text:
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redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}

Let $(T^2,g)$ be a two-dimensional Riemannian torus. By $c_v$ we denote the
unique geodesic $c_v:\Br \to T^2$ with the initial condition $\dot c_v(0)=v \in ST^2$.
The geodesic flow on the unit tangent bundle $ST^2$ is given by $\phi^t(v)=\dot c_v(t)$.
We also consider geodesics on $\Br^2$, where $\Br^2$ is equipped with the lifted metric.
The topological entropy  of a continuous dynamical system represents the exponential
growth rate of orbits segments distinguishable with arbitrarily fine but finite precision.
It therefore describes the total exponential orbit complexity by a single number.
Note that due to a theorem of A. Katok \cite{K80} for $C^{1+ \alpha}$-flows, $\alpha>0$,
on 3-dimensional spaces positive topological entropy and the existence of a horse-shoe
are equivalent. We consider in this paper the questions which consequences on the behavior
of geodesics on $T^2$ we can expect under the assumption of zero topological entropy and which
geometrical restrictions on their behavior forbid high complexity of the geodesic flow.
More precisely, we formulate necessary and sufficient conditions for zero topological
entropy.
In order to state the main theorems we have to define the asymptotic direction of
geodesic rays, i.e.\ geodesics  $c:[0, \infty) \to \Br^2.$ Note that unlike in most
text books in our context a geodesic ray does not have to be minimal.
Given a geodesic $c:\Br \to \Br^2$ then we associate to $c$ two geodesic rays given
by $c^+:=c|_{[0, \infty)}: [0, \infty) \to \Br^2$ and
$c^-:[0, \infty) \to \Br^2$ with $c^-(t):=c(-t)$.\\
\\
In the following it will be useful to distinguish the following types of geodesic rays
(see also \cite{W}).\\

\begin{dfn} [types of geodesic rays] ~\\
Given a complete Riemannian metric on $\Br^2$. A geodesic ray $c: [0, \infty) \to \Br^2$
is called
\begin{itemize}
\item
{\it bounded}, if the set $c([0, \infty))$ is bounded.
\item
{\it escaping}, if $\lim\limits_{t \to \infty} \|c(t)\|= \infty$, where $\| \;\cdot \;  \|$
denotes the Euclidean norm  on $\Br^2$.
\item
{\it oscillating}, if $c$ is neither bounded nor escaping.
\end{itemize}
A geodesic $c: \Br \to \Br^2$ is bounded (escaping or oscillating) if both of its geodesic
rays $c^+$ and $c^-$ are bounded (escaping or oscillating).
\end{dfn}

\begin{remo}
It will turn out that mixed cases of geodesics will not be relevant.
Furthermore, their Liouville measure is zero, as proved by M. Wojtkowski in \cite{W}.\\
\end{remo}

\begin{dfn}[asymptotic direction and rotation number] ~\\
Let $c : [0, \infty) \rightarrow \Br^2$ be an escaping geodesic ray on $\Br^2$.
Then, if the limit exists,
\begin{align*}
    \delta(c):= \lim_{t \rightarrow \infty} \frac{c(t)}{\|c(t)\|} \in S^1 \enspace \quad
\end{align*}
is called the {\it asymptotic direction} of $c$.
 Let  $$\pi: S^1= \{(x,y) \mid x^2 + y^2 =1 \} \to \mathbb{P}_1(\Br) \cong \Br \cup \{\infty\},$$
 be the canonical projection onto $\mathbb{P}_1(\Br)$
 defined by
 $$
 \pi(s) = \left\{ \begin{array}{cc}
 \frac{y}{x}, & \text{if} \; x \not=0\\
 \infty, & \text{otherwise} \enspace .
 \end{array} \right.
 $$
 Then we call
\begin{align*}
    \rho(c): = \pi \circ \delta(c) \in  \mathbb{P}_1(\Br) \cong \Br \cup \{\infty\} \enspace \quad
\end{align*}
the {\it rotation number} of $c$. \\

Let $c: [0, \infty) \to T^2$ be a geodesic ray such that its lift $\tilde c: [0, \infty) \to \Br^2$
provides an escaping geodesic ray for which $\delta(\tilde c)$ exists.
 Then we define $\delta(c):=\delta(\tilde c)$ and $\rho(c):=\rho(\tilde c)$.
\\
A rotation number $\rho(c)$ is called {\it rational} if
$ \rho(c)  \in \Bq \cup \{\infty\}$. A direction $\delta(c) \in S^1$ is called {\it rational}
if $\pi(\delta(c)) \in \Bq \cup \{\infty\}$.\\
\end{dfn}

\begin{remo}~\\
\begin{enumerate}
\item[(a)]
The definition
and the existence of the asymptotic direction are independent of the chosen lift.
\item[(b)]
A first definition of a rotation number for minimal geodesics induced by the slope of the
accompanying Euclidean lines goes back to G.~A.~Hedlund (see \cite{Hedlund}) and H.~M.~Morse
(see \cite{Morse}). In~\cite{B1988} V.~Bangert presents a definition of the rotation
number for minimal geodesics, which coincides with our definition restricted to minimal geodesics.
\end{enumerate}
\end{remo}

Now we are able to state our first theorem.\\

\noindent{\bf Theorem~I.}
{\it Let $g$ be a Riemannian metric on $T^2$ with vanishing topological
entropy. Then, on the universal covering $\Br^2$, every geodesic $c$ is escaping without
self-intersections and the directions $\delta(c^+)$ and $\delta(c^-)$ exist. Furthermore,
$\delta(c^+)=-\delta(c^-)$ and $c$ lies in a bounded Euclidean strip $S(c)$, where $S(c)$
is a strip in $\Br^2$ bounded by two Euclidean lines. }\\
\\
Consider the asymptotic direction and the rotation number for a geodesic $c_v: \Br \to T^2$
as functions on the unit tangent bundle defined as $\delta(v):=\delta(c_v^+)$ and
$\rho(v):= \rho(c_v^+)$, if they exist. Obviously $\delta(c_v^-)=\delta(-v)$ and $\rho(c_v^-)=\rho(-v)$.\\

\noindent{\bf Theorem~II.}
{ \it Let $g$ be a Riemannian metric on $T^2$ with vanishing topological entropy.
Then the asymptotic direction $\delta: ST^2 \to S^1$ and
the rotation number $\rho : ST^2 \to \Br \cup \{\infty\}$
are surjective continuous functions invariant under the geodesic flow.
}\\

\begin{remo}
If the constant of motion $\rho$ would be differentiable with non-zero differential
almost everywhere, the system would be integrable in the sense of Liouville-Arnold.\\
\end{remo}

>From now on we represent $T^2$ as $\mathbb{R}^2/\mathbb{Z}^2$, where $\mathbb{Z}^2$ acts
on $\mathbb{R}^2$ via
$$
\tau_{(m,n)}(x,y) =(x+m,y+n) \enspace .
$$
The non-trivial covering transformations $\tau_{(m,n)}$ will be called translation elements.
Sometimes we will denote $\tau_{(m,n)}$ only by $\tau$.  A translation element $\tau$
is called primitive if it is not a nontrivial power of another translation element.
Two translation elements $\tau$ and $\eta$ are called equivalent if there exist
$k, \ell \in \mathbb{Z}\setminus\{0\}$ such that
$$
\tau^k=\eta^{\ell} \enspace.
$$
By $[\tau]$ we denote the corresponding equivalence classes.
\begin{dfn}
 A geodesic
$c: \Br \to \Br^2$ is called an axis if there exists a nontrivial translation element $\tau$ such
that $\tau c(t)=c(t+l)$ for some $l \in \Br$ and all $t \in \Br$. By $\delta(\tau)$ we
will denote the asymptotic direction of the axes of $\tau$. An axis $\alpha$ of a translation
element $\tau^k$ with $k \geq 2$ is called a non-primitive axis if $\alpha(\Br) \not= \tau \alpha(\Br)$.
A geodesic $c: \Br \to T^2$ is called prime-periodic if its lift is an axis of a primitive translation
element $\tau$.
\end{dfn}

\begin{dfn}
For a geodesic ray $c: [0, \infty) \to \Br^2$ we define
$$
I(c) := \{ [\tau] \mid \# \{c([0, \infty)) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \eta c([0, \infty)) \} = \infty
\text{ for some $\eta \in [\tau]$}\}.
$$
\end{dfn}
We will now present a further characterization of the behavior of geodesics on the universal
covering in the case of vanishing topological entropy. By intersections we always mean
transversal intersections.\\

\noindent{\bf Theorem~III.}
{\it Let $(T^2,g)$ be a Riemannian torus with zero topological entropy.
Then $\# I(c) \leq 1$, for each geodesic ray  $c: [0, \infty) \to \Br^2$.
Furthermore, geodesics $c$ with irrational rotation number intersect their translates
$\tau c$ only a finite number of times. Geodesics $c$ with rational rotation number
intersect their translates $\tau c$ an infinite number of times at most for $\tau$ with
$\delta(\tau)=\delta(c^+)$.
}\\
\\
In a previous version of this paper we have had a weaker formulation of the following theorem.
We originally showed applying hyperbolic dynamics and the Curve Shortening flow that the
assumption of the theorem implies zero topological entropy. V.~Bangert showed to us how to
use variational arguments to even conclude flatness. We also note that we obtained flatness
under the stronger assumption that no geodesic on the universal covering intersects its
translate transversally.\\

\noindent{\bf Theorem~IV.}
{\it Let $g$ be a Riemannian metric on $T^2$. Then flatness of the metric $g$ is equivalent
to the condition that no axis $c: \Br \to \Br^2$ on the universal covering intersects
any of its translates.}\\

\begin{remo}~\\
\begin{enumerate}
\item[1)]
The assumption of Theorem~IV is equivalent to the fact that all axes $c: \Br \to \Br^2$ are axes of
primitive elements.
\item[2)]
As the theory of minimal geodesics and the Curve Shortening flow extend to symmetric Finsler metrics,
all Theorems generalize to symmetric Finsler metrics.
\item[3)]
M.~L.~Bialy and L.~Polterovich ~\cite{BP} and independently V.~Bangert~\cite{B1988} present a
formalism for minimal geodesics on $T^2$ based on a special class of orbits for monotone twist maps.
For this setting they formulate the notion of rotation number. However, in general it is not
possible to extend their formalism to non-minimal geodesics and arbitrary orbits of twist maps. \\
The study of monotone twist maps with variational methods and many results about their
properties mentioned in \cite{B1988} and \cite{BP} go back to J.~N.~Mather \cite{Mather} and
independently to S.~Aubry and P.~Y.~Le~Daeron \cite{Aubry}.
In \cite{A-int} S.~B.~Angenent studies orbits of monotone twist maps and their intersection
properties in the case that these maps have vanishing topological entropy. He formulates
analogious results to parts of the statements of Theorem~I and Theorem~III.
\item[4)]
A different approach to understand the relation between the complexity of the geodesic flow
and the behavior of geodesics similar to the one in this paper is presented by S.~V.~Bolotin
and P.~H.~Rabinowitz in \cite{BR}.
\end{enumerate}
\end{remo}

The paper is organized as follows: In the second section we introduce the notion of the
topological entropy and discuss the for us relevant properties of the Curve Shortening
flow. In section three we show that vanishing topological entropy implies the non-existence
of self-intersections of lifted geodesics and the non-existence of contractible geodesics
on $T^2$. Section four deals with geometric conditions which imply via the Curve
Shortening flow positive topological entropy. In the case of vanishing topological entropy
we exclude the existence of oscillating geodesic rays. Under the same assumption, in section
five we prove the existence and regularity of the rotation number. In section six we study
the intersection properties of geodesics on the universal covering with their translates.
In the last section we provide a characterization for flatness of Riemannian metrics on $T^2$.
In particular, we show that no axis on the universal covering intersects its translates
iff the torus is flat.

