text:
\section{Introduction}
Let $H\subset \PSL{2}$ be a Fuchsian Schottky group. This means that there exist $\ell_1,\dots,\ell_{2g}$ geodesics in the hyperbolic plane $\mathbb{H}^2$ with disjoint closures, and a collection of generators $A_1,\dots,A_g\in H$ which identify these geodesics pairwise \cite{OuterCircles}. By a classical ping pong argument, $H$ is free on its generators and the quotient $\mathbb{H}^2/H$ is a hyperbolic surface with non-empty boundary. The homeomorphism type of the surface depends on the combinatorial type of the pairings. If we allow the geodesics to be asymptotic, we obtain the converse: every complete hyperbolic metric on the interior of a compact surface with boundary arises as an example of this construction.

Schottky groups have been generalized in multiple directions. One application of the basic ping pong argument which has seen frequent use was pioneered by Tits \cite{Pingpong}. He used what is now generally known as the Ping Pong Lemma as a tool in the proof of the famous Tits alternative. Assuming that every element in a symmetric generating set of a linear group has a unique largest eigenvalue and that they are in sufficiently generic position, he concludes that high powers of these elements generate a free group.\\
Related constructions have been used more recently to obtain nonabelian free groups acting on homogeneous spaces. Benoist used a quantitative version of transversality to define what he calls \emph{$\epsilon$-Schottky groups} acting on projective spaces \cite{BenPropre}. Using their powerful machinery of \emph{Morse actions}, Kapovich, Leeb and Porti define Schottky groups in the more general setting of semisimple Lie groups acting on symmetric spaces of noncompact type \cite{KLP}. Again, they consider a suitable notion of transversality for geodesics and show that groups generated by sufficiently high powers of axial isometries are free. In fact, they prove that such a group is Morse (or \emph{Anosov}).

A somewhat more geometric approach to generalizations of Fuchsian Schottky groups is to start out with a collection of disjoint subsets of a space $X$ and then find a set of generators pairing these subsets according to some prescribed combinatorics. Following this idea, Schottky groups were generalized to projective linear groups acting on $\mathbb{CP}^n$ (\cite{CompSchott},\cite{CompSchott2}), affine groups acting on $\mathbb{R}^3$ \cite{crookedplanes}, and the conformal Lorentzian group $\mathrm{SO}(3,2)$ acting on the Einstein Universe \cite{CrookedSurf}. We introduce a new generalization which is based on the notion of cyclic orders and follows the latter approach.

The moduli space of representations $\rho:\Gamma \rightarrow G$ of a discrete group $\Gamma$ is called the $G$-\emph{character variety} of $\Gamma$. The study of character varieties and their connections to geometric structures has received a lot of attention lately. To name one example which is relevant for the paper at hand, when $\Gamma$ is the fundamental group of a closed surface and $G$ is $\PSL{2}$, there is a discrete bounded invariant called the Euler number which distinguishes connected components of the character variety \cite{BillThesis}. In this case, the component where this invariant is maximal coincides with the Teichmüller space of the surface.

When the surface is not closed, $\Gamma$ is free and the invariant can still be defined through the use of bounded cohomology \cite{BIW}. The subset of the character variety where the Euler number is maximal is again exactly the set of representations which are holonomies of complete hyperbolic structures. The Toledo invariant generalizes this situation to representations into any Lie group $G$ of Hermitian type. The space of maximal representations in this general setting was studied by Burger, Iozzi, Wienhard, Strubel, Pozzetti (\cite{BIW},\cite{Strubel},\cite{BurgerPozzetti}). In particular, in \cite{BIW} it is shown that maximality is equivalent to the existence of an equivariant, left-continuous boundary map $\xi:\partial \mathbb{H}^2 \rightarrow \mathcal{S}$, where $\mathcal{S}$ is the Shilov boundary of the symmetric space associated to $G$.

The main theorem of this paper generalizes the Schottky group construction to all maximal representations into Lie groups of Hermitian type.

In order to define the notion of Schottky group in this generality, we use \emph{partial cyclic orders} \cite{PCO}. Several notions of ordering have been successfully applied to the study of representations spaces (\cite{OrderPreserving},\cite{calegari}, \cite{GuiPositivity},\cite{GuiPositivity2}). Partial cyclic orders generalize the cyclic order on the boundary of the hyperbolic plane. The Shilov boundary of a Hermitian symmetric space admits such a partial cyclic order, defined using the generalized Maslov index introduced in  \cite{Clerc}. In the Schottky groups we introduce, pairs of points in a partially cyclically ordered space play the role of endpoints of hyperbolic geodesics. The requirement for the geodesics to be disjoint is replaced by a cyclic order condition on these pairs of points, and hyperbolic isometries are replaced by order preserving transformations. This gives a new very explicit description of maximal representations in the case of surfaces with non-empty boundary.
\begin{thmn}[\ref{SchottkyMaximal}]
Let $\Gamma = \pi_1(\Sigma)$ be the fundamental group of a compact surface with boundary, $G$ a Hermitian Lie group, and $\rho:\Gamma\to G$ a representation. Then $\rho$ is maximal if and only if it admits a Schottky presentation.
\end{thmn}

