text:
\section{Non minimality of cylindrical vortices}

\subsection{The 1-dimensional case}
\label{dimension-1}

\hspace{0.45cm}
In this section we prove that 1-dimensional cylindrical vortices cannot be minimal. To do
this, we need to distinguish three cases: when the linear part of the skew dynamics is the
identity everywhere, when it coincides with $-\mathrm{Id}$ everywhere, and when it combines
$\mathrm{Id}$ and $-\mathrm{Id}$.
The first case was settled by Besicovitch \cite{besi}. The second case follows by
slightly modifying Besicovitch's arguments. Finally, the third case can be reduced to
the second one.

We start by (recalling and) slightly modifying Besicovitch's proof.\footnote{We
do this in order to avoid the use of the fact that if $F$ is minimal then,
{\em a-priori}, it must have a dense orbit.} Consider a cylindrical cascade
\[F \!: (x,v) \mapsto  \big( T(x), v + \rho(x) \big),\]
and denote by $\Pi$ the projection of $X\times \mathbb{R}$ on $\mathbb{R}$. Obviously,
if (the $\Pi$-projection of) the orbit of a point is bounded either from
above or from below, then $F$ cannot be minimal. Assume next that all the
orbits are unbounded from above and from below. We will show that, in
this case, all the orbits are proper as maps from $\mathbb{Z}$ into
$X \times \mathbb{R}$ (compare \S \ref{almost-integrability}).
Indeed, if the orbit of a point
$(x,v) \in X \times \mathbb{R}$ is (unbounded from above
and from below and) not proper, then by examining all
possible cases one easily convinces that we may choose three
sequences of integers $n_j < n'_j < n''_j$ such that either
\begin{equation}\label{eq:enes}
\begin{split}
\lim_{j \to \infty} \Pi (F^{n_j}(x,v)) =
\lim_{j \to \infty} \Pi (F^{n''_j}(x,v)) \mbox{ belongs to } [-\infty,+\infty),\\
\Pi (F^{n'_j}(x,v)) >  j, \quad \textrm{and} \quad
\Pi (F^{n'_j}(x,v)) =  \max_{n_j\le n \le n''_j} \Pi (F^{n}(x,v)).
\end{split}
\end{equation}
or
\begin{equation}\label{eq:enes2}
\begin{split}
\lim_{j \to \infty} \Pi (F^{n_j}(x,v)) =
\lim_{j \to \infty} \Pi (F^{n''_j}(x,v)) \mbox{ belongs to } (-\infty,+\infty],\\
\Pi (F^{n'_j}(x,v)) <  -j, \quad \textrm{and} \quad
\Pi (F^{n'_j}(x,v)) =  \min_{n_j\le n \le n''_j} \Pi (F^{n}(x,v)).
\end{split}
\end{equation}
\noindent Let us first consider the case (\ref{eq:enes}).
Set, for every  $n_j - n'_j \le n \le n''_j - n'_j$,
\[
 z_{j,n} := F^{n'_j+n} \big( x,v - \Pi (F^{n'_j}(x,v)) \big).
\]
We have:
\begin{eqnarray}
\label{eq:compacto} z_{j,0}  =  \big( T^{n_j'} (x_0),0 \big),
&& \textrm{ for every } j\in \mathbb{N}, \\
\label{eq:bounded} z_{j,n} = F^{n}(z_{j,0}), \quad \Pi
(z_{j,n}) \le  0,&& \textrm{ for every } n_j - n'_j \le n \le n''_j - n'_j,
\quad \textrm{and }\\
\label{eq:Z} n_j - n'_j \rightarrow -\infty, \quad
n''_j - n'_j \rightarrow +\infty, && \textrm{ as } j\rightarrow +\infty.
\end{eqnarray}
Due to \eqref{eq:compacto}, passing to a subsequence, we may assume that $z_{j,0}$
converges to a point $z_0 := (x_0,0)$. By \eqref{eq:bounded}, $z_{j,n}$ converges
to $z_n := F^n (z_0)$, for every  $n\in \mathbb{Z}$. Finally, using \eqref{eq:bounded}
and \eqref{eq:Z}, one may easily see that the orbit $\{z_n\}_{n \in \mathbb{Z}}$ remains
bounded from above by zero, which contradicts our assumption. Similarly, in
case (\ref{eq:enes2}), one may conclude the existence of a point whose orbit
remains bounded from below by zero, which again contradicts our assumption.

\vspace{.2cm}

Consider now a cylindrical vortex of type
\[F \!: (x,v) \mapsto  \big( T(x), -v + \rho(x) \big).\]
Set $G := F^2$. This is a fibered transformation over the non-necessarily minimal map
$T^2$. Clearly, if $G$ has a bounded orbit, then the same holds for $F$, which violates
our hypothesis of minimality for $F$. If not, then the arguments given so far show
that either:

\noindent -- there is a point $z_0^* = (x_0,v_0)$ whose $G$-orbit is bounded from above, or

\noindent -- there is a point $z_0^* = (x_0,v_0)$ whose $G$-orbit is bounded from below, or

\noindent -- all the $G$-orbits are proper.

In the first case, we let
$$h := \sup_{n \in \mathbb{Z}} \Pi \big( G^n (x_0,v_0) \big).$$
With obvious notation, for every $w \in \mathbb{R}$,
we have
\begin{eqnarray*}
orb_{F} (x_0,w) &=& orb_{G} (x_0,w) \hspace{0.2cm} \bigcup \hspace{0.2cm} F \big( orb_G (x_0,w) \big)\\
&\subset& X \times (-\infty, w - v_0 + h] \hspace{0.2cm} \bigcup \hspace{0.2cm} F( X \times (-\infty,w - v_0 + h])\\
&\subset& X \times \big( -\infty, w - v_0 + h + \| \rho \| \big] \hspace{0.2cm} \bigcup \hspace{0.2cm}
X \times \big[ -w + v_0 - h - \| \rho \|, + \infty \big).
\end{eqnarray*}
Taking $w_0 := -1 + v_0 - h - \| \rho \|$, we see that the $F$-orbit of $(x_0,w_0)$ avoids
$X \times (-1,1)$. In particular, $F$ is not minimal. The second case can be
treated similarly. Finally, since $F$ is a proper map,
if the $G$-orbit of $(x,v_0)$ is proper, then its $F$-orbit
$orb_{F} (x_0,v_0) = orb_{G} (x_0,v_0) \cup  F \big( orb_G (x_0,v_0) \big)$
is also proper, and $F$ cannot be minimal neither.

\vspace{.4cm}

Finally, consider a general 1-dimensional cylindrical vortex
\[F: (x,v) \mapsto  \big( T(x), \Psi(x) v + \rho(x) \big),\]
where linear part $\Psi(x)$ can be either $\mathrm{Id}$ or
$-\mathrm{Id}$ at each
point.\footnote{For a nice discussion of the measure-theoretical
cohomological properties of such a map, see \cite{OO}.} Denote
by $Y \subset X$ the preimage of $\{-\mathrm{Id}\}$ under $\Psi$. This
is a clopen set. Assuming that it is nonempty, the map $F_Y$ induced
from $F$ by the first-return map $T_Y$ to the set $Y$ has the form
\[ F_Y(x,v) = \big( T_Y (x), -v + \tilde\rho(x) \big),\]
where $\tilde \rho \!: Y \to \mathbb{R}$. Since $T$ is a minimal homeomorphism,
the same must hold for $T_Y$. Therefore, $F_Y$ is a cylindrical vortex
of the type considered in the second case, thus it cannot be minimal.
Now, every point in $Y \times \mathbb{R}$ that does not have a dense orbit
for $F_Y$ also fails to have a dense orbit under $F$, since
$Y^c \times \mathbb{R}$ is a closed set. Therefore, $F$ is not minimal,
and this concludes the proof of the Main Theorem for $\ell = 1$.

\subsection{The case of higher dimension}
\label{higher-dimension}

\hspace{0.45cm}
In \cite{birk}, to each local planar homeomorphism fixing the origin, Birkhoff associated
a forward-invariant compact set touching the boundary of the definition domain. In the
holomorphic case, this construction was refined in \cite{perez-marco} by P\'erez-Marco,
who constructed a completely invariant compact set touching the boundary.
In this section, we construct a Birkhoff/P\'erez-Marco like set
{\em at infinity} (a B/P-M set, for short) for cyclindrical vortices of dimension
$\ell \geq 2$. To do this, we follow the strategy developed by the third-named author in
\cite{ponce}, where he extends the construction of P\'erez-Marco to fibered holomorphic
maps. Next, we use the B/P-M sets to show the non-minimality of these vortices. It is
worth mentioning that this last idea is not completely new, as Besicovitch's work
\cite{besi} already includes a remark relating forward non-minimality of
cylindrical cascades to the existence of Birkhoff invariant sets.

Given $\ell \geq 2$, let $I \!: \mathbb{R}^{\ell} \to \mathbb{R}^{\ell}$ be an affine Euclidean isometry, that is,
\[ I v = \Psi v + \rho \]
for certain $\Psi \in O (\mathbb{R}^{\ell})$ and $\rho\in \mathbb{R}^{\ell}$. This isometry extends
continuously to the one-point (Alexandrov) compactification $\mathbb{R}^{\ell} \cup \{\infty\}$ by
letting $I \infty = \infty$. We will show that the infinity is not an isolated point as
an invariant object. More precisely, we will show that given an open, bounded set
$U$, there exists a closed set containing $\infty$ that is completely invariant
under $I$, touches the boundary $\partial U$, and is contained in $\mathbb{R}^{\ell}\setminus U$.
We introduce a terminology for this. We say that $K$ is a B/P-M set for $I$
avoiding $U$ if the following conditions hold:
  \begin{enumerate}
  \item $K \subset \mathbb{R}^{\ell}\setminus U$.
  \item $K\cup \{\infty\}$ is compact and connected for the one-point
        compactification topology of $\mathbb{R}^{\ell}$.
  \item $I(K) = I^{-1}(K)=K$.
  \item $K\cap \partial U \neq \emptyset$.
  \end{enumerate}

\begin{prop}\label{p1}
For any open, bounded set $U \subset \mathbb{R}^{\ell}$ and any affine isometry
$I: \mathbb{R}^{\ell} \to \mathbb{R}^{\ell}$, there exists a B/P-M set $K$ for $I$ avoiding $U$.
\end{prop}

To prove this proposition, we need an elementary lemma.

\begin{lema}
If the claim of Proposition \ref{p1} holds for $I^2$ and any
open, bounded set $U \subset \mathbb{R}^{\ell}$, then it also holds for $I$.
\end{lema}

\noindent{\it Proof.} Set $\mathcal{U} := U \cup I(U)$. This is
an open, bounded set. By hypothesis, there exists a B/P-M set
$\mathcal{K}$ for $I^2$ avoiding $\mathcal{U}$. Letting
$K := \mathcal{K} \cup I^{-1}(\mathcal{K})$, it is not
hard to check that $K$ is a B/P-M set for $I$ avoiding $U$.
$\hfill\square$

\vspace{0.5cm}

\noindent{\it Proof of Proposition \ref{p1}.} We start with the case $\ell=2$.
Due to the lemma above, we may assume that $\Psi$ preserves orientation, and
hence we only need to consider two cases:
\begin{itemize}
    \item $\Psi= \mathrm{Id}_{\mathbb{R}^2}$: Take two parallel lines in the direction of
$\rho$ touching the boundary of the
open set $U$, and such that $U$ lies in between the band determined by these two lines. Then
choose $K$ as being the union of the two semi-planes that form the complement of this band.
(See Figure 1.)
    \item $\Psi$ equals the counterclockwise rotation $R_{\alpha}$ of angle $\alpha \neq 0$:
The isometry
$I$ fixes the point $v_0 := (\mathrm{Id} - R_{\alpha})^{-1}\rho$, and $I$ corresponds to the rotation of
angle $\alpha$ centered at this point. Let $B$ be the smallest open ball centered at this
point and containing $U$. We then may choose $K := B^c$. (See Figure 2.)
\end{itemize}

\begin{figure*}
\centerline{\includegraphics[height=6cm]{figures.eps}}
\end{figure*}

When $\ell=3$, in a suitable orthonormal basis, the
(orientation-preserving) isometry $I$ may written in the form
    \begin{displaymath}
     I v= \left[ \begin{array}{cc}\tilde{\Psi}& \\
& 1 \end{array}\right] v+\left[\begin{array}{c}a\\ b
\\ 0 \end{array}\right]+\left[\begin{array}{c}0\\ 0 \\ c \end{array}\right]
    \end{displaymath}
for some linear isometry $\tilde \Psi$ of $\mathbb{R}^2$. Let $\tilde U$ be the
orthogonal projection of $U$ onto $\mathbb{R}^2$, and let $\tilde K$ a B/P-M set
for $\tilde{I}=\tilde \Psi+[a\ b]^{T}$ avoiding $\tilde U$. (Such a set
is known to exist because the case $\ell = 2$ has already been settled.)
Then $K := \tilde K\times \mathbb{R}$ is a B/P-M set avoiding $U$.

