
$(Q_3L)^{\perp}\cap\cl{R}$ is the zero locus of a system of three
quadratic forms in at least 15 variables.  There is therefore a
$K_v$-point whenever $v$ is a finite place above a prime $p\ge 37$, by
Heath-Brown \cite[Corollary 1]{HBQs}. (The reader should note
that one only needs to know that some bound of the form $p\ge p_0$
suffices. This is a corollary of the famous Ax--Kochen Theorem
\cite{AK}.  However in the present situation one can now provide an
explicit value of $p_0$.  Indeed the work of Schuur \cite{SS} could
also be used here.)  Thus there are points
$[\b{x}_v]\in\cl{R}\cap(Q_3L)^{\perp}$ for every $v\in T$. Indeed, since $L$
is chain-admissible it is certainly admissible, so that
$\cl{R}\cap(Q_3L)^{\perp}$ must be nonsingular.  If then follows that the
$K_v$-points on $\cl{R}\cap(Q_3L)^{\perp}$ will be Zariski-dense. In
particular we can choose a $K_v$-point $[\b{x}_v]$ in $\cl{R}\cap(Q_3L)^{\perp}-\cl{R}_1$,
with $\cl{R}_1$ as in Lemma \ref{good4}. Thus, if we set
$L_v=<L,[\b{x}_v]>$, then $L_v$ is admissible so that $\cl{Q}_1\cap\cl{Q}_2\cap
L_v$ is nonsingular.  In particular $[\b{x}_v]$ will be a smooth point of
$\cl{Q}_1\cap\cl{Q}_2\cap L_v$.

Since $[\b{x}_v]\in\cl{R}\cap(Q_3L)^{\perp}$ it follows
in particular that $[\b{x}_v]\in\cl{Q}_3\cap(Q_3L)^{\perp}$.
We now claim that $[\b{x}_v]$ cannot be a singular point of
$\cl{Q}_3\cap(Q_3L)^{\perp}$. For otherwise the vectors
\[Q_3\b{x}_v,Q_3\b{e}_0,\ldots,Q_3\b{e}_3\]
would be linearly dependent. Since $Q_3$ is nonsingular it would
follow that
\[\b{x}_v\in<\b{e}_0,\ldots,\b{e}_3>,\]
so that $[\b{x}_v]\in L$.  However $[\b{x}_v]$ was chosen to lie in $\cl{R}$,
so that it would in particular be a $K_v$-point of $\cl{Q}_1\cap\cl{Q}_2\cap L$.
We would therefore have
a contradiction, since $T$ was defined to be the set of places where
$\cl{Q}_1\cap\cl{Q}_2\cap L$ had no local points.

Thus $[\b{x}_v]$ is a smooth point of
$\cl{Q}_3\cap(Q_3L)^{\perp}$.  This variety certainly has at
least one smooth $K$-point, since the quadratic form $Q_3$ was
constructed to split off at least $(n-5)/2>4$ hyperbolic planes over $K$.
Thus we can use weak approximation on $\cl{Q}_3\cap(Q_3L)^{\perp}$ to
produce a sequence of $K$-points $[\b{x}^{(m)}]$, such that
\[\lim_{m\rightarrow\infty}\b{x}^{(m)}=\b{x}_{v}\]
for every $v\in T$. For large enough $m$ a continuity argument shows that
if $L^{(m)}=<L,[\b{x}^{(m)}]>$ then $L^{(m)}\in F_4-Z_4$.
We also see
that $\cl{Q}_1\cap\cl{Q}_2\cap L^{(m)}$ approximates $\cl{Q}_1\cap\cl{Q}_2\cap L_v$
in the sense of Lemma \ref{approxV}.  Moreover $[\b{x}_v]$ is a
smooth point of $\cl{Q}_1\cap\cl{Q}_2\cap L_v$, whence Lemma \ref{approxV}
provides $K_v$-points $P^{(m)}$ on $\cl{Q}_1\cap\cl{Q}_2\cap L^{(m)}$ as
soon as $m$ is large enough, for all $v\in T$.  The lemma then follows
on choosing $L'=L^{(m)}$ with a suitably large $m$.
\bigskip

As the final step in our argument we state the
following result.
\begin{lemma}\label{fin}
There is an admissible 7-plane $L''$ containing $L'$, defined over
$K$, and such that
 $\cl{Q}_1\cap\cl{Q}_2\cap L''$ has points over every completion of $K$.
\end{lemma}

Before presenting the proof of the lemma, we show how it suffices for
our theorem. Since $L''\in F_7$ is admissible, the variety $\cl{Q}_1\cap\cl{Q}_2\cap
L''$ is contained in $\cl{R}$, and is a nonsingular intersection of two
quadrics in $\mathbb{P}^7$.  By construction it has points over $K_v$ for
every place $v$.  Indeed the subvariety $\cl{Q}_1\cap\cl{Q}_2\cap L$ has a
$K_v$-point $[\b{y}_v]$ for every $v\in S$, and has $K_v$-points for
all $v\not\in T$, while $\cl{Q}_1\cap\cl{Q}_2\cap L'$ has $K_v$-points for
any remaining places $v\in T$. We may therefore apply the
author's result \cite[Theorem 1]{HB8} described in the introduction,
which shows that $\cl{Q}_1\cap\cl{Q}_2\cap L''$ satisfies the Hasse principle
and weak approximation.  This allows us to conclude that
$\cl{Q}_1\cap\cl{Q}_2\cap L''$ has $K$-points arbitrarily close  to
$[\b{y}_v]$ for each $v\in S$.  Our theorem therefore follows.
\bigskip

It remains to establish Lemma \ref{fin}. We have already observed that
either $\cl{Q}_1\cap\cl{Q}_2\cap L$ has points everwhere locally, if $T$ is
empty, or $\cl{Q}_1\cap\cl{Q}_2\cap L'$ does.  The linear spaces $L$ and
$L'$ are chain-admissible, and are defined over $K$. It is therefore
enough to show that if $M$ is any chain-admissible linear space of
dimension $t\in[3,6]$, defined over $K$, then there is a chain
admisssible space $M'\supset M$ of dimension $t+1$, also defined over
$K$. Once this is proved we can use this repeatedly to go from $L$ or
$L'$ to $L''$.

