text:
\section{\em Introduction}

Let $M$ be a closed connected
 $C^{\infty}$ manifold and $g$ a $C^{r}$ ($r\geq 2$) Riemannian metric on $M$.
Let $\phi_{t}:SM\rightarrow SM$ denote the geodesic flow of $g$, acting on the
unit sphere bundle $SM$. Denote the topological entropy of the
flow $\phi_{t}$ with respect to a compact subset $K$ of $SM$ by $h_{top}(K)$.
Then $h_{top} \df h_{top}(SM)$ is the topological entropy of $\phi_t$.

Given $p$ and $q$ in $M$ and $T>0$, define $n_{T}(p,q)$ as the number of
geodesic segments joining $p$ and $q$ with length $\leq T$. Already in 1962
Berger and Bott \cite{BB} observed that there are significant relationships
between integrals of this function and the dynamics of the geodesic flow. As
was pointed out in \cite{BB}, it is not hard to
see that, for each $T>0$, the counting function $n_{T}(p,q)$ is finite and locally constant on an
open full measure subset of $M\times M$, and integrable on $M \times M$. More generally, if $N$ is a compact submanifold of $M$, define $n_T(N,q)$ to be the number of geodesic segments with length $\leq T$ that join a point in $N$ to $q$ and are initially orthogonal to $N$. The function $n_T(N,q)$ enjoys properties similar to those of $n_T(p,q)$.

For $M$ and $N$ that are $C^\infty$, it was shown in \cite{PP} that Yomdin's theorem \cite{Y} can be used to prove that
\begin{equation}
\limsup_{T\rightarrow \infty}\frac{1}{T}\log \int_{M}n_{T}(N,q)\,dq
\leq h_{top}(S^\perp N),
\label{PPineq}
\end{equation}
where $S^\perp N$ is the set of unit vectors with footpoint in $N$ that are orthogonal to $N$. It was also shown in \cite{PP} that when $N$ is the diagonal in the product manifold $M \times M$, the above inequality reduces to
\begin{equation}
\limsup_{T\rightarrow \infty}\frac{1}{T}\log \int_{M\times M}n_{T}(p,q)\,dpdq
\leq h_{top} .
\label{2}
\end{equation}
Another proof of (\ref{2}) is given in \cite{M}.
 On the other hand, for any $C^{r}$ Riemannian
metric ($r\geq 2$), Ma\~{n}\'{e} shows in \cite{M} that
 \begin{equation}
\liminf_{T\rightarrow\infty}\frac{1}{T} \log \int_{M\times M}n_{T}(p,q)\,dpdq
\geq h_{top} .
\label{3}
\end{equation}

Ma\~{n}\'{e} thereby  obtains the first purely Riemannian characterization of
$h_{top}$ for an arbitrary $C^{\infty}$ Riemannian metric:  combining
(\ref{2}) and (\ref{3}) gives

\begin{equation}
\lim_{T\rightarrow \infty}\frac{1}{T}\log \int_{M\times M}n_{T}(p,q)\,dpdq
 = h_{top}.
\label{4}
\end{equation}

Ma\~n\'e's result extends  earlier work of Manning and Freire-Ma\~n\'e.
Suppose $\widetilde p$ is a lift of a point $p \in M$ to the universal cover
$\widetilde M$ of $M$ and $B_T(\widetilde p)$ is the ball of radius $T$ about
$\widetilde p$ in $\widetilde M$ (with the metric lifted from $M$). It follows
easily from the results in \cite{BB} that
\begin{equation}
 \int_M n_{T}(p,q)\,dq \geq  {\hbox {\rm Vol}}\,B_T(\widetilde p),
\label{4.5}
\end{equation}
with equality if $M$ has no conjugate points. Manning showed that,
for any $\widetilde p \in \widetilde M$, \ $T^{-1} {\hbox {\rm Vol}}\,B_T(\widetilde p)$
converges to a limit $\lambda$ that is independent of $\widetilde p$. From
(\ref{4}) and (\ref{4.5}), it is easy to obtain the inequality $h_{top} \geq
\lambda$ for any Riemannian manifold, which was  first proved by Manning in \cite{Ma}.
One also sees that $h_{top} = \lambda$ if $M$ has no conjugate points, which was first proved by Freire and Ma\~n\'e in \cite{FM}.

Besides the natural appeal of a formula like (\ref{4}),
there are other reasons to be interested in relations between the topological
entropy of the geodesic flow and the growth rate of the
average number of geodesic segments between two points in the manifold. The
function $n_{T}(p,q)$ also counts the number of critical points of the energy
functional on the path space $\Omega^{T}(p,q)$ given by all the curves joining
$p$ and $q$ with length $\leq T$. Therefore using Morse theory, $n_{T}(p,q)$
can be bounded from below by the sum of the Betti numbers of $\Omega^{T}(p,q)$
(provided of course that $p$ and $q$ are not conjugate).  By averaging over
$M$ and using results of Gromov \cite{Gr2}, one can obtain in this fashion
remarkable relations between the topology of $M$ and the vanishing of
$h_{top}$; we refer to \cite{Gr1,P,P1,PP} for a detailed description of these
relations.

The present paper continues the study of relationships between $h_{top}$ and
the exponential growth rate of $n_T(p,q)$. The strongest relationship is
for metrics with no conjugate points: Ma\~n\'e showed in \cite{M} that in this
case
\[ \lim_{T \to \infty}\frac1T \log n_T(p,q) = h_{top} \quad\hbox{\rm  for all $p,q \in M$.}\]
 For any $C^\infty$ Riemannian manifold, taking $N$ to be the single point $p$ in (\ref{PPineq}) gives the following inequality, that was first proved by G.P. Paternain  in \cite{P}:
 \begin{equation}
 \limsup_{T\rightarrow\infty} \frac{1}{T} \log\int_{M}n_{T}(p,q)\,dq
 \leq h_{top}(S_pM)  \leq h_{top}\quad\hbox{\rm for all $p \in M$.}  \label{5}
 \end{equation}
 As Ma\~{n}\'{e} indicates in \cite{M}, a simple argument using the Borel-Cantelli Lemma shows
that for every $p \in M$ one has
\begin{equation}
 \limsup_{T \to \infty}\frac{1}{T}\log n_{T}(p,q) \leq
\limsup_{T\rightarrow\infty} \frac{1}{T} \log\int_{M}n_{T}(p,q')\,dq'
\quad\hbox{\rm for a.e. $q \in M$.} \label{5.1}
\end{equation}
It is immediate from  (\ref{5}) and (\ref{5.1}) that  for all $p\in M$ one has
 \begin{equation}
\limsup_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)\leq h_{top}
\quad \hbox{for a.e. $q\in M$.}    \label{5.2}
\end{equation}
This inequality motivated Ma\~{n}\'{e} to pose the following questions in
\cite{M}. \bigskip

\noindent{\bf Question I.} {\em Is it true that
 \begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)= h_{top}
\quad\hbox{\rm  for a.e. $(p,q)\in M\times M$?}
\label{6}
\end{equation}}

\noindent{\bf Question II.} {\em Is it true that equation (\ref{6}) holds for generic Riemannian
metrics when~${\hbox {\rm dim}}\, M = 2$?}

\medskip

The main purpose of the present paper is to give a negative answer to Question II.
Of course this also answers Question I negatively.

Because of inequalities (\ref{5}) and (\ref{5.1}) it is natural to consider modified versions of the first question. \bigskip

\noindent{\bf Question I$'$.} {\em Is it true that
 \[\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M} n_{T}(p,q)\,dq
  =   h_{top} \quad\hbox{ for all $p \in M$?}\]}

\noindent{\bf Question I$''$.} {\em Is it true that
 \[\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M} n_{T}(p,q)\,dq
  =   h_{top} \quad\hbox{for almost every $p \in M$?}\]}

We shall also give negative answers to Questions I$'$ and I$''$.
Because of (\ref{5.1}), a negative answer to Question I$''$ implies a negative answer to Question I. Our main example will be a surface for which Question I$''$ has a negative answer. The important features of this surface will be stable under small perturbations of the metric; we thereby obtain an open set of metrics for which the answer to Question I$''$ is negative. This implies a negative answer to Question II.

Let us describe the contents of the paper in more detail.
 In Section 2 we obtain a slight improvement of inequality (\ref{5}) that is
needed for our main example.
  We also prove an interesting related result, namely, that for
the case of surfaces, the set of points $p\in M$ for which inequality (\ref{5}) is an equality has non-empty interior.
This result will imply an alternative proof of Ma\~{n}\'{e}'s inequality (\ref{3}) for the two dimensional case.

In Section 3, we observe that the Weinstein examples, described in \cite{BeBe}, give rise to many manifolds
 (such as ${\bf CP}^{k}$)
 for which there exists a point $p$ such that $\displaystyle\int_{M}n_{T}(p,q)\,dq$ grows {\bf linearly} in $T$, even though $h_{top}>0$.
  It follows from (\ref{5.1}) that $\limsup_{T \to \infty} T^{-1} \log n_T(p,q) = 0$
for a.e. $q \in M$.
 All these examples have dimension
$\geq 3$. We also construct a metric on the two-sphere with $h_{top}>0$ which has a point $p$ such that all
the geodesics leaving from $p$ are simple, closed and have the same period, so again $\displaystyle\int_{M}n_{T}(p,q)\,dq$ grows {\bf linearly} in $T$.

Section 4 contains our main example that gives negative answers to
Ma\~{n}\'{e}'s Questions I and II and also to Question I$''$.
 We start with a one parameter family of
surfaces of revolution $C_{d}$, $0< d<\infty$, which have  the
properties shown in Figure 1.

\begin{figure}

\vspace{8cm}

\caption{Main example.}

\end{figure}

Each $C_d$ is diffeomorphic to the two sphere and contains a region $R$ bounded by a geodesic circle of latitude $\alpha$ which is shorter than any other geodesic circle of latitude in $R$.
Inside $R$ is a circle of latitude $\gamma_0$ that is a hyperbolic closed geodesic. The region $R$ contains geodesics that are both forward and backwards asymptotic to  $\gamma_{0}$.
This means that $\gamma_{0}$ has a homoclinic connection.
Attached to $R$ there is a flat cylinder of length $d$ which we close smoothly with a cap $D$. Let $P$ be the center of the cap and $Q$ the center of the region $R$.

We perturb the metric in a small part of $R$ (shaded in Figure 1) so as to break the homoclinic
connection of $\gamma_{0}$ and obtain a horseshoe with entropy $h > 0$.
 Moreover we arrange the perturbation so that the meridians of $C_{d}$ are preserved, i.e., all geodesics leaving from $P$ in the new metric are simple, closed, have the same period and they coincide outside $R$ with the meridians of the old metric.
 A careful application of KAM theory (Lemma \ref{key}) shows that  there is
$d_0 > 0$ such that for any $d\geq d_0$ the geodesic flow for the perturbed metric has an invariant torus whose projection to $C_{d}$ becomes singular on two curves, one near $P$, the other near $Q$, as shown in Figure 1.