Assume now that the proposition holds for $\ell = k-1$ and $\ell=k$, and consider the case
$\ell = k+1$. In a suitable orthonormal basis, the isometry may be written in the form
\begin{displaymath}
   I v = \left[ \begin{array}{cc}\tilde{\Psi}& \\ & \hat \Psi \end{array}\right]v
+ \left[\begin{array}{c}a_1\\ \vdots  \\ a_{k-1}\\ 0\\ 0 \end{array}\right]
+\left[\begin{array}{c}0\\ \vdots \\ 0\\ a_{k}\\ a_{k+1} \end{array}\right]
\end{displaymath}
where $\tilde \Psi, \hat \Psi$ are linear isometries of $\mathbb{R}^{k-1}$ and $\mathbb{R}^{2}$, respectively.
Let $\tilde U$ be the orthogonal projection of $U$ onto $\mathbb{R}^{k-1}$, and let $\tilde K$ be a
B/P-M invariant for $\tilde{I}=\tilde \Psi + [a_1 \ldots a_{k-1}]^{T}$ avoiding $\tilde U$.
Then $K := \tilde K\times \mathbb{R}^2$ is a B/P-M set for $I$ avoiding $U$. $\hfill\square$

\vspace{0.35cm}

In what follows, we give an analog of the preceding proposition for fibered isometries.
More precisely, we consider a cylindrical vortex $F \!: X \times \mathbb{R}^{\ell} \to X \times \mathbb{R}^{\ell}$
over a minimal homeomorphism $T \!: X \to X$. We say that an open, bounded set
$U \subset X \times \mathbb{R}^{\ell}$ is a {\em tube} for $F$ if for every $x\in X$, the
fiber $U_{x}$ is nonempty. A closed set $K$ is a B/P-M set for $F$ avoiding
$U$ if the following conditions hold:
\begin{enumerate}
  \item $K\subset \left(X\times \mathbb{R}^{\ell}\right)\setminus U$.
  \item For each $x\in X$, the fiber $K_x \cup \{\infty\}$ is compact and connected
for the one-point compactification topology of the fiber $\mathbb{R}^{\ell}$.
  \item $F(K) = F^{-1}(K) = K$.
  \item There exists $x^*\in X$ such that $K_{x^*} \cap \partial U_{x^*} \neq \emptyset$.
\end{enumerate}

We first consider a fibered dynamics over the finite
space $X := \mathbb{Z}_p = \mathbb{Z} / p\mathbb{Z} $ with the basis homeomorphism
$T(j)=j+1$. In other words, given a finite family $\{ I_0, I_1, \dots, I_{n-1} \}$
of affine isometries of $\mathbb{R}^{\ell}$, with $\ell \geq 2$, we consider the cylindrical vortex
    \begin{eqnarray*}
    F \!: \mathbb{Z}_p \times \mathbb{R}^{\ell}&\longrightarrow& \mathbb{Z}_p \times \mathbb{R}^{\ell}\\
    (j, v)&\longmapsto&\left(j+1, I_j v \right).
    \end{eqnarray*}

\begin{prop}\label{ewpp1}
For every tube $U \subset X \times \mathbb{R}^{\ell}$, there
exists a B/P-M invariant set $K$ for $F$ avoiding $U$.
\end{prop}

\noindent{\it Proof.} For simplicity, we only deal with the case
$p = 2$, leaving the (slightly more elaborate) general case to the
reader. Let $U_0,U_1$ be the two (nonempty) open, bounded subsets of
$\mathbb{R}^{\ell}$ such that $U = (\{ 0 \} \times U_0) \bigcup (\{ 1 \} \times U_1)$,
and let $\mathcal{U} := U_0 \cup I_0^{-1}(U_1)$. By Proposition \ref{p1}, there
exists a B/P-M invariant set $K_0$ for $I_1\circ I_0$ avoiding $\mathcal{U}$.
Set $K_1 := I_0(K_0)$ and $K := (\{0\} \times K_0) \bigcup (\{1\} \times K_1)$.
Properties 1. and 2. in the definition above are obvious, whereas
Property 3. follows from
\begin{eqnarray*}
    (I_1\circ I_0)^{-1}(K_0)&=&K_0,\\
    I_0^{-1}\left(I_1^{-1}(K_0)\right)&=&K_0,\\
    I_1^{-1}(K_0)&=&I_0(K_0) \hspace{0.2cm} = \hspace{0.2cm} K_1.
\end{eqnarray*}
Since $K_0\cap \partial\mathcal{U}\neq \emptyset$, either
$K_0\cap \partial U_0\neq \emptyset$ or $K_0\cap \partial I_0^{-1}(U_1)\neq \emptyset$.
Since the second condition is equivalent to $K_1\cap \partial U_1\neq \emptyset$, in
both cases Property 4. above holds. $\hfill\square$

\vspace{0.35cm}

We can now proceed to the general case covered by the Main Theorem. Slightly more generally,
we will say that $T \!: X \to X$ is {\it approximated by homeomorphisms
having periodic orbits} if there exists a
sequence of homeomorphisms $T_n \!: X \to X$ that converges to $T$ uniformly on $X$ so that
each $T_n$ has a periodic point. Our Main Theorem
follows directly from the next two propositions.

\begin{prop}
Let $X$ be a locally homogeneous compact metric space. If $T\!: X\to X$
is a minimal homeomorphism, then $T$ is approximated by homeomorphisms
having periodic orbits.
\end{prop}

\noindent{\it Proof.} Given $x\in X$, let $V_n$ be a decreasing sequence
of neighborhoods of $x$ converging to $\{ x \}$. Let $p_n \in \mathbb{N}$ be
the first-return time of $x$ to $V_n$ under $T$, and let
$y_n := T^{p_n} x$. Define $T_n := h_{V_n, x, y_n} \circ T$. Clearly,
$x$ is periodic for $T_n$ with period $p_n$. Finally, since
$V_n \to \{x\}$, we have the (uniform) convergence
$T_n\to T$. $\hfill\square$

\begin{prop}\label{dem-for}
Let $U \subset X \times \mathbb{R}^{\ell}$ be a tube. If $T \!: X \to X$ is approximated
by homeomorphisms having periodic orbits, then there exists a B/P-M set for
$F$ avoiding $U$.
\end{prop}

\noindent{\it Proof.} Let
$x_n^0, x_n^1 := T(x_n^0), \ldots, x_n^{p(n)-1} := T(x_n^{p(n)-2}) $
be a periodic orbit of period $p(n)$ of $T_n$, which we identify with
$\mathbb{Z}_{p(n)}$. Let $F_n \!: \mathbb{Z}_{p(n)} \times \mathbb{R}^{\ell} \to \mathbb{Z}_{p(n)}
\times \mathbb{R}^{\ell}$ be the cylindrical vortex defined by
\[
F_n (x_n^{j},v) = \left(x_n^{j+1}, \Psi(x_n^j)v + \rho(x_n^j)\right).
\]
Proposition \ref{ewpp1} yields a B/P-M invariant set $K_n$ for $F_n$ avoiding
$U$. Let $\hat K_{n} \subset X \times \overline{\mathbb{R}}^{\ell}$ be the compact
set resulting from $K_n$ by attaching the section $X \times \{\infty\}$
to it, that is, $\hat{K}_n := K_n \cup (X \times \{\infty\})$. Taking
an appropriate subsequence, we may assume that there exists a connected
compact set $K \subset X \times \overline{\mathbb{R}}^{\ell}$ that is the limit
of $K_n$ for the Hausdorff topology on compact sets. Since $\partial U$ is
a compact set and, for each $n\in \mathbb{N}$, one has $K_n\cap \partial U\neq \emptyset$,
the intersection $K \cap \partial U$ is nonempty. We denote $\hat K :=
K\cup (X\times\{\infty\})$. Since $T_n \to T$ uniformly,
we have the uniform convergence $F_n \to F$. Thus,
$F_n(\hat K_n) \to F (\hat K)$ and $F^{-1}_n (\hat K_n) \to F^{-1}(\hat K)$.
This implies that $F(K) = F^{-1}(K) = K$, which closes the proof. $\hfill\square$

\subsection{Two possible generalizations}

\hspace{0.45cm} Recall that the main theorem of \cite{CNP} applies not only to
cocycles of isometries of a Hilbert space but also to cocycles of isometries of a
CAT(0) proper space. It is very likely that our Main Theorem here holds in this
context as well. Indeed, there is a general description of isometries of such
a space that allows to give an analog of Proposition \ref{p1} for ``higher
dimensional'' CAT(0)-spaces ({\em e.g.} spaces that are not quasi-isometric
to the real line). In this situation, this would certainly allow to perform
a similar procedure to get a B/P-M set and show the non-minimality, whereas for
the ``1-dimensional case'', it should not be very difficult to adapt the arguments
of \S \ref{dimension-1}. We do not carry out the details of all of this here
because we do not see any interesting application except for spaces for which
the arguments apply without major modifications ({\em e.g.} hyperbolic spaces
$\mathbb{H}^n$). In this direction, it is worth pointing out that, though
no cylindrical vortex of isometries of $\mathbb{H}^2$ over an irrational
rotation is minimal, the action on the product of the circle and the
boundary of the Poincar\'e disk appears to be minimal in many
cases \cite{avila-krikorian}.

\vspace{0.1cm}

Perhaps more interesting is trying to settle the infinite dimensional case of the
Main Theorem. Indeed, most of the arguments of \S \ref{higher-dimension} strongly
depend on the fact that the fiber space, namely $\mathbb{R}^{\ell}$, is locally compact
(the same would apply to proper CAT(0)-spaces). This turns natural the following

\vspace{0.4cm}

\noindent{\bf Question.} Does there exist a {\em minimal} cylindrical vortex with
infinite-dimensional fiber~?

\vspace{0.34cm}

Recall that, by \cite[Exercise 5.3.15]{book}, no isometry of a Hilbert space
can be minimal (it cannot be even topologically transitive ``at large scale'').
However, isometries without fixed points but having recurrent points do exist:
see \cite{c-t-v} and \cite[Exercise 5.2.26]{book}. The situation might be
compared with that of general linear maps: in finite dimension, topological
transitivity is impossible, while in infinite dimension, even topological
mixing may hold \cite{book-linear}.

\section{Minimal invariant sets}

\subsection{Almost-reducibility and proper orbits}
\label{almost-integrability}

\hspace{0.45cm}
The next proposition is folklore but difficult to find in the literature
stated in this way (compare \cite{krieger,OP}); we include the proof just
for the convenience of the reader. For the statement, notice that for a
cylindrical cascade $F \! : (x,v) \to \big( T(x), v + \rho(x) \big)$
and each $n \in \mathbb{N}$,
$$F^n (x,v) = \big( T^n (x), v + \rho_n(x) \big),$$
where $\rho_n$ denotes the Birkhoff sum
$$\rho_n (x) := \sum_{j=0}^{n-1} \rho \big( T^j(x) \big).$$

\begin{prop} \label{caso-cascadas}
The following properties are equivalent:
\begin{enumerate}
\item There exists a family of continuous sections $\varphi_k \!: X \to \mathbb{R}$
that is almost invariant under the skew action. In other words, the associate
cohomological equation can be solved in reduced cohomology:
$$\lim_{k \to +\infty} \left( \sup_{x \in X}
\big| \rho(x) - \big[ \varphi_k (T(x)) - \varphi_k (x) \big] \big| \right) = 0.$$
\item We have the convergence
$$\lim_{n \to +\infty} \left( \sup_{x \in X}
\Big| \frac{\rho_n (x)}{n} \Big| \right) = 0.$$
\item The union of proper orbits has zero $\mu$-measure
for every $T$-invariant probability measure $\mu$.
\item For every $T$-invariant (ergodic) probability measure $\mu$ on $X$,
$$\int_{X} \rho(x) \hspace{0.015cm} d \mu(x) = 0.$$
\end{enumerate}
\end{prop}

\noindent{\it Proof.} We give the proof of more implications than necessary
because they will useful for the further discussion of general cylindrical vortices.