As an application of this description, we identify a domain of discontinuity for generalized Schottky subgroups of $\Sp{2n}$ acting on $\RP{2n-1}$. The key idea, suggested to us by Anna Wienhard and Olivier Guichard, is to associate certain ``halfspaces'' in $\RP{2n-1}$ to pairs of points in the Shilov boundary $\Lag{2n}$ which are used in the definition of the Schottky group. This allows us to construct a fundamental domain and analyze its orbit according to the same general strategy that one follows in the classical case of a Fuchsian Schottky group in $\PSL{2}$ acting on $\RP{1}$.

\begin{thmn}[\ref{Thm:dod}]
Let $\rho:\Gamma\to\Sp{2n}$ be a generalized Schottky group modeled on an infinite area hyperbolic surface. Then it induces a cocompact properly discontinuous action on the complement of a Cantor set of projectivized Lagrangian $n$-planes in $\RP{2n-1}$. This action admits a compact fundamental domain in $\RP{2n-1}$ bounded by finitely many smooth algebraic hypersurfaces.
\end{thmn}

In the situation of this theorem, the representation $\rho$ is Anosov and the domain of discontinuity coincides with the ones described in \cite{AnosovReps} and \cite{KLP2}.

The paper is structured as follows: In section \ref{parcyc} we introduce the notion of a partial cyclic order and some topological conditions on partially cyclically ordered sets.

Section \ref{Sec:pingpong} describes the notion of generalized Schottky group and shows the existence of an equivariant left-continuous limit curve under some topological assumptions.

In section \ref{Sec:symspaces}, we show that Shilov boundaries satisfy all the required hypotheses from the previous section and prove the main theorem as a corollary.

In the last section, we focus on the real symplectic group $\Sp{2n}$. We use our description of maximal representations as Schottky groups to construct fundamental domains bounded by algebraic hypersurfaces in $\RP{2n-1}$.

We thank William Goldman, Anna Wienhard, Olivier Guichard, and Brice Lousteau for helpful discussions and suggestions. We are grateful to Florian Stecker for his comments on a preliminary version.
\section{Partial cyclic orders} \label{parcyc}
A partial cyclic order is a relation on triples which is analogous to a partial order, but generalizing a cyclic order instead of a linear order. The definition we use was introduced in 1982 by Nov{\'a}k \cite{PCO}.
\begin{Def}
A \emph{partial cyclic order} (PCO) on a set $C$ is a relation $\cycle{}{}{}$ on triples in $C$ satisfying, for any $a,b,c,d\in C$ :
\begin{itemize}
 \item if $\cycle{a}{b}{c}$, then $\cycle{b}{c}{a}$ (\emph{cyclicity}).
 \item if $\cycle{a}{b}{c}$, then not $\cycle{c}{b}{a}$ (\emph{asymmetry}).
 \item if $\cycle{a}{b}{c}$ and $\cycle{a}{c}{d}$, then $\cycle{a}{b}{d}$ (\emph{transitivity}).
\end{itemize}
If in addition the relation satisfies:
\begin{itemize}
 \item If $a,b,c$ are distinct, then either $\cycle{a}{b}{c}$ or $\cycle{c}{b}{a}$ (\emph{totality}),
\end{itemize}
 then it is called a \emph{total cyclic order}.
\end{Def}
Let $C,D$ be partially cyclically ordered sets.
\begin{Def}
A map $f : C \rightarrow D$ is called \emph{increasing} if $\cycle{a}{b}{c}$ implies $\cycle{f(a)}{f(b)}{f(c)}$.
An automorphism of a partial cyclic order is an increasing map $f:C\rightarrow C$ with an increasing inverse. We will denote by $G$ the group of all automorphisms of $C$.
\end{Def}
Any subset $X\subset C$ such that the restriction of the partial cyclic order is a total cyclic order on $X$ will be called a \emph{cycle}. We will also use the term cycle for (ordered) tuples $(x_1,\ldots,x_n) \in C^n$ if the cyclic order relations between the points in $C$ agree with the cyclic order given by the ordering of the tuple.
\begin{Def}
Let $a,b\in C$. The \emph{interval} between $a$ and $b$ is the set $(a,b):=\{x \in C ~|~ \cycle{a}{x}{b}\}$. The set of all intervals generates a natural topology on $C$ under which automorphisms of the partial cyclic order are homeomorphisms. We call this topology the \emph{interval topology} on $C$. We call $C$ \emph{first countable} when its interval topology is first countable.

The \emph{opposite} of an interval $I=(a,b)$ is the interval $(b,a)$, also denoted by $-I$.
\end{Def}
\begin{Ex}
The circle $S^1$ admits a total cyclic order. The relation on triples is : $\cycle{a}{b}{c}$ whenever $(a,b,c)$ are in counterclockwise order around the circle. The automorphism group of this cyclic order is the group of orientation preserving homeomorphisms of the circle.
\end{Ex}
\begin{Ex}
We can define a product cyclic order on the torus $S^1 \times S^1$. Define the relation to be $\cycle{x}{y}{z}$ whenever $\cycle{x_1}{y_1}{z_1}$ and $\cycle{x_2}{y_2}{z_2}$. This is not a total cyclic order. Some intervals in this cyclically ordered space are shown in figure \ref{fig:kthorderintervals}.
\end{Ex}
\begin{Ex}\label{Ex:linearOrder}
Every strict partial order $<$ on a set $X$ induces a partial cyclic order in the following way: define $\cycle{a}{b}{c}$ if and only if either $a<b<c$, $b<c<a$, or $c<a<b$. The cyclic permutation axiom is automatic and the two other axioms follow from the antisymmetry and transitivity axioms of a partial order.
\end{Ex}
The key topological property that we will need in the next section is a notion of completeness that we can associate to a space carrying a PCO.
\begin{Def}
A sequence $a_1,a_2,\dots \in C$ is \emph{increasing} if and only if $\cycle{a_i}{a_j}{a_k}$ whenever $i<j<k$.