For each $d$, the torus and its image under the flip $v\rightarrow -v$ separate the unit sphere bundle of $C_{d}$ into two invariant
sets, $W_{1}$ and $W_{2}$.
This forces the geodesics which pass sufficiently close to $P$ or $Q$ to cross
the piece of the surface around $\gamma_{0}$ fairly perpendicular to
$\gamma_{0}$; intuitively this separating surface keeps the geodesics that leave from points sufficiently close to $P$ or $Q$ away from the horseshoe. This property allows us to
estimate their Liapunov exponents and make them small compared with $h$, by
choosing $d$ large enough.
  If $W_{2}$ is the set that contains the horseshoe, then the topological
entropy on $W_{2}$ is at least $h$ and the topological entropy on $W_{1}$ can
be made smaller than $h$, since the Lyapunov exponents in $W_1$ can be made small, as we explained above. The results in Section 2 will imply
that if $p$ is sufficiently close to $P$ or $Q$ , then

\[
\limsup_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)
 \leq
\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M}n_{T}(p,q')\,dq'\leq h_{top}(W_{1}), \]
for almost every $q\in M$.
Thus there exists a positive measure set $U\subset M\times M$ such that for
$(p,q)\in U$
\begin{equation}
\limsup_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)\leq
\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M}n_{T}(p,q)\,dq'<h \leq
h_{top}.
\label{Upq}
\end{equation}
 This gives negative answers to Question I and I$''$.
 A careful look at the details in
Section 4 will show that for small perturbations of this example
(\ref{Upq}) still holds for a positive measure set of $p$ and $q$. Thus Question II also has a negative answer.

{\sl Acknowledgements:}
The second author is grateful to Detlef Gromoll for suggesting several years ago the study of the relationship between the exponential growth rate of $n_{T}(p,q)$ and $h_{top}$.
He is also grateful to the University of Maryland and Northwestern University for their hospitality while this work began.
The first author thanks the Universidad de la Rep\'ublica, Uruguay, for hospitality while this work was completed.

\section{\em Some properties of the function $n_{T}(p,q)$}

We begin this section by proving a result (cf. Proposition \ref{p1} below) that we shall use in our main example. In what follows we will always assume that the Riemannian metric is $C^{\infty}$.

In \cite{M}, Ma\~{n}\'{e} shows the following application of the Borel-Cantelli Lemma:

\begin{Lemma}Let $(X,{\cal A},\mu)$ be a probability space and $f_{n}:X\rightarrow (0,+\infty)$ a sequence of integrable functions. Then:
\[\limsup_{n\rightarrow\infty}\frac{1}{n}\log f_{n}(x)\leq\limsup_{n\rightarrow\infty}\frac{1}{n}\log\int_{X}f_{n}\,d\mu,\]
for $\mu$-a.e. $x\in X$.  \label{Borel}
\end{Lemma}

As in the Introduction, denote by $SM$ the unit sphere bundle of $M$ and let $\pi:SM\rightarrow M$ be the canonical projection.
For $p\in M$, set $S_{p}= \pi^{-1}(p)$.
If $K\subset SM$ is a closed set, let $h_{top}(K)$ denote the topological entropy of the geodesic flow $\phi_{t}$ with respect to the set $K$; with this notation, $h_{top}(SM)=h_{top}$.
Combining Lemma (\ref{Borel}) above with Corollary 2.2 from \cite{PP} gives the following result; we include the proof for the convenience of the reader.

\begin{Proposition}Let $K$ be any closed set in $SM$ and suppose there exists $p\in M$, such that $S_{p}\subset K$, then
\[\limsup_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)\leq
\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M}n_{T}(p,q')\,dq'\leq h_{top}(K),\]
for almost every $q\in M$.     \label{p1}
\end{Proposition}

\PROOF It is known (cf. \cite{BB,P,PP}) that for any $p\in M$,
\begin{equation}
\int_{M}n_{T}(p,q)\,dq\leq\int_{0}^{T}{\hbox {\rm Vol}}(\phi_{t}S_{p})dt,   \label{7'}
\end{equation}
where ``Vol" stands for the $n-1$ dimensional Riemannian volume ($n= {\hbox {\rm dim}}\, M$).
Now we use Yomdin's Theorem as stated in \cite{Gr1} to obtain:
\[\limsup_{T\rightarrow\infty}\frac{1}{T}\log {\hbox {\rm Vol}}(\phi_{T}S_{p})\leq h_{top}(S_{p}).\]
Since $S_{p}\subset K$, $h_{top}(S_{p})\leq h_{top}(K)$ and by combining the last inequality with (\ref{7'}) we obtain
\[\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M}n_{T}(p,q)\,dq\leq h_{top}(K),\]
and by using Lemma \ref{Borel} we conclude:
\[\limsup_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)\leq
\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M}n_{T}(p,q')\,dq'\leq h_{top}(K),\]
for almost every $q\in M$.
~\hfill~ $\diamond$ \vspace{7mm}

For brevity, let $\sigma_{p}$ be:
 \[\sigma_{p}\df\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{M}n_{T}(p,q)\,dq.\]
 Then (\ref{5}) can be writen as $\sigma_{p}\leq h_{top}$ for all $p\in M$.
Our main example in Section 4 possesses the following property: there exists an open set $V\subset M$ such that for any $p\in V$, $\sigma_{p}<h_{top}$.
To complete the description of the possible behavior of the
correspondence $p\rightarrow \sigma_{p}$ when ${\hbox {\rm dim}}\,M=2$, we now show that in this case
 the set of points for which $\sigma_{p}=h_{top}$ always has
non-empty interior.

\begin{Theorem}Suppose ${\hbox {\rm dim}}\,M = 2$. Consider the set $\Omega$ of
points $p\in M$ such that
\[\lim_{T\rightarrow\infty}\frac{1}{T}log\int_{M}n_{T}(p,q)\,dq=h_{top}.\]
Then $\Omega$ has non-empty interior.  \label{teo}
\end{Theorem}

\PROOF For each vector $v\in S_{p}$ consider the Jacobi equation along
the geodesic $\gamma_v$ defined by $v$:
\begin{equation}
y''(t)+K(t)y(t)=0,   \label{jacobi}
\end{equation}
where $K(t)$ is the Gaussian curvature of $M$ at $\gamma_v(t)$.
Let $y_{v}(t)$ be the solution of
(\ref{jacobi}) that satisfies: $y_{v}(0)=0$ and $y'_{v}(0)=1$.
On account of the results in \cite{BB} we have:
\begin{equation}
\int_{M}n_{T}(p,q)\,dq=\int_{0}^{T}ds\int_{S_{p}} | y_{v}(s) | \,dv.
\label{one}
\end{equation}
On the other hand by definition, if $l$ denotes length, we can write
\begin{equation}
l(\phi_{T}S_{p})=\int_{S_{p}}\sqrt{y_{v}^{2}(T)+(y_{v}')^{2}(T)}\,dv.
\label{two}
\end{equation}
Equations (\ref{one}) and (\ref{two}) clearly imply
\begin{equation}
\limsup_{T\rightarrow\infty}\frac{1}{T}\log \int_{M}n_{T}(p,q)\,dq\leq\limsup_{T\to\infty}\frac{1}{T}\log l(\phi_{T}S_{p}).
  \label{three}
\end{equation}

Since $y'_{v}(0)=1$, it follows from (\ref{jacobi}) that
\[y_{v}'(T)=1-\int_{0}^{T}K(s)y_{v}(s)\,ds;\]
 if $L$ is the maximum of the absolute value of the curvature on $M$, we obtain
\[ |  y_{v}'(T) | \leq 1+L\int_{0}^{T} |  y_{v}(s) |  \,ds.\]
Therefore using (\ref{two}) we can write
\begin{eqnarray*}
l(\phi_{T}S_{p}) &\leq& \int_{S_{p}} \left( |  y_{v}(T) |
+1+L\int_{0}^{T} |  y_{v}(s) |  \,ds \right) \,dv  \\
&=& \int_{S_{p}} |  y_{v}(T) | \,dv+l(S_{p})+L\int_{0}^{T}\int_{S_{p}}
| y_{v}(s) |  \,dvds. \end{eqnarray*}
{}From the last inequality and (\ref{one}) we obtain
\begin{equation}
\liminf_{T\rightarrow\infty}\frac{1}{T}\log l(\phi_{T}S_{p})\leq \liminf_{T\rightarrow\infty}\frac{1}{T}\log \int_{M}n_{T}(p,q)\,dq.  \label{four}
\end{equation}

Recall now from \cite[page 220]{N}, that there exists an arc $\alpha\subset SM$ such that
\begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{T}\log l(\phi_{T}\alpha)=h_{top}.  \label{six}
\end{equation}
Moreover, if $\mu$ is an ergodic measure of maximal entropy, $\alpha$ could be {\sl any} arc transversal to the Pesin stable manifolds of $\mu$.
Let $v_{0}$ be a Pesin point of $\mu$ and let $W^{s}(v_{0})$ denote the weak stable manifold through $v_{0}$.
 By a well known property of the geodesic flow ($W^{s}(v_{0})$ is $\phi_{t}$-invariant), there exists $r>0$ such that $T_{\phi_{r}v_{0}}W^{s}(v_{0})\cap T_{\phi_{r}v_{0}}S_{\pi(\phi_{r}v_{0})}=\{0\}$, therefore the curve $S_{\pi(\phi_{r}v_{0})}$ is transversal to $W^{s}(v_{0})$ at the point $\phi_{r}v_{0}$.
 Hence (\ref{three}), (\ref{four}) and (\ref{six}) imply:
\begin{equation}
\lim_{T\rightarrow\infty}\frac{1}{T}\log \int_{M}n_{T}(p,q)\,dq=h_{top},                                                              \label{seven}
\end{equation}
where $p=\pi(\phi_{r}v_{0})$.
 On the other hand, $\pi:W^{s}(v_{0})\rightarrow M$ is a local diffeomorphism at $\phi_{r}v_{0}$, since $T_{\phi_{r}v_{0}}W^{s}(v_{0})\cap T_{\phi_{r}v_{0}}S_{\pi(\phi_{r}v_{0})}=\{0\}$.
Therefore every unit circle $S_{p'}$, with foot point $p'$ in a neighborhood of $p=\pi(\phi_{r}v_{0})$, is transversal to $W^{s}(v_{0})$ and thus (\ref{seven}) has to hold for an open set around $p$.
~\hfill~ $\diamond$ \vspace{7mm}

We finish this section by showing how Theorem \ref{teo} implies Ma\~{n}\'{e}'s inequality (\ref{3}) for the two dimensional case.

\begin{Corollary}Suppose ${\hbox {\rm dim}}\,M = 2$. Then:
\[\liminf_{T\rightarrow\infty}\frac{1}{T} \log \int_{M\times M}n_{T}(p,q)\,dpdq\geq h_{top}.\]
\end{Corollary}

\PROOF For $p\in M$, set $I_{p} = \int_{M}n_{T}(p,q)\,dq$.
Let $\Omega$ be the set from Theorem \ref{teo} and let $m$ denote its
measure (by Theorem \ref{teo}, $m>0$).
Then
\[ \int_{M\times M}n_{T}(p,q)\,dpdq=\int_{M}I_{p}(T)\,dp\geq \int_{\Omega}I_{p}(T)\,dp,\]
and by Jensen's inequality
\[\log \int_{M\times M}n_{T}(p,q)\,dpdq\geq\frac{1}{m}\int_{\Omega}\log I_{p}(T)\,dp.\]
Hence
\[\liminf_{T\rightarrow\infty}\frac{1}{T} \log \int_{M\times M}n_{T}(p,q)\,dpdq\geq\frac{1}{m}\int_{\Omega}\sigma_{p}\,dp=h_{top}.\]
~\hfill~ $\diamond$ \vspace{7mm}

\section{\em Simple examples}

We consider in this section the following modified version of Question I:\medskip

\noindent{\bf Question I$'$.} {\em Is it true that
\[\limsup_{T\rightarrow\infty}\frac{1}{T}\log \int_{M}n_{T}(p,q)\,dq=h_{top}
\quad\hbox{ for all $p$?}\]}

Suppose $M^{n}=D\cup_{S^{n-1}} D_{N}$, where $D$ is an $n$-dimensional disk with center $p$ and $D_{N}$ is a disk bundle over a manifold $N$, such that its associated sphere bundle $\partial D_{N}$ is diffeomorphic to $S^{n-1}$.
 A typical example of such a manifold is ${\bf CP}^{k}$; if we remove a disk from it, we obtain a disk bundle over ${\bf CP}^{k-1}$.
Let $g_{N}$ be any Riemannian metric on $N$, then it is shown in \cite{BeBe} that one can construct a metric $g$ on $M$ so that:
\begin{itemize}
\item all geodesics leaving from $p$ (the center of $D$) return to $p$ exactly at the same time.
\item $(N,g_{N})\rightarrow (M,g)$ is a totally geodesic isometric embedding.
\end{itemize}

Choose  a metric $g_{N}$ for which the topological entropy of the geodesic flow of $N$ is positive. Since $N$ is totally geodesic, this implies that $h_{top}$ of the geodesic flow of $M$ is also positive.
On the other hand since every geodesic leaving from $p$ returns to $p$ at exactly the same time, it follows that ${\hbox {\rm Vol}}(\phi_{T}S_{p})$ is uniformly bounded for all $T$ and by equation (\ref{7'}), $\displaystyle\int_{M}n_{T}(p,q)\,dq$ grows linearly with $T$.
  This gives a negative answer to Question I$'$.
In fact, it also shows that for some $p$ the growth of $\displaystyle\int_{M}n_{T}(p,q)\,dq$ could be only linear, even though $h_{top}>0$.