We recall that $\cl{Q}_3$ contains at least one 7-plane defined over $K$,
whence $M$ will be contained in such a 7-plane, $M^*$, say. We can
choose a basis $\b{e}_1,\ldots\b{e}_n$ for $K^n$ so that $M$ is
spanned by $[\b{e}_1],\ldots,[\b{e}_{t+1}]$ and $M^{*}$ by
$[\b{e}_1],\ldots,[\b{e}_8]$, and such that
\[Q_3(\sum_1^n
X_i\b{e}_i)=X_1X_9+\ldots+X_8X_{16}+Q(X_{17},\ldots,X_n)\]
for a suitable nonsingular form $Q$.  Then $(Q_3M)^{\perp}$ is spanned by
$[\b{e}_1],\ldots,[\b{e}_8]$ and $[\b{e}_{t+10}],\ldots,[\b{e}_n]$,
and one therefore sees that $[\b{e}_8]$ will be a smooth $K$-point of
$\cl{Q}_3\cap(Q_3M)^{\perp}$. Having shown that there is at least one
smooth $K$-point on $\cl{Q}_3\cap(Q_3M)^{\perp}$ we deduce that the
$K$-points are Zariski-dense, so that there is a $K$-point
$P\in\cl{Q}_3\cap(Q_3M)^{\perp}-E_t$, in the notation of Lemma \ref{good4}.
We can then complete the proof of Lemma \ref{fin} by taking $M'=<M,P>$.

redpajama_set_name: RedPajamaArXiv


People are either Yankee haters or lovers. There's no in between. Fortunately, I grew up in southern Connecticut, so I was born a real Yankee. Last month, I was lucky enough to be in attendance when the Yankees captured their 27th World Series title after beating the Philadelphia Phillies 7–3 in Game 6. Woot!
My buddy Joe Lewis called a few days earlier to say he had scored tix to Game 6, which meant we had to hope for the Yankees to lose a game in Philadelphia—not an easy thing for a Yankee fan to do. But once they did just that it was time to scramble to change my schedule, including booking flights with some help from .
Joe had to work late so we got a late start to the game—we left an hour before first pitch. We bypassed traffic by taking the 4 train from 86th street station. But boarding looked like something out of India: There were so many people trying to get in each car that people just plowed their way on. It was insane and I was the last to make it on so I was pinned up against the door. Seriously, it was so tight I couldn't even lift my arm to scratch my itchy nose or take my hot winter jacket off. People were sweating and it was eerily quiet for a rambunctious crowd—I think fans were just praying the car wouldn't break down or there'd be a terrorist attack—at least I was.
About 20 minutes and four stops later we made it to 161st Street. There's always something special about going to a ballgame in the Bronx, but it's even sweeter in October and November. I could never get sick of the smell of hot roasted chestnuts, pretzels, and the autumn air.
We waited in line for about five minutes to get through the security check and have our tickets scanned. We had paper tickets, which looked like airline boarding passes since we printed them up online from StubHub.com. In case you are wondering, like I was: Fans can bring cameras but not bags. You can also bring a bottle of water, if it hasn't been opened. Mine had been opened but the security girl let me slide after she shook it to make sure it wasn't anything other than water. That saved me $5.
Our seats were field level between first base and right field, 29 rows from the grass. It was ideal because if it rained we were covered and we had just a very short walk to the concession stands and bathroom (there always seemed to be a line). It got cold quick so I kept getting Yankee dogs and hot chocolate to warm my hands and belly.
There were a just few ballsy Philly fans spread out—I don't know what they were thinking not showing up with an army, but I guess they knew better than me. The five in our section didn't get razzed as bad as I expected. No one got beat up, had their stupid red hats confiscated, or was treated to a complimentary beer shower. That had to be because the tickets are so darn expensive that it kept the thugs out. This brings up a valid point. The negative of having such expensive tickets is that it actually keeps the uneducated folks out, which probably takes away the home field advantage. Think about it.
The House that George Built doesn't have the same intimidating feel as the House that Ruth Built. Obviously, someone's got to start paying off the $1.5 billion price tag. But if I was running the joint and the Yankees made it to the World Series I would give the first five rows to all the gang bangers or make it Yankee Fan Prisoner Appreciation Night. Then you can pretty much guarantee an advantage.
Seriously, I was surprised at the stark contrast with Game 6 in 1996. Thanks to my brother I was at that game when the Yankees won the World Series—their first championship in 18 years. Remember Wade Boggs riding on the back of a police horse?! I do, like it was yesterday. Maybe Yankee fans have become spoiled over their success of the past dozen years but without a doubt Game 6 in 1996 was much louder and rowdier than in 2009. A decade ago I thought the upper deck was going to collapse from fans jumping up and down the WHOLE game. This game we only stood up when there were two strikes or a base hit. In 1996 when Charlie Hayes caught the final out everyone hugged one another, and I'm talking EVERYONE. I was hugging people in the streets of the Bronx that you would normally cross the street to avoid or immediately yell help if you saw them get too close to you. Last month, I high fived a few fans sitting around me but that was it. No hugs. I felt unloved.
Either way it was still an incredible experience, and the Bronx Bombers are back on top. The "Let's Go Yankees!" chant is still in my head, and that's not a bad thing.
The following morning I flew to West Palm Beach (PBI) on jetBlue. It took 20 minutes to get to La Guardia (LGA) from the Upper East Side and there was no line at check-in. Thirty seconds later I had my boarding card (there wasn't a printer where I was staying) and cleared security in eight minutes. There was a long line 40 people deep but the TSA did a good job keeping it moving.
JetBlue's gate at LGA is pretty dismal and there's not a lot of seating, so I stood in the slow-moving line at the lone newsstand and then again at the Au Bon Pain, where the prices weren't ridiculously inflated: A large yogurt parfait, a banana, and a bottle of water came to just $7. That's good for New York.
We boarded on-time. boards families and people who need more time first, along with those passengers sitting in the exit row (I hadn't seen the latter get priority before). General boarding is by rows, the last few rows are first. Unfortunately for boarding purposes, I was in row 6. I asked the agent if it was okay if I boarded early so I could find space for my carry-on and he said no problem. I guess you just have to ask nicely (I'm sure it also depends on the agent).
I almost plopped down the $25 fee for the seats with the extra legroom but it turned out to be lucky they were all sold out because the middle seat in my row was vacant, giving me all the room I needed. JetBlue doesn't offer any real food for free or for sale but they do have friendly flight attendants passing out free snacks like blue potato chips and crackers. Flight time was just 2 hours and 27 minutes and the time flew by since jetBlue has free satellite radio and live TV (bring your headset or pay $2 to get one of jetBlue's).
The most awkward moment came before takeoff when the flight attendant made his pre-flight announcement. He started off by saying he was sorry for any Phillies fans on board but we were in New York, so CONGRATS to the Yankees fans! His energy got me all pumped and brought me back to 1996, so I stood up and started waving my jacket while yelling "Let's Go Yank…" Then I realized no one else was chiming in and that I looked like one of those lunatic prisoners who escaped from the Prisoner Appreciation game.
*PLEASE tell us what you think of this week's newsletter.