\section{Topological entropy and curve shortening}

The topological entropy is invariant under topological conjugations and measures as
described in the introduction the exponential orbit complexity  by a single non-negative
number. The precise meaning becomes apparent in the following definition of topological
entropy introduced by R.~E.~Bowen~\cite{BO}.

\begin{dfn}[Topological entropy] \label{topent}
Let $(Y,d)$ be a compact metric space, $\phi^t: Y \to Y$ a continuous flow and
$d(\cdot,\cdot)_{T}$ the dynamical metric defined by
$d(v,w)_{T}:= \max_{0 \leq t \leq T}d(\phi^tv, \phi^tw)$ for all $v,w \in Y$.
We fix $\varepsilon>0$. A subset $F\subset Y$ is called $(\phi,\varepsilon)$-{\it separated}
set of $Y$ with respect to $T$, if for $x_1 \not=x_2 \in F$ it holds $d(x_1,x_2)_{T}> \varepsilon.$ \\
The topological entropy of $\phi^t$ is defined as
$$
h_{top}(g) = h_{top}(\phi)=\lim_{\varepsilon \to 0} \limsup_{T \to \infty}
\left(\frac{1}{T} \log r_T(\phi, \varepsilon)  \right)
. $$
Here $r_T(\phi, \varepsilon)$ denotes the maximal cardinality of any
$(\phi,\varepsilon)$-separated set of $Y$ with respect to $T$.
\end{dfn}
For more details and properties of the topological entropy see for example
\cite{KH} or \cite{Wo}.
\\
\\
To prove that certain geometric constellations imply
positive topological entropy of the geodesic flow on $ST^2$ we will
use the Curve Shortening flow on $T^2$. In the following we will give the precise
definition of this flow and state the properties relevant in this context.
\\
\\
Let $(M,g)$ be a Riemannian surface and
$$\Gamma=\{\gamma\;|\; \gamma: S^1 \to M \;\text{smooth immersed closed curve
}\}$$
a family of (parameterized) immersed smooth closed curves on $M$.
We consider a continuous local semi-flow $\Psi^t: \Gamma \to \Gamma$ with
$\Psi^t(\gamma)=: \gamma_t$ and $t \in [0, T_{\gamma})$ defined by
\begin{eqnarray} \label{csf}
\frac{\partial \gamma_t}{\partial t} = k_t N_t,
\end{eqnarray}
where $k_t$ denotes the geodesic curvature of $\gamma_t$ and $N_t$ its unit normal vector.
This evolution equation defined in~(\ref{csf}) is called the {\it Curve Shortening flow}.
\\
\\
From the following theorem we will derive the existence of closed geodesics:

\begin{theorem}[M.~A.~Grayson, see \cite{GO}]
\label{GO}
Let $M$ be a smooth Riemannian surface which is convex at infinity, i.e.\ the convex hull
of every compact subset is compact. Let $\gamma_0:S^1 \to M$ be a smooth curve, embedded
in $M$. Then, $\gamma_t:S^1 \to M$ exists for $t \in [0, T)$, for some $T>0$, satisfying
the evolution equation (\ref{csf}).\\
If $T$ is finite, then $\gamma_t$ converges to a point. If $T$ is infinite, then the
curvature of $\gamma_t$ converges to zero in the $C^{\infty}$ norm, i.e.\ there exists
a subsequence $t_n$ such that $\gamma_{t_n}$ converges to a closed geodesic.
\end{theorem}

The assumption that $M$ is convex at infinity ensures that the set of limit curves exists
since an evolving curve $\gamma_t$ cannot leave a compact set. In our special case of the
torus we will directly show that the embedded curve $\gamma_t$ stays in a compact set on
the universal covering $\Br^2$ of $T^2$ or in a compact set on some unbounded cylinder
$C$ for all $t \in [0,T)$. Then, by excluding that $\gamma_t$ converges to a point we
conclude the existence of a closed geodesic.
\\
As the length $l_t$ of the curve $\gamma_t$ fulfills $$\frac{d l_t}{dt} = - \int k_t^2(s) ds,$$
the length is a decreasing function of $t$. This is why this flow is called Curve Shortening.\\
An important fact is that embedded curves never become singular, unless they shrink
to a point, as proved by M.~A.~Grayson~\cite{GO}.

Furthermore we will apply the following consequence of the maximum-principle for parabolic
differential equations several times:

\begin{theorem}[S.~B.~Angenent, see \cite{A-II}]  \label{2seg}
Let $\gamma_0, \eta_0:[0,1] \to M$ be two curve-segments and $\{\gamma_t\;|\; 0<t<T\}$,
$\{\eta_t\;|\; 0<t<T\}$ solutions of Curve Shortening which satisfy
$$\partial \gamma_t \cap \eta_t = \partial \eta_t \cap \gamma_t= \emptyset$$
for all $t \in [0,T)$. Then the number of intersections of the solutions $\gamma_t$ and
$\eta_t$ is a finite and nonincreasing function of $t \in (0,T)$. The solutions intersect
only transversally except at a discrete set of times $\{t_j\}\subset (0,T)$, and at each
$t_j$ the number of intersections of $\gamma_t$ and $\eta_t$ decreases.
\end{theorem}

\section{Contractible Closed Geodesics and Self-intersections}

In 1998 J.~Denvir and R.~S. MacKay \cite{DM} showed the following result as a
conclusion of their study on geodesically convex surfaces of negative Euler characteristic.

\begin{theorem}\label{scc}
Let $g$ be a Riemannian metric on $T^2$ with a simple closed contractible geodesic $c$.
Then $g$ has positive topological entropy.
\end{theorem}

This theorem allows us to conclude the following lemma:

\begin{lemma}\label{2-loops}
Let $g$ be a Riemannian metric on $T^2$ and $c:\Br \to \Br^2$ a lift of a geodesic
with two self-intersections such that $c(t_1)=c(t_2)$ and $c(t_3)=c(t_4)$ with
$t_1<t_2<t_3<t_4$. Then $g$ has positive topological entropy.
\end{lemma}

\begin{proof} We use similar ideas as S.~Angenent in~\cite{A-pre}.
Let $\gamma_0:S^1 \to \Br^2$ be a simple closed contractible and smooth curve such
that $\gamma_0$ and $c([t_1, t_4])$ do not intersect on $\Br^2$ and such that
$c[t_1, t_4]$ lies in the bounded connected component of $\Br^2 \setminus \gamma_0(S^1)$.
We apply the curve-shortening flow to $\gamma_0$ and $c$. Note that $c([t_1, t_4])$
is a constant solution. Assume, that $\gamma_t$ shrinks to a point. Then the solution
$\gamma_t$ has to pass $c([t_1, t_4])$ under the flow. Consider $\tilde t \geq 0$ with
$\gamma_{\tilde t}(S^1) \cap c([t_1, t_4]) \not= \emptyset$ such that $\tilde t$ is
the first time at which the segments meet. By Theorem \ref{2seg} new intersections
under the flow can only appear in the endpoints $c(t_1)$ or $c(t_4)$. Consider a small
$\varepsilon>0$ such that $t_1 + \varepsilon < t_{2}$ and $t_4 - \varepsilon>t_3$.
As $c(t_1)=c(t_2)$ and $c(t_3)=c(t_4)$ the curve $\gamma_{\tilde t}$ meets the
geodesic segment $c([t_1 + \varepsilon, t_4 - \varepsilon])$ also at
$s \in (t_1 + \varepsilon, t_4 - \varepsilon)$, in contradiction to Theorem \ref{2seg}.
Hence, as $\gamma_t$ cannot pass $c([t_1, t_4])$ it will not shrink to a point.\\
Furthermore, $\gamma_t$ will stay in a bounded set ${\mathcal{K}} \subset \Br^2$
for all $t$: We consider two non-equivalent primitive translation elements $\tau, \eta$
and two corresponding axes $\alpha_{\tau}$ and $\alpha_{\eta}$. For $k \in \Bn$ large
enough the curve $\gamma_0$ lies in the open strip between $\tau^k \alpha_{\eta}$ and
$\tau^{-k}\alpha_{\eta}$, and in the open strip between $\eta^k \alpha_{\tau}$
and $\eta^{-k}\alpha_{\tau}$.
\begin{figure}[h!]
\begin{center}
\psfrag{a}{$\tau^k\alpha_{\eta}$}
\psfrag{b}{$\eta^k\alpha_{\tau}$}
\psfrag{d}{$\tau^{-k} \alpha_{\eta}$}
\psfrag{c}{$\eta^{-k} \alpha_{\tau}$}
\psfrag{e}{$c$}
\psfrag{K}{$\mathcal{K}$}
\psfrag{f}{$\gamma_0$}
\includegraphics[scale=0.36]{figure1.eps}
\end{center}
\caption{Illustration of the two loops of $c$ and $\gamma_0$ in the proof of Lemma~\ref{2-loops}.}
\end{figure}
The intersection ${\mathcal{K}}$ of the closures of these strips defines a region from
which $\gamma_t$ cannot escape, because since $\gamma_0$ does not intersect any of
the bounding minimal geodesics, by Theorem~\ref{2seg} $\gamma_t$ will not intersect
them for all $t>0$.\\
Now we apply M.~A.~Grayson's Theorem~\ref{GO}: Since $\gamma_t$ will not shrink to a
point and stays in a compact set for all times, there exists a simple closed geodesic
on $\Br^2$. By Theorem \ref{scc} $g$ has positive topological entropy.\\
\end{proof}