Observe that this construction requires ${\hbox {\rm dim}}\,N \geq 2$, and hence ${\hbox {\rm dim}}\,M \geq 3$, so it is natural to ask if similar examples can be constructed in the case of surfaces.
We shall show below that there exist metrics on $S^{2}$ with $h_{top}>0$ and the additional property that there exists a point $p$ such that all the geodesics leaving from $p$ are simple, closed and with the same period.
As a consequence, $\displaystyle\int_{M}n_{T}(p,q)\,dq$ grows {\sl linearly} with $T$.

Let $C$ be a surface of revolution diffeomorphic to $S^{2}$.
On $C$ the geodesic flow is completely integrable with an integral of motion --- the Clairaut integral --- given by $r\sin\phi$, where $r$ is the radial distance of a point on the surface to the axis of revolution and $\phi$ is the angle a geodesic makes with the meridians.
We assume that $C$ contains a circle of latitude $\gamma_{0}$ where the function $r$ has a nondegenerate minimum. This means that $\gamma_0$ is a hyperbolic closed geodesic. We also assume that the other circles of latitude where $r$ has the same value $r_0$ as on $\gamma_0$ are not critical points of $r$. Thus $\gamma_0$ is the only closed geodesic on which $r = r_0$.
The closed orbit of the geodesic flow of $C$ corresponding to the geodesic $\gamma_{0}$  has a homoclinic connection.
This means that the weak stable and weak unstable manifolds of this orbit coincide.
They lie in the set of vectors tangent to the geodesics for which the value of the Clairaut integral is $r_0$.
These geodesics are asymptotic to $\gamma_{0}$ as $t\rightarrow \pm\infty$.
One of them is shown in Figure 2.

Let $P$  and $Q$ denote the poles where the axis of revolution intersects $C$.
We parametrize $C$ with geodesic polar coordinates $(\theta,l)$, $\theta\in [0,2\pi)$ and $l\in [0,R]$.
In these coordinates $C$ is determined by the profile function $r(l)$, which is required to be $C^{\infty}$, that gives the radius of a circle of latitude at a distance $l$ from the point $Q$ at the bottom of the surface.
For technical reasons we assume that there exists $l_1\in (0,R)$ and $b>0$, such that for $l\in (l_1-b,l_1+b)$ we have $r(l)\equiv r_1$, where $r_1$ is a constant such that $r_1>r_0$.
We will also assume that the value $l_0$ of $l$ on $\gamma_{0}$ satisfies $l_0<l_1-b$ and that $r$ is strictly increasing for $l\in (l_0,l_1-b)$.
Let $F$ denote the flat region given by those $p\in C$ such that $l(p)\in [l_1-b,l_1+b]$.
A typical shape for $C$ is shown in Figure 2.
\begin{figure}
\vspace{8cm}
\caption{Surface of revolution $C$.}
\end{figure}

Recall that the meridians of $C$ are all simple closed geodesics with the same period.

\begin{Lemma}There exist arbitrarily small smooth perturbations of the Riemannian metric of $C$ with support in $F$ such that:

(1) The meridians are preserved, i.e. all geodesics leaving from $P$ are simple, closed and with the same period, and coincide with the meridians outside $F$.

(2) $\gamma_{0}$ possesses a {\bf transverse} homoclinic orbit.\label{meridians}

\end{Lemma}

\PROOF The assumptions made above (in particular the requirement $r_1 > r_0$) and the properties of the Clairaut integral ensure that there are geodesics that pass through the flat region $F$ and are both forwards and backwards asymptotic to $\gamma_0$; indeed any vector in $F$ that makes angle~$\phi$ with the meridians will be tangent to such a geodesic if $r_1\sin\phi = \pm r_0$. Let $\gamma_{su}$ be such a geodesic and let $p$ be a simple point of $\gamma_{su}$ that lies in $F$.
Parametrize $\gamma_{su}$ so that $\gamma_{su}(0) = p$.
Consider the frame $\{E_{1}, E_{2}\}$ in $T_{p}C$, where $E_{1}=\gamma_{su}'(0)$ and $E_{2}$ is the unit vector tangent to the meridian through $p$ pointing towards $P$.
Let $\{E_{1}(t), E_{2}(t)\}$ denote the frame in $T_{\gamma_{su}(t)}C$ (not orthonormal) obtained by parallel transport along $\gamma_{su}$.
Consider the map $f:{\bf R}^{2}\rightarrow C$ given by
\[f(t,x)=\exp_{\gamma_{su}(t)}(xE_{2}(t)).\]
For a small enough $\delta > 0$, the map $f$ is a diffeomorphism form $\Delta = (-\delta,\delta)^2$ to a neighborhood~$U \subset F$ of the point~$p$; the set~$U$ is shaded in Figure 2.
 For each fixed $t$, the curves $x\rightarrow f(t,x)$ are meridians parametrized by arc length.
In these coordinates the metric of $C$ satisfies:
\[g_{11}(t,x)=1,\;\;\;g_{12}(t,x)= a,\;\;\;g_{22}(t,x)=1,\]
where $a$ is a constant satisfying $-1 < a < 1$.

Let $\alpha : {\bf R}^2 \rightarrow {\bf R}$ be a smooth function with support inside $\Delta$
and let $g^\alpha$ be the metric defined by:
\[g^\alpha_{11}(t,x) = 1 - \alpha(t,x)x^{2},\;\;\;g^\alpha_{12}(t,x)= a,\;\;\;g^\alpha_{22}(t,x)=1.\]
 A simple computation shows that the Christoffel symbols of $g^\alpha$ satisfy:
 \[\Gamma^{1}_{22}\equiv \Gamma^{2}_{22}\equiv 0,\]
\[\Gamma^{1}_{11}(t,0)=\Gamma^{2}_{11}(t,0)=0.\]
Therefore the curve, $t\rightarrow f(t,0)$, and the curves, $x\rightarrow f(t_{0},x)$ for each fixed~$t_0$, are still geodesics, and thus the perturbation $g^\alpha$ preserves the geodesic $\gamma_{su}$ and the meridians.

A further computation shows that the Gaussian curvature of $g^\alpha$ at the point $\gamma_{su}(t) = f(t,0)$ is given by:
\begin{equation}
K_\alpha(t,0)=\frac{-\alpha(t,0)}{1-a^{2}}.  \label{gauss}
\end{equation}
This equation and Donnay's arguments from \cite{D} imply that for a suitable choice of the function~$\alpha$ the stable and unstable manifolds of $\gamma_{0}$ must have a {\sl transverse} intersection at the point $\gamma_{su}'(0)$, which concludes the proof of the lemma.

For the convenience of the reader, we give a brief sketch of Donnay's idea.
Let $H^-$ and $H^+$ be the projections to $C$ of the strong stable and strong unstable manifolds associated to the geodesic $\gamma_{su}$.
Recall that the strong stable and strong unstable manifolds are given by the unit normal vectors to $H^-$ and $H^+$ that point to the same side as the tangent vector to $\gamma_{su}$.
 The geodesic curvatures at $\gamma_{su}(t)$ of $H^-$ and $H^+$ are solutions $u^-$ and $u^+$ of the Riccati equation
     \begin{equation}
u'(t) + u^2(t) + K(t) = 0, \label{riccati}
     \end{equation}
where $K(t)$ is the curvature at $\gamma_{su}(t)$. Before the perturbation we have $u^-_{old} \equiv u^+_{old}$. Let $t_1$ be the time when $\gamma_{su}$ enters and $t_2$ the time when $\gamma_{su}$ leaves the support of the perturbation of the metric. Then $u^-_{new}(t) =  u^-_{old}(t)$ for $t \geq t_2$ and  $u^+_{new}(t) =  u^+_{old}(t)$ for $t \leq t_1$. All that one needs to arrange is that $u^+_{new}(t_2) \neq u^-_{new}(t_2) = u^+_{old}(t_2)$. It is clear from (\ref{gauss}) and (\ref{riccati}) that this will be the case if the function $\alpha$ is chosen suitably.

~\hfill~ $\diamond$ \vspace{7mm}

\section{\em The main example}

As mentioned in the introduction, we begin with a one parameter family of surfaces of revolution $C_d$,  $0<d<\infty$, which have
the properties shown in Figure 1.
Each $C_d$ contains a region $R$ whose geometry is the same for all $d$. The boundary of $R$ is  a geodesic circle of latitude $\alpha$, which is shorter than any other geodesic circle of latitude in $R$: this ensures that any geodesic which enters $R$ must leave after a finite time. The region $R$ contains a flat cylinder $F$ and a hyperbolic closed geodesic $\gamma_{0}$; the distance of the surface from the axis of revolution is strictly decreasing as one moves from $F$ to $\gamma_{0}$. Attached to $R$ is a flat cylinder of length $d$ which is smoothly closed with a cap $D$; the geometry of $D$ is independent of $d$. The boundary of $D$ is a closed geodesic, which we shall denote by $\beta$. We choose $D$ so that $\beta$ is the only circle of latitude in $D$ that is a geodesic: this ensures that any geodesic which enters $D$ must leave after a finite time. Let $P$ be the center of the cap $D$ and $Q$ the center of the region $R$.

We parametrize $C_d$ with geodesic polar coordinates $(\theta,l)$ where $\theta \in {\bf R}/2\pi$ and $l$ is the geodesic distance of a point from $Q$. In these coordinates $C_d$ is determined by the profile function $r(l)$ which gives the distance from the axis of revolution of the circle of latitude which is at geodesic distance $l$ from $Q$. We  choose $r(l)$ to be $C^\infty$. Let  $l_\alpha$ be the value of $l$ on $\alpha$. Let $\rho = r(l_\alpha)$. Finally we introduce a third coordinate $\phi \in (-\pi/2,\pi/2]$ on the unit tangent bundle of $C_d \setminus \{P,Q\}$, which is the angle measured in the counterclockwise direction between a vector and the meridian  passing through the point where the vector is based (we shall take $\phi = \pi/2$ for the tangent vectors to the circle of latitude where this description is ambiguous).