\noindent ${\it 1. \to 2.}$ Given $\varepsilon > 0$, let $k = k(\varepsilon)$
be such that for all $x \in X$,
$$\big| \rho(x) - [ \varphi_k (T(x)) - \varphi_k (x) \big| \leq \varepsilon.$$
For each $n \in \mathbb{N}$,
$$\big| \rho_n (x) - [ \varphi_k (T^n(x)) - \varphi_k (x) ] \big|
\leq n \varepsilon,$$
thus
$$\Big| \frac{\rho_n (x)}{n} \Big|
\leq \frac{2 \| \varphi_k \|_{\infty}}{n} + \varepsilon.$$
Passing to the limit, this yields, with an uniform rate,
$$\lim_{n \to + \infty} \Big|\frac{\rho_n(x)}{n}\Big| \leq \varepsilon.$$
Since this holds for all $\varepsilon > 0$, we have the desired uniform
convergence to zero.

\noindent ${\it 2. \to 1.}$ For each $k \in \mathbb{N}$, let
$$\varphi_k (x) : = - \frac{\rho_1 (x) + \rho_2 (x) + \cdots + \rho_k(x)}{k}.$$
We have
$$\varphi_k (T(x)) - \varphi_k(x) = \frac{1}{k} \sum_{j=1}^{k}
\big[ \rho_j(x) - \rho_j(T(x)) \big] =
\frac{1}{k} \sum_{j=1}^{k} \big[ \rho(x) - \rho (T^{j}(x)) \big]
= \rho(x) - \frac{\rho_k(T(x))}{k},$$
and the last expression converges uniformly to $\rho(x)$.

\noindent ${\it 1. \to 4.}$ For every $T$-invariant probability measure
$\mu$ and each $k \in \mathbb{N}$, we have
$$\int_X \big[ \varphi_k (T(x)) - \varphi(x) \big] d \mu = 0.$$
Thus,
$$\int_X \rho \hspace{0.05cm} d \mu
= \int_X \lim_k \big[ \varphi_k \circ T - \varphi \big] d \mu
= \lim_k \int_X \big[ \varphi_k (T(x)) - \varphi(x) \big] d \mu = 0.$$

\noindent ${\it 4. \to 1.}$ Let $\mathbb{L}$ be the closure of the subspace of
$C(X)$ spanned by the functions of the form $\varphi \circ T - \varphi$, where
$\varphi \in C(X)$. If $\rho$ does not belong to $\mathbb{L}$, the Hahn-Banach
separation theorem provides us with a linear functional $I$ that restricted to
$\mathbb{L}$ is zero and $I (\rho) = 1$. Such an $I$ comes from integration
with respect to a signed probability measure on $X$. Since $I(\mathbb{L}) = \{0\}$,
this measure is $T$-invariant. Finally, the Hahn decomposition theorem yields
an invariant probability measure for which the integral of $\rho$ is nonzero.

\noindent ${\it 2. \to 3.}$ This implication directly follows from a classical
lemma due to Kesten \cite{kesten} (it can be also derived from a well-known
lemma of Atkinson \cite{atkinson-lema}).

\noindent ${\it 3. \to 4.}$ This follows directly from the Birkhoff ergodic theorem.
$\hfill\square$

\vspace{0.4cm}

Notice that $F$ is reducible if and only if the cohomological equation
$$\rho (x) = \varphi (T(x)) - \varphi (x)$$
has a continuous solution $\varphi \!: X \to \mathbb{R}$. We will hence
say that the cylindrical cascade $F$ is {\em almost-reducible} if the
equivalent conditions of the preceding proposition hold.

The situation for a cylindrical vortex
$$F \!: (x,v) \to \big( T(x), \Psi(x) v + \rho(x) \big)$$
is less transparent. First, in order to introduce a drift-like condition, notice that
if we define $\rho_n \!: X \to \mathbb{R}$ (and $\Psi_n \!: X \to O (\mathbb{R}^{\ell})$) by
$$F^n (x,v) =: \big( T^n(x), \Psi_n (x) v + \rho_n(x) \big),$$
then for all $m,n$ in $\mathbb{Z}$, we have
$$\rho_{m+n}(x) = \Psi_n (T^m(x)) \big( \rho_m(x) \big) + \rho_n (T^m (x)).$$
In particular,
$$\big\| \rho_{m+n}(x) \big\| \leq \big\| \rho_m(x) \big\|
+ \big\| \rho_n (T^m (x)) \big\|.$$
By the sub-additive ergodic theorem \cite{krengel},
for every $T$-invariant ergodic probability measure $\mu$, the value of
$$\frac{\|\rho_n (x)\|}{n}$$
converges to a limit (drift) $D = D(\mu)$ for $\mu$-almost
every point $x \in X$.

The main difference here is that Proposition \ref{caso-cascadas} does not extend
to cylindrical vortices, even to those with $\Psi \equiv \mathrm{Id}$. More precisely,
the equivalence between conditions 1. and 2. still holds with an analogous (direct)
proof. (See \cite{BN} for a more general version of this fact.)
Nevertheless, all the arguments relying on ergodic type theorems fail.
As a matter of example, let us consider Yoccoz' example from \cite{yoccoz}.
This is an irrational rotation of the 2-torus
$T \!: (x,y) \mapsto (x + \alpha,y + \beta)$ together with two continuous
functions $\hat{\rho} = \hat{\rho} (x)$ and $\check{\rho} = \check{\rho} (y)$,
both of zero integral, such that for almost every $(x,y) \in \mathbb{T}^2$,
$$\lim_{n \to \pm \infty} \big[ |\hat{\rho}_n (x)| + |\check{\rho}_n (y)| \big] = \infty.$$
Then letting $\Psi \equiv \mathrm{Id}$ and $\rho := (\hat{\rho},\check{\rho})$, almost
every orbit of the induced cylindrical vortex $F$ on $\mathbb{T}^2 \times \mathbb{R}^2$
is proper. However, for {\em every} point $(x,y) \in \mathbb{T}^2$, we have
$$\limsup_{n \to + \infty} \frac{\| \rho_n (x,y) \|}{n} =
\limsup_{n \to + \infty} \frac{|\hat{\rho}_n (x)| + |\check{\rho}_n(y)|}{n} = 0,$$
where the convergence is uniform in $(x,y)$. In particular, $D = D(\mathrm{Leb}) = 0$.

\subsection{On a theorem of Matsumoto and Shishikura}

\hspace{0.45cm}
One of the major interests on proper orbits is that they are nontrivial minimal
invariant closed sets. It easily follows from Denjoy-Koksma's inequality that for any
almost-reducible cylindrical cascade over an irrational circle rotation, there is no
such orbit provided that the function $\rho$ has finite total variation (without the
last assumption, proper orbits may appear; see \cite{sch} and \cite{besi}; see also
\cite{KS} for recent simpler examples). Actually, as it is shown by Matsumoto and
Shishikura in \cite{MS}, no nonempty, proper, minimal invariant closed set can
appear in this situation. A slight extension of this result is our next

\vspace{0.1cm}

\begin{prop} \label{ms-general}
Let $F$ be an almost-reducible, 1-dimensional cylindrical vortex over
an irrational rotation of the circle. If the corresponding function $\rho$ has finite
total variation, then $F$ admits no nonempty, proper, minimal invariant closed set.
\end{prop}

\noindent{\it Proof.} Let $F \!: (x,v) \to \big( x + \alpha, -v + \rho(x) \big)$
be a cylindrical vortex satisfying the hypothesis. (If $\Psi \equiv \mathrm{Id}$, then
Matsumoto-Shishikura's result applies.) Since $F$ is assumed to be almost-reducible,
the same must hold for $F^2$. Notice that $F^2$ is a cylindrical cascade to which
Matsumoto-Shishikura's theorem applies.

Let $C \neq \emptyset$ be a nonempty minimal closed $F$-invariant subset of $\mathbb{T}^1 \times \mathbb{R}$.
Given $(x,v) \in C$, denote by $C_{(x,v)}$ the closure of its (full) orbit under $F^2$. We
will show below that either $C$ is $F^2$-minimal or $C_{(x_0,v_0)}$ is $F^2$-minimal for
some $(x_0,v_0)$. Before showing this, notice that the previous remark yields either
$C = X \times \mathbb{R}$ or $C_{(x_0,v_0)} = X \times \mathbb{R}$, respectively. Since
$C_{(x_0,v_0)} \subset C$, in the second case we still have
$C = \mathbb{T}^1 \times \mathbb{R}$, as we wanted to check.

Assume that no $C_{(x,v)}$ is $F^2$-minimal, and let $C^*_{(x,v)} \subsetneq C_{(x,v)}$
be a nonempty closed $F^2$-invariant set. Then the set
$C^*_{(x,v)} \cup F(C^*_{(x,v)})$ is nonempty, closed, $F$-invariant and
contained in $C$. By minimality, it coincides with $C$. Now choose $(y,w)$
(depending on $(x,v)$) in $C_{(x,v)} \setminus C^*_{(x,v)}$. We must have
$(y,w) \in F(C^*_{(x,v)}) \subset F(C_{(x,v)})$. Thus, the closed set
$C_{(x,v)} \cap F(C_{(x,v)})$ is nonempty. Since it is $F$-invariant,
it must coincide with $C$, and by minimality, this easily implies that $C_{(x,v)} = C$.
Finally, since this holds for every $(x,v) \in C$, this shows that $C$
is $F^2$-minimal, which concludes the proof. $\hfill\square$

\subsection{An interesting family of cylindrical vortices}
\label{an-example}

\hspace{0.45cm}
Following \cite[Example 3]{CNP}, given two rationally independent
angles $\alpha,\beta$ and a continuous function $\rho \!: \mathbb{T}^1
\rightarrow \mathbb{C}$, we consider the cylindrical vortex
$F \!: (x,z) \mapsto \big( x+\alpha, e^{2\pi i\beta}z + \rho(x) \big)$
on $\mathbb{T}^1 \times \mathbb{C}$.

\vspace{0.15cm}

\begin{lema} The map $F$ has zero drift and is conservative
(that is, it admits no wandering open domain).
\end{lema}

\noindent{\em Proof.} The function $\rho_n$ defined so that
\begin{equation}
F^n(x,z) = \big(x + n \alpha, e^{2\pi i n \beta} z + \rho_n (x) \big)
\label{iterados}
\end{equation}
may be rewritten in the form
$$\rho_n (x)
= \sum_{k=0}^{n-1} e^{2 \pi i (n-k-1)\beta} \rho(x + k\alpha)
= e^{2\pi i n \beta} \sum_{k=0}^{n-1} e^{- 2\pi i (k+1) \beta} \rho(x + k\alpha).$$
Up to the factor $e^{2\pi in \beta}$, this coincides with the $n^{th}$ Birkhoff sum
$S_n^T (\chi)(x,0)$ at the point $(x,0) \in \mathbb{T}^2$ of the function
$\chi (x,y) = e^{2\pi i (y-\beta)} \rho(x)$ with respect to the dynamics of
the rotation $T \!: \mathbb{T}^2 \to \mathbb{T}^2$ of angle
$(\alpha,-\beta)$. Indeed,
\begin{equation}\label{suma-de-Birkhoff}
S_n^T (\chi)(x,y) = \sum_{k=0}^{n-1} e^{2\pi i (y - (k+1) \beta)} \rho(x + k\alpha),
\qquad \rho_n (x) = e^{2\pi in \beta} S_n^T (\chi) (x,0).
\end{equation}
In particular,
\begin{equation}\label{dividido}
\left| \frac{\rho_n(x)}{n} \right| = \left| \frac{S_n^T (\chi)(x,0)}{n} \right|.
\end{equation}
Since $\alpha,\beta$ are rationally independent, the map $T$ is uniquely ergodic. Since
$\chi$ is continuous, the right side member of (\ref{dividido}) uniformly converges to
$$\int_{\mathbb{T}^2} \chi(x,y) \hspace{0.04cm} dx \hspace{0.04cm} dy =
\int_{\mathbb{T}^1}\int_{\mathbb{T}^1} e^{2\pi i(y-\beta)} \rho(x)
\hspace{0.04cm} dx \hspace{0.04cm} dy =
\int_{\mathbb{T}^1} e^{2\pi i(y-\beta)} dy \int_{\mathbb{T}^1} \rho(x) dx = 0.$$
In particular, the drift of $F$ is zero.