In the sequel we will make use of the following standard recurrence theorem
(see for instance \cite{Wo}, page 157).
\begin{theorem}
Let $X$ be a manifold and $\mu$ a probability measure invariant under a continuous
flow $\phi^t: X \to X$. Then almost all $p \in X$ are recurrent, i.e.\ there exists a
sequence $t_n$ with $t_n \to \infty$ as $n \to \infty$ such that
$$
\lim\limits_{n \to \infty} \phi^{t_n}p =p \enspace .
$$
Furthermore, if $p$ is recurrent all points on the orbit $\phi^t(p)$ are recurrent
as well. If $\mu$ is positive on open sets, the set of recurrent points is dense.
\end{theorem}

\begin{theorem}\label{no-intersections}
Let $g$ be a Riemannian metric on $T^2$ with zero topological entropy. Then no lift
of a geodesic has a
self-intersection.
\end{theorem}

\begin{proof}
Let $c:\Br \to \Br^2$ be a lift of a geodesic and assume that $c$ has a self-intersection.
Then there exist $t_0 < t_1 \in \Br$ with $c(t_0)= c(t_1)$. Let $\dot c(t_0)=w$ denote
the initial condition of $c$ in $t_0$. Assume first that $c$ is recurrent. Then there
exists an increasing sequence of times $(t_n)_{n \in \Bn}$ and a sequence $\tau_n$ of
translation elements such that $D\tau_n\dot c(t_n)=w_n \to w$ for $n \to \infty$, see
Figure~\ref{int}. For $n$ large enough and $t_n$ much larger than $t_1$ there exist $t$
near $t_n$ and $s>0$ such that $\tau_n c(t)=\tau_n c(t+s)$. Hence, $c(t)=c(t+s)$.
As $t_0<t_1<t<t+s$ by Lemma~\ref{2-loops} the metric $g$ has positive topological entropy,
in contradiction to the assumption.
\begin{figure}[h!]
\begin{center}
\psfrag{c}{$c$}
\psfrag{ct}{$\tau_n c$}
\psfrag{w}{$w$}
\psfrag{wn}{$w_n$}
\includegraphics[scale=0.56]{figure2.eps}
\end{center}
\caption{Illustration of $c$ and $\tau_n c$ in the proof of Theorem~\ref{no-intersections}.} \label{int}
\end{figure}
Let $c$ be an arbitrary geodesic. As recurrent geodesics are dense,
there exists a sequence of recurrent geodesics $c_n$ with $\dot c_n(t_0)=w_n$ such
that $w_n$ converges to $w$. Since $c$ has a self-intersection, i.e.\ $c(t_0)=c(t_1)$,
the continuous dependency implies that for $n$ large enough the recurrent geodesic $c_n$
has a self-intersection. Then by Lemma~\ref{2-loops} the metric $g$ has positive
topological entropy in contradiction to the assumption.
\end{proof}

The following theorem is due to V.~Bangert in~\cite{B1981}
\begin{theorem} \label{bounded}
Let $g$ be a complete Riemannian metric on $\Br^2$. Then the existence of a bounded
geodesic ray $c:[0, \infty) \to \Br^2$ implies the existence of a simple closed geodesic.
\end{theorem}

\begin{remo}
We note that this result can be also obtained combining Curve Shortening with
arguments from topological dynamics. For more details about this approach
see~\cite{Glas} where a slightly weaker result has been obtained.
\end{remo}

\begin{cor} \label{bounded2}
Let $g$ be a Riemannian metric on $T^2$. Assume there exists a bounded geodesic ray $c$
on the universal covering $\Br^2$. Then the metric $g$ has positive topological entropy.
\end{cor}

\section{Central Geometric Argument}

In this section we present a fundamental geometric constellation on the universal
covering of $T^2$ and prove that it implies positive topological entropy for the
Riemannian metric $g$ on $T^2$. In the further sections we will use this argument
several times.

\begin{dfn}
A continuous curve $c: I \to \Br^2$, for $I \subset \Br$, is called a broken geodesic
if there exists a finite set $W \subset I$ such that $c$ is geodesic on $I\setminus W$.
We call $W$ the set of vertices.
\end{dfn}

\begin{lemma}[Fundamental Lemma] \label{fundamentallemma}
Let $g$ be a Riemannian metric on $T^2$ and $\alpha: \Br \to \Br^2$ a
minimal axis of the translation element $\tau$. Let $c_1:[0, a] \to \Br^2$ and
$c_2:[0, b] \to \Br^2$ be two geodesic segments with endpoints on $\alpha$ and
$c_1((0, a)) \cap \alpha(\Br)=\emptyset$, $c_2((0, b)) \cap \alpha(\Br)=\emptyset$.
Assume that there exists a translation element $\eta$, with
$\eta\alpha(\Br) \cap \alpha(\Br)= \emptyset$ such that
$\eta \alpha(\Br) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c_1([0, a]) \not= \emptyset$ and
$\eta^{-1} \alpha(\Br) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c_2([0, b]) \not= \emptyset$. Then the metric $g$ has
positive topological entropy.
\end{lemma}