Define $S_1$ and $S_4$ to be the sets of unit vectors with footpoints on $\alpha$ that point out of and into $R$ respectively. Define $S_3$ and $S_2$ to be the sets of unit vectors with footpoints on $\beta$ that point out of and into $D$ respectively. The coordinates $\theta$ and $\phi$ allow us to identify $S_i$ with $ ({\bf R}/2\pi) \times (-\pi/2,\pi/2)$ for $i = 1,\dots,4$.

For a vector $v \in S_1$, let us  follow the geodesic $\gamma_v$ defined by $v$ starting from time $t = 0$. We have chosen the geometry of $R$ and $D$ so that $\phi_t(v) = \gamma'_v(t)$ moves from $S_1$, through $S_2$, $S_3$ and $S_4$ in succession and then returns to $S_1$.
This process defines maps from
$S_{i}$ to  $S_{i+1}$ for $i=1,\dots,4$~mod $4$,
which induce maps
\[\psi_{i}:({\bf R}/2\pi) \times (-\pi/2,\pi/2)\rightarrow ({\bf R}/2\pi) \times (-\pi/2,\pi/2),\;\;\;\;i=1,\dots,4.\]
Elementary trigonometry shows that
\begin{equation}
\psi_i(\theta,\phi) = (\theta + \frac{d\tan \phi}{\rho}, \phi) \qquad\hbox{for $i=1,3$.}     \label{cylinder}
\end{equation}

The properties of the Clairaut integral imply that
\begin{equation}
\psi_i(\theta,\phi) = (\theta + a_i(\phi), -\phi), \qquad\hbox{for $i=2,4$,} \label{minus}
\end{equation}
where $a_2$ and $a_4$ are $C^\infty$ functions  satisfying
\begin{equation}
a_2(0)= \pi = a_4(0).
\label{aa}
\end{equation}
 Notice that $a_2$ and $a_4$ {\bf do not} depend on $d$.

As explained in the introduction, we make a small perturbation of the surfaces of revolution $C_d$.
Recall from Section 3 (cf. Lemma \ref{meridians}) that there exist arbitrarily small, smooth perturbations  of the metric, such that:

(1) The support of the perturbation is confined to the interior of the flat cylinder $F \subset R$.

(2) The meridians are preserved, i.e. all geodesics leaving from $P$ are simple, closed and with the same period, and coincide with the meridians outside $F$.

(3) $\gamma_{0}$ possesses a transverse homoclinic orbit.

We can also assume that the perturbed metric has the following property:

 (4) There is a constant $\phi_0 > 0$ such that every geodesic that enters $R$ through $\alpha$ with $|\phi| < \phi_0$ must exit $R$ after a finite time.

To see this, note that we can choose $\phi_0 > 0$, $T_0 > 0$ and $\varepsilon_0 > 0$ so that every geodesic of the original metric that enters $R$ with $|\phi| \leq 2\phi_0$ must exit the $2\varepsilon_0$ neighborhood of $R$ by time $T_0$ at the latest. If the perturbation is small enough, every geodesic of the new metric that crosses $\alpha$ with $|\phi| \leq \phi_0$ must exit the  $\varepsilon_0$ neighborhood of $R$ by time $2T_0$ at the latest.

The above properties ensure that the orbits of the geodesic flow that start in $S_1$ with $|\phi| < \phi_0$ still pass through $S_2$, $S_3$ and $S_4$ in succession and then return to $S_1$. The transitions from $S_i$ to $S_{i+1}$, for $i = 1,2,3$, are exactly the same as for the unperturbed metric and do not change $|\phi|$. Property (4) means that every orbit that leaves $S_4$ with $|\phi| < \phi_0$ must pass through $S_1$.  Let $\tilde{\psi}_{4}:({\bf R}/2\pi) \times (-\phi_0,\phi_0)\rightarrow ({\bf R}/2\pi) \times (-\pi/2,\pi/2)$ be the map induced by the transition from $S_{4}$ to $S_{1}$ for the perturbed surface.
Notice that $\tilde{\psi}_{4}$ is determined by the metric in $R$ and is independent of $d$.

The next result will be crucial for our construction:

\begin{Lemma}There exists $d_{0}>0$ and a perturbation of the metric in $R$ satisfying properties (1)--(4) above, such that for any $d > d_{0}$, the map
\[ \tilde\Psi_d \df \tilde \psi_4 \circ \psi_3 \circ \psi_2 \circ \psi_1 : ({\bf R}/2\pi) \times (-\phi_0,\phi_0) \to ({\bf R}/2\pi) \times (-\pi/2,\pi/2)\]
has a homotopically nontrivial invariant circle in the set
\( ({\bf R}/2\pi) \times  [-1/d,1/d] . \) \label{key}
\end{Lemma}

\PROOF
For $d>0$ we define a scaling map $\varphi_{d} : ({\bf R}/2\pi) \times {\bf R} \to ({\bf R}/2\pi) \times {\bf R}$ by $\varphi_{d}(\theta,\phi) \df (\theta,d\phi)$.
Observe that $\varphi_{d}$ maps the region $| \phi| \leq 1/d$ diffeomorphically onto the set \[ S\df ({\bf R}/2\pi) \times [-1,1].\] In order to show that $\tilde\Psi_d$ has an invariant circle in the region $|\phi| < 1/d$, it suffices to show that the map \[\tilde k_d \df \varphi_d \circ \tilde\Psi_d \circ \varphi_d^{-1} : S \to ({\bf R}/2\pi) \times {\bf R},\] which is well defined for $d > 1/\phi_0$, has an invariant circle.

Let $\Psi_d =  \psi_4 \circ \psi_3 \circ \psi_2 \circ \psi_1$ and let $k_d = \varphi_d\circ \Psi_d \circ \varphi_d^{-1}$.

Our first step is to study the limiting behaviour of $k_d$ and $\tilde k_d$ as $d \to \infty$ using:

\begin{Lemma}Let $f=(g,h):({\bf R}/2\pi) \times (-\phi_0,\phi_0)\rightarrow ({\bf R}/2\pi) \times (-\pi/2,\pi/2)$ be a $C^{\infty}$ map with the property
\[h(\theta,0)=0,\]
for all $\theta\in {\bf R}/2\pi $.
For $d>1/\phi_0$ define
\[f_{d}=\varphi_{d}\circ f\circ\varphi_{d}^{-1}:S\rightarrow ({\bf R}/2\pi) \times {\bf R}.\]
Then $f_{d}$ converges in the $C^{\infty}$ topology as $d \to \infty$ to the map
\[(\theta,\phi)\rightarrow (g(\theta,0),\phi\frac{\partial h}{\partial \phi}(\theta,0)).\]    \label{technical}
\end{Lemma}

\PROOF
Since $h(\theta,0) = 0$, there is a $C^\infty$ function $H(\theta,\phi)$ such that
\begin{equation}
 h(\theta,\phi) = \phi H(\theta,\phi).
\label{key*}
\end{equation}
We have
\begin{equation}
f_d(\theta,\phi) = (g(\theta,\phi/d),dh(\theta,\phi/d)).
\label{key**}
\end{equation}
It is easy to see that as $d \to \infty$ the first component of $f_d$ converges in the $C^\infty$-topology to the function
\[  (\theta,\phi) \to g(\theta,0).\]
It follows from (\ref{key*}) and (\ref{key**}) that the second component of $f_d$ is
\[   (\theta,\phi) \to \phi H(\theta,\phi/d),\]
which converges in the $C^\infty$-topology to the function
\[   (\theta,\phi) \to \phi H(\theta,0) = \phi \frac{\partial h}{\partial \phi}(\theta,0).\]
~\hfill~ $\diamond$ \vspace{7mm}

Now observe that $k_{d}$ is the composition of the four maps  $\varphi_{d}\circ\psi_{i}\circ\varphi^{-1}_{d}$, $i=1,\dots,4$.
It is easily seen from (\ref{cylinder}) that for $i = 1$ and $3$
\[\varphi_{d}\circ\psi_{i}\circ\varphi^{-1}_{d}(\theta,\phi)=(\theta+\frac{d}{\rho}\tan\frac{\phi}{d},\phi),\]
which converges in the $C^{\infty}$ topology to the map
\[(\theta,\phi)\rightarrow (\theta+\frac{\phi}{\rho},\phi).\]

It follows from equations (\ref{minus}), (\ref{aa}) and Lemma \ref{technical} that for $i = 2$ and $4$ the map $\varphi_{d}\circ\psi_{i}\circ\varphi^{-1}_{d}$ converges to the map
\begin{equation}
(\theta,\phi)\to (\theta+\pi,-\phi).
\label{eins}
\end{equation}

A simple calculation now shows that as $d \to \infty$ the map $k_d$ converges in the $C^\infty$ topology to the map
\[ k_\infty  (\theta,\phi) = ( \theta + \frac{2\phi}{\rho} , \phi ). \]

Observe that $k_\infty$ is a nondegenerate twist map. By Moser's twist theorem \cite{Mo}, there is a neighbourhood $\cal U$ of $k_\infty$ in the $C^\infty$ topology such that every map in $\cal U$ possesses a homotopically nontrivial invariant circle.

Now consider the perturbed surface.
Let $\tilde{\psi}_{4}=(\tilde{g},\tilde{h})$.
It follows immediately from Lemma \ref{technical} that $\varphi_{d}\circ\tilde{\psi_{4}}\circ\varphi^{-1}_{d}$ converges to the map
\begin{equation}
(\theta,\phi)\to (\tilde{g}(\theta,0),\phi\frac{\partial \tilde{h}}{\partial \phi}(\theta,0)).
\label{zwei}
\end{equation}

We now know that $\tilde k_d$ is the composition of $\varphi_{d}\circ\tilde\psi_{4}\circ\varphi^{-1}_{d}$ and $\varphi_{d}\circ\psi_{i}\circ\varphi^{-1}_{d}$, $i=1,2,3$, and each of these maps approaches a limit as $d \to \infty$. Hence there is a map $\tilde k_\infty$ such that $\tilde k_d$ converges to $\tilde k_\infty$ in the $C^\infty$ topology as $d \to \infty$. If the perturbation of the metric is  small enough, $\tilde\psi_4$ will be close  to $\psi_4$,  the map in (\ref{zwei}) will be close to the map in (\ref{eins}), and $\tilde k_\infty$ will be in $\cal U$. It follows that for such a perturbation, $\tilde k_d$ belongs to $\cal U$, and thus has a homotopically nontrivial invariant circle, for all large enough $d$. This completes the proof of Lemma \ref{key}.

~\hfill~ $\diamond$ \vspace{7mm}

{From} now on we shall assume that $d \geq d_0$ so the conclusion of Lemma \ref{key} holds.
Let $\tilde{C}_{d}$ denote the perturbation of $C_{d}$ we constructed above.
Observe that the invariant circle of the map $\tilde\Psi_d$ in the region $|\phi | \leq 1/d$, gives rise to an invariant torus $T_{d}$ for the geodesic flow of $\tilde{C}_{d}$.
The projection map $\pi\mid_{T_{d}} : T_{d} \to \tilde{C}_{d}$ becomes singular on two simple closed curves which project to simple closed curves $\Gamma_{P,d}$ and $\Gamma_{Q,d}$ as indicated in Figure 1.
If $d$ is large enough, $\Gamma_{P,d}$ will lie in $D$ and $\Gamma_{Q,d}$ will lie in $R$.
Henceforth we assume that all $d\geq d_{0}$ have this property.
We shall also drop the subscript $``d"$ in order to simplify the notation.