Equivalently, the map $a:\mathbb{N} \rightarrow C$ defined by $a(i)=a_i$ is increasing, where the cyclic order on $\mathbb{N}$ is given by $\cycle{i}{j}{k}$ whenever $i<j<k$, $j<k<i$ or $k<i<j$ (as in Example \ref{Ex:linearOrder}).
\end{Def}
\begin{Def}
A partially cyclically ordered set $C$ is \emph{increasing-complete} if every increasing sequence converges to a unique limit in the interval topology.
\end{Def}
The following is a natural equivalence relation for increasing sequences.
\begin{Def}
Two increasing sequences $a_n$ and $b_m$ are called \emph{compatible} if they admit subsequences $a_{n_k}$ and $b_{m_l}$ making the combined sequence $a_{n_1},b_{m_1},a_{n_2},b_{m_2},\ldots$ increasing.
\end{Def}
\begin{Lem} \label{Lem:compatible_seq}
Let $C$ be an increasing-complete partially cyclically ordered set, and let $a_n$ and $b_m$ be compatible increasing sequences. Then their limits agree.
\begin{proof}
Any increasing sequence has a unique limit, and any subsequence of an increasing sequence therefore has the same unique limit.\\
The combined sequence (see the previous definition) is increasing, hence its unique limit must agree with the unique limits of both subsequences $a_{n_k}$ and $b_{m_l}$.
\end{proof}
\end{Lem}
To complete this list of definitions related to PCOs, we finish with two further restrictions on a set with a PCO which will be useful in section \ref{Sec:limitcurves}.
\begin{Def}
A partially cyclically ordered set $C$ is \emph{proper} if for any increasing quadruple $(a,b,c,d)\in C^4$, we have $\overline{(b,c)} \subset (a,d)$. Here, ``bar'' denotes the closure in the interval topology.
\end{Def}
\begin{Def}
Two points $a,b \in C$ in a partially cyclically ordered set $C$ are called \emph{comparable} if there exists a point $c\in C$ with either $\cycle abc$ or $\cycle acb$.
\end{Def}
\begin{Def}
A PCO set $C$ is \emph{full} if whenever $(a,b)$ is non-empty for some pair $a,b$, then $(b,a)$ is also non-empty. Equivalently, whenever $a,b$ are comparable then both intervals they bound are non-empty.
\end{Def}
\begin{Rem}
The motivation for the term ``full'' stems from the following construction. Assume we have a non-empty interval $(a,b)$. Then we can find a point $c\in(a,b)$ and another point $d\in(b,a)$. But then the point $d$ also lies in the interval $(c,a)$, so by fullness, there is a point inside $(a,c)$ as well. Continuing in this fashion, we can subdivide all resulting intervals further and further, and thereby construct a countably infinite subset $X\subset(a,b)$ with the following two properties: Firstly, $X$ is a cycle. Secondly, for any pair $x_1,x_2$ of distinct elements of $X$, the intersection $(x_1,x_2)\cap X$ is nonempty.
\end{Rem}
\section{Generalized Schottky groups}
\label{Sec:pingpong}
Throughout this section, $C$ denotes a partially cyclically ordered set and $G=\Aut(C)$.

\subsection{Definition of generalized Schottky group}
Let $\Sigma$ be the interior of a compact surface with boundary, of Euler characteristic $
\chi<0$. Then, the fundamental group $\pi_1(\Sigma)$ is free on $g=1-\chi$ generators. Let $\Gamma\subset \PSL{2}$ be the holonomy of a finite area hyperbolization of $\Sigma$. In this section, we construct free subgroups of $G$ using $\Gamma$ as a combinatorial model.

It is well known that there is a presentation for $\Gamma$ of the following form : $\Gamma$ is freely generated by $A_1,\dots,A_g \in \PSL{2}$ and there are $2g$ disjoint open intervals $I_1^+,\dots,I_g^+,I_1^-,\dots,I_g^-\subset S^1$ such that $A_j(-I_j^-)=I_j^+$. Moreover, we have that $\bigcup\limits_i \overline{I_i^+} \cup \bigcup\limits_i \overline{I_i^-} = S^1$ (figure \ref{fig:Pairings}).

\begin{figure}
    \centering
    \includegraphics[width=.6\textwidth]{pairings.pdf}
    \caption{A combinatorial model for the once punctured torus.}
    \label{fig:Pairings}
\end{figure}

The cyclic ordering on $S^1$ gives a cyclic ordering to the intervals in the definition.

We call a $k$-th order interval the image of any $I_j^+$(respectively $I_j^-$) by a reduced word of length $k-1$ not ending in $A_j^{-1}$(respectively $A_j$). There are exactly $(2g)(2g-1)^{k-1}$ $k$-th order intervals. There is a natural bijection between words of length $k$ and $k$-th order intervals. We use this bijection to index $k$-th order intervals : $I_W$ is the interval corresponding to the word $W$. For any fixed $k$, the $k$-th order intervals are all pairwise disjoint, and so they are cyclically ordered. This induces a cyclic ordering on words of length $k$ in $\Gamma$. The union of all closures of $k$-th order intervals is all of $S^1$.