To show the conservativity of $F$ notice that, due to Atkinson's lemma \cite{atkinson-lema},
if we fix $x \in \mathbb{T}^1$ and $\varepsilon > 0$, there exist $\bar{x} \in \mathbb{T}^1$,
$y \in \mathbb{T}^1$, and $n \in \mathbb{N}$, such that
$S_n^T (\chi) (\bar{x},y) \leq \varepsilon$ and
$$\mathrm{dist} (x,\bar{x}) \leq \varepsilon, \quad \mathrm{dist} (\bar{x},
\bar{x} + n \alpha) \leq \varepsilon, \quad
\mathrm{dist} (y,0) \leq \varepsilon, \quad \mathrm{dist} (y,y-n \beta) \leq \varepsilon.$$
This implies that both $n\alpha$ and $n\beta$ are
$\varepsilon$-close to zero. Together with (\ref{iterados}), (\ref{suma-de-Birkhoff}), and
$$\mathrm{dist} (x,\bar{x}) \leq \varepsilon, \qquad S_n^T (\chi) (\bar{x},y) \leq \varepsilon,$$
this implies that, for any $z \in \mathbb{C}$, the $n^{\mathrm{th}}$-iterate under $F$
of the $(\varepsilon,\varepsilon)$-neighborhood of $(x,z)$ is
$(\varepsilon,\varepsilon(\|z\|+1))$-close to it. Therefore,
$F$ has no wandering open domain. $\hfill\square$

\vspace{0.5cm}

As shown by the proof above, the dynamics of $F$ is closely related to the
cylindrical cascade $G$ on $\mathbb{T}^2 \times \mathbb{C}$ defined by
$$G \big( (x,y), z \big) \mapsto
\big( (x+\alpha,y-\beta), z + e^{2\pi i(y-\beta)} \rho(x) \big).$$
Besides (\ref{suma-de-Birkhoff}), both maps are related in that $F$ is a
factor of $G$. Indeed, letting $\Pi \!: \mathbb{T}^2 \times \mathbb{C}
\to \mathbb{T}^1 \times \mathbb{C}$ be the {\em proper map}
defined by
$$\Pi \big( (x,y), z \big) := (x, e^{-2\pi iy}z),$$
we have
$$F \circ \Pi = \Pi \circ G.$$
These relations allow showing the next

\vspace{0.1cm}

\begin{lema} \label{implica}
The map $F$ is reducible if and only if $G$ is.
\end{lema}

\noindent{\em Proof.} In view of either the second relation in (\ref{suma-de-Birkhoff})
or the fact that $F$ is a factor of $G$ by a proper map, this follows as a direct
application of the main result of \cite{CNP}. A direct argument proceeds as follows.

Recall that for $G$ being reducible we mean that there exists a
continuous function $\varphi \!: \mathbb{T}^2 \to \mathbb{C}$ such that,
for all $(x,y) \in \mathbb{T}^2$,
$$\varphi(x+\alpha,y-\beta) = \varphi(x,y) + e^{2\pi i(y-\beta)} \rho (x).$$
If this holds, then defining the continuous function
$\varphi_* \! : \mathbb{T}^1 \to \mathbb{C}$ by
$$\varphi_* (x) := \int_{\mathbb{T}^1} e^{-2\pi iy} \varphi(x,y) dy,$$
we obtain
\begin{multline*}
\varphi_* (x)
= \int_{\mathbb{T}^1} \big[ e^{-2\pi iy} \varphi(x+\alpha,y-\beta) - e^{-2\pi i\beta} \rho(x) \big] dy\\
= e^{-2\pi i \beta} \int_{\mathbb{T}^1} e^{-2\pi i(y-\beta)} \varphi(x+\alpha,y-\beta) dy
\hspace{0.2cm} - \hspace{0.2cm} e^{-2\pi i\beta} \rho(x)
= e^{-2\pi i \beta} \big[ \varphi_* (x + \alpha) - \rho(x) \big].
\end{multline*}
Hence
$$\varphi_* (x+\alpha) = e^{2\pi i\beta} \varphi_* (x) + \rho(x),$$
which shows that $F$ is reducible.

Conversely, assume that $F$ is reducible, that is, there exists a
continuous function $\varphi_* \! : \mathbb{T}^1 \to \mathbb{C}$ such that
$$\varphi_* (x+\alpha) = e^{2\pi i\beta} \varphi_* (x) + \rho(x).$$
If we multiply by $e^{2\pi i(y-\beta)}$ both sides of this equality, we obtain
$$e^{2\pi i(y-\beta)} \varphi_* (x + \alpha)
= e^{2\pi iy} \varphi_* (x) + e^{2\pi i(y-\beta)} \rho(x).$$
Therefore, if we define $\varphi \!: \mathbb{T}^2 \to \mathbb{C}$
by $\varphi(x,y) := e^{2\pi iy} \varphi_* (x)$, we have
$$\varphi(x+\alpha,y-\beta) = \varphi(x,y) + e^{2\pi i(y-\beta)} \rho (x),$$
thus showing that $G$ is reducible. $\hfill\square$

\vspace{0.6cm}

A more elaborate relation between $F$ and $G$ is given by the next

\vspace{0.1cm}

\begin{prop}\label{trans-top}
The map $F$ is topologically transitive if and only if $G$ is.
\end{prop}

\vspace{0.1cm}

To show (the difficult implication of) this proposition, we will strongly use a deep
theorem due to Atkinson that characterizes the failure of topological transitivity
by the existence of reducible linear factors \cite{atkinson}. In our case,
this may be stated as follows:

\vspace{0.15cm}

\begin{teo} {\em {\bf [Atkinson]}} Assuming that $G$ is conservative
and non-reducible, a necessary and sufficient condition for its topological
transitivity is that there is no $\theta \in \mathbb{T}^1$ such that the
1-dimensional cylindrical cascade
\begin{equation}\label{factor}
\big( (x,y), t \big) \mapsto
\big( (x+\alpha,y-\beta), t + \big\langle e^{2\pi i\theta}, e^{2\pi i(y-\beta)} \rho(x) \big\rangle \big)
\end{equation}
is reducible, where $\langle \cdot , \cdot \rangle$ stands
for the inner product of vectors in $\mathbb{R}^2 \sim \mathbb{C}$.
\end{teo}

\vspace{0.4cm}

\noindent{\em Proof of Proposition \ref{trans-top}.}
Since $F$ is a factor of $G$, the map $F$ is topologically transitive whenever $G$ is.

To prove the converse implication, assume first that (\ref{factor}) is reducible
for some $\theta \in \mathbb{T}^1$, that is, there exists a continuous function
$\varphi \!: \mathbb{T}^2 \rightarrow \mathbb{R}$
such that, for all $(x,y) \in \mathbb{T}^2$,
$$\langle e^{2\pi i\theta}, e^{2\pi i(y-\beta)} \rho(x) \rangle = \varphi(x+\alpha,y-\beta) - \varphi(x,y).$$
Then, letting
$$\varphi_{\vartheta}(x,y) := \varphi(x,y-\vartheta),$$
we have
$$\langle e^{2\pi i\theta'}, e^{2\pi i(y-\beta)} \rho(x) \rangle =
\varphi_{\theta'-\theta} (x+\alpha,y-\beta) - \varphi_{\theta' - \theta} (x,y).$$
In particular, both cocycles \hspace{0.2cm}
$\mathrm{Re} (e^{2\pi i(y-\beta)} \rho(x)) = \langle 1, e^{2\pi i(y-\beta)} \rho(x)\rangle$ \hspace{0.2cm}
\hspace{0.2cm} and \hspace{0.2cm} $\mathrm{Im} (e^{2\pi i(y-\beta)} \rho(x)) = \langle i, e^{2\pi i(y-\beta)} \rho(x)\rangle$
are reducible. Hence, the cocycle $e^{2\pi i(y-\beta)} \rho(x)$ is reducible, which is
impossible since $G$ is topologically transitive.

Now, if no cylindrical cascade (\ref{factor}) is reducible, then in order to apply
Atkinson's theorem for concluding that $G$ is topologically transitive, we need
to show that $G$ is conservative whenever $F$ is topologically transitive. To do
this, we first notice that the non-wandering set of $G$ is invariant under both
$G$ and the translations along the fibers
$$\big( (x,y), z \big) \mapsto \big( (x,y), z + t \big), \quad t \in \mathbb{C}.$$
Therefore, this set is either empty or the whole space $\mathbb{T}^2 \times \mathbb{C}$.
To exhibit a non-wandering point of $G$ we proceed as follows. Since $F$
is topologically transitive, we may chose a point $(x_0,z_0)$ in
$\mathbb{T}^1 \times \mathbb{C}$ having a dense orbit under $F$. If
we denote $(x_n,z_n) := F^n (x_0,z_0)$, this implies in particular that
there exists a strictly monotone sequence of integers $(n_k)$ such that
$(x_{n_k},z_{n_k}) \to (x_0,z_0)$ as $k \to \infty$. Since $\Pi$ is a
proper map, the sequence of subsets
$\Pi^{-1} (x_{n_k},z_{n_k}) \subset \mathbb{T}^2 \times \mathbb{C}$
remains inside a compact set. In particular, there must be a point
$\big( (\tilde{x}_0,\tilde{y}_0), \tilde{z}_0 \big) \in
\Pi^{-1} (x_0,z_0)$ such that the sequence
$\big( (\tilde{x}_{n_k}, \tilde{y}_{n_k}), \tilde{z}_{n_k}) :=
G^{n_k} \big( (\tilde{x}_0,\tilde{y}_0), \tilde{z}_0 \big)$
accumulates at some point
$\big( (\tilde{x}_{\infty},\tilde{y}_{\infty}), \tilde{z}_{\infty} \big)$.
As it is easy to check, every such an accumulation point is non-wandering
for $G$. $\hfill\square$

\vspace{0.65cm}

The construction of a map
$$F \!: (x,z) \mapsto \big( x+\alpha, e^{2\pi i\beta}z + \rho(x) \big)$$
that is topologically transitive will be the main issue of the next section.
Let us close this section by pointing out
that we do not know whether there exists a cylindrical
vortex $F$ of the form above that is neither reducible nor
topologically transitive. This is related to the existence of Yoccoz-like
examples (see \cite{yoccoz}) associated to functions of a particular
form, which seems to be a difficult problem. Indeed, the previous
arguments easily show the following

\vspace{0.15cm}

\begin{prop} If $F$ is neither reducible nor topologically transitive,
then $G$ is non-conservative. In particular, the function
$(x,y) \mapsto e^{-2\pi i(y-\beta)}\rho(x)$ does not satisfy
the Denjoy-Koksma property.
\end{prop}

\subsection{A ``concrete'' example}
\label{an-example2}

\hspace{0.45cm} Quite surprisingly, to perform our construction we will need
a lemma about the density of a certain set obtained by an arithmetic type
construction.\footnote{We strongly believe that a much more general result
should be true; in particular the set $\mathrm{FS}(m,n)$ analogous to that defined
further one should be dense for all $m > n$. Nevertheless, we were unable
to produce a conceptual proof of this seemingly interesting fact.} Given
$q \in \mathbb{N}$, let $t(q) := [\sqrt[3]{q}]$ and $r(q) := [\sqrt{q}]$.
If $q$ and $t(q)$ are coprime, define $p(q) \in \{1,\ldots,q-1\}$ as the
inverse (mod $q$) of $t(q)$. Otherwise, set $p(q) := 0.$ Now, let
$$\mathrm{FS}_q (2,3) :=
\left\{ \Big( \frac{s p(q)}{q}, \frac{s p(q) r(q)}{q} \Big), \hspace{0.2cm}
1 \leq s \leq 2 t(q) \right\},$$
where each coordinate is reduced modulo $\mathbb{Z}$ (thus, $\mathrm{FS}_q (2,3)$
is though of as a subset of $[0,1]^2$). Finally, let us consider the
{\em set of frequencies}
$$\mathrm{FS}(2,3):= \bigcup_{q \in \mathbb{N}} \mathrm{FS}_q (2,3).$$

\begin{lema} \label{F-SET}
The set $\mathrm{FS}(2,3)$ is dense in $[0,1]^2$.
\end{lema}

\noindent{\em Proof.} For a fixed $m \in \mathbb{N}$, let
$q := m^6 + 2 m^4 + m^2 + 1 = (m^3 + m)^2 + 1$. One readily
checks that \hspace{0.2cm} $t(q) = m^2$ \hspace{0.2cm} and \hspace{0.2cm} $r(q) = m^3 + m.$ \hspace{0.2cm}
Since \hspace{0.2cm} $t(q) (m^4 + 2m^2 + 1) = m^6 + 2m^4 + m^2 = q -1,$ \hspace{0.2cm}
we have
$$t(q) (q - m^4 - 2m^2 - 1) \equiv 1 \hspace{0.2cm} (\mathrm{mod} \hspace{0.2cm} q).$$
Hence,
$$p(q) = q - m^4 - 2m^2 - 1 = m^6 + m^4 - m^2.$$
In particular, modulo $\mathbb{Z}$,
$$\frac{p(q)}{q} = - \frac{m^4 + 2m^2 + 1}{m^6 + 2m^4 + m^2 + 1}.$$
Similarly, we have the equality
$$\frac{p(q) r(q)}{q} =
- \frac{(m^4 + 2m^2 + 1)(m^3 + m)}{m^6 + 2m^4 + m^2 + 1}
= -\frac{m^7 + 3m^5 + 3m^3 + m}{m^6 + 2m^4 + m^2 + 1}
= -m - \frac{m^5 + 2m^3}{m^6 + 2m^4 + m^2 + 1}.$$
Hence, modulo $\mathbb{Z}$,
$$\Big(\frac{s p(q)}{q}, \frac{s p(q) r(q)}{q} \Big) =
\Big( - \frac{s(m^4 + 2m^2 + 1)}{m^6 + 2m^4 + m^2 + 1},
- \frac{s(m^5 + 2m^3)}{m^6 + 2m^4 + m^2 + 1} \Big).$$