\begin{proof}
We choose $k \geq 1$ such that
$$\eta^k \alpha(\Br) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c_1([0, a]) \not= \emptyset \quad \text{but} \quad
\eta^{k+1} \alpha(\Br) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c_1([0, a]) = \emptyset.$$ Analogously we choose
$l \geq 1$ such that $$\eta^{-l} \alpha(\Br) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c_2([0, b])\not= \emptyset \quad
\text{but} \quad \eta^{-(l+1)} \alpha(\Br) \cap c_2([0, b])= \emptyset.$$
W.l.o.g.\ let $l \geq k$. Then $$\alpha(\Br) \cap \eta^{l+1} c_2([0, b])
= \emptyset \quad \text{but} \quad
\eta \alpha(\Br) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \eta^{l+1} c_2([0, b]) \not= \emptyset.$$
\begin{figure}[h!]
\begin{center}
\psfrag{a}{$\alpha$}
\psfrag{ka}{$\eta^{k} \alpha$}
\psfrag{1a}{$\eta\alpha$}
\psfrag{l1a}{$\eta^{(l+1)} \alpha$}
\psfrag{c2}{$c_1$}
\psfrag{c1}{$\eta^{(l+1)} c_2$}
\includegraphics[scale=0.3]{figure3.eps}
\end{center}
\caption{Illustration of the translated segments $c_1$ and $\eta^{(l+1)} c_2$
in the proof of the Fundamental Lemma~\ref{fundamentallemma}.}
\end{figure}
By $S$ we denote the geodesic strip bounded by $\alpha$ and $\eta^{l+1} \alpha$.
In the next step we use the segments $c_1$ and $\eta^{l+1}c_2$ to construct
suitable broken geodesics $\sigma_1$ and $\sigma_2$ which intersect. We distinguish
two cases:
\begin{enumerate}
\item[Case 1)] There exists $\tilde n \in \Bz$ such that $\tau^{\tilde n} c_1$
and $\eta^{l+1}c_2$ intersect. In this case we will denote
$\sigma_1: = \tau^{\tilde n} c_1$ and $\sigma_2: = \eta^{l+1}c_2$.
\item[Case 2)]
For all $\tilde n \in \Bz$ the segments $\tau^{\tilde n} c_1$ and
$\eta^{l+1}c_2$ do not intersect. Using $c_1$ and $\eta \alpha$ we will construct
a broken geodesic segment $\sigma_1$ with endpoints on $\alpha$ which
intersects $\eta^{l+1}c_2$, see Figure~\ref{fig-ref4}.
\begin{figure}[h!]
\begin{center}
\psfrag{a}{$\alpha$}
\psfrag{c}{$c_1$}
\psfrag{tc}{$\tau^{n} c_1$}
\psfrag{ec}{$\eta^{l+1} c_2$}
\psfrag{ka}{$\eta^{k} \alpha$}
\psfrag{1a}{$\eta\alpha$}
\psfrag{l1a}{$\eta^{l+1} \alpha$}
\includegraphics[scale=0.3]{figure4.eps}
\end{center}
\caption{The construction of intersecting broken geodesic segments $\sigma_1$
and $\sigma_2$ in the proof of the Fundamental Lemma~\ref{fundamentallemma}.} \label{fig-ref4}
\end{figure}
We choose $n \in \Bz$
such that on $\eta \alpha(\Br)$ the set of intersections
$\eta^{l+1}c_2 \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \eta \alpha(\Br)$ lies
between the finite sets
$M_+=\tau^n c_1 \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \eta \alpha(\Br)$ and
$M_-= c_1 \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \eta \alpha(\Br)$.\\
Let $\eta \alpha([t_1,t_2])$ be the smallest segment of $\eta \alpha$
such that
$$\eta \alpha([t_1,t_2]) \cap M_+ \not= \emptyset
\not= \eta \alpha([t_1,t_2]) \cap M_-.$$
Let $A$ be the union of the bounded connected components of
$$S \setminus \left(c_1([0, a])
 \cup \tau^n c_1([0, a])
 \cup \eta \alpha([t_1,t_2]) \right).$$
We consider the broken geodesic $\sigma_1([0, \tilde a])$ endowed
with a new para\-metrization consisting of $\eta \alpha([t_1,t_2])$
and segments of $\tau^n c_1$ and $c_1$, such that it separates
$\bar A$ from $S \setminus \bar A$. By construction $\sigma_1([0,\tilde a])$
and $\sigma_2([0,b]) = \eta^{l+1} c_2([0, b])$ intersect.
\end{enumerate}
Let $m \in \mathbb{N}$ be large enough such that
\begin{eqnarray} \label{m2}
(\sigma_1 \cup \sigma_2) \cap \tau^{mi}(\sigma_1 \cup \sigma_2) = \emptyset
\quad \text{for all $i \in \Bz \setminus\{0\}$.}
\end{eqnarray}
For fixed $j \in \Bn$ and a finite sequence of $j$ elements
$(a_0,...,a_{j-1})=a(j)$ with $a_k \in \{1,2 \}$ we consider
the set
$$
A(a(j)):= \bigcup\limits_{k =0}^{j-1} \tau^{mk} \sigma_{a_k}\enspace.
$$
For each $a(j)$ let $\gamma_0^{a(j)} : \mathbb{R} \to S \setminus A(a(j))$
be a smooth curve such that $\tau^{mj}\gamma_0^{a(j)}=\gamma_0^{a(j)}$,
see Figure \ref{fig-ref5}.
\begin{figure}[h]
\begin{center}
\psfrag{S}{$S$}
\psfrag{c1}{$\tau^m \sigma_1$}
\psfrag{c2}{$\tau^m \sigma_2$}
\psfrag{tc1}{$\tau^{2m}\sigma_1$}
\psfrag{tc2}{$\tau^{2m} \sigma_2$}
\psfrag{ttc1}{$\tau^{3m} \sigma_1$}
\psfrag{tal}{$\alpha$}
\psfrag{ta}[c]{$\eta^{l+1} \alpha$}
\psfrag{t2c2}{$\tau^{3m} \sigma_2$}
\psfrag{g0}{$\gamma_0^{(a)}$}
\psfrag{1}{$\gamma_0^{(a)}$}
\psfrag{a}{$\alpha$}
\psfrag{C}[c]{$C_3$}
\psfrag{ga}{$\gamma_0$}
\includegraphics[scale=0.24]{figure5.eps}
\end{center}
\caption{Simplified illustration of the translates of $\sigma_1$,
$\sigma_2$, and $\gamma_0^{(a)}$ in the proof of the Fundamental
Lemma~\ref{fundamentallemma}.} \label{fig-ref5}
\end{figure}
Furthermore, we can choose these curves such that the length of
the segment of $\gamma_0^{a(j)}$ joining  $\gamma_0^{a(j)}(0)$
and $\tau^{jm}\gamma_0^{a(j)}(0)$ is less than $jb_2$ and larger than
$jb_1$ for some universal constants $0<b_1 < b_2$. Consider for each
$j \in \Bn$ the cylinder
$$
C_j: = \Br/\langle \tau^{mj} \rangle
$$
where $\langle \tau^{jm} \rangle$ is the subgroup of $\Bz^2$ generated
by $\tau^{jm}$. In particular, the projection of $S$ onto $C_j$ defines
an annulus. By assumption the curves $\gamma_0^{(a(j))}$ project to closed
curves $\gamma_0^{(a(j))}: \mathbb{R} \to C_j \setminus  B(a(j))$, where
$B(a(j))$ is the projection of $A(a(j))$ onto $C_j$. By construction the
interior angles at the vertices $w_i$ of the connected component of
$C_j \setminus  B(a(j))$ containing $\gamma_0^{(a(j))}$ are all less than $\pi$.
Applying the curve shortening flow to the smooth curve $\gamma_0^{(a(j))}$
we obtain smooth curves $\gamma_t^{a(j)}: \mathbb{R} \to C_j \setminus  B(a(j))$
which stay in $C_j$ and do not intersect $ B(a(j))$. This follows from
Theorem~\ref{2seg} and the fact that due to their homotopy class $\gamma_t^{a(j)}$
never become singular. Since the curves $\gamma_0^{a(j)}$ are not contractible,
M.~A.~Grayson's Theorem~\ref{GO} implies that the curvature of $\gamma_t^{a(j)}$
converges to zero. This yields the existence of a closed geodesic on $C_j$.  \\
Let $D_i$ be the bounded connected component of $S \setminus \sigma_i$ for
$i \in \{1,2\}$. Let $B(x, \varepsilon) \subset D_1 \cap D_2$ be a geodesic
ball with radius $\varepsilon>0$. Obviously, by construction the curves
$\gamma_0^{a(j)}$ do not intersect $\tau^{mk} B(x, \varepsilon)$ for
$0 \leq k \leq j-1$ for all sequences $a(j)$ and $\gamma_t^{a(j)}$ will not
intersect $\tau^{mk} B(x, \varepsilon)$ for all $0 \leq k \leq j-1$ and
$t \in (0, T)$ by the properties of the Curve Shortening flow.\\
By this construction for fixed $j$ and different sequences $a(j)$ we get $2^j$
different closed geodesics on $C_j$ of length between $b_1j$ and $b_2j$.
We project the constructed closed geodesics onto $C_1$. Identifying $\alpha$
and $\eta^{l+1} \alpha$ we get a torus $\tilde T^2$ on which we still have at
least $\frac{2^j}{j}$ different closed geodesics. By construction the initial
conditions of these closed geodesics form a $(\phi, \varepsilon)$-separated set
which grows exponentially for $j \to \infty$. By Definition~\ref{topent} this implies
positive topological entropy for the Riemannian metric on $\tilde T^2$.
As $\tilde T^2$ is a finite cover of $T^2$ also the Riemannian metric $g$ on
$T^2$ has positive topological entropy.
\end{proof}

\begin{remo}
For the proof of the Fundamental Lemma it is not necessary that $\alpha$ is minimal.
It suffices the condition $\alpha \cap \eta^{k+1} \alpha= \emptyset$ for the
special $k\in \Bz$ used in the proof.
\end{remo}

\begin{cor} \label{closed}
Let $(T^2,g)$ be a Riemannian torus. The existence of a contractible closed
geodesic on $T^2$ implies the existence of non-primitive axes for all translation
elements $\tau$.
\end{cor}

\begin{proof}
Considering an arbitrary translation element and replacing in the proof of
the Fundamental Lemma~\ref{fundamentallemma} the segments $\sigma_1$ and
$\sigma_2$ by a closed geodesic we can use a similar construction, as presented
starting in (\ref{m2}), to produce a non-primitive axis.
\end{proof}

Using the recurrence of the geodesic flow on $T^2$ we obtain the following
refinement of the Fundamental Lemma.

\begin{theorem}\label{1g}
Let $(T^2,g)$ be a Riemannian torus. Let $c:\Br \to \Br^2$ be a geodesic and
$\alpha$ a minimal axis such that $c(a), c(b) \in \alpha (\Br)$ for $a < b$.
If there exists a translation element $\tau$ with $\tau(\alpha( \Br)) \cap \alpha( \Br)= \emptyset$
and $\tau(\alpha( \Br)) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c( [a,b])\not= \emptyset$, then $g$ has positive
topological entropy.
\end{theorem}

\begin{proof}
Let $c:\Br \to \Br^2$ be a geodesic and $\alpha$ a minimal axis such that
$c(a), c(b) \in \alpha (\Br)$ for $a < b$ and $\tau$ the translation element
with $\tau(\alpha( \Br)) \cap \alpha( \Br)= \emptyset$  and
$\tau(\alpha( \Br)) \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; c( [a,b])\not= \emptyset$.
We will now prove that this implies positive topological entropy.\\
Consider the two halfplanes given by the connected components of
$ \Br^2 \setminus \tau(\alpha( \Br))$ and denote by $A$ the one containing
$\alpha(\Br)$ and by $A'$ the other one. By assumption there exists $t_0 \in [a,b]$
such that  $c(t_0) \in A'$.
\\
We can assume that $c[b, \infty)$ does not intersect $\tau(\alpha(\Br))$ since
this would immediately imply positive topological entropy by the Fundamental
Lemma~\ref{fundamentallemma}. We can also assume that $c$ is recurrent, since
the recurrent geodesics are dense. In particular,
there exists an increasing sequence $(t_n)_{n \in \Bn}$ and a sequence $\tau_n$
of translates such that $D\tau_n \dot c(t_n) \to \dot c(t_0)$ for $n \to \infty$.
But then for $n$ large enough $c([b, t_n])$ must intersect the translates
$\tau_n \tau \alpha( \Br) $ and  $\tau_n \alpha( \Br) $. Then, $c([b, t_n])$
fulfills the assumption of the Fundamental Lemma~\ref{fundamentallemma} and
hence, $g$ has positive topological entropy and there exist non-primitive axes
of $\tau$.
\end{proof}

\begin{cor} \label{2g}
Let $(T^2,g)$ be a Riemannian torus.
Let $c:[0 , \infty) \to \Br^2$ be a geodesic ray on the universal covering
$\Br^2$ which intersects a minimal axis $\alpha$ and disjoint translate
infinitely often. Then $g$ has positive topological entropy.
\end{cor}

\begin{lemma} \label{osc}
Let $g$ be a Riemannian metric on $T^2$. Then the existence of an
oscillating geodesic ray $c$ on $\Br^2$ implies positive topological
entropy for the metric $g$.
\end{lemma}

\begin{proof}
Due to the definition of an oscillating geodesic ray there exist sequences
$t_n, s_n$ tending to infinity and a compact set $K \subset \Br^2$ such that
$c(s_n) \in K$ and $\|c(t_n)\| \to \infty$ for $n \to \infty$. W.l.o.g.\ we can
assume that
$$
\lim\limits_{n \to \infty} \frac{c(t_n)}{\|c(t_n)\|} = z
$$
for some $z \in S^1$. Let
$\alpha: \Br \to \Br^2$ be a minimal axis such that $\pm \delta(\alpha) \not= z$.
Furthermore, we can choose $\alpha$ and a disjoint translate $\alpha'$ in such a
way that $K$ and $\{ c(t_n) \mid n \ge n_0 \}$ are contained in the two
different halfplanes of $\Br^2 \setminus (\alpha(\Br) \cup \alpha'(\Br))$
for sufficiently large $n_0$. This implies that the geodesic ray
$c:[0 , \infty) \to \Br^2$ will intersect $\alpha$ and $\alpha'$ infinitely
often. By Corollary~\ref{2g} the metric $g$ has positive topological entropy.
\end{proof}