The following easy Lemma, which is a reformulation of transivity, motivates our definition of generalized Schottky groups in $G$.
\begin{Lem}
Let $(a,b,c)\in C^3$ be a cycle. Then we have $(b,c) \subset (b,a)$. In particular, the intervals $(a,b)$ and $(b,c)$ are disjoint.
\begin{proof}
Let $x\in(b,c)$, so we have $\cycle bxc$. By transitivity, together with $\cycle bca$, this implies $\cycle bxa$.
\end{proof}
\end{Lem}
We now define generalized Schottky subgroups of $G$ by asking for a setup of intervals similar to the $\PSL{2}$ case and requiring generators to pair them the same way.
\begin{Def}
Let $\xi_0$ be an increasing map from the set of endpoints of the intervals $I_1^+,\dots,I_g^+$, $I_1^-,\dots,I_g^-$ into $C$. For $I_i^\pm=(a_i^\pm,b_i^\pm)$, define the corresponding interval $J_i^\pm=(\xi_0(a_i^\pm),\xi_0(b_i^\pm))\subset C$. Next, assume there exist $h_1,\dots,h_g\in G$ which pair the endpoints of $J_i^\pm$ in the same way that the $g_i$ pair the $I^\pm_i$, so that $h_i(-J_i^-)=J_i^+$. We call the image of the induced map $\Gamma \to G$ sending $A_i$ to $h_i$ a \emph{generalized Schottky group}, and the intervals $J_i^\pm$ used to define it a set of \emph{Schottky intervals} for this group.
\end{Def}
\begin{Rem}\
\begin{enumerate}
    \item A generalized Schottky group will in general have many possible choices of a set of Schottky intervals. We will only use this term when a specific choice of both generators and intervals corresponding to these generators is fixed.
    \item Since the cyclic ordering is a property of $\RP{1}$ which is not shared by $\mathbb{CP}^1$, the Schottky groups defined here do \emph{not} generalize the more well known $\mathbb{CP}^1$ Kleinian case.
    \item The requirement that the combinatorial model be a finite-volume hyperbolization is artificial. It is helpful in order to avoid having to separate our analysis into several cases. We could use a model where the intervals have disjoint closures and the construction would work in the same way. In section \ref{sec:FundamentalDomains} we will use presentations of this type to describe domains of discontinuity in $\RP{2n-1}$.
    \item Our use of the term "Schottky" differs slightly from most references in that we allow for the closures of the ping pong subsets to intersect. This is sometimes called "Schottky-type".
\end{enumerate}
\end{Rem}

With this setup, we can define $k$-th order intervals in $C$ in the same way as above but starting with the intervals $J_i^\pm$ and their images under words in the $h_i$ (see figure \ref{fig:kthorderintervals}). As above, denote by $J_W$ the interval corresponding to $W$. Note that since $\xi_0$ is increasing, the $k$-th order intervals in $C$ are also cyclically ordered, where the ordering is the same as the ordering of the corresponding intervals in $S^1$.
\begin{figure}
    \centering
    \includegraphics[width=.6\textwidth]{kthorderintervals.pdf}
    \caption{Some first, second and third order intervals for a generalized Schottky group acting on $S^1 \times S^1$.}
    \label{fig:kthorderintervals}
\end{figure}
\begin{Prop}
The group generated by $h_1\dots h_g$ is free on those generators.
\begin{proof}
Define $J_i = J_i^+ \cup J_i^-$. Note that $J_i\cap J_j = \emptyset$ whenever $i\neq j$. Moreover, for any $n\neq 0$, $h_i^n (J_j) \subset J_i$ and so the proposition follows from the ping pong lemma.
\end{proof}
\end{Prop}