Let us consider all possible $s$ of the form
$s (j,k) := jm + k$, where $j,k$ range from $1$ to $m$. Notice that all
these values satisfy the restriction $1 \leq s \leq 2 t(q) = 2 m^2$, and
therefore the associated pairs
$$\Big( - \frac{(jm + k)(m^4 + 2m^2 + 1)}{m^6 + 2m^4 + m^2 + 1},
- \frac{(jm + k) (m^5 + 2m^3)}{m^6 + 2m^4 + m^2 + 1} \Big)$$
belong to $\mathrm{FS} (2,3)$. Now, easy computations show that the
pair above coincides with
$$\Big( -\frac{j m^5 + km^4 + 2jm^3 + 2km^2 + jm + k}{m^6 + f_4 (m)},
-j - \frac{k m^5 + 2km^3 - jm^2 - j}{m^6 + f_4 (m)} \Big)$$
where $f_4$ is a degree-4 polynomial. Since both $j,k$ are
(positive and) smaller than or equal to $m$, for $m$ large-enough,
the pair above is very close (modulo $\mathbb{Z}$) to
$(-\frac{j}{m},-\frac{k}{m}) \sim (1-\frac{j}{m},1-\frac{k}{m})$
with an error that converges to zero as $m \to \infty$ (independently
of $j,k$). As a consequence, every pair of rational numbers in
$[0,1]^2$ is contained in the closure of $\mathrm{FS} (2,3)$,
which shows that this set is dense. $\hfill\square$

\vspace{0.5cm}

To construct our desired topologically transitive cylindrical vortex
$$F \!: (x,z) \mapsto \big( x+\alpha, e^{2\pi i\beta}z + \rho(x) \big),$$
we will perform a sequence of approximations inspired in
the classical Anosov-Katok's method \cite{katok}.
More precisely, we will construct a sequence of skew maps
$$F_k: (x, z) \mapsto
\big( x + \alpha_k, e^{2\pi i\beta_k}z + \rho_k (x) \big),
\quad \alpha_k \in \mathbb{Q}, \hspace{0.2cm} \beta_k \in \mathbb{Q},$$
over periodic rotations so that they converge uniformly on compact
sets. The main point consists in prescribing a sequence of sections
$\varphi_k \!: \mathbb{T}^1 \rightarrow \mathbb{C}$ whose images become
more and more dense in larger and larger regions of $\mathbb{C}$
and that are invariant under $F_k$, that is,
$$\rho_k (x)
= \varphi_k (x + \alpha_k) - e^{2\pi i\beta_k} \varphi_k (x).$$
The construction of the sequence $(\varphi_k)$ is made so that
\hspace{0.2cm} $\varphi_k := \varphi_{k-1} + \psi_k,$
\hspace{0.2cm} where $\psi_k$ has the form
$$\psi_{k}(x) := \ell_k \sin(2\pi t_k x) e^{2\pi i r_{k} x}.$$
To perform this construction, we will need to define inductively the
sequences $(\ell_k)$, $(r_k)$, $(t_k)$, $(\alpha_k)$ and $(\beta_k)$.
The choice is irrelevant for $k = 1$ (we just need to impose the condition
$\ell_1 \geq 1$). Now, assuming that these values
have been already defined for $k \in \mathbb{N}$, we let $C_k \geq 1$ be
a Lipschitz constant for $\varphi_k$ and $D_k \geq 1$ be the supremum
of the norm of $\varphi_k$. According to Lemma \ref{F-SET}, we
may choose $(\alpha_{k+1}, \beta_{k+1}) :=
(\frac{s_{k+1} p_{k+1}}{q_{k+1}}, r_{k+1} \alpha_{k+1})$,
where $p_{k+1}$ is the inverse (modulo $q_{k+1}$) of
$t_{k+1} := [\sqrt[3]{q_{k+1}}]$ and $r_{k+1} := [\sqrt{q_{k+1}}]$,
in such a way that the following conditions are satisfied:
\begin{enumerate}
\item \hspace{0.2cm} $1 \leq s_{k+1} \leq 2 t_{k+1}$.
\item \hspace{0.2cm} $q_{k+1} > q_k^3$, \hspace{0.2cm} $q_{k+1}^{5/12} \geq 2^{k+1} q_k$,
\hspace{0.2cm} $q_{k+1}^{1/12} \geq 2 k q_{k}^{1/12}$ \hspace{0.2cm} and \hspace{0.2cm}
$q_{k+1} \geq \big[ 2 (k+1) \sum_{j=1}^{k} \ell_j \sqrt{q_j} \big]^3.$
\item \hspace{0.2cm} $\big| \alpha_{k+1} - \alpha_k \big| \leq
\frac{1}{2^{k + 2} q_k C_k}$.
\item \hspace{0.2cm} $\big| \beta_{k+1} - \beta_k \big| \leq
\frac{1}{2^{k + 2} q_k D_k}$.
\end{enumerate}
Finally, we let $\ell_{k+1} := q_{k+1}^{1/12}$.

\vspace{0.1cm}

For each $k,n$ in $\mathbb{N}$,
we define $(\rho_{k})_n \!: \mathbb{T}^1 \rightarrow \mathbb{C}$ by
$$(\rho_{k})_n (x):= \varphi_{k} (x + n \alpha_k)
- e^{2\pi in\beta_k} \varphi_k (x).$$
Then we have
$$F_k^n (x, z) =
\big( x + n \alpha_k, e^{2\pi in\beta_k} z + (\rho_{k})_n (x) \big).$$
The next lemma yields a quantitative estimate for the convergence
of the maps $F_k$ as well as some of their iterates.

\vspace{0.1cm}

\begin{lema} \label{explicit-estimate}
For each $k \in \mathbb{N}$ and $1 \leq n \leq q_k$,
\begin{equation}\label{necesaria}
\big| (\rho_{k+1})_n - (\rho_{k})_n \big| \leq
\frac{\ell_{k+1} q_k s_{k+1}}{q_{k+1}} +
C_k q_k |\alpha_{k+1} - \alpha_k| + D_k q_k |\beta_{k+1} - \beta_k|.
\end{equation}
\end{lema}

\noindent{\em Proof.} Notice that \hspace{0.2cm} $(\rho_{k+1})_n (x) - (\rho_{k})_n (x)$ \hspace{0.2cm} equals
\begin{small}
\begin{eqnarray*}
&& \hspace{-1cm}
\varphi_{k+1}(x + n\alpha_{k+1}) - e^{2\pi in\beta_{k+1}} \varphi_{k+1}(x) -
\big[ \varphi_{k}(x + n \alpha_k) - e^{2\pi i n \beta_k} \varphi_{k}(x) \big] \\
& = & \varphi_{k+1}(x + n \alpha_{k+1}) - \varphi_k (x + n \alpha_k) +
e^{2\pi i n \beta_k} \varphi_k (x) - e^{2\pi i n \beta_{k+1}}
[\varphi_{k}(x) + \psi_{k+1}(x)] \\
& = & \big[ \psi_{k+1}(x + n \alpha_{k+1}) - e^{2\pi i n \beta_{k+1}} \psi_{k+1}(x) \big]
+ \big[ \varphi_{k} (x + n \alpha_{k+1}) - \varphi_k (x + n\alpha_k) \big]
+ \varphi_k (x) \big[ e^{2\pi in\beta_k} - e^{2\pi in\beta_{k+1}} \big].
\end{eqnarray*}
\end{small}Thus, the value of $\big| (\rho_{k+1})_n (x) - (\rho_{k})_n (x) \big|$
is smaller than or equal to
\begin{small}
$$\ell_{k+1} \big| \sin ( 2\pi t_{k+1}(x + n\alpha_{k+1}))
e^{2\pi i r_{k+1} (x + n \alpha_{k+1})} - e^{2\pi in\beta_{k+1}}
\sin(2\pi t_{k+1} x) e^{2\pi i r_{k+1} x} \big| +
C_k n |\alpha_{k+1} - \alpha_k| + D_k n |\beta_{k+1} - \beta_k|.$$
\end{small}The first term of this expression is bounded from above by
\begin{footnotesize}
\begin{eqnarray*}
&& \hspace{-1.6cm} \big| \ell_{k+1} e^{2\pi ir_{k+1}(x + n \alpha_{k+1})}
\big[ \sin(2\pi t_{k+1}(x + n \alpha_{k+1})) -
\sin(2\pi t_{k+1} x) \big] +
\ell_{k+1} \sin(2\pi t_{k+1} x) \big[ e^{2\pi i r_{k+1} (x + n \alpha_{k+1})}
- e^{2\pi i(n \beta_{k+1} + r_{k+1} x)} \big] \big|\\
\hspace{0.6cm} &\leq&  \ell_{k+1} \big( \{ t_{k+1} n \alpha_{k+1} \} +
|e^{2\pi inr_{k+1}\alpha_{k+1}} - e^{2\pi in\beta_{k+1}}| \big).
\end{eqnarray*}
\end{footnotesize}Since $p_{k+1} t_{k+1} \equiv 1 \hspace{0.2cm} (\mathrm{mod} \hspace{0.2cm} q_{k+1})$,
we have
$$\{ t_{k+1} n \alpha_{k+1} \}
= \left\{ \frac{n s_{k+1} t_{k+1} p_{k+1}}{q_{k+1}} \right\} =
\left\{ \frac{n s_{k+1}}{q_{k+1}} \right\}.$$
Moreover, since $\beta_{k+1} = r_{k+1} \alpha_{k+1}$, the term
$|e^{2\pi inr_{k+1}\alpha_{k+1}} - e^{2\pi in\beta_{k+1}}|$
vanishes. We thus obtain
$$\big| (\rho_{k+1})_n (x) - (\rho_{k})_n (x) \big| \leq
\frac{\ell_{k+1} n s_{k+1}}{q_{k+1}} +
C_k n |\alpha_{k+1} - \alpha_k| + D_k n |\beta_{k+1} - \beta_k|,$$
which shows the lemma. $\hfill\square$

\vspace{0.65cm}

The next lemma deals with the density of the invariant curve
$x \mapsto (x,\varphi_k (x))$ inside a large
region of $\mathbb{T}^1 \times \mathbb{C}$.

\vspace{0.1cm}

\begin{lema} \label{last}
For every $k \ge 2$, the graph of $\varphi_k$ is $(\frac{1}{t_k},
\frac{\ell_k t_k}{r_k} + \frac{4\pi}{t_k} \sum_{j = 1}^{k-1} \ell_j r_j)$-dense
in the cylinder $\mathbb{T}^1 \times \mathrm{Ball} \big( 0, \ell_k -
\sum_{j=1}^{k-1} \ell_j \big)$.
\end{lema}

\noindent{\em Proof.} We first claim that the graph of $\psi_k$ is
$(\frac{1}{t_k},\frac{\ell_k t_k}{r_k})$-dense
in $\mathbb{T}^1 \times \mathrm{Ball} (0, \ell_k)$. Indeed, given $(x,z)$ in
this cylinder, there must exist $\tilde x \in
\big[ x - \frac{1}{2 t_k}, x + \frac{1}{2 t_k} \big]$ such that
$\psi_k (\tilde x)= |z|$. The claim then follows by noticing that
$\frac{1}{r_k} \leq \frac{1}{t_k}$ and that for every $|s|< \frac{1}{r_k}$,
$$\big| |\psi_k (\tilde x + s)| - |\psi_k (\tilde x)| \big|
\le \ell_k \big( |\sin(2\pi t_k (\tilde x + s))| -
    |\sin(2\pi t_k \tilde x)| \big)
\le \ell_k \frac{t_k}{r_k}.$$

Now, to deal with the graph of $\varphi_k$, we begin by noticing that
$$|\psi_j'| \leq 2\pi(\ell_j t_j + \ell_j r_j) \leq 4\pi \ell_j r_j.$$
Thus, on each interval $[m/t_k, (m+1)/t_k] \subset \mathbb T^1$,
the oscillation of $\varphi_{k-1}$ is at most
$\frac{4\pi}{t_k} \sum_{j = 1}^{k-1} \ell_j r_j.$
Since
$$|\varphi_{k-1}| \leq \sum_{j=1}^{k-1} \ell_j,$$
this proves the lemma.
$\hfill\square$

\vspace{0.6cm}

The next lemma deals with the density of a
certain $F_k$-orbit along the graph of $\varphi_k$.