Paul McShane and John Swift will be available for Reading after hamstring problems for Saturday's Championship clash with Middlesbrough at the Madejski Stadium.
Striker Jon Dadi Bodvarsson will miss out however, still dogged by a calf problem.
Reading boss Jaap Stam feels Saturday's clash will come too soon for long-term absentee Yann Kermorgant, despite the forward closing in on a return.
Kermorgant continues to fight for full fitness after surgery on a hip and groin issue in pre-season.
Middlesbrough boss Garry Monk is likely to revert to his strongest line-up after making 10 changes for the midweek Carabao Cup defeat at Premier League Bournemouth.
Central defender Ben Gibson is among those who will hope for a swift return, in his case for the first time since signing a new five-year contract at the Riverside Stadium.
Defender Ryan Shotton and midfielder Marvin Johnson were cup-tied at the Vitality Stadium and are available once again, although neither has featured since the 2-2 draw with Brentford on September 30, Shotton's only appearance for the club to date.
Boro are without a win in five league games and have taken just three of the last 15 points on offer to slip to 13th place in the table.


text:
Home News Content Blog Blog Items Intelligence Bin Laden Issues Warning on Iraq, Israel
Bin Laden Issues Warning on Iraq, Israel
Written by AP
In New Audiotape to Unify al Qaeda
CAIRO, Egypt -- Osama bin Laden warned Iraq's Sunni Arabs against fighting al Qaeda and promised to expand the terror group's holy war to Israel in a new audiotape Saturday, threatening "blood for blood, destruction for destruction."
Most of the 56-minute tape dealt with Iraq, apparently al Qaeda's latest attempt to keep supporters in Iraq unified at a time when the U.S. military claims to have al Qaeda's Iraq branch on the run.
The tape did not mention Pakistan or the assassination of Benazir Bhutto, though Pakistan's government has blamed al Qaeda and the Taliban for her death on Thursday.
But Mr. bin Laden's comments offered an unusually direct attack on Israel, which has warned of growing al Qaeda activity in Palestinian territory. The terror network is not believed to have taken a strong role there so far.
"We intend to liberate Palestine, the whole of Palestine from the (Jordan) river to the sea," he said, threatening "blood for blood, destruction for destruction."
"We will not recognize even one inch for Jews in the land of Palestine as other Muslim leaders have," Mr. bin Laden said.
In Iraq, a number of Sunni Arab tribes in western Anbar province have formed a coalition fighting al Qaeda-linked insurgents that U.S. officials credit for deeply reducing violence in the province. The U.S. military has been working to form similar "Awakening Councils" in other areas of Iraq.
White House spokesman Tony Fratto said Mr. bin Laden's tape shows that al Qaeda's aim is to block democracy and freedom for all Iraqis.
"It also reminds us that the mission to defeat al Qaeda in Iraq is critically important and must succeed," Mr. Fratto said. "The Iraqi people -- every day, and in increasing numbers -- are choosing freedom and standing against the murderous, hateful ideology of AQI. And we stand with them."
In the audiotape, Mr. bin Laden denounced Abdul-Sattar Abu Risha, the former leader of the Anbar Awakening Council, who was killed in a September bombing claimed by al Qaeda.
"The most evil of the traitors are those who trade away their religion for the sake of their mortal life," Mr. bin Laden said.
Mr. bin Laden said U.S. and Iraqi officials are seeking to set up a "national unity government" joining the country's Sunnis, Shiites and Kurds.
"Our duty is to foil these dangerous schemes, which try to prevent the establishment of an Islamic state in Iraq, which would be a wall of resistance against American schemes to divide Iraq," he said.
The authenticity of the tape could not be independently confirmed. But the voice resembled that of Mr. bin Laden. The tape was posted on an Islamic militant Web site where al Qaeda's media arm, Al-Sahab, issues the group's messages.
Copyright © 2007 Associated Press
86 Wounded in Wednesday Clashes, Red Cross Says
'The Last Ones': Listening to survivors of the concentration camps

redpajama_set_name: RedPajamaCommonCrawl


text:
@implementation AddPromotionsViewController

- (void)viewDidLoad {
    [super viewDidLoad];
    if (!(appDelegate.venueStatus == Staff && [appDelegate.activeVenue[@"promo_perm"] integerValue] == 0)) [self setRightBarButton:@"Publish" actionSelector:@selector(addPromotion)];

    self.tableData = @[@"Title", @"", @"Passcode", @"Starts", @"Ends", @"Limit", @"Audience"];
    textFields = [NSMutableDictionary dictionary];
    pickerValues = [NSMutableDictionary dictionary];

    [self.doneButton setTitleTextAttributes:
     @{NSForegroundColorAttributeName: [UIColor blackColor],
       NSFontAttributeName: [UIFont systemFontOfSize:14.0]}
                                   forState:UIControlStateNormal];

    appDelegate.audience = appDelegate.activePromotion == nil ? 0 : [appDelegate.activePromotion[@"level"] integerValue];
}

- (void)viewDidAppear:(BOOL)animated {
    [super viewDidAppear:animated];
    [self.tableView reloadRowsAtIndexPaths:@[[NSIndexPath indexPathForRow:0 inSection:6]] withRowAnimation:UITableViewRowAnimationNone];
}