Let $\Theta: S\tilde{C}_{d} \to S\tilde{C}_{d}$ denote the flip:
\[\Theta(p,v)=(p,-v).\]
Then $\Theta T_{d}$ is also an invariant torus of the geodesic flow, and the projection map $\pi\mid_{\Theta T_{d}}$ becomes singular above the curves $\Gamma_{P}$ and $ \Gamma_{Q}$.

Let $V_{P}$ and $V_{Q}$ be the open neighborhoods of $P$ and $Q$ that are bounded by $\Gamma_{P}$ and $\Gamma_{Q}$ respectively.
Clearly, $T_{d}$ and $\Theta T_{d}$ separate the unit sphere bundle of
$\tilde{C}_{d}$ into two invariant sets, $W_{1}$ and $W_{2}$.
One of them, which we shall call $W_{1}$, contains $S(V_{P}\cup V_{Q})$.
If $v$ is in $W_{1}$ the geodesic $\gamma_{v}$ oscillates between $D$ and $R$; in particular $\gamma_{v}$ enters both $D$ and $R$.
It is clear from this that the set $W_{2}$ contains the unit vectors tangent to $\gamma_{0}$, and thus the transverse homoclinic orbit associated with $\gamma_{0}$.

\begin{Lemma} There exists $t_{0}>0$, such that if $\gamma$ is any geodesic leaving from a point in $V_{P}\cup V_{Q}$, then $\gamma$ spends time at most $t_{0}$ inside $D\cup R$ (in each passage). The time $t_0$ is independent of $d$. \label{time}
\end{Lemma}

\PROOF The tangent vectors to $\gamma$ belong to the invariant set $W_1$ defined above.
Observe that $\overline{W}_1$ is compact and does not contain the tangent vectors to $\alpha$ and $\beta$, the boundaries of $R$ and $D$ respectively.

Both regions $R$ and $D$ have the property that their boundary is a closed geodesic and that any geodesic which crosses the boundary and enters the region must exit after finite time.
Since geodesics always cross transversally, it is easy to see that the time between entry and exit is a continuous function of the tangent vector at the time of entry.
Let $t_{0}$ be the supremum of the longest geodesic segment in $D\cup R$ that is part of a geodesic whose tangent vectors are in $\overline{W}_1$.
It is easy to see from the above that $t_{0}<\infty$ and each passage of $\gamma$ through $D\cup R$ takes time at most $t_{0}$.

~\hfill~ $\diamond$ \vspace{7mm}

\begin{Lemma}Given $\varepsilon>0$, there exists $d(\varepsilon)$ such that for $d>d(\varepsilon)$, the geodesic flow of $\tilde{C}_{d}$ satisfies:
\[h_{top}({W_{1}})<\varepsilon.\]  \label{entropy}
\end{Lemma}

\PROOF On account of the variational principle for entropy and the relationship between Liapunov exponents and entropy, it suffices to show that given $\varepsilon>0$, there exists $d(\varepsilon)$ such that for $d>d(\varepsilon)$ and all $v\in W_{1}$ we have:
\[\limsup_{t\rightarrow +\infty}\frac{1}{t}\log \parallel d\phi_{t}(v)\parallel<\varepsilon.\]

We follow an argument of Manning \cite{Manning}.
Let $K$ denote the Gaussian curvature of $\tilde{C}_{d}$ and let $Y$ denote any Jacobi field along the geodesic $\gamma_{v}$ defined by $v$.
Also let $L$ be an upper bound for $ K^2$ in the region $R\cup D$. We may assume that $L > 1$ and $\varepsilon < 1$.

Consider the function:
\[ y_{\varepsilon}(t) \df \varepsilon^{2} \langle  Y(t),Y(t) \rangle  + \langle  Y'(t),Y'(t) \rangle. \]
Using the Jacobi equation we deduce:
\[ y_{\varepsilon}'(t) = 2\varepsilon^{2} \langle  Y'(t),Y(t)\rangle  + 2\langle  -K(t)Y(t),Y'(t) \rangle. \]
Let  $t$ be such  that $\gamma_{v}(t)\in R\cup D$. Since $\varepsilon < L$, we have
\[ |y_{\varepsilon}'(t)| \leq 2\varepsilon^{2} \| Y'(t)\|\| Y(t)\| + 2L^{2}\| Y'(t)\|\| Y(t)\| \leq 4L^{2}\| Y'(t)\|\| Y(t)\| \]
\[ = \frac{2L^{2}}{\varepsilon} (2\varepsilon \| Y'(t)\|\| Y(t)\|) \leq \frac{2L^{2}}{\varepsilon} (\varepsilon^{2} \langle  Y(t),Y(t)\rangle  + \langle  Y'(t),Y'(t) \rangle) \]
\begin{equation}
= \frac{2L^{2}}{\varepsilon}y_{\varepsilon}(t).  \label{cinco}
\end{equation}

On the other hand, if $t$ is such that $\gamma_{v}(t)$ belongs to the flat cylinder, we have:
\[  y_{\varepsilon}'(t) = 2\varepsilon^{2}\langle  Y'(t),Y(t) \rangle  , \]
and thus
\begin{equation}
 |y_{\varepsilon}'(t)| \leq 2\varepsilon^{2}\| Y'(t)\|\| Y(t)\|
\leq \varepsilon(\varepsilon^{2} \langle  Y(t),Y(t) \rangle  + \langle  Y'(t),Y'(t) \rangle) = \varepsilon y_{\varepsilon}(t).  \label{seis}
\end{equation}
Next observe that
\[ \limsup_{t\rightarrow + \infty}\frac{1}{t}\log \| d\phi_{t}(v)\| \leq \limsup_{t \to +\infty} \frac{1}{t} \log\sqrt{\frac{y_{\varepsilon}(t)}{\varepsilon^{2}}}\]
\begin{equation}
=\limsup_{t\rightarrow+\infty}\frac{1}{2t}\log y_{\varepsilon}(t)=\limsup_{t\rightarrow+\infty}\frac{1}{2t}\int_{0}^{t}\frac{y_{\varepsilon}'(s)}{y_{\varepsilon}(s)}\,ds.  \label{siete}
\end{equation}

Recall that the geodesic $\gamma_{v}$ oscillates between $D$ and $R$.
In each passage through the flat cylinder between $D$ and $R$, $\gamma_{v}$ spends time at least $d$ and by Lemma \ref{time}, it spends time at most $t_{0}$ during each visit to $R\cup D$.
By combining equations (\ref{cinco}), (\ref{seis}) and (\ref{siete}) we deduce:
\[\limsup_{t \to +\infty}\frac{1}{t}\log \| d\phi_{t}(v) \| \leq \frac{L^{2}t_{0}}{d\varepsilon} + \frac{\varepsilon}{2}.\]
If we take $d > d(\varepsilon) \df 2L^{2}t_{0} / \varepsilon^{2}$, we obtain
\[ \limsup_{t \to +\infty}\frac{1}{t}\log \| d\phi_{t}(v)\| < \varepsilon. \]

~\hfill~ $\diamond$ \vspace{7mm}

We are ready to prove the main result of this section:

\begin{Theorem}For all $d$ sufficiently large and $p\in V_{P}\cup V_{Q}$,
\[\limsup_{T\rightarrow\infty}\frac{1}{T}\log n_{T}(p,q)\leq
\limsup_{T\rightarrow\infty}\frac{1}{T}\log\int_{\tilde{C}_{d}}n_{T}(p,q)\,dq< h_{top},\]
for a.e. $q\in \tilde{C}_{d}$.
\end{Theorem}

\PROOF Since $\pi^{-1}(V_{P}\cup V_{Q}) \subset W_1$, we see from  Proposition \ref{p1} that it suffices to show that for all $d$ sufficiently large,
\[h_{top}({W_{1}})<h_{top}.\]

Let $h$ denote the entropy of the horseshoe associated with $\gamma_{0}$; obviously $h$ is independent of $d$ and $h_{top}(\phi_{t})\geq h$.
The theorem now follows from the last lemma.

~\hfill~ $\diamond$ \vspace{7mm}

This theorem gives negative answers to Questions I, I$'$ and I$''$ in the Introduction.

Now observe that the tori $T_d$ and $\Theta T_d$ that were considered above will survive under any sufficiently small (in the $C^\infty$ topology) perturbation of the metric that we constructed above. These tori will still separate the unit tangent bundle and the conclusions of Lemma 4.4 and Theorem 4.5 carry over. It is clear from this that Question II in the introduction has a negative answer.

Finally we observe that since $h_{top}(S_p) \leq h_{top}(W_1)$ for any $p \in V_p \cup V_q$, we have $h_{top}(S_p) < h_{top}$ for any $p \in V_p \cup V_q$.

redpajama_set_name: RedPajamaArXiv


You will most likely be tested for Group B Strep at the end of your pregnancy, typically between weeks 34-36. If you are positive for GBS during that test you will most like receive antibiotics in labor via your IV or saline lock. You may also be given antibiotics if you have given birth to a baby who contracted GBS.
In some birth situation, you may be given antibiotics just in case, an example might be for a cesarean section or c-section. Since the risk of infection is increased due to the nature of surgical birth, an antibiotic is given to help prevent that infection. There may be other reasons that you need antibiotics as well, talk to your doctor or midwife for a discussion given your personal medical history. It was once proposed that all women be given antibiotics at the end of their pregnancy to prevent some types of infections. There is no data showing this is beneficial and may actually cause harm, so this is not what we mean with prophylactic. It is also not beneficial to give antibiotics to every mother before breaking her water to try to prevent chorioamnionitis.
If you are in preterm labor, it may be caused by an infection. Because of this fact, using antibiotics may stop your labor or at least help with the infection. Even if you are unable to stop the labor, you can help reduce the risks of infection to you.
Typically antibiotics are given after 18 hours of your membranes (water) being ruptured to try to prevent a uterine infection or chorioamnionitis (an infection of the amniotic sac) if your Group B Strep status is unknown. This may be sooner or later depending on your hospital or practitioner's protocol. This is true if your water broke on its own or if it was done as a way to induce labor or speed labor up.
If you are running a fever in labor, they may give you antibiotics, even before they know why you are running a fever to try to prevent transmission of whatever you may have to the baby. This depends on your medical history and the protocol at your place of birth and with your practitioners.
It used to be that if you had a mitral valve prolapse (MVP) you would be asked to do prophylactic antibiotics to prevent heart problems. It was also used in cases of rheumatic heart disease. This is no longer considered the protocol.
Gabbe, S, Niebyl, J, Simpson, JL. Obstetrics: Normal and Problem Pregnancies. Fifth Edition.
Ray A, Ray S. Cochrane Database Syst Rev. 2014 Oct 1;10:CD010626. doi: 10.1002/14651858.CD010626.pub2. Antibiotics prior to amniotomy for reducing infectious morbidity in mother and infant.
Thinkhamrop J, Hofmeyr GJ, Adetoro O, Lumbiganon P, Ota E. Cochrane Database Syst Rev. 2015 Jun 20;6: CD002250. doi: 10.1002/14651858.CD002250.pub3. Antibiotic prophylaxis during the second and third trimester to reduce adverse pregnancy outcomes and morbidity.