\vspace{0,1cm}

\begin{lema} \label{densidad-orbita}
The set $\{ F_k^n (0,0) \!: 1 \leq n \leq q_k \}$ is
$(\frac{1}{t_k}, \frac{\ell_k t_k}{r_k} +
\frac{4\pi}{t_k} \sum_{j = 1}^{k-1} \ell_j r_j +
\frac{16\pi}{q_k^{2/3}} \sum_{j=1}^{k} \ell_j r_j)$-dense
in the cylinder $\mathbb{T}^1 \times \mathrm{Ball} \big( 0, \ell_k - \sum_{j=1}^{k-1} \ell_j \big)$.
\end{lema}

\noindent{\em Proof.} Since $\varphi_k$ is an invariant section for
$F_k$, the set of points we are dealing with coincides with
\begin{equation}\label{set}
\left\{ \left( n \frac{s_k p_k}{q_k}, \varphi_{k}
\Big( n \frac{s_k p_k}{q_k} \Big) \right)
\!: 1 \leq n \leq q_k \right\}.
\end{equation}
Since $p_k$ and $q_k$ are coprime and $s_k \leq 2 t_k \leq 2 q_k^{1/3}$,
the projection on the first coordinate of this set consists of at
least \hspace{0.2cm} $q_k^{2/3} / 2$ \hspace{0.2cm} points uniformly distributed on $\mathbb{T}^1$.
Therefore, the distance between two consecutive points of the set
(\ref{set}) is less than or equal to
$$\int_{\frac{2j}{q_k^{2/3}}}^{\frac{2(j+1)}{q_k^{2/3}}} \sqrt{1 + \varphi_k'(x)^2}
\hspace{0.2cm} d x \hspace{0.2cm} \leq \hspace{0.2cm} \frac{4 \max | \varphi_k'|}{q_k^{2/3}}
\hspace{0.2cm} \leq \hspace{0.2cm} \frac{16\pi \sum_{j=1}^{k} \ell_j r_j}{q_k^{2/3}},$$
and the claim of this lemma follows from that of the preceding one.
$\hfill\square$

\vspace{0.6cm}

To close the construction, notice that for each $1 \leq n \leq q_k$, the properties
of the inductive construction together with the estimates (\ref{necesaria}) and
$s_{k+1} \leq 2 q_{k+1}^{1/3} < q_{k+1}^{1/2}$ yield
\begin{equation}\label{rapidez}
\big| (\rho_{k+1})_n - (\rho_k)_n \big| \leq
\frac{q_{k+1}^{1/12} q_k q_{k+1}^{1/2}}{q_{k+1}} +
C_k q_k |\alpha_{k+1} - \alpha_k| + D_k q_k |\beta_{k+1} - \beta_k|
\leq \frac{q_k}{q_{k+1}^{5/12}} + \frac{1}{2^{k+1}} \leq \frac{1}{2^k}.
\end{equation}
Letting $n := 1$, this shows that $(\rho_k)$ is a Cauchy sequence,
hence it converges to a continuous function $\rho \!: \mathbb{T}^1 \rightarrow \mathbb{C}$.
Moreover, from Property 3. it follows immediately that $(\alpha_k)$ converges to
some angle $\alpha \in [0,1]$. Similarly, by Property 4, $(\beta_k)$
converges to a certain angle $\beta \in [0,1]$.

Checking that the limit map $F \!: (x,z) \mapsto \big( x+\alpha, e^{2\pi i\beta}z +
\rho(x) \big)$ is topologically transitive is not very difficult. Indeed, from
Property 2. it follows that
$$\frac{1}{t_k} \sum_{j=1}^{k-1} \ell_j r_j \leq \frac{2}{q_k^{1/3}}
\sum_{j=1}^{k-1} \ell_j q_j^{1/2} \leq \frac{1}{k}.$$
Moreover,
$$\frac{1}{q_k^{2/3}} \sum_{j=1}^{k} \ell_j r_j \leq
\frac{1}{t_k} \sum_{j=1}^{k-1} \ell_j r_j + \frac{\ell_k r_k}{q_k^{2/3}}
\leq \frac{1}{k} + \frac{\ell_k}{q_k^{1/6}} = \frac{1}{k} + \frac{1}{q_k^{1/12}}.$$
The last two inequalities combined with Lemma \ref{densidad-orbita} imply that
$\{ F_k^n (0,0) \!: 1 \leq n \leq q_k \}$ is $(\varepsilon_k, \delta_k)$-dense
in the cylinder $\mathbb{T}^1 \times \mathrm{Ball} \big( 0, \ell_k - \sum_{j=1}^{k-1} \ell_j \big)$
for certain sequences $(\varepsilon_k)$, $(\delta_k)$ converging to zero as $k$
goes to infinite. Since $F_k^n (0,0) = \big( n \alpha_k, (\rho_k)_n (0) \big)$,
using (\ref{rapidez}) and the estimate (valid for $1 \leq n \leq q_k$)
$$\big| n \alpha_k - n \alpha_{k+1} \big| \leq \frac{q_k}{2^{k+2} q_k C_k}
\leq \frac{1}{2^{k+1}},$$
we conclude that the orbit of $(0,0)$ under the limit map $F$ is dense
in $\mathbb{T}^1 \times \mathrm{Ball} \big( 0, \ell_k - \sum_{j=1}^{k-1} \ell_j \big)$
for every $k \in \mathbb{N}$. Finally, since Property 2. yields
$$\sum_{j=1}^{k-1} \ell_j \leq (k-1) \ell_{k-1} =
(k-1) q_{k-1}^{1/12} \leq \frac{q_k^{1/12}}{2} = \frac{\ell_k}{2},$$
we have that \hspace{0.2cm}
$\ell_k - \sum_{j=1}^{k-1} \ell_j \geq \frac{\ell_k}{2}$ \hspace{0.2cm}
goes to infinite together with $k$, thus showing that the $F$-orbit
of $(0,0)$ is dense in the whole space $\mathbb{T}^1 \times \mathbb{C}$.

It remains to show that $\alpha$ and $\beta$ are rationally independent.
Actually, we do not know whether this is always true, but we can ensure
it provided that the sequence $(q_k)$ satisfies a supplementary condition.

\begin{lema} If the sequence $(q_k)$ satisfies
\begin{itemize}
\item[{\rm 5}. ] $\sum_{j=k}^{\infty}|\alpha_{j+1} - \alpha_j| +
\sum_{j=k}^{\infty}|\beta_{j+1} - \beta_j| < \frac{1}{k q_k},$
\end{itemize}
then $\alpha$ and $\beta$ are rationally independent.
\end{lema}

\noindent{\em Proof.} Since $p_k \leq q_k$, if we consider the
representatives of $\alpha_k$ and $\beta_k$ in $[0,1]$, then we have
$$\alpha_k = \frac{p_k}{q_k}, \quad \beta_k = r_k \alpha_k =
\frac{r_k p_k}{q_k} - n_k, \quad \mbox{where } n_k \in \mathbb{Z}.$$
Assume that $(p, q, r) \neq (0,0,0)$ is 3-tuple of integers such that
\hspace{0.2cm} $p \alpha + q \beta + r = 0.$ \hspace{0.2cm} On the one hand,
\begin{eqnarray*}
|p \alpha_k + q \beta_k + r|
&=&    |p (\alpha_k - \alpha) + q (\beta_k - \beta) + p \alpha + q \beta + r|\\
&=&    |p (\alpha_k - \alpha) + q (\beta_k - \beta)|\\
&\leq& |p| \sum_{j \geq k} |\alpha_{j+1} - \alpha_j| +
|q| \sum_{j \geq k} |\beta_{j+1} - \beta_j|.
\end{eqnarray*}
On the other hand, if
$$p \alpha_k + q \beta_k + r
= p \!\hspace{0.2cm} \frac{p_k}{q_k}
+ q \Big( \frac{r_k p_k}{q_k} - n_k \Big) + r$$
equals zero, then
$$p_k (p + q r_k) = q_k (q n_k - r).$$
Since $p_k$ and $q_k$ are coprime, this implies that
$q_k$ must divide $p + q r_k$. Nevertheless, since $r_k \leq \sqrt{q_k}$
and $(p,q) \neq (0,0)$, this is impossible for a large-enough $k$.
Therefore, for a large-enough $k \in \mathbb{N}$, the value of
$p \alpha_k + q \beta_k + r$ is nonzero, and hence
$$|p \alpha_k + q \beta_k + r|
= \left| \frac{p p_k + q (r_k p_k - n_k q_k) + r q_k}{q_k} \right|
\geq \frac{1}{q_k}.$$
Therefore,
$$\frac{1}{q_k} \leq |p| \sum_{j \geq k} |\alpha_{j+1} - \alpha_j|
+ |q| \sum_{j \geq k} |\beta_{j+1} - \beta_j|,$$
which contradicts Property 5. for $k$ larger than $|p|$ and $|q|$.
$\hfill\square$

\section{Reducibility v/s arithmetic properties of the rotation angles}

\hspace{0.45cm} Although the preceding construction provides us with a
pair of angles $(\alpha, \beta)$ that are rationally independent, it also
suggests that $\beta$ is very fast approximated by multiples of $\alpha$.
As we next show, this is the case of every non-reducible smooth cylindrical vortex.
What follows is inspired (and may be deduced almost directly) from \cite{herman}.

We say that a pair $(\alpha, \beta) \in \mathbb{T}^1 \times \mathbb{T}^1$ satisfies
a type-$1$ Diophantine condition, and write $(\alpha, \beta) \in \mathcal{CD}_1$,
if there exist $C > 0$ and $\tau \geq 0$ such that for every $n \in \mathbb{Z}$,
$$\big| e^{2\pi i(n\alpha-\beta)} - 1 \big| \geq \frac{C}{n^{1+\tau}}.$$
For a fixed irrational $\alpha \in \mathbb{T}^1$, denote by $\mathcal{CD}_1^{\alpha}$ the set of
$\beta \in \mathbb{T}^1$ such that $(\alpha, \beta)$ belongs to $\mathcal{CD}_1$.
Standard arguments show the next

\vspace{0.1cm}

\begin{lema} The set $\mathcal{CD}_1^{\alpha}$ is a countable union
of closed sets with empty interior and has full Lebesgue measure.
\end{lema}

\vspace{0.1cm}

The next proposition is nothing but a straightforward
application of the classical baby-KAM theorem.

\begin{prop}
If $\rho:\mathbb{T}\to \mathbb{C}$ is a $C^{\infty}$-function, then
for every $(\alpha, \beta) \in \mathcal{CD}_1$ the cylindrical vortex
$$F \!:(x, z)\longmapsto \left(x+\alpha, e^{2\pi i\beta}z+\rho(x)\right)$$
is reducible.
\end{prop}

\noindent{\em Proof.} Recall  that reducibility is equivalent to the
existence of a continuous solution to the cohomological equation
\begin{equation}\label{ceq}
\varphi(x + \alpha) - e^{2\pi i\beta} \varphi(x)=\rho(\theta).
\end{equation}
At the level of Fourier series expansion, this is equivalent to that,
for all $n \in \mathbb{Z}$,
$$\hat \varphi_n = \frac{\hat\rho_n}{e^{2\pi in\alpha} - e^{2\pi i\beta}},$$
where $\hat\varphi_n$ and $\hat\rho_n$ stand for the Fourier coefficients
of $\phi$ and $\rho$, respectively. Since $\rho$ is a $C^{\infty}$-function,
$\hat\rho_n$ decreases faster than $1/n^k$ for any $k \geq 1$. The type-1
Diophantine condition on $(\alpha, \beta)$ allows us to conclude that
the coefficients $\hat\varphi_n$ defined by the previous equality
also decrease faster than $1/n^k$ for any $k \geq 1$. Therefore,
they correspond to the coefficients of a $C^{\infty}$-function
which solves our cohomological equation. $\hfill\square$

\vspace{0.55cm}

A better result relating the differentiability classes of
$\rho$ and the solution $\varphi$ can be stated as follows:

\vspace{0.1cm}

\begin{prop} {\bf [Herman]} Let $\rho$ be a  $C^{r}$-function. If
$(\alpha,\beta) \in \mathcal{CD}_1$ is such that the associated
$\tau \geq 0$ satisfies $r > \tau + 1$ and $s := r-1-\tau$ is
not an integer, then the solution $\varphi$ to the equation
(\ref{ceq}) is a $C^{s}$-function.
\end{prop}

\vspace{0.1cm}

Finally, concerning the case of ``Liouville pairs'',
we have the following

\vspace{0.1cm}

\begin{prop}
If $(\alpha, \beta)$ satisfies no type-1 Diophantine condition, then there exists a
$C^{\infty}$-function $\rho \!: \mathbb{T}^1 \to \mathbb{C}$ such that the equation
(\ref{ceq}) has no measurable solution. Moreover, $\rho$ can be chosen so that the
coefficients $\frac{\hat\rho_n}{(e^{2\pi in\alpha} - e^{2\pi i\beta})}$ do not correspond
to those of any distribution.
\end{prop}

\vspace{0.35cm}

\begin{small}

\noindent{\bf Acknowledgments.} We would like to thank to Jan Kiwi for helping
us in dealing with the sets $\mathrm{FS}(m,n)$, to Raph\"ael Krikorian for
fruitful discussions on the subject of this work, and to Roman Hric for
useful information concerning minimal homeomorphisms and for pointing
out to us the existence of \cite{G}.