\section{Existence and Regularity of the Rotation Number}
In the proof of Theorem~I we will need the following property of minimal geodesics:
\begin{remo}
As already mentioned, the study of minimal geodesics on $T^2$ goes back to
H.~M.~Morse~\cite{Morse}, G.~A.~Hedlund~\cite{Hedlund}, and V.~Bangert~\cite{B1988}.
Central results are that for each $r \in \Br \cup \{\infty\}$ there exists
a minimal geodesic $c:\Br \to \Br^2$ with asymptotic direction $r$ and that
furthermore there exists a constant $D>0$, such that for each minimal geodesic
$c: \Br \to \Br^2$ there exists a Euclidean line $l_c$, and for each Euclidean
line $l_c$ there exists a minimal geodesic $c$ such that
$$d (l_c, c(t)) \leq \frac{D}{2}, \quad \text{for all $t \in \Br.$}$$
Furthermore, we can assume that $D$ is larger than the diameter of the
fundamental domain. As shown by V.~Bangert~\cite{B1988}, for irrational rotation
numbers the set of minimal geodesics with this rotation number is totally ordered,
i.e., all these minimal geodesics have pairwise no intersections with each other.
For the set of minimal geodesics with a fixed rational rotation number the
subset of axes is ordered. Two minimal axes with the same rotation number bounding
a strip containing no further minimal axes are called neighboring minimals.
\end{remo}

\begin{proof}[Proof of Theorem~I]
Combining Corollary~\ref{bounded2}, Lemma~\ref{osc}, and Theorem~\ref{no-intersections}
we conclude that in the case of vanishing topological entropy all geodesic rays
are escaping and have no self-intersections on the universal covering $\Br^2$.\\
If for $c^+$ the asymptotic direction does not exist, there exist two
accumulation points $z_1 \not= z_2 \in S^1$ for the quotient
$\frac{c^+(t)}{\|c^+(t)\|}$ as $t$ tends to $\infty$. Choose a rational direction
$z \in S^1$ which lies in the connected component of $S^1 \setminus\{z_1,z_2\}$,
which is not larger than a half-circle. Let $\alpha$ be a minimal axis of the
primitive translation element $\tau$ with $\delta(\tau)=z$. Then the geodesic
$c:[0, \infty) \to  \Br^2$ must intersect $\alpha(\Br)$ and a given disjoint
translate $\eta(\alpha(\Br))$ with $[\eta] \not= [\tau]$ (which is a minimal axis
of $\tau$ as well) infinitely often. But then, by Corollary~\ref{2g} the metric
$g$ has positive topological entropy in contradiction to the assumption.\\
Assume that $\delta(c^+) \not= - \delta(c^-)$.
Choose $z \in S^1$ rational such that $\pm z$ both lie in one
connected component of $S^1 \setminus \{\delta(c^+), \delta(c^-)\}$. Let
$\alpha$ be an axis with $\delta(\alpha)=z$ which intersects $c$. Then $\alpha$
divides $\Br^2$ in two halfplanes. There is precisely one of those halfplanes
denoted by $A$ for which there exists $t_0 >0$ such that $c(t) \in A$ for $|t| \ge t_0$.
Choose a disjoint translate $\tau \alpha$ which is contained in $A$ as well.
Then there  exist $a <b$ such $c(a) , c(b) \in \tau(\alpha( \Br))$ and
$c[a,b] \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \alpha(\Br) \not= \emptyset$ which contradicts Theorem~\ref{1g}.\\
We will now show that there exists a Euclidean strip $S(c)$ with direction
$\delta(c^+)$ such that $c(\Br) \subset S(c)$. We already know that the
asymptotic directions exist and that $\delta(c^+)=-\delta(c^-)$.
Assume that $c$ is not contained in a Euclidean strip. Consider an arbitrary
minimal geodesic with asymptotic direction $\delta(c^+)$. As minimal geodesics
are accompanied by Euclidean lines with the same direction and as $c$ is not
bounded by any strip with direction $\delta(c^+)$, c also intersects an infinite
number of translates of this geodesic. Consider $D>0$ as introduced in the
previous remark. Choose a translation element $\tau$, a time $t_0 >0$ and a
minimal geodesic $b$ with asymptotic direction $\delta(c^+)$ such that
$c[0 , t_0]$ crosses $b$ and $\tau b$, $c(0)$ and $c(t_0)$ lie in different
halfplanes of $\Br^2 \setminus (b(\Br) \cup \tau b(\Br))$ and such that there
fits a Euclidean strip $R$ with direction $\delta(c^+)$ and width $4D$ between
$b$ and $\tau b$, see Figure~\ref{fig-ref6}.
\begin{figure}[h!]
\begin{center}
\psfrag{a}{$c(0)$}
\psfrag{f}{$R$}
\psfrag{b}[r]{$c(t_0)$}
\psfrag{c}{$S$}
\psfrag{d}{$b$}
\psfrag{e}{$\tau b$}
\psfrag{g}{$c$}
\psfrag{j}{$\alpha$}
\psfrag{i}{$\eta \alpha$}
\includegraphics[scale=0.55]{figure6.eps}
\end{center}
\caption{Illustration of the argumentation in the second part of the proof of Theorem~I.} \label{fig-ref6}
\end{figure}
Then there exists a Euclidean strip $S$ with width $3D$ and a rational
asymptotic direction $z$ near $\delta(c^+)$ on $S^1$ but $z \not= \delta(c^+)$
such that $c(0)$ and $c(t_0)$ lie in different halfplanes of $\Br^2 \setminus S$.
As $S$ has width $3D$, by the previous remark there exists a minimal axis $\alpha$
and a disjoint translate $\eta \alpha$ with asymptotic direction $\delta(\alpha)=z$
and $(\alpha(\Br) \cup \eta \alpha(\Br)) \subset S$. Furthermore, the minimal axes $\alpha $
and $\eta\alpha$ intersect $c^+$. Since $\delta(\alpha^+)=z \not= \delta(c^+)$,
the ray $c^+$ has to intersect $\alpha $ and $\eta\alpha$ a second time which
implies positive topological entropy by Theorem~\ref{1g}.
\end{proof}
Now we like to prove regularity properties of the rotation number. We begin with
the following lemma which holds for all Riemannian metrics.

\begin{lemma} \label{surj}
For all $r \in \Br \cup \{\infty\}$ and $x \in T^2$ there exists $v=v(r) \in ST^2$
such that the corresponding geodesic $c_v:\Br \to T^2$ fulfills $c_v(0)=\pi(v)=x$,
$\dot c_v(0)=v$ and for the lift of $c_v$ on the universal covering it holds $\rho(c_v^+)=r$.
Let $E_x=\{v \in S_xT^2\;|\; \rho(v) \text{ exists }\}$. Then, for fixed $x \in T^2$ the map
$$f_x: E_x \to \Br \cup \{\infty\} \quad \text{with} \quad v \mapsto \rho(v)$$
is surjective.
\end{lemma}

\begin{proof}
We fix $r \in \Br \cup \{\infty\}$ and $x \in T^2$. By the remark at the beginning
of this section there exists a minimal geodesic $\gamma$ with rotation number $r$.
We assume that $x \notin \gamma(\Br)$, otherwise we set $\gamma=c$ after a
reparameterization such that $c(0)=x$ and so we choose $v=\dot c(0)$. \\
We consider minimal geodesic segments $c_t$ connecting $x$ and $\gamma(t)$ on the
universal covering such that $c_t(0)=x$ with $\dot c_t(0)=:w_t \in S_x \Br^2$.
As $S_x\Br^2$ is compact there exists a sequence $t_n$ with $t_n \to \infty$ such
that $w_{t_n}$ converges to $w$. By its minimality the segment $c_t$ intersects
$\gamma(\Br)$ only once in $\gamma(t)$.
\\
Let $\tau \gamma$ be a translate of $\gamma$ such that $x$ lies in the bounded strip
$S$ between $\gamma(\Br)$ and $\tau \gamma(\Br)$. Repeating the minimality arguments
we conclude that $c_t$ and $\tau \gamma(\Br)$ have no intersections for all $t$.
By the continuous dependence on the initial conditions also the limit geodesic ray
$c^+$ with $c^+(0)=x$ and $\dot c^+(0)=w$ will not intersect $\tau \gamma(\Br)$ and
$\gamma(\Br)$.\\
Hence, the geodesic ray $c^+$ lies in the geodesic strip $S$ with rotation number $r$.
This implies the existence of the rotation number and even that $\rho(c^+)=r$.
We extend $c^+$ on $T^2$ to the geodesic $c: \Br \to T^2$ with $c(0)=x$ and $\dot c(0)=w$.
The choice $v=w$ fulfills the required properties.
\end{proof}

\begin{proof}[Proof of Theorem~II]
By Lemma~\ref{surj} the rotation number is surjective. We have to show its continuity.
As the topological entropy vanishes, by Theorem~I the asymptotic directions $\delta(v)$
and $\delta(-v)$ exist and it holds $\delta(v)=- \delta(-v)$ for all $v \in S \Br^2$.
Here, we think of the asymptotic directions as a function on the unit tangent bundle
of $\Br^2$. Assume that $\delta$ is not continuous at $v_0 \in S\Br^2$. Let $c$ be
the geodesic with the initial condition $\dot c(0)=v_0$ and let $\delta(c^+)=z$.
As $\delta$ is not continuous at $v_0$, there exists a sequence $v_n \to v_0$ such
that $\delta(v_n)$ does not converge to $z$.
Since $S^1$ is compact there exists a subsequence $v_{n_k}$ with
$\delta(v_{n_k}) \to \tilde z \not= z$. We will again denote this subsequence by $v_n$
and the geodesics corresponding to $v_n$ by $c_n$, i.e.\ $\dot c_n(0)=v_n$.\\
Consider the two Euclidean rays $r: [0, \infty) \to \Br^2$ and $\tilde r: [0, \infty) \to \Br^2$
given by $r(t) = tz $ and $\tilde r(t) = t \tilde z$. Choose  a minimal axis
$\gamma: \Br \to \Br^2$ and a disjoint translate $\gamma': \Br \to \Br^2$ such that
the two halfplanes $H_1$ and $H_2$ which are two of the three connected components of
$\Br^2 \setminus (\gamma(\Br) \cup \gamma'(\Br))$ have the following properties:
First, there exists an $\varepsilon>0$ such that $B(c(0), \varepsilon) \subset H_1$.
Second, the sets $H_2 \cap r[0, \infty)$ and $H_1 \cap \tilde r[0, \infty)$ are
unbounded, see Figure~\ref{fig-ref7}.
\begin{figure}[h!]
\begin{center}
\psfrag{a}{$\tilde r$}
\psfrag{b}{$r$}
\psfrag{c}{$c$}
\psfrag{d}{$c_n$}
\psfrag{e}{$\gamma$}
\psfrag{f}{$\gamma'$}
\psfrag{g}{$H_1$}
\psfrag{h}{$H_2$}
\psfrag{c0}{$c(0)$}
\psfrag{cnt}[c]{$c_n(t_1)$}
\psfrag{cns}{$c_n(t_2)$}
\includegraphics[scale=0.7]{figure7.eps}
\end{center}
\caption{Illustration of the geodesics $c$ and $c_n$ in the proof of Theorem~II.} \label{fig-ref7}
\end{figure}
Since $\delta(c^+)=z$ and $B(c(0), \varepsilon) \subset H_1$, the continuous
dependence on the initial conditions implies the existence  of  $n_0 \in \mathbb{N}$
such that for all $n \ge n_0$
we have $c_n(0) \in H_1$ and $c_n(t_1) \in H_2$ for a suitable $t_1 > 0$. Moreover,
$\delta(v_{n}) \to \tilde z$ as $n \to \infty$ yields the existence of  $n_1 \ge n_0$
such that the set $H_1 \cap c_n[0 , \infty)$ is unbounded for each $n \ge n_1$.
In particular, for a given $n \ge n_1$ we will find $t_2 \ge t_1$ such that $c_n(t_2) \in H_1$.
Such a geodesic $c_n$  crosses the pair of minimal axes $\gamma$ and $\gamma'$ at least
twice which implies by Theorem~\ref{1g} positive topological entropy in contradiction to
the assumption in Theorem~II. Hence, the asymptotic direction is continuous and its
continuity implies the continuity of the rotation number.
\end{proof}