The endpoints of $k$-th order intervals in $C$ satisfy the same cyclic order relations as the corresponding endpoints in $S^1$, and we can extend $\xi_0$ to an increasing equivariant map defined on the countable dense set of all endpoints of Schottky intervals in $S^1$. We denote this set by $S^1_\Gamma$.
\subsection{Limit curves}\label{Sec:limitcurves}
\begin{Lem} \label{Lem:closures_pos}
Let $C$ be a partially cyclically ordered set which is full and proper, and $(x_1,\ldots,x_6) \in C^6$ a cycle. Let $I_1 = (x_1,x_2), I_2=(x_3,x_4),I_3=(x_5,x_6)$ and $a_i \in \overline{I_i}$ be arbitrary points in the closures of the intervals. Then we have $\cycle {a_1}{a_2}{a_3}$.
\begin{proof}
Since $C$ is full, we can choose auxiliary points $y_i$ such that the $12$-tuple $(y_1,x_1,x_2,y_2,y_3,x_3,x_4,y_4,y_5,x_5,x_6,y_6)$ is a cycle. This allows us to conclude that $\overline{(x_i,x_{i+1})} \subset (y_{i},y_{i+1})$ for odd $i$ as $C$ is proper. Since $(y_1,\ldots,y_6)$ is a cycle, transitivity implies the lemma.
\end{proof}
\end{Lem}
\begin{Lem}
\label{Lem:nested_sequences}
Let $P_n \rightarrow P$ be an increasing sequence in a proper, increasing complete, PCO set $C$. Assume $Q_n$ is another sequence with $Q_n\in \overline{(P_n,P_{n+1})}$ for all $n$. Then $Q_n$ converges to $P$ and is $3$-increasing in the following sense: whenever $i+2<j<k-2$, we have $\cycle{Q_i}{Q_j}{Q_k}$.
\begin{proof}
For every $n\geq 2$, $Q_n\in (P_{n-1},P_{n+2})$ by properness, which already implies that $Q_n$ is $3$-increasing. Now, consider the following sequence: \[P_1,Q_2,P_4,Q_5,\dots,P_{3n+1},Q_{3n+2},\dots\]
It is increasing, and admits a subsequence which is also a subsequence of $P_n$. Since increasing sequences have unique limits, this sequence must converge to $P$. The increasing subsequence $Q_{3n+2}$ therefore converges to $P$. Using the same argument, we see that $Q_{3n+1}$ and $Q_{3n}$ also converge to $P$, so in fact the sequence $Q_n$ converges to $P$.
\end{proof}
\end{Lem}
We now come to the main theorem of this section, which explains how to construct a boundary map for generalized Schottky groups, under some topological assumptions.\\
The boundary map we construct will be left-continuous as a map from $S^1$ to some first countable topological space $C$. To avoid confusion, let us fix the definition here: In a small neighborhood $U$ of a point $a\in S^1$, the cyclic order induces a linear order. A sequence $a_n \in U$ converges to $a$ from the left if $a_n<a$ and $a_n\xrightarrow{n\to\infty}a$. The function $f$ is left-continuous at $a$ if $f(a_n)\to f(a)$ for all sequences $a_n$ converging to $a$ from the left.\\
It is worth noting that to check for left-continuity at a point $a$, it is in fact sufficient to check the convergence of $f(a_n)$ for increasing sequences $a_n$ converging to $a$ from the left. The reason is the following: Assume $a_n$ is a sequence converging to $a$ from the left such that $f(a_n)$ does not converge to $f(a)$. Then it has a subsequence such that $f(a_{n_k})$ stays bounded away from $f(a)$. But since $a_{n_k} \to a$ from the left, we can pick a further subsequence which is increasing and still a counterexample to left-continuity.
\begin{Thm}    \label{Thm:bdry_map_gen_Schottky}
Let $\rho: \Gamma \to G$ be the map defining a generalized Schottky group. Assume that $C$ is first countable, increasing-complete, full and proper. Then there is a left-continuous, equivariant, increasing boundary map $\xi: S^1 \to C$.
\begin{proof}
We construct the map $\xi$ as follows: Recall that $S^1_\Gamma \subset S^1$ denotes the domain of $\xi_0$ and is a dense subset. For $x\in S^1$, pick any increasing sequence $x_n \in S^1_\Gamma$ converging to $x$ and set
\[ \xi(x) = \lim_{n\to\infty} \xi_0(x_n). \]
First of all, let us show that this value is well-defined. Since $x_n$ is an increasing sequence, the increasing map $\xi_0$ maps it to an increasing sequence in $C$ which therefore has a unique limit. Furthermore, this limit does not depend on the choice of $x_n$: Let $y_m$ be another increasing sequence converging to $x$. Then the two sequences $\xi_0(x_n)$ and $\xi_0(y_m)$ are compatible, so they have the same limit by Lemma \ref{Lem:compatible_seq}.

We then verify that $\xi$ is equivariant. Let $x\in S^1$, $\gamma\in\Gamma$, and $x_n\to x$ an increasing sequence, so we have $\xi(x) = \lim\limits_{n\to\infty} \xi_0(x_n)$. Then $\gamma(x_n)$ is an increasing sequence converging to $\gamma(x)$, so by continuity of $\rho(\gamma)$ and equivariance of $\xi_0$, we have the following equalities:
\[ \rho(\gamma)(\xi(x)) = \lim_{n\to\infty} \rho(\gamma)(\xi_0(x_n)) = \lim_{n\to\infty} \xi_0(\gamma(x_n)) = \xi(\gamma(x)). \]

Next, we show that it is left-continuous. Assume $x_n \in S^1$ is a sequence converging to $x$ from the left. As explained above, without loss of generality we can take $x_n$ to be an increasing sequence. We pick points $y_n \in S^1_\Gamma$ such that $y_n \in (x_{n-1},x_n)$. Then $y_n$ is increasing and we have $x_n\in (y_n,y_{n+1})$. Furthermore, $y_n$ also converges to $x$, hence
\begin{equation}    \label{bdry_map_eq1}
 \xi(x) = \lim \xi_0(y_n).
\end{equation}
Now, for each $n$, let $\{a_k(n)\}_{k\in\mathbb{N}}\subset S^1_\Gamma$ be an increasing sequence converging to $x_n$, so
\begin{equation}    \label{bdry_map_eq2}
 \xi(x_n) = \lim_{k\to\infty} \xi_0(a_k(n)).
\end{equation}
Then $a_k(n) \in (y_n,y_{n+1})$ for large $k$, so
\begin{equation}    \label{bdry_map_eq3}
 \lim\limits_{k\to\infty} \xi_0(a_k(n)) \in \overline{(\xi_0(y_n),\xi_0(y_{n+1}))}
\end{equation}
because $\xi_0$ is increasing. Now Lemma \ref{Lem:nested_sequences} applies and, combined with \eqref{bdry_map_eq1}, \eqref{bdry_map_eq2} and \eqref{bdry_map_eq3}, tells us that $\xi(x_n)$ converges to $\xi(x)$.