\noindent -- D. Coronel was funded by the Fondecyt Post-doctoral Grant 3100092.

\noindent -- A. Navas was funded by the Math-AMSUD Project DySET.

\noindent -- M. Ponce was funded by the Fondecyt Grant 11090003 and the Math-AMSUD Project DySET.

\end{small}

\begin{footnotesize}

redpajama_set_name: RedPajamaArXiv


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Is the British nation state still needed? Debate on this has at last moved into the glare of media concern. Yet this very development has generated a phoney war to derail growing opposition to the EU.
Democracy and the EU
We find ourselves with our laws made by a foreign body, daily asserting extra powers for itself, trying to force us into a single economy with a single currency and soon to establish a single armed force.

redpajama_set_name: RedPajamaCommonCrawl


Recommended via the typical middle kingdom criteria for English Language Arts and Literacy as an exemplary informational text.
For the captains of males like Andrew Carnegie, John D. Rockefeller, J. P. Morgan, and Henry Ford the Gilded Age is a time of huge cash. expertise boomed with the discovery of trains, phones, electrical lighting, harvesters, vacuum cleaners, and extra. yet for thousands of immigrant employees, it's a time of huge struggles, with adults and kids alike operating 12 to fourteen hours an afternoon below severe, harmful stipulations. The disparity among the wealthy and the bad used to be dismaying, which triggered a few humans to motion. In An Age of Extremes, you will meet mom Jones, Ida Tarbell, gigantic invoice Haywood, Sam Gompers, and different movers and shakers, and get swept up within the enthusiasm of Teddy Roosevelt. you will additionally watch the us take its maximum position at the international level because the Revolution, because it enters the bloody battlefields of Europe in global conflict I.
Master storyteller pleasure Hakim has excited thousands of younger minds with the nice drama of yank heritage in her award-winning sequence A historical past people. instructed by means of the typical middle country criteria for English Language Arts and Literacy as an exemplary informational textual content, A historical past people weaves jointly fascinating tales that deliver American historical past to existence. Hailed through reviewers, historians, educators, and fogeys for its interesting, thought-provoking narrative, the books were well-known as a break-through instrument in instructing background and significant interpreting abilities to teenagers. In ten books that span from Prehistory to the twenty first century, teenagers won't ever contemplate American historical past as dull again.
In the event you have been requested to come back up with an inventory of an important occasions, innovations, humans, or discoveries some time past a hundred or 1,000 years, what could you opt? TIME for children mammoth publication of while, the latest unencumber from the highly winning mammoth ebook sequence, is a riveting new source for more youthful curious readers.
Pour le jeune chief, tous les hommes sont des managers en devenir, appelés à gérer au mieux et en permanence leur photo. Pour réussir, il faut incarner avec perfection plusieurs rôles, plaisants ou non. Le supervisor devient alors un imposteur. Passionné par le training en administration et advertising and marketing, Brou Ulrich Ahui rend hommage à tous ces grands Leaders du monde.
For the 1st name during this sequence, Künstler's work carry historical past to lifestyles with brilliant, high-action portrayals of the first occasions that received americans their freedom from Britain: the Boston Tea get together, the Siege of Yorktown, Paul Revere's journey, and the signing of the assertion of Independence. The epic artistic endeavors faithfully chronicle those moments from background and inspire young children to seem time and again for specific details—from the variety of stars on George Washington's flag to the fashion of a soldier's uniform.
During this biographical publication, younger readers will find out about the courageous lifetime of Harriet Tubman. Readers will become aware of how Tubman bravely escaped from her existence as a slave to move the Mason-Dixon line into freedom and the way she helped 1000's of different slaves into freedom in the course of the Underground Railroad with assistance from abolitionists.


Semenov, Yu. P., S P Korolev Space Corporation Energia, RKK Energia, 1994.
Dual English/Russian language picture book of the history of the Energia Corporation. Many unique photos and drawings of Korolev's rockets and spacecraft. Republished by Apogee books in 2000.


I remember coming away from the TAG F2F presentation at the F2F with the impression that SPDY was optimized for a relatively narrow set of use cases compared to the full breadth of current HTTP applications,  that SPDY would serve well in those cases (a high-performance 'upgrade' option for those servers that matched the use case for which it was designed), but that SPDY was also far from being a HTTP replacement.  I think if SPDY were *only* useful for the "top 100 sites on the internet", it would be still worth developing and bringing to standard.
This wasn't based on a technical analysis of SPDY and counter-examples, but rather that (as it seemed in the Q&A session after the presentation) that the analysis and optimization against the HTTP-NG failure cases hadn't really been done.
To be clear, I think SPDY could succeed and be an important optimization of the Internet, and I'm all for further development, standardization, analysis and further work, I just don't imagine we are close to calling it HTTP 2.0.
Now, maybe this is just my personal recollection? Did I miss something or mis-remember? The minutes of the discussion at the F2F are sparse.
Also, since a great deal of the HTTP-NG work was done by W3C before the project was abandoned, perhaps there might be some organizational memory that would help.... a review of "why HTTP-NG failed" would be helpful; my memory is a little fuzzy about it, although at the time, I was disappointed that the project was abandoned.
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text:
class Asset {
public:
	Asset();
	Asset(std::string urlOfAsset);
	virtual ~Asset();

	virtual AssetType GetAssetType() = 0;
	virtual void LoadAsset() = 0;

	inline std::string GetUrl() const                       { return this->url; }
	inline void        SetUrl(const std::string urlOfAsset) { this->url = urlOfAsset; }

private:
	std::string url;

	static AssetType assetType;
	friend class AssetLibrary;
};

#endif // ASSET_H

redpajama_set_name: RedPajamaGithub


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Home / Categories / Literature / Apple Inc / Computers / Retail / Steve Jobs / Apple is an American High-tech Company
Apple is an American High-tech Company
Apple Inc , Computers , Retail , Steve Jobs
Category: Literature
3 Products and Services
Apple is an American high-tech company headquartered in Cupertino, California. It is considered one of the Big Four of technology, including Amazon, Google, and Facebook (Ritholtz, 2017). According to statistics of 2018, there are more than 500 Apple retail stores in the world and had approximately 123,000 full-time employees. At the beginning of its founding, Apple focused on developing and selling personal computers. By 2014, it was committed to designing, developing and selling consumer electronics, computer software, online services, and personal computers.
Apple had a history of the instability of the company. Apple was founded on April 1, 1976, by Steve Jobs, Steve Wozniak and rarely mentioned third partner Ronald Wayne. They cobbled together their combination of a microprocessor, a video screen, and keyboard. The inventors called it the Apple I personal computer kit, designed by Wozniak, which seems simple but opened a new page of microcomputers. Millionaire Mike McCulla provided $250,000 during the founding of Apple. On January 3, 1977, Apple was incorporated as Apple Computer, Inc. without Wayne, who left Apple and sold his shares to Jobs and Wozniak for $800 after having co-founded Apple (Simon, 2010).
Apple II, also designed by Wozniak, was introduced in 1977 at the first West Coast Computer Faire. It included a floppy-disk drive, it differs from its major competitors who still used the cassette-tape drive. Apple II also included graphics and colors and it became the first kind of widespread microcomputer made by Apple Inc. In order to compete with IBM and Microsoft in the business and enterprise computing markets, Apple introduced Apple III in May 1980 (Staff, 2015). Unfortunately, thousands of Apple III computers were recalled due to quality issues. The first GUI computer, Apple Lisa was introduced in 1983, but because of its high price, Lisa failed to enter the commercial market. In 1984, Apple launched Macintosh, the first personal computer without a programming language.
The first Mac was launched through a famous Super Bowl advertisement, imitating after Orwell's 1984. The sales of Macintosh remained difficult until the launch of laser printers and PageMaker software. In 1985, a power struggle broke up out between Jobs and John Scully, who had been hired two years ago. In the same year, Jobs resigned from Apple and founded NeXT Inc. Wozniak also left the company and he continues to represent the company at events or in interviews. Apple has launched a series of Mac products including Mac Plus, Mac SE, Mac II, Mac Classic and Mac LC from 1986 to 1990. At the same time, Apple tried its first portable Macs: the failed Macintosh laptop in 1989 and more popular one PowerBook in 1991 which marked the beginning of real Apple notebooks. It was targeted at the professional market and received numerous awards.
Year 1989 to 1991 was "first golden age" of the Macintosh. On February 19, 1987, Apple registered the "Apple.com" domain name, becoming one of the top 100 companies register the .com address on the early Internet. Soon afterward, Apple has launched several new brands to sell identical machines at different prices for different markets. These are the Quadra, Centris and the Performa series. However, the performance of these new products in the market is not good due to there are too many models, and customers do not understand the differences between different models, which confused many consumers and damaged Apple's reputation. In the 1990s, Apple also tried other unsuccessful consumer-oriented products, including digital cameras, portable CD audio players, speakers, video consoles, eWorld online services, and TV appliances.
None of these products worked, and Apple's market share and share price continued to decline. In 1997, Apple bought the NeXT for its NeXTSTEP operating system and brought Steve Jobs back. With Jobs coming back, Apple began to restructure. On November 10, 1997, Apple introduced the Apple Store website, which linked to new order-based production strategies. On August 15, 1998, Apple launched an all-in-one computer: iMac. The iMac design team was led by Ive, who later designed the iPod and the iPhone (Wilson, 2007). In 1998, Apple acquired Macromedia's key GRIP software project, marking its entry into the digital video editing market and it was renamed Final Cut Pro when it was listed on the retail market in April 1999 (GR Reporter). In October 1999, Apple released the consumer video-editing product iMovie (Matt & Mark, 2002). iTunes was purchased and renamed in 2000 and was released by Casady & Greene in 1998, then called SoundJam MP. On May 19, 2001, the first official Apple Retail Stores opened in Virginia and California. On 23 of the same year, Apple launched the iPod portable digital audio player.
The product has achieved amazing sales, with sales exceeding 100 million in six years (Cantrell, 2006). iTunes Store has quickly become a market leader in online music services. Two years later, the iTunes Store became the largest music retailer in the world. Jobs announced that Apple would start producing Intel-based Macs in 2006. On January 9, 2007, Jobs announced that Apple Computer (Inc.) would later be called "Apple Inc." because the company had shifted its focus from computers to consumer electronics (Forbes, 2007). The iPhone, iPad and Apple TV have quickly become fashionable and mainstream, they have achieved extensive success.
With the popularity of the iPhone, Apple has become the third largest mobile phone supplier in the world. From 2011 to 2012, Apple released the iPhone 4S and 5, including improved cameras, a smart assistant named Siri, and cloud-synced data with iCloud and iPad 3, 4 and mini, which all achieved success and good reputation. In January 2016, 1 billion Apple devices were announced to be in use worldwide (Statt, 2016). Apple may have received some criticism or some scandals in the past, but it is constantly improving its products and services. From 1976 to 2019, Apple has occupied an important position in the consumer electronics market for more than 20 years, and it is still one of the most famous brands and has truly become the most valuable technology company in the world.
According to their online store, the company's hardware products include the Mac personal computer, the iPod portable media player, the iPhone smartphone, the iPad tablet computer, the Apple Watch, the Apple TV, the HomePod smart speaker, Software, and electric vehicles. The online services include App Store, Apple Card, Apple Music, iBooks Store, iCloud, iTunes Store. Apple offers various customer services including web support, App and live chat and phone support. Apple sells its products directly through Apple retail stores and Apple online store or through a variety of indirect distribution channels. Such as the third-party wholesalers, retailers. You can even find Apple products at Best Buy, Wal-Mart and Target. That's why Apple retail stores increased dramatically and become so successful.
Apple is an American high-tech company. (2020, Mar 17). Retrieved from https://papersowl.com/examples/apple-is-an-american-high-tech-company/
Apple Inc	,	Computers	,	Retail	,	Steve Jobs
Apple Strategy Analysis	Pages: 8 (2386 words)
Steve Jobs and Hitler: Personal Branding	Pages: 4 (1054 words)
IOS Gaming 10 Years on	Pages: 2 (718 words)
Apple, Inc. is an American Multinational Company	Pages: 3 (832 words)
Life and Career of Steve Jobs	Pages: 2 (483 words)
Apple Inc. Company History	Pages: 5 (1633 words)
Apple Computer, Inc.
Executive Summary Apple Inc. is a multinational corporation that designs and manufactures consumer electronics, computer software, and personal computers. Apple was founded by two college dropouts Steve Jobs and Steve Wozniak in April of 1976. In 1977, Apple was incorporated in Cupertino, California. The company's products include Macintosh, iPhone, iPad, Apple Watch, Air Pods, Apple [...]
Topics: Apple Inc, Computers, Retail, Steve Jobs
Mount Rushmore National Memorial
The heads of Washington, Jefferson, Roosevelt, and Lincoln are forever sculpted in stone at the Mount Rushmore National Memorial. The likeness of each former president was sculpted by Gutzon Borglum. Borglum chose each man to represent a different facet of U.S. culture. (National Park Service [NPS], 2018). Washington is best known as the First President [...]
Topics: Apple Inc, Entrepreneurship, Retail, Steve Jobs, United States, Walmart, Walt Disney
Foundation and Development of Apple Inc
Apple Inc., formerly Apple Computer Inc., is located in Cupertino, California, a multinational technology company. It was originally created by Steve jobs, Steve Woznick, and Ronald Wayne on April 1, 1976, and on January 3, the following year, it was officially called Apple Computer (Yahoo! Finance, n.d). The birthplace of Apple Computer is in the [...]
Company Analyses – Apple Inc.
Apple Inc was incorporated in 1977 in California. The company engages in the designing, manufacturing, and manufacturing mobile communications, media devices, and personal computers. It sells variety of software, services, accessories, networking solutions, and third-party digital content and applications. The products and services of the company are iPhone, iPad, Mac, Apple Watch, Apple TV, IOS, [...]
Topics: Apple Inc, Computers, Retail, Smartphone, Steve Jobs
Steve Wozniak and Apple Inc.
""Never trust a computer you can't throw out a window."" A quote from Stephen Gray "Woz" or Steve Wozniak. He was born on August 11, 1950 in San Jose, California. Woz is a man of many talents. "American electronics engineer, cofounder, with Steve Jobs of Apple computer, and designer of the first commercially successful personal [...]
Topics: Apple Inc, Computer engineering, Computers, Steve Jobs
Essay about Apple AirPods
Apple Inc., a company out of California, is one of the most popular technology companies in the industry leading in phone manufacturing and the creation of other personal technology like computers, headphones, television, tablets, watches, and music/music players. Looking through a corporate standpoint, the company operates with a mission that states, ""Apple designs Macs, the [...]
Topics: Apple Inc, Computers, Steve Jobs