\section{Further Conditions for Vanishing Topological Entropy}
In this section we show that vanishing topological entropy implies strong restrictions
of the intersections of geodesics on the universal cover with their translates.

\begin{proof}[Proof of Theorem~III]
As $h_{top}(g)=0$, by Theorem~I the asymptotic direction $\delta(c)$ exists for all
$c$ and for each $c$ there exists a Euclidean strip $S(c)$ bounding $c$.
Assume there exists a geodesic ray $c:[0, \infty) \to \Br$ with $\#I(c) \geq 2$.
Consider a translation element $\tau$ with $\#\{c \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \tau c\}= \infty$ and
$\delta(\tau) \not=\pm \delta(c)$, e.g.\ it does not leave $S(c)$ invariant.
Let $k \in \Bz$ be large enough such that the distance between $S(c)$ and
$\tau^k(S(c))$ is larger than $4D$ with $D>0$ introduced in the remark at the
beginning of the previous section. We denote the Euclidean strip between $S(c)$
and $\tau^k (S(c))$ by $E$. For each $l \in \{0,\dots, k\}$ we denote the connected
component of $\Br^2 \setminus \tau^l c(\Br)$ which contains $\tau^{k+l} c(\Br)$ by
$A_{\tau^l c}$ and by $B_{\tau^l c}$ the other one. \\
As $c(\Br) \subset S(c)$ it follows that $\tau^k c(\Br) \subset \tau^k(S(c))$.
Consider $t_0 \in [0, \infty)$. As $\#\{c \;\;\makebox[0pt]{$|$}\makebox[0pt]{$\cap$}\;\; \tau c\}= \infty$ there exist
$t_1,\tilde t_1>t_0$ with $c(t_1)=\tau c(\tilde t_1)$ such that $\tau c$ passes in
$c(t_1)$ from $B_c$ to $A_c$. Analogously there exist $t_2,\tilde t_2>\max\{t_1, \tilde t_1\}$
with $\tau c(t_2)=\tau^2 c(\tilde t_2)$ such that $\tau^2 c$ passes in $\tau c(t_2)$
from $B_{\tau c}$ to $A_{\tau c}$. By this construction we get a broken geodesic segment
$$ \alpha[t_0, t_{k}]:=c[t_0,t_1] \cup \tau c[\tilde t_1, t_2] \cup
\tau^2 c[\tilde t_2, t_3]\cup \dots \cup \tau^{k-1} c[\tilde t_{k-1}, t_{k}]$$
connecting $c$ and $\tau^k c$. Consider $s_0>t_{k}$ and analogously to the previous
construction there exist $s_1,\tilde s_1>s_0$ with $\tau^k c(s_1)=\tau^{k-1} c(\tilde s_1)$
such that $\tau^{k-1} c$ passes in $\tau^k c(s_1)$ from $A_{\tau^k c}$ to $B_{\tau^k c}$.
Analogously we get the broken geodesic segment
$$ \beta_1[s_0, s_{k}]:=\tau^k c[s_0,s_1] \cup \tau^{k-1} c[\tilde s_1, s_2]
\cup \tau^{k-2} c[\tilde s_2, s_3]\cup \dots \cup  \tau c[\tilde s_{k-1}, s_{k}].$$
Gluing these two geodesic segments we obtain the broken geodesic segment
$$V_1:=\alpha([t_0, t_{k}]) \cup \tau^k c([\tilde t_{k}, s_0]) \cup \beta_1([s_0, s_{k}])$$
which connects $S(c)$, $\tau^k(S(c))$, and $S(c)$. We call the unbounded connected
component of $\Br^2 \setminus\{S(c) \cup V_1\}$ which is not a Euclidean halfplane the
exterior of $V_1 \cup S(c)$. By construction the exterior angles of $V_1 \cup S(c)$ are
smaller than $\pi$. Analogously we construct a broken geodesic $V_2$ connecting $\tau^k(S(c))$,
$S(c)$, and $\tau^k(S(c))$ such that the exterior angles of $V_2 \cup \tau^k S(c)$ are
smaller than $\pi$. Consider two minimal axes $\xi_1$ and $\xi_2$ of $\tau$ bounding a
geodesic strip $G$ such that $V_1,V_2 \subset G$. As the distance of $S(c)$ and
$\tau^k(S(c))$ is larger than $4D$ consider three ordered translates of minimal axes
denoted by $\alpha_1$, $\alpha_2$, and $\alpha_3$ with a rational asymptotic direction
near $\delta(c^+)$ such that $\alpha_i \cap G \subset E$ for all $i \in \{1,2,3\}$.
Then, there exist subsegments $c_1$ and $c_2$ of $V_1$ and $V_2$ with endpoints on
$\alpha_2$ intersecting $\alpha_1$ or $\alpha_3$, respectively. These broken geodesic
segments $c_1$ and $c_2$ fulfill by construction the assumptions of the segments in the
Fundamental Lemma~\ref{fundamentallemma}.
\begin{figure}[h]
\begin{center}
\psfrag{a}{$V_1$}
\psfrag{b1}{$ $}
\psfrag{b2}{$ $}
\psfrag{g}{$V_2$}
\psfrag{E}{$E$}
\psfrag{G}{$G$}
\psfrag{x1}{$\xi_1$}
\psfrag{x2}{$\xi_2$}
\psfrag{sc}[c]{$S(c)$}
\psfrag{a1}{$\alpha_1$}
\psfrag{a2}{$\alpha_2$}
\psfrag{a3}{$\alpha_3$}
\psfrag{tsc}[c]{$\tau^k (S(c))$}
\includegraphics[scale=0.35]{figure8.eps}
\end{center}
\caption{Illustration of the broken geodesic segments $V_1$ and $V_2$ in the proof of Theorem~III.}
\end{figure}
Hence, $h_{top}(g)>0$ in contradiction to the assumption.
Furthermore, by construction we conclude, that geodesics $c$ with irrational
rotation number intersect their translates $\tau c$ only a finite number of times.
Geodesics $c$ with rational rotation number intersect their translates $\tau c$ an
infinite number of times at most for $\tau$ with $\delta(\tau)=\delta(c^+)$.
\end{proof}

\section{Characterization of Flatness}

In order to prove Theorem~IV we first prove the following lemma due to V.~Bangert
which is of independent interest.

\begin{lemma}\label{lemmaflat}
Let $(T^2, g)$ be a Riemannian torus. Suppose there exists a primitive translation element $\tau$
such that its minimal axes do not foliate $\mathbb{R}^2$. Then there exists $k \geq 2$ and a
non-primitive axis of $\tau^k$.
\end{lemma}

\begin{proof}
If the minimal axes of $\tau$ do not foliate $\mathbb{R}^2$, then there exist two neighboring
minimal axes $\tilde c_1, \tilde c_2: \mathbb{R} \to \mathbb{R}^2$ of $\tau$. Consider the cylinder
$\mathbb{R}^2 \diagup \langle \tau \rangle =C$ and the projections $c_1, c_2:\mathbb{R} \to C$
of $\tilde c_1$ and $\tilde c_2$ which are closed geodesics of equal length $\ell$. By assumption
all closed curves between $c_1$ and $c_2$ in the same homotopy class have length strictly larger
than $\ell$. Choose locally convex neighborhoods $C_1$ and $C_2$ about $c_1$ and $c_2$. (Take for
instance minimal geodesic loops homotopic to $c_1$ and $c_2$, respectively, close to $c_1$ and $c_2$.)
There exists $\varepsilon>0$ such that for all closed curves $\gamma$ homotopic to $c_1$ with
$L(\gamma) \leq \ell + \varepsilon$ we obtain that
$$\gamma \text{ is contained in } C_1 \cup C_2 \enspace. \qquad \qquad (\ast)$$
Using the methods in \cite{B2}, pp. 87/88 we obtain a constant $A>0$ such that for all $n>0$
there exists a closed geodesic $a_n$ between $c_1$ and $c_2$ with the following properties:
\begin{enumerate}
\item[(i)]
$a_n$ is homotopic to $c_1:[0, n\ell] \to C$
\item[(ii)]
$L(a_n) \leq n \ell + A$
\item[(iii)]
$a_n$ is not contained in $C_1 \cup C_2$
\end{enumerate}
Choose $n \geq 2$ large enough such that $\frac{A}{n} < \varepsilon$. Then there does not exist a curve
$b_n$ homotopic to $c_1:[0, \ell] \to C$ whose $n$-th iterate is equal to $a_n$. Otherwise using
property (ii),
$$L(b_n)=\frac{1}{n}L(a_n) \leq \ell + \frac{A}{n} \leq \ell + \varepsilon.$$
But by ($\ast$) this would imply that $b_n$ and hence $a_n$ is contained in $C_0 \cup C_1$ which
contradicts property (iii). Hence, $a_n$ is a non-primitive axis of $\tau^n$.
\end{proof}

\begin{proof}[Proof of Theorem~IV]
Obviously the flatness of the metric on $T^2$ implies that no periodic geodesic (and in fact no
geodesic) intersects its translates. Conversely, assume that no axis intersects its translates.
Then Lemma~\ref{lemmaflat} implies that for each primitive translation element $\tau$ the
associated minimal axes foliate $\mathbb{R}^2$. Since all axes are minimal the result of
N.~Innami in \cite{I} implies that $(T^2,g)$ does not have conjugate points. Then by E.~Hopf's
theorem \cite{H} the metric $g$ is flat.
\end{proof}

{\it Acknowledgement.} The authors like to thank Sigurd Angenent for explaining to us the essential features of the Curve Shortening flow which is a crucial technique in this paper. We are also grateful to Victor Bangert for explaining to us his variational methods which led to Theorem~IV.

redpajama_set_name: RedPajamaArXiv


Stunning pair of vintage Opaline Murano table lamps in SOFT LAVENDER by Alfredo Barbini. The glass has defined ribs and loads of controlled bubbles.
The lamps are 28 inches to the top of the socket. As shown, the top of shade is 33 inches high. Lampshades are for display only and not available for sale.