- (void)addPromotion {
    NSMutableDictionary *params = [@{@"venue_id": appDelegate.activeVenue[@"id"],
                                     @"title": ((UITextField *)textFields[@0]).text,
                                     @"description": ((UITextField *)textFields[@1]).text,
                                     @"passcode": ((UITextField *)textFields[@2]).text,
                                     @"start": @([(NSDate *)pickerValues[@3] timeIntervalSince1970]),
                                     @"end": @([(NSDate *)pickerValues[@4] timeIntervalSince1970]),
                                     @"maximum": ((UITextField *)textFields[@5]).text ? ((UITextField *)textFields[@5]).text : @0,
                                     @"level": @(appDelegate.audience)} mutableCopy];
    if (appDelegate.activePromotion[@"id"] != nil) {
        params[@"promotion_id"] = appDelegate.activePromotion[@"id"];
    }
    [self.view makeToastActivity];
    [Request post:@"promotions/set" params:params callback:^(id response) {
        [self.view hideToastActivity];
        if (response) {
            [self.navigationController popViewControllerAnimated:YES];
        } else {
            UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Error" message:nil delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
            [alert show];
        }
    }];
}

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath {
    UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:[NSString stringWithFormat:@"CellToAddPromo%ld", (long)indexPath.section]];
    cell.textLabel.text = self.tableData[indexPath.section];
    if (indexPath.section == 0) {
        UITextField *textField = (UITextField *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UITextField *textField = [[UITextField alloc] initWithFrame:CGRectMake(110, 8, 185, 30)];
            textField.delegate = self;
            textField.backgroundColor = [UIColor clearColor];
            textField.text = appDelegate.activePromotion != nil ? appDelegate.activePromotion[@"title"] : @"";
            textField.tag = indexPath.section + 1;
            textFields[@0] = textField;
            [cell.contentView addSubview:textField];
        }
    } else if (indexPath.section == 1) {
        UITextField *textField = (UITextField *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UITextView *textField = [[UITextView alloc] initWithFrame:CGRectMake(10, 35, 280, 100)];
            textField.delegate = self;
            textField.backgroundColor = [UIColor clearColor];
            textField.text = appDelegate.activePromotion != nil ? appDelegate.activePromotion[@"description"] : @"";
            textField.tag = indexPath.section + 1;
            textField.font = [UIFont systemFontOfSize:16];
            textFields[@1] = textField;
            [cell.contentView addSubview:textField];
            UILabel *label = [[UILabel alloc] initWithFrame:CGRectMake(10, 6, 280, 25)];
            label.backgroundColor = [UIColor clearColor];
            label.font = [UIFont fontWithName:cell.textLabel.font.fontName size:label.font.pointSize];
            label.text = @"Promotion Details";
            [cell.contentView addSubview:label];
        }
    } else if (indexPath.section == 2) {
        UITextField *textField = (UITextField *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UITextField *textField = [[UITextField alloc] initWithFrame:CGRectMake(110, 8, 185, 30)];
            textField.delegate = self;
            textField.backgroundColor = [UIColor clearColor];
            textField.text = appDelegate.activePromotion != nil ? appDelegate.activePromotion[@"passcode"] : @"";
            textField.tag = indexPath.section + 1;
            textFields[@2] = textField;
            [cell.contentView addSubview:textField];
        }
    } else if (indexPath.section == 3) {
        UITextField *textField = (UITextField *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UILabel *textField = [[UILabel alloc] initWithFrame:CGRectMake(110, 6, 185, 30)];
            textField.backgroundColor = [UIColor clearColor];
            if (pickerValues[@(indexPath.section)] == nil) {
                if ([@"0" isEqualToString :[appDelegate.activePromotion[@"start"] stringValue]]) {
                    textField.text = @"";
                } else {
                    textField.text = appDelegate.activePromotion != nil ? [NSDateFormatter localizedStringFromDate:[NSDate dateWithTimeIntervalSince1970:[appDelegate.activePromotion[@"start"] integerValue]] dateStyle:NSDateFormatterShortStyle timeStyle:NSDateFormatterShortStyle] : @"";
                }
            } else {
                textField.text = [NSDateFormatter localizedStringFromDate:pickerValues[@(indexPath.section)] dateStyle:NSDateFormatterShortStyle timeStyle:NSDateFormatterShortStyle];
            }
            if ([@"0" isEqualToString : textField.text]) textField.text = @"";
            textField.tag = indexPath.section + 1;
            textFields[@3] = textField;
            [cell.contentView addSubview:textField];
        }
    } else if (indexPath.section == 4) {
        UITextField *textField = (UITextField *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UILabel *textField = [[UILabel alloc] initWithFrame:CGRectMake(110, 6, 185, 30)];
            textField.backgroundColor = [UIColor clearColor];
            if (pickerValues[@(indexPath.section)] == nil) {
                if ([@"0" isEqualToString :[appDelegate.activePromotion[@"end"] stringValue]]) {
                    textField.text = @"";
                } else {
                    textField.text = appDelegate.activePromotion != nil ? [NSDateFormatter localizedStringFromDate:[NSDate dateWithTimeIntervalSince1970:[appDelegate.activePromotion[@"end"] integerValue]] dateStyle:NSDateFormatterShortStyle timeStyle:NSDateFormatterShortStyle] : @"";
                }
            } else {
                textField.text = [NSDateFormatter localizedStringFromDate:pickerValues[@(indexPath.section)] dateStyle:NSDateFormatterShortStyle timeStyle:NSDateFormatterShortStyle];
            }
            textField.tag = indexPath.section + 1;
            textFields[@4] = textField;
            [cell.contentView addSubview:textField];
        }
    } else if (indexPath.section == 5) {
        UITextField *textField = (UITextField *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UITextField *textField = [[UITextField alloc] initWithFrame:CGRectMake(110, 8, 185, 30)];
            textField.backgroundColor = [UIColor clearColor];
            textField.keyboardType = UIKeyboardTypeNumberPad;
            textField.text = appDelegate.activePromotion != nil ? [appDelegate.activePromotion[@"maximum"] stringValue] : @"";
            textField.tag = indexPath.section + 1;
            textFields[@5] = textField;
            [cell.contentView addSubview:textField];
        }
    } else if (indexPath.section == 6) {
        UILabel *textField = (UILabel *)[cell viewWithTag:indexPath.section + 1];
        if (!textField) {
            UILabel *textField = [[UILabel alloc] initWithFrame:CGRectMake(110, 6, 185, 30)];
            textField.backgroundColor = [UIColor clearColor];
            textField.text = [self levelName:appDelegate.audience];
            textField.tag = indexPath.section + 1;
            textFields[@6] = textField;
            [cell.contentView addSubview:textField];
        }
    }

    return cell;
}

- (NSString *)levelName:(int)level {
    switch (level) {
        case 0:
            return @"Everyone";
        case 1:
            return @"Explorers (Top 20%)";
        case 2:
            return @"Scouts (Top 5%)";
        case 3:
            return @"Shnerglers (Top 1%)";
        default:
            return @"";
    }
}