Envox Worldwide, a provider of IP-based voice solutions, has announced a partnership with Carousel Industries, a communication systems integrator and reseller that specializes in the design, implementation, and support of integrated communication solutions. Under the terms of this partnership, Carousel will develop and deploy voice solutions based on the entire Envox Worldwide product portfolio. Together, the two companies will work to ensure that enterprise contact centers can optimize their efficiency.
The Envox Worldwide portfolio includes: the Envox Communications Development Platform, an open, standards-based platform with integrated application development and management components; Envox VoiceXML Studio, a graphical development environment for the creation of VoiceXML-based voice solutions; Envox CT Connect, standards-based CTI software designed to allow application developers and systems integrators to create voice self-service, contact center, and unified communications solutions for IP and TDM networks; Envox CT ADE, a collection of utilities designed to enable developers to create and deploy CT applications; and Envox OnDemand, a speech-enabled contact center voice service for enterprises and service providers.


* Balloon can be inflated with helium or balloon pump.
* Please note when using helium confetti can pool at the base of the balloon. To get confetti to stick to sides of the balloon- rub hands around the perimeter.
Colors not perfect for your event? Check out our Custom confetti option for the perfect blend of confetti.


Dr. Fiona Robertson is retiring from Seattle University this year after 27 years of service. She has always been known as a high energy and demanding professor and has been a favorite instructor of many students over the years. She frequently taught the courses in financial markets and institutions and risk management. As a teacher, Fiona was one of the first in Albers to incorporate service learning into her classes since she could see how it aligned with the SU mission.
Since 2006, Fiona has served as Chair of the Department of Finance and presided over several US News and World Report Top 25 rankings and the establishment of distinctive programs such as the Redhawk Fund, in which students manage over $450,000 of university endowment funds. The position of department chair is not an easy one, as the chair has to represent the department to the dean and the dean to the department, and Fiona has done a fine job of striking a good balance between the two. I really appreciate her willingness to take on this role for the past eight and half years! Her service has not just been limited to her role as department chair, but over the years she has been very active in the Colleagues program and in representing Albers in the NAEF Scholarship process.
Congratulations to Fiona on her 27 years of dedicated service to SU and Albers. Through the many students she has taught, she has created an impressive legacy to take pride in. She will be dearly missed by her colleagues in Albers and around the SU campus.


Why choose us for our flexible service offering?
Total Gas & Power have been working with brokers and consultants to support the gas and electricity needs of large UK customers since 1987. As a result, we have developed a team of highly experienced sales and account managers in our Indirect Team to help Partners secure the best products for larger customers – from fixed to fully flexible. We are also able to offer basket deals for groups of smaller customers when appropriate.
If you are interested in becoming a Partner of our Indirect Team, please complete the enquiry form. We will be in touch within 3-5 working days to discuss what we require from a potential Partner. Assuming all criteria are met and we have an agreement in place, we need the Partner to sign our Code of Conduct which includes all details including initial payment terms. Following documents being signed, there is a 90 day trial put in place, and after 90 days a review will take place with the new Partner to discuss performance, and gain feedback on the initial service received from Total Gas & Power.


If upon arriving at work this morning, someone asked you how your day was going, and your response was, "Well, I'm breathing aren't I?"
That may be a sign that things in your life are out of balance. Our breath is actually a great indicator of our current mood, as well as a key to discovering clues about our, overall, emotional and physical health. Below you will find out what specific types of breath have to say about you and what you can do about it.
When we catch ourselves taking short, shallow chestal breaths, there is probably a good chance that we are experiencing a good amount of stress in our professional or personal lives. While this will always be true to some degree, it is especially important to remedy this type of irregular breathing. The problem here, is that the primitive portion of our brain has not completely evolved to domestic life. It does not discriminate between the stress of actual, physical danger versus the type of stress we feel before going for a big job interview. Rapid, shallow breathing sends a message to our brain that we are in distress and calls in for back up; releasing the hormones and neurotransmitters epinephrine, adrenaline and cortisol. If we were in eminent danger, these might be life-saving. But, in daily life, where many of the stressors we deal with are permanent fixtures, having our bodies flooded with large amounts of these chemicals poses serious health risks and can lead to issues like Adrenal Burn-out and Chronic Fatigue.
There are a variety of breathing techniques available that easily fall under the category of Breath Therapy. Most are very easy to follow and come stamped with expert approval. By training our minds to become conscious and active toward the usually passive process of respiration, we now have the ability to slow everything down and breathe a little deeper. Here are a couple of links for some great exercises to get you started.
If, at any point, rapid, shallow breathing persists or falls out of the range that could reasonably be considered as stress-related, this may be caused by serious medical conditions, not limited to Asthma, COPD, Emphysema, Pneumonia, Lung Infections or partial lung collapse. In this case, please consult your physician.
In the event of sour, rancid breath and a possible bacterial infection, to prevent your friends from dropping like the flies that are beginning to gather around you, it would be best to consult with your medical physician as soon as possible.
There are many potential factors that can affect the quality and rhythm of our breath. The key is to be mindful of any changes and to seek help when there is cause for concern. When issues and disturbances are within normal parameters, we can employ these tools to help us get our breath and our sniffer, back up to snuff.
The information provided herein should not be used during any medical emergency or for the diagnosis or treatment of any medical condition. A licensed physician should be consulted for diagnosis and treatment of any and all medical conditions.


text:
Watches of Switzerland valuation hits £647m after IPO
Barbara Santamaria
today May 30, 2019
Watches of Switzerland, the British retail group behind luxury chains including Mappin & Webb and Goldsmiths, made its debut on the London Stock Exchange on Thursday with shares rising by 15% during the morning.
The company had set a share price at 270p each, valuing the business at £647 million on its stock market debut. The stock started trading at 295p and rose to as much as 315p on Thursday morning.
The group said 34% of its share capital was floated immediately in the initial public offering under the symbol WOSG. A further 8 million shares could be made available by main shareholder Jewel HoldCo S.à.r.l. if it decides to pursue its so-called over-allotment option, which would increase the total of shares being traded to 37.4% of the company.
"Today marks the next phase in our growth story," Brian Duffy, group CEO, said: "I am delighted by the reaction we have received from the market to our business and the significant opportunities that lie ahead.
"On behalf of all our colleagues, I look forward to welcoming our new shareholders to the business and to working with them as we continue to execute on our growth strategy."
The group, which sells luxury watch brands such as Rolex, Cartier and Omega, has previously stated that the money raised from the float will be used to cut debt and fund its expansion in the US and UK.
In the year to 27 January, the business made revenues of £773 million, up 22.5% on the previous year. UK like-for-like sales grew 10% during the period.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: JNA - Ubuntu - XGetInputFocus OS: Ubuntu 16.04
JNA: 4.2.X
JDK: 1.8.0u111
I'm trying to get the currently focused window programmaticaly on a JavaFX application.
if (Platform.isLinux()) {
    final X11 x11 = X11.INSTANCE;
    final XLib xlib = XLib.INSTANCE;

    X11.Display display = x11.XOpenDisplay(null);
    X11.Window window = new X11.Window();
    Pointer pointer = Pointer.NULL;

    xlib.XGetInputFocus(display, window, pointer); // << ERROR
    X11.XTextProperty name = new X11.XTextProperty();
    x11.XGetWMName(display, window, name);
    System.out.println(name.toString());
}

public interface XLib extends StdCallLibrary {

    XLib INSTANCE = (XLib) Native.loadLibrary("/usr/lib/x86_64-linux-gnu/libX11.so", XLib.class);

    int XGetInputFocus(X11.Display display, X11.Window focus_return, Pointer revert_to_return);
}

But it doesn't work and throws this exception :
java.lang.IllegalArgumentException: Unrecognized calling convention: 63
at com.sun.jna.Native.invokeInt(Native Method)
at com.sun.jna.Function.invoke(Function.java:390)
at com.sun.jna.Function.invoke(Function.java:323)
at com.sun.jna.Library$Handler.invoke(Library.java:236)
at com.sun.proxy.$Proxy1.XGetInputFocus(Unknown Source)
at application.Main.start(Main.java:33)

Is this line correct ?
XLib INSTANCE = (XLib) Native.loadLibrary("/usr/lib/x86_64-linux-gnu/libX11.so", XLib.class);

I tested with an older version of JNA (4.1.X) and the error changed for :
Unrecognized calling convention: 1

Debug log with -Djna.debug_load=true
Looking in classpath from sun.misc.Launcher$AppClassLoader@73d16e93 for /com/sun/jna/linux-x86-64/libjnidispatch.so
Found library resource at jar:file:/home/puglic/eclipse/jna-4.2.2.jar!/com/sun/jna/linux-x86-64/libjnidispatch.so
Looking for library 'X11'
Adding paths from jna.library.path: null
Trying libX11.so
Found library 'X11' at libX11.so
Looking for library 'X11'
Adding paths from jna.library.path: null
Trying libX11.so
Found library 'X11' at libX11.so
java.lang.IllegalArgumentException: Unrecognized calling convention: 63

A: Your XLib definition uses StdCallLibrary, which only makes sense on 32-bit windows systems.  It should simply be Library, or derive from the JNA contrib-defined version.
You're basically asking for a calling convention that does not exist.

redpajama_set_name: RedPajamaStackExchange


The Bonneville County Sheriff's Office says Leah Ball has been located and is safe. No further details were released.
The Bonneville County Sheriff's Office is seeking the public's help in locating 33-year-old Leah J. Ball of Idaho Falls. Leah's family reported her as missing earlier today after not hearing from her since Sunday, April 14.
Ball was last seen Sunday evening at Buffalo Wild Wings in Idaho Falls and her vehicle was located near the Shilo Inn on Lindsay Blvd. Family has not been able to contact her and has been worried about her wellbeing.
Leah is 5′ 08″ tall with red hair and blue eyes. It is unknown what she was last wearing.
Anyone with information on the welfare of Leah Ball is asked to contact the Sheriff's Office via dispatch at (208) 529-1200.