In my discussion of federal political advocacy last issue, I mentioned that the politicians all seem to be very aware of and interested in our industry's response to the need for flood coverage. I'm always quick to mention the terms "overland water" and "sewer backup" in the discussion because people need to know that the changing dynamics of these coverages seem to be merging them into one. In order to drive flood coverage towards premium adequacy, you can't have one without the other in some markets. It's a saleable perspective: if you are in an area where you should be concerned about a sewer backup, you should be also concerned about overland water. Most people will agree that the proximate cause of most sewer backup is actually overland water. Combining the two coverages eliminates "the chicken and the egg" claims discussion that generates public confusion and protest when one side of the street is covered. This random coverage is an ongoing issue because many companies are still not providing any coverage for "overland water" perils and are offering only the old "sewer backup" endorsement. This competing coverage is now both a public issue and a broker E&O nightmare.
The failure of the insurers to develop a standard of coverage that is set out in common wording and is used by all companies referencing this protection will likely be a major issue for our business. No company or insurance adviser is immune from the confusion that will arise with the next catastrophic loss in a major urban centre. If we as brokers are advising clients to take coverage with one company or another but are confused by the variances in the wording and the special limits for "overland water" or "flood" coverage, we will not be able to describe the differences sufficiently for the client to understand the coverage and make an informed choice. Consequently, we will be held responsible for any shortfall in the coverage that causes the client an uninsured loss. Of course, the insurers will suffer the same challenges and may even be found liable to the public for failure to set a standard of coverage, but any liability charges would be years after the headlines sullying our reputations have long passed.
The political perspective on disaster relief programs may further jeopardize public good will. Some provincial and federal politicians have suggested that the programs in place for uninsured losses from flood can now be reviewed and the budgets for them limited so that these funds can be freed up for other needed programs. Well, the reduction may not be that simple. As most brokers will tell you, "overland water" coverage is excluded for those living within 300 meters of running water and the new forms won't likely even cover "sewer backup" for these folks. When these exclusions become apparent after the next weird weather event, don't be surprised if you're asked to explain what happened. I can't help but note that, as I write, the community of Dawson Creek is suffering from just this kind of thing. I know everyone's tired of hearing me quip that we are just three days of rain away from a repeat of the 2013 flooding that devastated Calgary and High River. I recently toured some of the work underway to mitigate the Highwood River's potential to create havoc again. While mitigation is happening, we're far from prepared to deal with the kind of flood we saw just three years ago.
The issue of "broker" companies entering into the direct-writing business is always foremost in discussions between brokers. Of particular concern is the manner in which the competition takes place and the use of the data accumulated from the customers for further sales and marketing by both the brokers and the company. The other main concern remains the name by which the competition takes place. It should be clear to anyone that competing with your brokers by using the same name as that used to write the business the broker sends the company is seen as a breach of trust, a violation of the exclusivity anticipated in the contract between the broker and the company and just downright wrong! To further belabour the point, the public is also very confused when they see the same name on the insurance contract that is on the broker's wall and advertising. They expect to receive service and advice from the broker. Well, this confusion will of course sort itself out, eventually. If a "broker" company wants to be a "direct writer" under the same name as it is a "broker" company, then it will damage the relationship. Their support from brokers may wane to their detriment. All things being equal, I believe they will either have to commit fully to participating in the marketplace one way or the other.
This is quite a move for an affinity-group insurance company. Would a membership in the Automobile Association come as part and parcel with an insurance application? I wonder as well if the agent will get a fee for this part of the transaction. Auto clubs have affected the insurance industry before. The roadside assistance endorsement SEF 35 was created to provide a direct competitive response by the insurance industry to the entry of automobile clubs into the automobile insurance markets. An actual Auto Club package that provided some of the auto club services such as mapping and hotel planning used to be sold by brokers for one market. Now most that has gone by the wayside—no need to spend an hour talking about a road trip with someone when you can simply Google a route (complete with turning directions) and immediately get a whole bunch of information on all the amenities you might need on your trip. As in most of the travel industry, technology has taken over the heavy lifting, leaving little money for the service providers, but the affinity group of auto-club members continues to survive. In Alberta, their operations seem quite healthy, and they compete actively in the insurance and travel marketplace. I wonder if they will be appointing Alberta brokers to represent them. If not, I'm pretty sure they'll be watching the success of the Ontario efforts closely. I know I will be.
Stampede Breakfast in Calgary, Wednesday, July 13, 7:00–11:00 a.m.
Come and join around 1500 of our closest friends for a regular old-fashioned western whoop up. There'll be lots to eat, world famous chuck-wagon drivers, line dancers, and a live band. It's our 23rd time, and I think we're starting to get it right. This has turned into an "all industry" event that you don't want to miss. It starts early enough that even the 15 people in Calgary who do any work that week can get in on the fun before they get to the office.
I'm working my way through the chapters of CAIB 2 covering business interruption and crime coverage and reviewing their content for relevancy and accuracy in our ever-changing marketplace. Reacquainting myself with the material is interesting: I find I learn more by teaching than I ever did by studying. As I'm plodding along with this project, the thought crossed my mind that those of you far more in tune with the current market than I am might have helpful contributions to these topics. If you, do I'd be happy to include them in my overview of the suggested changes to the instruction program. For example, we've come a long way from the old 3D wording as the basics for crime coverage, but all the new package stuff just builds on the old foundation. I wonder if tossing out the basics might hinder comprehension of the reasons the new package works the way it does. Many of you are far wiser on this topic than I, so I'd welcome your ideas. The email address below is my personal one. Feel free to drop me a note.
Real life is stranger than fiction—watching the news these days sure hammers that saying home. The events unfolding to the south of us make me pause and reflect. The aberration that occurred in Orlando leaves me simply overwhelmed with grief for the people killed and injured and their families. I can make no comment that makes any sense of this. It is just so sad. Something has to be done. While I can hope for change, the lack of it after the massacre at Sandy Hook leaves me with a fatalistic view, believing that it will continue to get worse before it gets better. Pray I'm wrong!


You've got a great client looking for new ideas. They have a budget and a decent timeline. You can't wait to get started. You can just feel the ideas coming to the surface. Then, without warning, crickets. What happened?! And, more importantly, what can you do about it?
During the creative process, our subconscious is constantly at work on new ideas, but we have a habit of crushing them before they can even be heard. We let our fear in: we worry that our idea is bad, we become preoccupied with the challenges we must overcome, and with the work it'll take to make the idea successful. And, no matter the effort, there's always a chance we will fail.
But what if we were to take a step back and look at a problem not as an expert, but as a beginner – like someone free of all the baggage and nay-saying a person with years of experience naturally brings to the ideation process? It's great to be an expert, but by adopting a beginner's mindset in the concept phase, we can free ourselves from constraints and instead choose to see endless possibilities.
So how can we achieve a beginner's mindset? One way we can get ourselves to think this way is to disrupt our routines. One of the biggest creative roadblocks is getting caught up in the inspiration-sucking schedule of our day-to-day. We need to consciously disrupt our daily patterns. It can be as simple as driving a new route to work, or trying a new sandwich for lunch. It could also be something more involved, like taking a class and learning a new skill. By taking these actions, big or small, we force ourselves to detach from our rigidly structured days and expectations, and become open to finding new and different inspiration.
We should also give ourselves permission to come up with all solutions, whether they are good or bad. We discount many ideas based upon past experiences. Maybe we discredit them for not being possible, or too difficult to bring to life. Instead, let's look at every ideation session as a new opportunity to creative problem-solve, with endless opportunities and no right or wrong answers. If we give ourselves the chance to be wrong, it frees our thinking and moves us away from safe ideas toward ones that are truly original. Chances are if you come up with lots of unusable ideas, you'll also come up with a few gems along the way.
Changing our environment can also help when it comes to embracing the beginner's mindset. For some, sitting in a conference room with a large group of people isn't the best environment to come up with the "world's most awesome idea." Our best thinking happens while relaxed, even if that means being outside the office. Acknowledge that your surroundings play a big part in the ideation process, and if the ideas aren't coming, change your scenery.
These are all small changes in our thinking and the way we approach our daily lives. But, by allowing ourselves to shift our perspective and mindset, we can kick-start our creativity. Just imagine what we could dream up and accomplish for our clients (as well as in our own lives) if we adopted this mentality. To see a world of endless possibilities, we just need to get out of our own way.


Valentine's Day may have passed but love is still in the air for two border collies. Dewey and Blaze may be famous online for their awesome tricks and stunts in dog shows but it's their latest video showcasing their puppy love that is garnering them even more fans.
Dewey is an expert, competing in the obedience category. The seven-year-old is highly intelligent and knows more than 100 commands and behaviours. Although, three-year-old Blaze is the daredevil of the two, always ready to try something new.
The pair, who live in Texas, also devote a lot of their time with people. They teach kids about dog safety, care, and training. In this clip, Dewey and Blaze teach us something new: how canines, who happen to be best buds, hug it out.
Next, Nana proves border collies are the smartest dog breed in the world.