The final property we need to check is that $\xi$ is increasing. Assume that we have $\cycle xyz$ for points $x,y,z \in S^1$. By density of $S^1_\Gamma$, we can find a cycle $(a_1,a_2,b_1,b_2,c_1,c_2) \in (S^1_\Gamma)^6$ such that $x\in (a_1,a_2), \ y\in (b_1,b_2), \ z\in (c_1,c_2)$. As in the proof of left-continuity, this implies that $\xi(x) \in \overline{(\xi_0(a_1),\xi_0(a_2))}$, and similar for the other two points. Using Lemma \ref{Lem:closures_pos}, we conclude $\cycle{\xi(x)}{\xi(y)}{\xi(z)}$.
\end{proof}
\end{Thm}
The very general construction described in this section applies to many examples. For instance, various notions of positivity in homogeneous spaces give rise to partial cyclic orders. More specifically, the Shilov boundary of Hermitian symmetric spaces admits a PCO satisfying all the above properties, and the rest of the paper is dedicated to this example. It is also possible, using techniques similar to Fock-Goncharov total positivity \cite{FocPositivity}, to construct a PCO on spaces of complete oriented flags. In this last case, for flags in $\mathbb{R}^3$, Schottky groups can be used to describe convex projective structures on surfaces with boundary \cite{NextPaper}.
\section{Hermitian symmetric spaces of tube type}\label{Sec:symspaces}
In this section, we show that the Shilov boundary of a Hermitian symmetric space of tube type $X$ admits a partial cyclic order invariant under the holomorphic isometry group of $X$. Moreover, we prove that Shilov boundaries satisfy the topological assumptions from theorem \ref{Thm:bdry_map_gen_Schottky}, so we have a boundary map for every generalized Schottky subgroup. Then, using the machinery of section \ref{Sec:pingpong}, we show that Schottky subgroups in this case correspond to maximal representations.

\subsection{The partial cyclic order on the Shilov boundary}
Consider a tube type domain $X=V+i\Omega$, where $V$ is a real vector space and $\Omega\subset V$ is a symmetric cone. Let $G$ be the group of holomorphic isometries of $X$.
\begin{Def} The \emph{Shilov boundary} $\mathcal{S}$ of $X$ is the unique closed $G$-orbit in the visual boundary of $X$. It can be identified with a subset of the topological boundary of the bounded domain realization of $X$. The action of $G$ extends smoothly to $\mathcal{S}$.
\end{Def}
The vector space $V$ admits a Jordan algebra structure associated to the symmetric cone $\Omega$ \cite{farautkoranyi}.
\begin{Def} The partial order $<_\Omega$ on $V$ is given by $x<_\Omega y$ if and only if $y-x\in\Omega$. Equivalently, $y-x = z^2$ for some $z\in V$.\end{Def}
\begin{Prop}[\cite{farautkoranyi}, Proposition X.2.3]
The \emph{Cayley Transform} $p:V \rightarrow V_\mathbb{C}$ defined by
\[p(v) = (v-i e)(v+i e)^{-1},\]
where $e$ is the identity of the Jordan algebra, maps the vector space $V$ into the Shilov boundary $\mathcal{S}$ and $\overline{p(V)}=\mathcal{S}$.
\end{Prop}
Using the notion of transversality, we can make the statement more precise.
\begin{Def}
Two points $x,y\in \mathcal{S}$ are called \emph{transverse} if the pair $(x,y)\in \mathcal{S}\times \mathcal{S}$ belongs to the unique open $G$-orbit for the diagonal action.
\end{Def}
The image of the Cayley transform is exactly the set of points $x\in \mathcal{S}$ which are transverse to a fixed point which we denote by $\infty$. \cite[Section 6.6.1]{AnnaThesis}

The next object we need to define is the generalized Maslov index. This index is a function on ordered triples of points in $\mathcal{S}$, invariant under $G$. It will be used in order to define a partial cyclic order on $\mathcal{S}$, extending the partial cyclic order induced by $<_\Omega$ on $p(V)\subset \mathcal{S}$.

The generalized Maslov index is defined in \cite{Clerc} using the notion of $\Gamma$-radial convergence. For our purposes we will use the following equivalent definition, given in the same paper.
\begin{Def}
Let $x,y,z\in \mathcal{S}$. Applying an element of $G$, we may assume $x,y,z\in p(V)$. Let $v_x,v_y,v_z\in V$ be the vectors which map respectively to $x,y,z$ under the Cayley transform $p$. Then, the \emph{generalized Maslov index} of $x,y,z$ is the integer
\[\mathsf{M}(x,y,z) := \mathsf{k}(v_y-v_x) + \mathsf{k}(v_z-v_y) + \mathsf{k}(v_x-v_z)\]
where $\mathsf{k}(v)$ is the difference between the number of positive eigenvalues of $v$ and the number of negative eigenvalues of $v$ in its spectral decomposition.