text:
Phantom Liberty
Speaking Simulator
Watch my ear pop off and my nose shoot sparks when I go on a date in Speaking Simulator
By Christopher Livingston
Luckily I don't think my date suspects that I am not a human. And why would she? I am an extremely normal human!
At one time or another, everyone's had a slip of the tongue. Maybe you called your boss by the wrong name (as I did the very first time I emailed Tim Clark and called him Tyler). Maybe you once accidentally called your teacher "Mom," or said "Wait a sex" instead of "Wait a sec." It happens to the best of us!
But a slip of the tongue isn't usually so bad that it knocks several of your teeth out of your mouth and down your shirt. Stumbling over your words doesn't typically cause an eyeball to pop out of its socket, sparks and smoke to spray from your nostrils, or motor oil to squirt out of the spot where your ear used to be, now that your ear is protruding from your head on a long steel spring. Unless you're playing Speaking Simulator, in which case these things are pretty common.
This is because you're really a robot trying to pass for human while at work, on a date, in church, or at the doctor's office, and the best way to pass for human is to speak, look, and act like a human by pressing buttons with your tongue, opening and closing your mouth to form words, smiling at the appropriate times (for the appropriate amount of time) and keeping a natural amount of eye contact while you speak. And always stay calm, even after one of your eyes falls out and sparks shoot out of your nostrils. Maybe no one noticed!
I got my hands (and mouth) on a preview build and can report that speaking in Speaking Simulator is extremely tricky. You control your robotic tongue with the WASD keys, using it to hit the buttons inside your robot mouth when they light up green—and trying to avoid hitting them when they're red. At the same time, you use the mouse to stretch your mouth in four directions to form the shapes needed to make sounds as you follow your script, which consists of extremely normal and genuine human sentences.
For example, when your boss calls you into his office for a chat, it's perfectly normal and human to address him by saying "Greetings, mighty supreme leader." (Especially if you mess it up and your nose flips open like a car trunk, spraying motor oil everywhere.) And while talking to a coworker who is currently sitting on a toilet in a bathroom stall, just say: "Like many humans, Carl, I enjoy the process of excretion, but not at the expense of productivity." This is perfectly normal human communication. You're doing great!
Below (and here on YouTube (opens in new tab)) you can watch me go on a date. As you can see, it all went perfectly smoothly. Sure, I accidentally tried to order coolant instead of a human beverage, but I think I covered it naturally.
Between missions to reassure humans that you're totally human and not a robot, you'll earn tokens that can be spent to upgrade your robot body by yanking teeth from your head and replacing them with microchips. These upgrades increase your human-simulating abilities, but they also make speaking even more challenging. Adding a smile function gives you a reassuring facial movement, but also a new lever to pull with your mouse when you're told to raise or lower your smile. Adding eye movement means having to quickly swivel your eyes with the mouse as you make and break eye contact—the same mouse you've been using to control your mouth.
Pretty quickly your eyes (your real ones, I mean) will be darting around the screen to address all the facial movements you need to make and buttons you need to press. Make eye contact, but not for too long or it's weird. Did you look away? That's also weird! Look back, quickly! Smile! Now stop smiling. Smile again! Press buttons! Make mouth-shapes! This all has to happen pretty much at once to get through Speaking Simulator's long sentences. It's funny but it's not even remotely easy. Robots might be great at multitasking, but I sure ain't.
Luckily, the humans you're speaking to aren't that quick on the uptake. Their suspicions rise the worse you perform, but very, very slowly, giving you time to reassure them by performing your actions before the timer drains. As you bite your tongue, miss-press buttons in your mouth, and various bits of your face fall apart when you take ten or fifteen seconds to complete simple words like "Grant" or "Yes," they'll still be pretty convinced you're human.
Funny and gruesome as this all is, I can definitely relate to the robot in Speaking Simulator (whose name is Blank Textfield). As an extremely self-conscious, shy, anxiety-riddled person, I spend most of my face-to-face conversations struggling to get my thoughts out properly and imagining the other person's suspicion meter rising, too. (I don't worry they're suspicious I'm a robot, I worry they're suspicious I'm an idiot). Talking to people has always been tough for me, there's plenty of miscommunication between my brain and my mouth, and I basically spend most of my conversations wondering just how badly the conversation is going. I dread talking to people, and before a social event I've definitely found myself thinking, "Relax. You'll be fine. Just pretend you're a human being."
So, I definitely empathize with your struggles, Blank Textfield. Hopefully with some more practice, we'll be able to get through a single conversation without awkwardness, embarrassment, or with any of our face-parts falling off. If you're interested in taking Speaking Simulator's mouth for a drive, it comes out on Steam on January 30 (opens in new tab).
Christopher Livingston
Chris started playing PC games in the 1980s, started writing about them in the early 2000s, and (finally) started getting paid to write about them in the late 2000s. Following a few years as a regular freelancer, PC Gamer hired him in 2014, probably so he'd stop emailing them asking for more work. Chris has a love-hate relationship with survival games and an unhealthy fascination with the inner lives of NPCs. He's also a fan of offbeat simulation games, mods, and ignoring storylines in RPGs so he can make up his own.
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text:
Defendants facing sexual battery charges have a high likelihood of being treated unfairly by the criminal justice system and the general public. Many people including judges, prosecutors, friends and even family can assume a defendant is guilty before looking at the actual evidence. A sexual battery charge carries a stigma that can sometimes feel unsurmountable, but this could not be further from the truth. A skilled and experienced lawyer can fight back against the state to protect your freedom and restore your reputation. Benjamin Herbst is a South Florida criminal lawyer who specializes in sexual battery defense. He has successfully defended dozens of clients who have been accused of rape and sexual battery, and has achieved multiple dismissals and charge reductions. Benjamin represents clients in the Tri-County jurisdictions of Miami-Dade, Broward and Palm Beach County, and is also an experienced Martin and St. Lucie County criminal lawyer. He is standing by to offer free consultation at (954) 543-0305, and has flexible office hours at his Fort Lauderdale and West Palm locations. Benjamin is also willing to meet at a convenient location anywhere from South Miami to Port St. Lucie.
Under Florida law a sexual battery means that there was some sort of penetration. Touching or rubbing typically will not fall under this statute, and could be charged as a different sex offense. In addition to penetration, the other main element that the state must prove in any Florida sexual battery case is that the act was done without the consent of the victim. Consent can be a more complicated issue, as the two sides can have completely different recollections of whether consent existed. In addition, the issue of consent is almost never supported by physical evidence, and rather comes down to the word of the victim. In contrast, the state will often be able to present physical evidence that penetration occurred. The problem is that law enforcement and prosecutors have a history of succumbing to pressure from the victim, and often file charges when consent is in doubt. Benjamin will not let charges based on the victim's word stand up in court. He has extensive experience cross examining victims in court, and will fight tooth and nail to expose any inconsistencies in the victim's story.
The basic definition of an act done without consent is when the victim says no or physically resists and the defendant ignores this and proceeds anyway. Under Florida law though, the definition of consent is more layered. The state does not actually have to prove the victim resisted or told the defendant to stop, as evidence of submission under duress or coercion would satisfy the consent element of sexual battery. The state may also try to prove the act was done without consent by arguing that the victim was mentally incapacitated or physically helpless. Mental incapacitation means that the victim was temporarily incapable of consenting due to the influence of drugs or alcohol. In these cases, the victim can actually consent to the sexual act in the moment but then later say it was done without consent. The state can technically prosecute an individual on this theory even if he or she did not know the mental state of the victim, which seems incredibly unjust. In this scenario the state would have to prove by circumstantial evidence that the defendant should have known the mental state of the victim. Benjamin has defended clients under this scenario and will do whatever it takes to show his client could not have reasonably known the condition of the victim.
Physically helpless is another common theory that the state may argue in order to prove there was no consent in a sexual battery case. Under Florida law a physically helpless victim is one that is either asleep or otherwise physically unable to communicate unwillingness to participate in an act. The theory of physically helpless often comes down to the word of the victim, as these types of cases rarely have objective witnesses who can testify to the condition of the victim at the time of the act. The state will try to produce evidence of the victim's condition before and after the act, but Benjamin will put this theory under intense scrutiny. The bottom line is that the state must prove the victim was physically helpless at the time of the act, and not an hour before or after. Benjamin has successfully represented defendants charged with sexual battery on a physically helpless individual in Florida, and is standing by to offer a free consultation how he can defend you.
Sexual battery can range anywhere from a life felony if it was committed by the use of force, with a weapon or on a child, to a second-degree felony if there were no aggravating factors involved. Sexual battery on a physically helpless individual is a first-degree felony under Florida law, and is perhaps the most common charge due to the involvement of drugs and alcohol. It is also a charge that is commonly filed against a completely innocent individual who had no intention of committing a crime. Benjamin understands that the consequences involved with anyone facing a sexual battery charge, and will not only fight to keep his clients out of jail but also will preserve their future by fighting to have the charges dropped. All sexual battery charges carry mandatory sex offense registry upon conviction, and this could be a lifetime sentence that could affect a person for decades. Benjamin has the experience defending sexual battery cases in South Florida to be able to argue all available defenses, and he will never back down from the state at trial. If you have been charged or are under investigation for any sex offense call Benjamin today at (954) 543-0305.
Sexual Battery | South Florida Criminal Lawyer The Herbst Firm


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text:
The world was in a mess... but their hair was perfect!!
$6.4 mil
Genre(s): Comedy, Family, Music
Leslie Ann Powers
Ruth Brown
Michael St. Gerard
Clayton Prince
Rachel Talalay
Tracey Turnblatt is obsessed with getting on the local music show The Corny Collins Dance Show in 1960's Baltimore. When she finally does get on the show, she is an instant hit and becomes a regular dancer. Tracey decides to use her new found fame to push for integration on the show. This infuriates Amber, the former star of the show, and her racist parents. Things come to a head when the two young women both decide to compete in a beauty contest.
Release Date March 4, 2014
Release Date November 5, 2002
Driving Miss Daisy
Tin Men
imdb: 6.4 R