text:
Home Politics How Hillary Must Change, for Us
How Hillary Must Change, for Us
If Hillary Clinton does indeed become America's next president, she must change the course she has been traveling in three major areas: (1) an economy benefiting Wall Street's wolves, (2) endless war which could lead to world war, and (3) pushing for a police state. Her history shows she won't want to change, in fact may feel she can't change in this oligarchy called America. But she must for us, the people. And only you, the voter, can make her change.
The Economy Benefiting Wall Street's Wolves
The millionaire candidate has reaped millions financially from Wall Street, including obscene sums for speeches to big-money audiences, and she won't reveal those speech transcripts. It's difficult to see her stepping away from the Wall Street wolves and their money factory, but she must break from them if she is going to govern and care for the American people.
This includes keeping her husband, former President Bill Clinton, away from any say involving the economy. She said during her summer campaign she was going to rely on him on economic matters. That would be a mistake.
Why? Because it was President Clinton's ties to Wall Street which led him to sign the two harmful pieces of legislation which led to the 2008 global meltdown: the Gramm-Leach-Bliley Act, which scrapped the economically vital Glass-Steagall Act; and the Commodity Futures Modernization Act of 2000 which quelled federal and state governments from regulating the $600 trillion over-the-counter derivatives industry.
Keep Bill away from the economy
The Glass-Steagall Act had been created following the Great Depression. It separated commercial banks, which handled checking and savings accounts and home mortgages, from the Wall Street wolves and their investment gambling which led to the Depression. After signing Gramm-Leach-Bliley, Clinton stated that Glass-Steagall "is no longer relevant." He couldn't have been more wrong.
Wall Street continues sucking trillions through derivatives. And it also has returned in recent years to the parasitic practice of subprime loans. Subprime mortgages helped cause the 2008 meltdown. Now Wall Street is not only involved in subprime mortgages, but has spread into subprime small-business loans, subprime auto loans, subprime credit cards, and subprime college-student loans. We've written in-depth about these issues here, here, and here.
Mrs. Clinton was a six-year member of Walmart's board of directors.
Hillary Clinton's love for Wall Street includes close ties to major corporations, such as Walmart, on whose board she served for six years. It also means her flipflop on major trade agreements like the Trans-Pacific Partnership – which she favored before her candidacy, then has opposed as a candidate – will see her flip again as president and support the TPP and TTIP with Europe. Critics shout that both these secret trade agreements are more the creation of corporate forts than trade pacts, giving corporations powers that transcend nations' powers.
You might notice that the National Democratic Party in its platform did not include opposition to TPP, a clear sign of where Mrs. Clinton is going. We've written about TPP and TTIP here and here.
Endless War Leading to World War
We've written in detail about President Obama's continuum of endless war ranging from killing innocents with drones to expanding previous presidents' efforts at world rule. See our column "Obama Widens Carter's, Bush's Global-Rule Policies".
Clinton wants military control of Eurasia.
Mrs. Clinton has been an ally in these global-rule policies. And she has been a stalwart supporter of Obama's "Asian Pivot," the U.S. effort to send 40% of America's military to encircle Eurasia as a threat to both Russia and China. In fact, she wrote about it in 2011 in a column headlined "America's Pacific Century".
This is a major problem. Russia and China are not Vietnam, Afghanistan, Iraq or Syria. They're ancient civilizations like the smaller nations, but they are not the smaller nations. They are major world powers who will not accept invasion nor threats of invasion and war, nor efforts to control their seas.
No doubt they've read Mrs. Clinton's column on the Asian Pivot. And both Russia and China continue to respond to it.
Russia and China have also seen Obama begin a 30-year, $1 trillion rebuilding of America's nuclear arsenal. It has led those two countries to expand their nuclear capabilities. We wrote about this in our column "Nukebuild: This Will Not End Well".
We've also written about Andrew J. Bacevich, a retired Army colonel, Vietnam veteran, military historian, academic and best-selling author who has hammered the U.S. four-decade Middle East military policy. He has expressed grave concerns about Mrs. Clinton, calling her a hawk. And when someone with Bacevich's credentials warns of the Democratic candidate's neo-con war obsession, you should listen. You can read that column here.
If you read the column, you'll also see Bacevich has no stomach for Trump either, comparing his personality to that of a "five year old".
Police-State Push
We wrote in late March of Hillary Clinton wanting to push America into a deeper police state in our column "Hillary Joins the Police-State Parade".
In her nomination acceptance speech, and other speeches, she prefers the politically acceptable term "security". But such a generality is dangerous, and can easily move to more surveillance, more militarizing of police, and less freedom for Americans.
So there it is. We see the direction Hillary Clinton has been traveling. She won't change this direction — or even if she might want to, the oligarchy, and its puppet Millionaire Congress, won't let her — unless you get organized (you can't do it alone), get educated to these issues, and get active in letting her know she needs to change course, for our own good, and the good of the world.
Of course, you could greatly help the cause at the November polls by cleaning the millionaires out of Congress and replacing them with public servants who really mirror the U.S. public, like John Adams called for in his landmark paper "Thoughts on Government".
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text:
Meet Your Elites: Dick Scaife
By The republican Mother
In my last post in this series, I delved into the background of Robert Mueller, who was educated in an exclusive prep school, and came from a family in "high" standing. While rummaging through the society page clippings of yesteryear, I found that two of the people who were in attendance of Mr. Meuller's inlaws' marriage ceremony in 1940 were Peter and William Standish. They were just boys at the time, but one, William grew up to be a federal judge for Western Pennsylvania. In the obituary for Peter, it was noted by one Richard Scaife how much fun they had playing together as boys. I'm just connecting this to show you how these people run in circles together that are not advertised by the media. I've been meaning to do a piece on Dick Scaife for awhile, but I knew it would be quite the undertaking, as no one funds things like he does. To the left, he is their George Soros, yet the vast majority of conservatives have never heard of him. He prefers it that way...
Dick Schaife is heir to the Mellon steel fortune. You may remember Andrew Mellon, Treasury Secretary during the 20s. Well, Dick is his great-nephew. The fortune came down to Scaife's mother Sarah, and upon her death, he inherited a huge fortune. His father, Alan was a member of the OSS, the forerunner of the CIA, and as researchers of the underground history of the 20th century would tell you - that is the source of so much of the foreign policy mess we're into today. Bear this in mind as we look at the organizations Mr. Scaife supports.
Now just because someone is an "Elite" doesn't mean that I want to go around gossiping about them, but I will say that Mr. Scaife hasn't had a tranquil home-life. He was expelled from Yale for drunkenness, according to his sister, their mother was a "gutter drunk", his first marriage ended in divorce, he had a falling out with his only sibling, Cordy (although they reportedly became friends again before she died), and his second marriage didn't last either. For pure entertainment, see this interview Mr. Scaife did at the Vanity Far. You WILL be entertained (catfights, dog-nappings, asparagus tongs, etc.) and will understand what a character this fellow is.
He is owner of the Pittsburgh Tribune Review, and as you can imagine has an influence on Pennsylvania politics. Wait a minute.. who is running for president that is also from Pennsylvania? That's right, Rick Santorum, former US Senator from PA, I wonder if they've met? (eyeroll) Mr. Saife also owns Newsmax, perhaps you've heard of it? He also was responsible for trying to put on that sham debate with Donald Trump.
As regular readers of my blog know, we focus on the strings between the puppet population and the puppeteers (banksters). Those strings would be the tax exempt foundations set up ingeniously by those who also set up our tax law. Mr. Scaife has a few foundations of his own. If you look them up, you can easily get a copy of the financial disclosure forms of his three foundations: The Sarah Scaife Foundation, named in honor of his mother, The Carthage Foundation, and the Allegheny Foundation. Before we begin, I want you to first keep something firmly in your mind: According to Dick Scaife, one of his grandmother's most important causes was population control and was friends with Margaret Sanger herself. In fact, about a year ago, Mr. Scaife wrote an op-ed in the Wall Street Journal, saying that conservatives should support Planned Parenthood because it does so much good. Does this explain to you why Rick Santorum took one for "the team" to fund Planned Parentood, when one of the biggest Pennsylvania money men apparently has some sort of family crusade to see Planned Parenthood succeed? So getting back to these foundations, are you ready to do this thing!!!??
The Sarah Scaife Foundation
Named in honor of his mother, who was deeply concerned about population control amongst minorities, white trash, and useless eaters, we see a jarring juxtoposition. For the interest of space, attention spans, and brevity, I'm just going to list the recipients of the largest portions of grant money. All figures come from the 2010 finanical statements, which showed the Sarah Scaife Foundation had a book value of roughly $180 million dollars.
Institute for Foreign Policy Analysis, Inc ($760,000)
I bet you've never heard of this outfit, but it employees a lot of former defense types to do a little rah! rah! over the latest defense sector products. According to their website :
The Institute for Foreign Policy Analysis is an independent, nonpartisan research organization specializing in national security, foreign policy, and defense planning issues. We help senior government policy makers, industry leaders, and officials in the public policy community make informed decisions in a dynamic and unpredictable global security environment. IFPA staff is a mix of scholars, business professionals, retired military officers, and foreign policy specialists who bring a uniquely broad array of talents and experiences to every task they undertake.
Call me nuts, but I thought that we had a defense department that handled that kind of thing for us. Turns out, that all the big wigs in the defense contracting sphere donate to this "nonprofit". You see, you get your "experts" to sign off on what you want to do beforehand and because your "experts" are nonpartisan and nonprofit, there's no way they could be biased or be giving fraudulent information. An example:
On April 8, 1992, Pfaltzgraff wrote a piece for the Wall Street Journal praising the Raytheon-produced Patriot missile, and attacking MIT professor's Dr. Theodore Postol's questioning of its effectiveness. Other members of the IFPA sung the praises of the Raytheon corporations Patriot missile as well. In the fiscal year 1991, Raytheon's Missile Systems Division had donated $60,000 to the IFPA.
Heritage Foundation ($600,000 from SS and $1,250,000 from Allegheny)
That's right, the good ol' Heritage Foundation is an operation of the Dick Scaife foundations. Now he's obviously not the only source of their funding, but is a long-time trustee, and his early hefty donations were crucial to getting it off the ground. What's sad is that pretty much all conservatives think that this is a "good, all-American organization". But just click over to their financial statement and as you look at how much they spend and the almost $200,000,000 they have in their net asset column, and wonder why they aren't more effective at getting real Constitutionalists in positions of power? Answer: They're paid NOT to. That's right, I've come to the conclusion, like former Dept. of Ed. official Charlotte Iserbyt that these organizations are MORE responsible than the leftists for the explosion of communism we've seen. They act like they're working for us, flyover country apple pie eaters, while really they're just talking our language just enough to make most conservatives, especially those that listen to Mark Levin (whom we'll get to later) think that Heritage has their back. When you look up Ron Paul in the Heritage search, you'll find nothing there put how "isolationism is wrong, so wrong". Well, Ron Paul supporters would agree with you; I don't know of a one that advocates isolationism, but in this 1984 newspeak world, the meanings of words are deliberately changed. Search the "Fed" on their website, and you'll just see tumbleweeds. But a FoxNews contributor for Heritage is on call 24/7 should anyone want to talk endlessly about some peripheral issue that is completely pointless when our nation is facing total economic ruin.
I liken the Heritage Foundation to that Kaa the snake charachter in The Jungle Book, the one that hypnotizes the prey. We're supposed to follow their talking points meanwhile the Elites take us completely over. What else explains the fact that during my brief little life, the Congress has been totally Republican twice, once also having the executive in their column, and we saw government drastically expanded? Education in America was completely nationalized (read socialized), and taxpayers began paying for everyone's pills (Medicare D, thanks to Team Player Santorum). Under the Republican's watch socialism marched forward, meanwhile the Heritage Foundation and like groups were placating us grassroots out here with inane drivel.
American Enterprise Institute ($500,00 )
Oh yes, they put on that foreign policy debate jointly with the Heritage Foundation. Their emphasis of course is a strong national defense. But be aware their true definition of national defense might go something like this:
American national defense - in order to maximize profits of our benefactors and to keep churning the vicious cycle of transferring money from the Federal Reserve Bank to large weapon-building firms through the pretense and secrecy of national security we hereby reserve the right to lie and distort about anything in the whole wide world that the average American could not possibly verify for themselves, making them putty in our psychological media matrix.