When $x,y$ are transverse to $z$, equivalently, we can map $z$ to $\infty$ using an element of $G$ and define
\[\mathsf{M}(x,y,\infty) = \mathsf{k}(v_y-v_x)\]
\end{Def}
\begin{Prop}
The Maslov index enjoys the following properties:
\begin{itemize}
\item $G$-invariance : $\mathsf{M}(gx,gy,gz) = \mathsf{M}(x,y,z)$.
\item Skew-symmetricity : $\mathsf{M}(x_1,x_2,x_3) = \sgn(\sigma)\mathsf{M}(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)}).$
\item Cocycle identity : $\mathsf{M}(y,z,w) - \mathsf{M}(x,z,w) + \mathsf{M}(x,y,w) - \mathsf{M}(x,y,z) = 0$.
\item Boundedness : $|\mathsf{M}(x,y,z)| \leq \mathrm{rk}(X)$
\end{itemize}
\end{Prop}
These properties allow us to define a partial cyclic order on the Shilov boundary.
\begin{Prop}
The relation $\cycle{x}{y}{z}$ if and only if $\mathsf{M}(x,y,z)=\mathrm{rk}(X)$ defines a $G$-invariant partial cyclic order on $\mathcal{S}$.
\begin{proof}
Since $M$ is skew-symmetric, the relation automatically satisfies the first two axioms of a partial cyclic order. To prove the third axiom, assume $\mathsf{M}(x,y,z)=\mathsf{M}(x,z,w)=\mathrm{rk}(X)$. By the cocycle identity,
\[\mathsf{M}(y,z,w) - \mathsf{M}(x,z,w) + \mathsf{M}(x,y,w) - \mathsf{M}(x,y,z) = 0\]
and so
\[\mathsf{M}(y,z,w) + \mathsf{M}(x,y,w) = 2rk(X)\]
which is only possible if $\mathsf{M}(Q,R,S)=\mathsf{M}(P,Q,S)=\mathrm{rk}(X)$.
\end{proof}
\end{Prop}
The partial cyclic order $\cycle{}{}{}$ is closely related with the causal structure on $\mathcal{S}$ introduced by Kaneyuki \cite{KanShilov}. Namely, whenever $\cycle{x}{y}{z}$, there is a future-oriented timelike curve going through $x,y,z$ in that order. Informally, $y$ is in the intersection of the future of $x$ and the past of $z$. The following two lemmas describe some immediate properties of cyclically ordered triples.
\begin{Lem}[\cite{AnnaThesis}, Lemma 5.5.4] \label{Lem:PosTriplesTransverse}
Let $x,y,z \in \mathcal{S}$ with $\cycle xyz$. Then $x,y,z$ are pairwise transverse.
\end{Lem}
\begin{Lem} \label{Lem:Intervals_V}
Assume $x,y\in V$. Then, $\cycle{x}{y}{\infty}$ if and only if $x<_\Omega y$.
\begin{proof}
The cone $\Omega$ coincides with the region where $\mathsf{k}(v)=\mathrm{rk}(X)$.
\end{proof}
\end{Lem}
\begin{Rem}
The interval topology on $\mathcal{S}$ is the same as the usual manifold topology.
\end{Rem}
\begin{Prop} \label{Thm:Shilov_complete}
The PCO defined by $\cycle{}{}{}$ on $\mathcal{S}$ is increasing-complete, full and proper.
\begin{proof}
We first show that it is increasing-complete. Let $x_1,x_2,\dots$ be an increasing sequence in $\mathcal{S}$. Let $g\in G$ be such that $gx_2 = \infty$. Then, since we have $\cycle{x_k}{x_{k+1}}{x_2}$ for all $k\geq 3$, the sequence $gx_3,gx_4,\dots$ is an increasing sequence transverse to $\infty$. Hence, there exist $v_3,v_4,\ldots \in V$ with $p(v_k)=gx_k$.

This new sequence is increasing with respect to $<_\Omega$. Moreover, it is bounded since we have $\cycle{gx_k}{gx_1}{gx_2}$ for all $k>2$, so $v_k<_\Omega v_1$ where $p(v_1)=gx_1$. The tail of the sequence is contained in $\overline{(v_3,v_1)}$ which is compact, so it has an accumulation point. If $w,w'$ are two accumulations points of the sequence, let $w_k$, $w'_k$ be subsequences converging respectively to each of them. Passing to subsequences if necessary, we can arrange so that $w_k<_\Omega w'_k$ for all $k$, and so $w'_k-w_k\in \Omega$. This implies $w'-w\in \overline{\Omega}$, and by the same argument we can also show $w-w'\in \overline{\Omega}$. Since $\Omega$ is a proper convex cone (in the sense of \cite{farautkoranyi}), its closure does not contain any opposite pairs, so $w=w'$.

Now we turn to fullness of the PCO. Whenever an interval $(x,y)$ is nonempty, its endpoints have to be transverse by Lemma \ref{Lem:PosTriplesTransverse}. We can therefore apply an element of $G$ to map $x$ to $\infty$ and $y$ inside $p(V)$. Then Lemma \ref{Lem:Intervals_V} shows that the interval $(y,x)$ is also nonempty.