- (BOOL)textView:(UITextView *)textView shouldChangeTextInRange:(NSRange)range replacementText:(NSString *)text {
    if ([text isEqualToString:@"\n"]) {
        [textView resignFirstResponder];
        return NO;
    }
    return YES;
}

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath {
    if (indexPath.section == 3 || indexPath.section == 4) {
        if (self.pickerView.superview == nil) {
            CGRect startFrame = self.pickerView.frame;
            CGRect endFrame = self.pickerView.frame;
            startFrame.origin.y = self.view.frame.size.height;
            endFrame.origin.y = startFrame.origin.y - endFrame.size.height;

            self.pickerView.frame = startFrame;

            [self.view addSubview:self.pickerView];

            [UIView beginAnimations:nil context:NULL];
            [UIView setAnimationDuration:0.40];
            self.pickerView.frame = endFrame;
            [UIView commitAnimations];

            publishButton = self.navigationItem.rightBarButtonItem;

            self.navigationItem.rightBarButtonItem = self.doneButton;
        }
    } else if (indexPath.section == 6) {
        UIViewController *promoAud = [self.storyboard instantiateViewControllerWithIdentifier:@"PromoAudience"];
        [self.navigationController pushViewController:promoAud animated:YES];
    }
}

- (CGFloat)tableView:(UITableView *)tableView heightForRowAtIndexPath:(NSIndexPath *)indexPath {
    UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:@"CellToAddPromo0"];

    return indexPath.section == 1 ? 140 : cell.bounds.size.height;
}

- (IBAction)doneAction:(id)sender {
    CGRect pickerFrame = self.pickerView.frame;
    pickerFrame.origin.y = self.view.frame.size.height;

    [UIView beginAnimations:nil context:NULL];
    [UIView setAnimationDuration:0.40];
    [UIView setAnimationDelegate:self];
    [UIView setAnimationDidStopSelector:@selector(slideDownDidStop)];
    self.pickerView.frame = pickerFrame;
    [UIView commitAnimations];
    self.navigationItem.rightBarButtonItem = publishButton;
    NSIndexPath *indexPath = [self.tableView indexPathForSelectedRow];
    pickerValues[@(indexPath.section)] = self.pickerView.date;
    [self.tableView deselectRowAtIndexPath:indexPath animated:YES];
    [self.tableView reloadRowsAtIndexPaths:@[indexPath] withRowAnimation:UITableViewRowAnimationNone];
}

- (BOOL)textFieldShouldReturn:(UITextField *)textField {
    [textField resignFirstResponder];
    return YES;
}

- (void)slideDownDidStop {
    [self.pickerView removeFromSuperview];
}

- (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView {
    return [self.tableData count];
}

- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section {
    return 1;
}

- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string {
    if (textField.tag == 1) {
        NSUInteger newLength = [textField.text length] + [string length] - range.length;
        return (newLength > 20) ? NO : YES;
    }
    return YES;
}

@end

redpajama_set_name: RedPajamaGithub


Some screenshots testifying to his words.
Selling an organ abroad, especially a kidney, might seem like an arduous task, but it turns out to be easier than thought: EstremeConsequenze even managed to strike a deal with a clinic in India, using everyday search engines on the web. We received a reply within a few hours of contacting them, and the price agreed upon was 120,000 euros, to be paid in two installments, travel excluded.
But it's one thing to try and sell an organ abroad (traffickers can be sentenced with up to 12 years imprisonment in Italy), quite another to understand if this phenomenon exists in Italy. And there is another side to this crime, one that stretches the imagination in its atrocity: the extraction of organs of people who have not given their consent. EstremeConsequenze is still investigating this barbaric aspect.
The founder of the association "Indiani d'Occidente", which has for years been exploring this issue, is Santa Rossi, who explains to EstremeConsequenze why light needs to be shed on such a seemingly distant problem: "We have always thought that problems like this are light years from us, and that this inhumane traffic only takes place in certain foreign countries, that at most, maybe a few Italians take part in 'transplant tourism'. But that's not the way things are. Indiani d'Occidente – my own personal battle – began 11 years ago when my husband died while on a waiting list to have a liver transplant. It all happened one afternoon, I remember every detail of that afternoon. It was May 10th, 2007.
"I was heading into the parking lot of the Cardarelli hospital in Naples, where my husband was a patient. A stranger grabbed me by the shoulder from behind and warned me not to turn around. He said I could help save my husband. That he had a new liver for him. He even knew my husband's blood type, he knew his entire medical history. He said he could help accelerate things and save him. All this for 250,000 euros. I spoke to my husband, and we came to a decision: not only would we refuse, but we would begin a battle to end this phenomenon. Sadly he died 20 days later, but this is a battle I have carried forward right to this day.
Organised crime has a finger in many pies: drugs, arms smuggling, the traffic of people. And in organ trafficking too. And what worries us is the trafficking linked to prostitution, mainly run by the Nigerian mafias. There are no scruples, unheard of violence is used. We know these women have to repay a debt. But we don't know how they do it. There are women forced into prostitution in Campania, who then seem to disappear off the face of the earth. What happens to them?
The Salerno branch of Libera has been fighting for years against criminal trafficking of all sorts, and for the first time they turn their attentions to this very specific and atrocious type of trafficking.
Organ trafficking was first mentioned by the World Health Organization in 1987, and on May 20th, 2010, the WHO reaffirmed its dedication to eradicating this phenomenon, with the publication of Resolution WHA63.22. According to WHO estimates, there are approximately one million organ transplants carried out per year (the majority are kidney transplants, which are estimated at making up 75% of the illegal trade), with a little over 10% of that total (120,000) being considered absolutely necessary.
In the document, the WHO Member States strongly condemn "the buying of human body parts for transplantation and the exploitation of the poorest and most vulnerable populations and the human trafficking that results from such practices".
Poverty, unemployment and the lack of socio-economic opportunities are the factors that create vulnerability to organ trafficking and to human trafficking aimed at organ extraction. The needy are easy prey for traffickers who persuade them to sell an organ in the desperate attempt to improve their sort. At the same time, desperate patients are willing to pay vast sums for an organ that could save their lives, unaware of the consequences that a transplant with a lucrative financial goal behind it could have on their health.
It is not only traffickers that oil the wheels of this market: unscrupulous medical staff are also fundamental players in the mechanism of organ trafficking, as transplant tourists travel to unauthorised clinics for their surgery.
There are a number of internationally agreed measures in place which define, condemn and penalise such pratices: the UN Palermo Protocols, against people trafficking, the Council of Europe Convention on Action against Trafficking in Human Beings and the Council of Europe Convention against trafficking in Human Organs.