And if you think I'm the head engineer of the crazy train for saying that, please review the map that took me so long to make connecting the top ten defense contractors with every single sector of the US economy in a big way. But to give you an idea of the types of people who are plugged into AEI, here is a map I did on them that just illustrates that these are not just good ol' boys who want the best for America. These are special interests, that include big oil, big pharma, and big banksters. Are these industries inherently evil? No, but when they become so large that they no longer wish to compete by making superior products and decide to just buy legislation that shuts down all competition, we have the definition of fascism. Also note the think tank overlap. All your favorites are there: Bilderberg, RAND, Trilaterals, the World Bank and CFR.
link for American Enterprise Institute - full of the Elites you've come to know and despise
But they just want us Americans to be safe! My eye! Many of them make a pretty penny every time America goes to war, and it's a known fact that banks do best in times of war. So why not have an outfit that works overtime to scare the crap out of the American people to let them do what they want, when they want and how they want to do it overseas. If you don't agree, you must be some kind of unpatriotic scumbag. That attitude is nationalism, friends, not patriotism.
Social Philosophy and Policy Foundation ($470,000)
A program at Bowling Green State University in Ohio that funds philosophers to study public policy. Translation: they study right and wrong without that troublesome Bible doing the job for them. This outfit is what you get to do with a degree in philosophy. I'm not saying that their work is all bad, their book list looks interesting. I'm just wary of more "experts" telling us what to think about xyz.
Center for Strategic and International Studies ($335,000 from S. Scaife and $100,000 from Allegheny)
Here we have a robust group advocating global collectivism. This is a big think tank, and if you aren't convinced yet that big corporations are all about it, you really need to check out their page for giggles. This right here made me lol:
Investing in America, Contributions by the Indian IT Sector
Write your own punchline.
If you check them out for yourself, you'll find that the Peace Corps is awesome, but we need to bomb, bomb, bomb Iran for the love of all that's good and greedy. It's like a globalist collectivist bonananza over there --- rah! rah! United Nations. But what about American Exceptionalism, the focus-group tested phrase to make regular Americans sacrifice their sons to gain global hegemony? Hey, if that's what the banksters have to sell to get you to buy their crap, then so be it.
Intercollegiate Studies Institute, ISI ($300,000)
This is a conservative college group that does the rah! rah! American exceptionalism thing combined with paying homage to good ol' Western Civ. What's your deal repubmom? Are you anti-American and anti Western Civilization? Yes, that's right, you've got me now.. Actually I am so pro-America that I'm taking a good deal of time going through this so you can realize that this is all a big show. If they really cared about America and progressives, then they would be using all these resources to call bs on the whole WWF politics and find out where the Fed money has disappeared to exactly, for instance. Let's find out why our elected officials don't represent us anymore. Ok, calming down..
One thing that made me crack up about this group was that in one of their journals, they had an article about forming the counter-establishment. And then (giggle, hee hee) on the back of the journal is a list of the trustees, which include Ed Meese, Richard DeVos (Amway fame), and others that are so Establishment it ain't funny. Oh, you can also buy Rick Santorum's book, It Takes a Family, through their store. So what we've got here is a group that is indoctrinating a group of young people to form the conservative antithesis to the progressive thesis. Meanwhile, total collectivism (synthesis) marches forward.
Other Groups of Note:
Landmark Legal Foundation - where Mark "the grate one" Levin has his Headquarters. How does Mark feel about his benefactor donating and advocating for Planned Parenthood and for globalist groups like the CSIS who advocate global inter-dependency? How the heck does that fit into the America is the awesomest country ever meme? (Donation amount $250,000)
Stanford University, Hoover Institution on War, Revolution and Peace - where Antony Sutton got his start investigating technology transfers between the US and USSR. He was drummed out of academia for exposing the fact that the source for all this evil was financed right here in the good ol' USA. It currently seems to be a mixed bag supporting some things conservatives would agree on, with a hint of globalism. (Donation amount $300,000)
Judicial Watch Inc ($175,000 and $67,000 through Allegheny)
David Horowitz Freedom Center ($250,000 from S. Scaife and $200,000 from Allegheny)
FreedomWorks Foundation ($125,000) Astroturfing the Tea Party
Foundation for the Defense of Democracies ($200,000) --all your fav. neocons are there in the war drum-line as it were..
The Carthage Foundation
This one is a smaller outfit, but the largest grantees are:
The Maldon Institute, Inc. ($200,000) Their mission statement:
The Maldon Institute's (TMI) mission is to provide and publish information on matters ignored or misrepresented by the media, or that is believed will amplify the knowledge necessary to understand a variety of international and domestic press and government reports. TMI seeks to publish truthful information where false reporting may have contributed to our loss of tranquility.
Sounds good. What's the catch? They do data-mining, but for whom? hmmm.
The Institute on Religion and Democracy, Inc. ($100,000) First off, this IS NOT a democracy. Sheesh! This seems to be a group reaching out to church members who have "had it" with liberalism in their church, but very subtly downplays the Just War Theory that has guided Christians for ages.
Allegheny Institute for Public Policy ($115,000 and $125,000 from Allegheny) seems like a good group, centered on Pittsburgh area politics, with a lot of good libertarian books in their bookstore, but something here smacks of public-private partnerships (the code word for fascism). Let's look at the mission statement:
The Allegheny Institute is a non-profit research and education organization. Our mission is to defend the interests of taxpayers, citizens and businesses against an increasingly burdensome and intrusive government. To that end, we will formulate and advocate public policies that roll back the size and scope of local government as well as create a more accountable government. Our efforts will be guided by the principles of free enterprise, property rights, civil society and individual freedom that are the bedrock upon which this nation was founded.
I'm all about limiting government, but I hope they don't mean rolling back the representation of local government. That recently happened in Knoxville where each county commission district went from two to one representative, cutting our county commission in half. But anyway, remember what Frederic C. Howe said, the real money is in going political, and I wouldn't be surprised if that dynamic is in play with this group.
Allegheny Foundation
This one mainly has a lot benefactors local to Mr. Scaife's hometown of Pittsburgh, but there are some groups that are national and many I've already mentioned. Here are some more heavy hitters:
America's Survival, Inc. ($200,000) Hold on to your hats, kids, because this group is really interesting. It has filed a complaint with the FEC complaining of Adam Kokesh's advocacy of Ron Paul citing that because he has a show on Russia Today, and everyone knows that Russians are communists, ergo it's all a communist plot. There's also a lot on their website about the worldwide communist conspiracy. The Catholic Church is totally in on it man! This group is headed up by Cliff Kincaid, who sounds the alarm against world government through the United Nations. This is all well and good, but how do we reconcile the fact that Mr. Scaife, a major benefactor of his organization, also gives to the CSIS, which is very UN friendly. Contradiction City in my opinion. This group is also all about George Soros being the evil mastermind. Little does the average person know that a pro-abort, pro-gay marriage guy like Dick Scaife is funding this outfit, which is also very anti-gay marriage. I wonder if he and Soros hang out? That would be the ultimate irony.
Counterterrorism and Security Education and Research Foundation ($125,000) Founded in 1999 to make people aware of terrorism, they provide scholarships for those wishing to study terrorism. Yes, they also have educational programs for the general public, security professionals and military/government personnel. Quick question: Why in the sam hill would our military and government personnel, after the billions, trillions etc. the USG has spent over the past 100 years on this very issue need the help of a tax-exempt foundation to tell them about terrorism, a thousands year old tactic? Don't turn me into Big Brother for using critical thinking skills there. Gee, whiz...
Young America's Foundation - ($100,000) Home to the Reagan Ranch, where conservative kids can go to camp. That's sounds great and wholesome to the uncynical conservative out there, which would have been me a mere five years ago. The new me says, "what's the angle?" This group hosts all your fav. speakers, like Star Parker, Herman Cain, SE Cupp, and even Rudy Giuliani. It's more of Obama, Obama, Obama, the source of all things evil. Teaching point: Machiavellians like to get you to focus on one thing, and like a lightning rod that attention span God gave you gets sucking into the ground on side issues. I can bet you some FRNs that Antony Sutton's work connecting the dots between the rise of communism and fascism and central banking aint' in their bookstore. Get me? And anyone who brings up these cold hard facts, will be ridiculed. They like the talking points written for them these cookie cutter conservatives. Nothing bothers them more than thinking for themselves. That's scary. Well, it IS SCARY, I won't sugar coat if for you, you're being played, conservatives. And building yourselves idols to Ronald Reagan won't help you.
So that was just a smattering of the stuff Dick Scaife funds. The way his foundations are set up, he has to give away 5% of his foundation money per year, so he'll be giving this money to someone, but isn't it a crazy mix? It's kind of hard to find a pattern here, but I see an overarching one: death. Stay with me here- I'm about to get spiritually serious:
By the admission of his sister, Mr. Scaife comes from a pretty sad home. In fact, he demolished his childhood home brick and brick and built a new house upon it. That screams something, doesn't it? His mother advocated for eugenics, as did his grandmother. He heavily funds the purveyors of wars. When asked in the Vanity Fair interview linked above, he was asked if any book had an influence on his life. He said, no. That means that he is not a Christian, for the Christian every word in the Bible is a precious word of life. I conclude that his mother did not spend any time on her knees praying that her son would remain safe in the cleft of the Rock. She did not plead the Blood for his soul. I view Mr. Scaife as I view a child living in spiritual darkness. Instead of jungles to keep him from the Word, he grew up in an encasement of money and ease. The result is still spiritual blindness. Jesus said it would be easier for a camel, (whose meaning in the Aramaic is actually a huge rope used on boats) to pass through the eye of a needle than for a rich man to enter heaven. Why? Because all his necessities are taken care of, blinding him from his spiritual needs. So often we seek the Lord because of material issues before we really mature into craving our spiritual issues to be fulfilled. Remember the rich man who could not bring himself to give away all that he had to follow the Lord, which would save his very soul. Remember that Satan tempted Jesus with everything in the world, no wonder then that so many of the world's wealthiest people seem to work against Christianity at every turn. We see this today in our "War on Terror", it is always the Christian minority in these countries that suffer. And Mr. Scaife's many fear factories will use this fact to solicit support from Christians to continue the cycle of violence. Wars and rumors of wars, yet another sign the Lord Jesus gave us about the end times. Doesn't it fit that these wars are financed and promoted by the moneychangers? Remember that this is the only group of people the Lord Jesus went violent on during his earthly ministry. Keep that in your hat and know that Mystery Babylon is built upon moneychanging and war.
Doing this post really made me feel sorry for this man, who needs prayer. Jesus said bless and not curse. How do you fight fire? With water or an oxygen-extinguishing foam. You fight evil with its opposite. Like in the David Wilkerson story, when the hood said, I'll cut you into a million pieces. Pastor Wilkerson replied, and every piece will cry out "Jesus Loves You!" That hood is an evangelist for the Lord today, because the gentleness of good outlasts the violence of evil.
Cross Posted at The Classic Liberal, Don't Tread on Us, and The republican Mother
They Say/We Say March 23, 2012 at 9:48 PM
Someone told me as a young one, wars ain't nothin but profit and population control.
Then one day I was thumbing through the encyclopaedia, and came across a artical that said the early tribes-maybe greek, council-ed around the campfire and some smaller clans said that they would leave because of over population. Then when the over population thing came up again, some of the council met in secret and decided that they would Raid the tribes and would kill off many and Keep Their Possession. The Birth of THE SPOILS OF WAR And POPULATION CONTROL. Move along nothing new here.
We can not fight this--the Bible says so.
BUT WE CAN use the G_d given Constitution and Bill Of Rights to Slow them down and Make a Mockery of them to the World, maybe not the world - but HERE in The Good 'Ole USA!
AT AMAZON, Deal of the Day, Ivation 70 Pint Energy Star Compressor Dehumidifier with Pump, Large Cap…
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Let's hear it for the Mac.
There's been plenty of ink shed on the meaning and impact of iPhone and iPad sales in Apple's most recent quarters. In and of itself, that's no surprise: The iPhone is the company's biggest product and makes up more than two-thirds of its revenue. The iPad, on the other hand, has been struggling, unlike most of Apple's other devices.
But through it all, the Mac has quietly kept on doing steady business, and I think the sturdy workhorse of Apple's lineup deserves some accolades. Not only has the Mac entered its fourth decade—impressive for any piece of technology—but it's seen tremendous success and even growth in an era where all anybody can talk about is smartphone this and tablet that.


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