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text:
Louisiana Joins Lawsuit Against Nation's Largest Generic Drug Manufacturers For Price Fixing
May 13, 2019 04:35PM ● By Press Release
Fighting for Lower Prescription Drug Prices and to Hold Drug Companies Accountable, AG Jeff Landry Joins Lawsuit Targeting Major Generic Drug Companies for Price Fixing
BATON ROUGE, LA – Louisiana Attorney General Jeff Landry joined Connecticut Attorney General William Tong and 42 other attorneys general in a lawsuit against Teva Pharmaceuticals and 19 of the Nation's largest generic drug manufacturers alleging a broad conspiracy to artificially inflate and manipulate prices, reduce competition, and unreasonably restrain trade for more than 100 different generic drugs.
The lawsuit, filed in U.S. District Court for the District of Connecticut, also names 15 individual senior executive defendants responsible for sales, marketing, pricing, and operations. The drugs at issue account for billions of dollars of sales in the United States, and the alleged schemes increased prices affecting the health insurance market, taxpayer-funded healthcare programs like Medicare and Medicaid, and individuals who must pay artificially-inflated prices for their prescriptions drugs.
"Louisiana patients rely on generic drugs every day and they have felt the burn from the alleged conspiracy," said General Landry. "I am proud to join General Tong in these efforts to ensure those who have harmed patients are brought to justice."
Landry, Tong, and the others AG's allege the drug manufacturers engaged in a broad, coordinated, and systematic campaign to conspire with each other to fix prices, allocate markets, and rig bids for more than 100 different generic drugs. The drugs span all types, including tablets, capsules, suspensions, creams, gels, ointments, and classes, including statins, ace inhibitors, beta blockers, antibiotics, anti-depressants, contraceptives, non-steroidal anti-inflammatory drugs, and treat a range of diseases and conditions from basic infections to diabetes, cancer, epilepsy, multiple sclerosis, HIV, ADHD, and more. In some instances, the coordinated price increases were over 1,000 percent.
Their lawsuit seeks damages, civil penalties, and actions by the court to restore competition to the generic drug market.
People, News, Business, Local Life Breaking News landry generic drugs lawsuit Attorney General


It's truly astounding to consider the breadth and diversity of the lineup for the 59th edition of this Emmy Award-winning holiday show. There is West African music and dance from Le Ballet Dembaya; Hawaiian from Daniel Ho & Halau Keali'i o Nalani; Korean from the Jung Im Lee Korean Dance Academy, and Filipino from Kayamanan Ng Lahi Pilippine Folk Arts. Infinite Flow – An Inclusive Dance Company is America's first professional wheelchair ballroom dance company while the Inner City Youth Orchestra of Los Angeles is the largest majority African American orchestra in the U.S. These are just a portion of the local choirs, music ensembles and dance companies, hailing from all around Southern California, that will be taking the stage at the Music Center's Dorothy Chandler Pavilion to celebrate the season with this free three-hour extravaganza. The entire three hour event also will be broadcast live on PBS SoCal KOCE as well as live stream on pbssocal.org/.


text:
EXCLUSIVE: Billy Slater explains why Cameron Smith has never won coveted award
By Ben Glover
Cameron Smith is often referred to as 'The GOAT' by his younger teammates.
His career has universal respect to the point that he has been described as "the greatest player I've ever seen" by an Immortal with his own GOAT claims, Andrew Johns.
Yet in an era blessed by greatness, with the likes of Johnathan Thurston also a worthy candidate for the 'Greatest of All-Time' debate, the resume of Melbourne's legendary hooker has one glaring omission.
Cameron Munster of the Storm is congratulated by Cameron Smith after scoring a try (Getty)
A coveted individual award that is peer voted has never been won by Smith.
The Rugby League Players' Association's Players' Champion award has only been around since 2004 but its list of winners in its 16-year history is a veritable who's who.
But for the surprising nod for cult winger Akuila Uate in 2011 every player to have taken home the award was considered among the game's best players when they won it.
Jarryd Hayne won it in his famous 2009 season. Darius Boyd won it in the same season that he led the Dragons to a premiership and was the best player on the field in the grand final.
Ben Barba won it for his insane 2012 season that was packed full of ridiculous highlights, propelled the Bulldogs to the grand final and also won him the Dally M Medal.
Thurston, Smith's obvious rival for the tag of 'best player' of their generation, won the award four times.
Cameron Smith and Johnathan Thurston speak after a State of Origin game in 2017. (Getty)
Smith has been a model of consistency throughout the last 16 years and for most of that time he has been the captain of the game's most successful club, its dominant state and Australia. It's hard to think of a time when a player, teammate or rival, has openly questioned his legacy or disrespected his contribution to the game or his talent.
Yet his absence from the Players' Champion honour board clearly says that he's never had a season that has convinced enough of his peers that he was indeed the game's best player.
Few observers have had a closer view of Smith's career than his great mate and fellow legendary player Billy Slater and the former Storm fullback attempted to explain why the 37-year-old may have gone under the radar in voting despite his commanding presence.
Cooper Cronk, Johnathan Thurston, Cameron Smith, Billy Slater (Getty)
"It's quite hard to believe isn't it? He's won every other award," Slater told Wide World of Sports for a panel show discussing the history of the Players' Champion award.
"I think it might have something to do with Cameron's best asset and that's his ability to make everyone else look good.
"He doesn't score tries very often and he doesn't make big explosive tackles, so therefore you don't sort of notice him a whole heap.
"He doesn't have five or six big wins in the game like tries or big hits or linebreaks. He has wins every single set of six. He's getting his side over the advantage line, he's kicking early, he's leading the line speed when they haven't got the ball, he's putting pressure on the kicker, so sometimes little Cameron Smith can go unnoticed and therefore he's at his most dangerous.
"I think that might have something to do with it with why he hasn't won this award but it's an intriguing sort of debate on why he hasn't, 'cause I regard him as one of the greatest players to ever play the game, certainly the greatest hooker.
"He's been ultra-consistent over probably a two-decade period but he's also stood up in those big games, the Origins, the Test matches and the grand finals."
Smith responds to retirement speculation
Slater's assessment suggests that Smith's skillset is so subtle that even those on the field with him can't fully appreciate his impact on the result.
It's an incredible observation that in some ways contradicts the value players put on the award, with Peter Sterling explaining why it is so revered within NRL dressing rooms.
"Players are the best judges," Sterling told Wide World of Sports.
"They're out there, they're alongside for the 80 minutes, they see things that other people don't see. The little things that you do, the sacrifices that are made on the field. Your teammates see that and acknowledge that and a lot of that can be missed by people who aren't as closely involved or sitting up in the stands.
"They see the flamboyant, spectacular stuff but not the nitty gritty things that ultimately add up to be the most important things."
The history of the award: RLPA Players' Champion
In his 19th season in first grade, Smith has led the Storm to within a game of the grand final for a sixth consecutive season.
He is still one of the greatest players in the game as he weighs up retirement but the absence of the Players' Champion trophy in his house must sting at least a little bit.
Told by Sterling that he could still win it this year, Slater replied with a smirk: "Or next year."
Even at 38, few would doubt that he could at least put himself back in the conversation.
The short list for this year's Players' Champion will be announced in the coming days, with Smith a possible finalist but trailing the likes of Panthers star Nathan Cleary and two-time winner James Tedesco for favouritism.
Wide World of Sports is a partner of the 2020 RLPA Players' Champion, with the winner to be announced in Nine's coverage of the grand final on Sunday, October 25.
cameron smith
RLPA
Billy Slater
Peter Sterling
EXCLUSIVE: Paul Gallen on why top prop David Klemmer missed NSW State of Origin squad
NRL investigating Melbourne Storm trainer over Suliasi Vunivalu injury stoppage


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Heidi Przybyla (NBC News) Wiki, Bio, Wedding, Family, Net Worth, Child, Education, Career
Richard Daniel July 19, 2019
Heidi Przybyla Biography (Childhood, Education, Early life)
Heidi Przybyla Married Life
Heidi Przybyla Career and Net Worth
Heidi Przybyla, a renowned personality, a spellbinding reporter of NBC news and an enchanting news analyst. She started off her career as a reporter in 1997, as it worked out as ladder for her, to rise higher and gradually she reach to the position of National Political Reporter. In her initial years she began as a reporter for Blooms berg news and the USA today.
On the off chance if you are interested to think about her life so here is the opportunity for you to become acquainted with by seeking after this article.
Birth Name Heidi Przybyla
Date of Birth November 18, 1973 (age 45 years), Alexandria, Virginia, United States
Husband Name Unknown
Education Michigan State University, University of Freiburg
Heidi Przybyla was born on 18th November, 1973 in Alexandria, Virginia, USA. She belongs to White ethnicity. She holds the American nationality. She rose up in America with her siblings however her family information is still undercover.
When talking about her education, the information that has been mentioned in various sources is related to her university education and there is no further information has ever been mentioned on any of the social media website. Above all she did her Bachelor's degree from Michigan State University with the major in International Relations and German. It has also been mentioned that she attended Albert-Ludwigs-University Freiburg in Breisgau, Germany, between 1993 and 1994.
She has not only spent all her life in the USA but also completed her education and kicked started her career there. She has been a prestigious journalist and earns a name to a great extent as a media analyst. But apart from Journalist profession, she had greatly made an appearance in various television advertisements.
Heidi Przybyla has always kept her low profile status. She hasn't ever revealed any information about her affairs and married life. She is a media personality for over so many years but no one knows about her personal family life. However, she has tweeted a few years back in which she ends her sentence with the word husband. It is assumed a hint about her being a married woman. She neither mentioned anything about him nor put any picture with her better half on any social media. Everything is still undercover and wraps up secretly.
Heidi started her career as a reporter at Washington Business Journal and then kicked started from Bloomberg News, where she began to maintain her high profile status. She has experienced a great deal of work form the above-mentioned media agencies. Later in 1997, she turned out as a White House reporter and afterward she was appointed by USA Today in 2015.
\Despite of this she has popularly come to be known as senior political Reporter. From then and till now she is doing a great job as she has greatly been furnished as successful senior and most prominent journalist of USA. She has earned a very handsome amount from her salary so far. As according to an estimate she acquired total assets of $1 million.
Richard Daniel
Richard Daniel is the content writer on Thecelebinfo.com. After joining this website he is putting a lot of efforts in searching and collecting information regarding Celebrities & Famous Personalities,
By Richard Daniel
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