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Catalyze partner highlights College Possible's impact
The College of Saint Benedict and Saint John's University are nestled in central Minnesota, two institutions only seven miles from each other along Interstate 94. These schools have a unique partnership – St. Ben's is a women's college and St John's, a men's university – that allows students to take classes, work and socialize on both campuses.
They also boast another unique partnership: St. Ben's and St. Johns, or CSB/SJU as students call it, have long been champions of first-generation and low-income students. College Possible's Catalyze program is one among the many programs the institutions support.
"While the demographics and socioeconomic backgrounds of CSB and SJU students continue to evolve, the need and desire to support first-generation students remains an essential part of the fabric of our institutions," says Tammy Moore, chief communications and marketing officer at CSB/SJU.
In fact, over the summer, The College of Saint Benedict Magazine dedicated an entire issue to its first-generation students. College Possible was featured among the institution's core student support services.
"Featuring first-generation students and the programs and people who support them is one way to provide our readers with a glimpse of the characteristics and needs of our evolving student body and also highlights our institutional commitment to inclusive excellence," says Moore.
CSB/SJU sees near-peer coaching as an effective way to meet those needs.
"CSB/SJU has excellent resources available to support student success," says Moore. "Many of these, however, require that the student identify and take steps to access the resource. College Possible coaches take the lead in reaching out to students to help them connect first to the coaching resource and then to other appropriate services.
"The near-peer, AmeriCorps model for College Possible is a cost-effective strategy for this high level of outreach to ensure that low-income students are well-supported and can effectively navigate the college experience."
Learn more about programs, such as College Possible, at CSB/SJU in The College of Saint Benedict Magazine.
If you are interested in learning more about the Catalyze program model and how it can improve student outcomes on your campus, subscribe to our newsletter or contact Carrie Carroll, Senior Partnerships Manager at CCarroll@CollegePossible.org.
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A smart future for identify verification
Innovative trials towards digital identity authentication in China
Alison Liu
Verifying applicant identity currently constitutes a significant use of time and manpower when processing travel document applications. In Guangdong, China, the immigration bureau is trialling two new methods of digital identity verification: the self-service 'Smart Kiosk', and mobile identity verification via the applicant's own mobile phone. These methods aim to facilitate the administrative process for both applicants and authorities. Whilst some questions still remain regarding how to implement these new systems, pilot schemes for both have shown promising results and it is likely that their use will be expanded in the future.
Due to uneven economic development and mobility of the population, the driving force for adopting new technologies in China varies from province to province, leading to the fragmentation of administrative processes in different provinces. However, as one of the most modernised and advanced provinces, Guangdong takes on the role of pioneer in promoting innovations in handling administrative affairs.
Guangdong is also the most populated province in China, with over 111 million residents in 2017 – the only province in China with a population over 100 million. Therefore, the local government faces numerous challenges in terms of population management, posing a heavy burden on the provincial immigration bureau when handling travel document applications. Traditional application processes can no longer meet expectations in terms of efficiency, and therefore the Guangdong provincial immigration bureau has become very proactive in seeking innovative ways of handling travel document applications.
Starting in early 2018, two pilot cases – the Smart Kiosk (see Figure 1) and Mobile ID
(see Figure 2) – have been trialled by the Guangdong provincial immigration bureau in an attempt to assess the feasibility of adopting digital identification using self-service kiosks and enabling automatic identity verification
by electronic means. Both cases have successfully passed the feasibility assessment, marking the first cornerstone in a
move towards a world of digital identities.
Pilot 1: The Smart Kiosk
Besides general self-services such as fingerprint collection, portrait shooting, application processing, electronic signature and e-Payment, several innovations have been implemented in the Smart Kiosk to ensure the integrity of the enrolment process and data security, including liveness detection, big data forecasting, risk assessment for early warning, and facilitation of the application process in one go. Figure 1: The Smart Kiosk: a cubicle with an inner private space for handling travel document applications, and a side module for issuance of travel permits for mainland travellers to Hong Kong and Macao.
Figure 2: Mobile ID: identity verification by mobile phone.
Liveness detection
Live-captured digital images such as the movement of eyelashes and eyelids, pupil contraction and expansion in response to different light sources, are auto-processed and compared in the background to ensure the integrity of the application.
Figure 3: The nationwide platform for big data forecasting and risk assessment for early warning.
Big data forecasting and risk assessment for early warning
A nationwide platform that synergises personal information from across multiple government segments provides new forecasting models and risk assessment data, which enable the authority to better plan for resource allocation and risk management (see Figure 3).
Facilitation of the application process in one go
Empowered by big data and intelligent handling, all self-services can be done in one go at the kiosk, including immediate issuance of a permit for travel to Hong Kong and Macao, minimising the effort spent on applicant enrolment processes and strengthening the security of personal information.
Observations on the Smart Kiosk pilot
The Smart Kiosk trial has been very successful in terms of its practicability for both the public and authorities. Firstly, the average time spent by an applicant on operating the kiosk for a single application has been reduced to less than 12 minutes in one go, compared to the traditional processing method which would take 80% more time to complete in multiple stages. Secondly, the artificial intelligence implemented in the kiosk has helped to significantly reduce the cost of additional resources, especially manpower, required on site to handle the traditional method of processing applications. It is expected that the number of applications processed by the Smart Kiosks will reach more than 120 per day per kiosk when they are running at full capacity, which would be equivalent to two additional members of staff on duty. Moreover, the Smart Kiosk is able to run 24/7, which adds a lot of value to the current provincial immigration services.
The Smart Kiosk has also successfully met expectations from the perspective of data security and accuracy. Currently all Smart Kiosks are implemented in secure locations at authority offices, where data security during all transactions is guaranteed by the local secure network. Moreover, thanks to the application of big data and intelligent handling, the Smart Kiosk facilitates the task of automatically updating personal data, while at the same time using machine learning to increase the confidence level of liveness detection. According to the assessment report on self-service applications handled by the Smart Kiosk, so far no false or invalid cases have been identified or reported.
Pilot 2: Mobile ID
The Mobile ID concept being trialled focuses on identity verification on the user's mobile phone, which aims to facilitate e-Government services, synergise resources, and allow implementation within the private sector. Beside the adoption of the latest mobile technology for use in hardware security and biometric verification, several innovative ideas have been implemented in day-to-day situations, such as CTID-based liveness detection, and cardless electronic identity verification. Figure 4: CTID-based liveness detection interface on Mobile ID.
CTID-based liveness detection
Cyber Trusted Identification, or CTID, is an electronic identity verification platform developed and operated by the First Research Institute of the Ministry of Public Security, which serves as a trusted portal for identity verification and ID card verification. Besides the live-capture of digital images for various biometric features, voice recognition is also being implemented as an additional facet of the identity verification process (see Figure 4).
Cardless electronic identity verification
The aim of this initiative is to reduce the risk of personal information being leaked during traditional identity verification, where a physical identity card is required to be presented and photocopied. The cardless electronic identity verification system will generate an encrypted certification and generate a dynamic QR code on the user's mobile phone. Figure 5: Cardless hotel check-in with Mobile ID.
This cardless electronic identity verification system has already been initiated in several situations, such as checking into a hotel (see Figure 5), pawn shop registration, patrol checking, access control and logistic registration. The level of uptake varies from city to city, depending on existing infrastructure.
Observations on the Mobile ID pilot
In the current digital world, demand is growing for convenient services that the customer can access on their own mobile phone in real time. The Mobile ID trial shows great potential in facilitating e-services in a more secure and efficient way by implementing identity verification on a mobile device. From a data security and information protection point of view, the Mobile ID system is very successful due to the fact that CTID is endorsed by various authorities and certified by the Ministry of Public Services, ensuring user trust and the security of data transactions.
However, there have proven to be some difficulties in implementing Mobile ID in practice. Firstly, there is a lack of motivation in some parts of the private sector as the implementation requires investment, whilst immediate profit cannot be expected. Secondly, many administrative processes in the public sector are too complex and layered, resulting in difficulties scaling the use of the new technology to its full potential. Last but not least, there are still limitations on the Mobile ID enrolment process that discourage uptake by the public. One example of this is that the cardholder is required to make a trip to the designated station with their physical identity card in order to activate the electronic verification function. There is still a long way to go.
Although both pilot cases have shown great potential in the feasibility assessment, the overall plan to integrate the schemes within the digital identity world is still only half way there, and a number of concerns need to be addressed before progressing forward.
Security at public or remote locations
The core concern in improving the mobility feature of the Smart Kiosk is how to ensure the kiosk itself and the associated data will not be tampered with when a kiosk is moved to a public or remote location, and how to ensure data integrity as well as personal information security in the absence of a secure private network environment and guarded facility. It is believed that additional hardware security features, such as trusted embedded elements, will have to be implemented in order to increase the level of security when handling travel document applications at remote locations. However, exactly what kind of hardware security features are adequate to the expected security requirements has yet to be evaluated.
Possibility to modernise and improve the Smart Kiosk
It is believed that e-Government services could be the first to benefit in terms of efficiency and cost reduction from adopting Mobile ID verification in the enrolment process. By adopting Mobile ID verification using the Smart Kiosk, the applicant might be able to access the same level of service without having to go to the immigration bureau office. Future travel permits may also be replaced by an electronic equivalent, which could be used along with Mobile ID verification, adding value to the authority as cost of printing documents can be reduced and card durability will be improved.
Possibility to expand the use of Mobile ID
Similarly to the implementation of e-Government services by the immigration bureau, the social security bureau is also pilot-testing the use of Mobile ID as an aid in liveness and remote identity verification for elders over 65+ years old to minimise the effort they must spend on travelling. The transportation department is also performing a feasibility evaluation on the adoptability of an e-Driving licence using Mobile ID verification. It is believed that more and more government sectors will take Mobile ID into consideration as a tool to facilitate existing e-Government services.
Previous articleBiometrics in counter-terrorism
Next articleIdentity currencies
Alison Liu holds a bachelor's degree in biological science and a master's degree in information management from the University of Pittsburgh. With over seven years' experience in various e-Travel document projects, Alison serves as an independent consultant for information utilisation.
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Q: Proper way to refer to references in natbib? I'm using the natbibpackage for managing my references in my latex document.
Is it correct to refer to a reference in the following way:
A survey can be found in Ref.~\citep{Radke:2005ug}

Which is typeset as: A survey can be found in Ref. (Radke et al, 2005)
Since my bibliography is not numbered I'm not sure if this is the correct way to do it. Or should I avoid the passive voice altogether?

A: The way you reference is typically governed by whatever journal, book och other publication series you intend to publish in. The "Ref." seems unnecessary to me and in this case I would have written the line "A survey can be found in Radke et al. (2005)
A survey can be found in \citet {Radke:2005ug}

But again, the way this is done varies. The natbib package provides numerous ways to "bend" the referencing so check the standard you are supposed to follow and the natbib documentation fo the way to do it.
Your first suggestion is often called passive referencing whereas my sugestion is referred to as active. Both are used but obviously in different situations.

redpajama_set_name: RedPajamaStackExchange


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Ruf SCR 2018 revealed at Geneva motor show
Combines a retro skin and carbon core with a 503bhp naturally aspirated flat-six and manual gearbox
by: Will Beaumont
Ruf has revealed the latest of its retro-styled, but all-new models at the Geneva motor show. Following the CTR Yellow bird that was shown last year, aside from the colour change, Ruf's newest model has lost the turbos and its iconic nomenclature and transformed into the naturally aspirated Ruf SCR 2018.
Although Ruf is best known for its heavily turbocharged 911-shaped cars, it has long been making naturally aspirated-powered cars too, most bearing the name SCR. The new SCR continues that convention and behind the rear axle sits a naturally-aspirated 4-litre flat-six engine that produces 503bhp at 8270rpm and 347lb ft of torque at 5760rpm. Although Ruf hasn't quoted an acceleration figure for its new car, it has revealed it has a top speed of 199mph.
> Read our review of the Ruf SCR 4.2
If the prospect of a high-revving atmospheric flat-six and a top speed just shy of 200mph didn't seem thrilling enough, power is sent to the rear wheels only via a six-speed manual gearbox.
Like the CTR 2017 that Ruf showed at last year's Swiss motor show, the new SCR is based around a full carbonfibre monocoque chassis. It is not, as its old-school 911 shape might suggest, based on a 90s Porsche chassis just with composite bodywork on top. Its predecessor, the SCR 4.2, was an engineering feat; Ruf took a 993 shell elongated the wheelbase, but kept the overall length of the car the same, then added radiators up front to cool the water-cooled 4.2-litre Mezger flat-six in the back.
> Find out more about Ruf's CTR 2017
The new SCR's underpinnings are even more impressive, though. There's not a hint of old 911 in its make-up, it has been designed, engineered, developed and built all by Ruf. Like last year's CTR, the SCR has a central carbonfibre tub reinforced with an integrated roll cage that supports the roof. Bonded to the main composite structure are the front and rear subframes that support double-wishbone suspension with pushrod actuated dampers. The rear also holds the engine and gearbox.
The outer body is then fully constructed from carbonfibre, the brakes are carbon ceramic and there are also fixed-back hound's-tooth Recaro seats, all helping keep the weight of the entire car down to an impressive 1250kg.
> Take a look behind the scenes of Ruf's factory
Unlike its close relative, the CTR 2017, the SCR does not have a big whale-tale fixed-rear wing. Instead there's a retractable rear wing, like on modern 911 Carreras and the 911 GT3 Touring.
The rear wing isn't the only departure from Ruf's typical look. Instead of the blocky, functional five-spoke alloys that are seen on most of the company's cars, the SCR 2018's wheels have five twin-spokes. They still look just as purposeful however, with their centre-lock attachment as well as being wrapped in Michelin Pilot Cup 2 tyres.
As Ruf has proved over the pas 37 years as an independent manufacturer, it's capable of creating truly impressive and exciting vehicles. The SCR 2018, with its blend of high-tech carbonfibre construction and involving manual gearbox and atmospherically fed engine, promises to be one of the most thrilling and engaging car's Ruf has ever created.
Geneva motor show 2018
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Put Poughkeepsie bus saga to a fitting end: Editorial
Poughkeepsie Journal Editorial Board
The City of Poughkeepsie bus issue – really a saga at this point – has been beset by befuddling, bizarre decisions that have made the city look ridiculous and confused. Recriminations, allegations and misinformation have been pervasive throughout the year, creating a needless divide among Common Council members -- and the public at large. The city's bumbling also cost it any chance of receiving $10 million in a state competition for downtown revitalization money – funds that were awarded to the City of Kingston. And the situation was only going to get worse unless enough Common Council members came to their senses and finished what they had started last year by passing a budget that funded bus service for only the first six months of the year.
OVERCOMING HURDLE:Council approves leasing Poughkeepsie city buses to Dutchess County
PRESSURE CONTINUED:Put these taxpayer-funded buses back to work: Editorial
Fortunately, reason apparently has prevailed, and Poughkeepsie's bus fleet -- idle since June 30 -- may be back on the road in the near future. It never should have come to this. Five months after Dutchess County took over the city system, the buses have been collecting dust in a garage on Howard Street. Four council members -- Democrats Randall Johnson, Lorraine Johnson, Chris Petsas and Ann Perry – refused to part with them, hoping against hope the city would reinstate the service. That belief was groundless; such a move would be far more costly and complicated than if the city had kept the service going in the first place.
On Monday, Perry rightly relented and voted with Democrats Mike Young, Matthew McNamara and Natasha Cherry and Republican Lee David Klein to allow Poughkeepsie to lease the buses to the county for $1 a year -- so long as they are being used for city transportation. For months, the Federal Transportation Administration had made it perfectly clear Poughkeepsie had to turn over these resources (including six hybrid and two diesel buses) to support public transit service in some way – or pay up $1.8 million for holding onto them. That money constitutes the value the FTA puts on these resources through their "useful life," meaning the buses are not near retirement age.
While this standstill has been a unmitigated waste of time and energy, it somehow managed to get stranger following the election of Ann Finney to the newly created at-large council position. Before taking office next month, Finney had the audacity to send a letter to the FTA suggesting the city would be restoring its bus service next year. Yet she didn't confer with all the members of the council – those re-elected in November and those who will be joining the board next year – before making this assertion.
The agreed-upon lease deal should put the focus where it belongs – how can the county improve bus service in the city?
Surely, Poughkeepsie officials can and will be part of that dialogue, but they have plenty of other problems to sort out. In a recent report, state Comptroller Thomas DiNapoli deemed the city "one of the most fiscally stressed municipalities in New York," and this comes after a series of findings regarding the shoddy budget practices that have left Poughkeepise in financial disarray. Remember, the city originally (and wisely) decided to fund the bus service for only half a year because Poughkeepsie is facing a massive budget deficit and must find ways to run a sharper, more effective government -- as should all forms of government.
For the overall financial health of the city, the bus change should serve as a model and not cause mayhem. And if the county is truly looking for a way to help the city, getting this service right is a sound and imperative way to prove it.


The Greenfield Library is delighted to display projects from Jason Lempieri's History of Industrial Design and Architectonics classes. Works on the main level of the library were created for the assignment Hidden History: Know Your ID. Location and information booklets are available at the Greenfield circulation desk.
Inspired to create a communal space on campus, Architectonics students have designed and built Living Room. This model is on display in the lower level of the library.