Finally, we show that the PCO is proper. Let $(x_1,x_2,x_3,x_4) \in \mathcal{S}^4$ be a cycle. Using an element of $G$, we can assume that $x_4$ is $\infty$, so that $x_1,x_2,x_3 \in p(V)$. Let $v_i \in V$ be the vector such that $p(v_i) = x_i$ for $i=1,2,3$. Now the cyclic relations $\cycle{x_1}{x_2}{\infty}$ and $\cycle{x_2}{x_3}{\infty}$ imply that both $v_2-v_1$ and $v_3-v_2$ lie in the cone $\Omega$. The interval $(x_2,x_3)$ is therefore given by $p\left( (v_2 + \Omega) \cap (v_3 - \Omega)\right)$. This implies the claim since $(v_2 + \Omega) \cap (v_3 - \Omega)$ is a relatively compact set in $V$ whose closure is contained in $v_1 + \Omega$, which is mapped onto $(x_1,\infty)$ by $p$.
\end{proof}
\end{Prop}
\subsection{Maximal representations}
In the previous section we defined a PCO on the Shilov boundary $\mathcal{S}$ of a Hermitian symmetric space of tube type on which the group of isometries $G$ acts by order-preserving diffeomorphisms. We recall that this action is transitive on transverse pairs. The Schottky construction described in section \ref{Sec:pingpong} therefore gives maps $\rho:\Gamma\rightarrow G$ where $\Gamma$ is the fundamental group of a surface with boundary. Maximal representations, defined and studied in \cite{BIW}, are a class of geometrically interesting representations and we will show in this section that they correspond to Schottky subgroups.

The key is the following characterization from \cite{BIW} (Theorem 8) :
\begin{Thm}
Let $h:\Gamma\rightarrow \PSL{2}$ be a complete finite area hyperbolization of the interior of $\Sigma$ and $\rho:\Gamma\rightarrow G$ a representation into a group of Hermitian type. Then $\rho$ is maximal if and only if there exists a left continuous, equivariant, increasing map
\[\xi : S^1 \rightarrow \mathcal{S}\]
where $\mathcal{S}$ is the Shilov boundary of the bounded symmetric domain associated to $G$.
\end{Thm}
Using this characterization and our earlier construction of a boundary map for generalized Schottky representations, we see that the two notions agree:
\begin{Thm}\label{SchottkyMaximal}
The representation $\rho:\Gamma\rightarrow G$ is maximal if and only if it admits a Schottky presentation.
\begin{proof}
Assume $\rho$ is Schottky. Proposition \ref{Thm:Shilov_complete} states that all the prerequisites of Theorem \ref{Thm:bdry_map_gen_Schottky} are fulfilled. Therefore, there exists a boundary map $\xi$ satisfying the conditions of the characterization above, so $\rho$ is maximal.

Conversely, if $\rho$ is maximal, then we have such a map $\xi$. Choosing a Schottky presentation for the hyperbolisation $h$, we get a Schottky presentation for $\rho$ by using the intervals $(\xi(a),\xi(b))$ where $(a,b)$ is some Schottky interval in the presentation for $h$. Equivariance and positivity of $\xi$ ensure that these intervals fit our definition of generalized Schottky groups.
\end{proof}
\end{Thm}
Theorem \ref{SchottkyMaximal}, as stated, assumes that $G$ is of tube type. However, this assumption is not necessary. This is because of the following observations. Let $X$ be a Hermitian symmetric space, and $\mathcal{S}$ its Shilov boundary. Then, in the same way as for tube type, the generalized Maslov index defines a partial cyclic order on $\mathcal{S}$. Let $x,y\in \mathcal{S}$ be transverse. Then, $x,y$ are contained in the Shilov boundary of a unique maximal tube type subdomain of $X$ \cite[Lemma 4.4.2]{AnnaThesis}. Moreover, this is also true of any increasing triple in $\mathcal{S}$ \cite[Proposition 5.1.4]{AnnaThesis}. This means that any increasing subset of $\mathcal{S}$ is contained in the Shilov boundary of a tube type subdomain, and so the proofs of this section generalize to arbitrary Hermitian symmetric spaces.
\section{Schottky groups in $\Sp{2n}$}

In this section, we consider the symplectic group $\Sp{2n}$, acting on $\mathbb{R}^{2n}$ equipped with a symplectic form $\omega$, and describe the construction of Schottky groups in detail.
\subsection{The Maslov index in $\Sp{2n}$}
\begin{Def}
Let $P,Q$ be transverse Lagrangians in $\mathbb{R}^{2n}$. We associate to them an antisymplectic involution $\Inv{P}{Q}$ defined using the splitting $\mathbb{R}^{2n}=P\oplus Q$:
\begin{align*} \Inv{P}{Q}: P\oplus Q & \to P\oplus Q \\
 (v,w) & \mapsto (-v,w)
 \end{align*}
We call this antisymplectic involution the \emph{reflection} in the pair $P,Q$. This generalizes the projective reflection in $\RP{1}$.

We will sometimes abuse notation and use $\Inv{P}{Q}$ to denote the induced transformation on Grassmannians.
\end{Def}
Using this involution, we associate a symmetric bilinear form to the pair $P,Q$ :
\begin{Def}
\[\B{P}{Q}(v,w) := \omega(v, \Inv{P}{Q}(w))\]
\end{Def}
This bilinear form is nondegenerate and has signature $(n,n)$.
\begin{Def} \label{Def:MaslovIndex}
Let $P,Q, R$ be pairwise transverse Lagrangians in $\mathbb{R}^{2n}$. The \emph{Maslov index} of the triple $(P,Q,R)$ is the index of the restriction of $\B{P}{R}$ to $Q$. We denote it by $\mathsf{M}(P,Q,R)$.
\end{Def}

Since $\Lag{2n}$ is the Shilov boundary for the bounded domain realization of the symmetric space of $\Sp{2n}$, it is an example of the general construction in section \ref{Sec:symspaces}. In fact, the Maslov index we just defined agrees with the more general version that we introduced before. Hence, the relation defined by $\cycle{P}{Q}{R}$ whenever $\mathsf{M}(P,Q,R)=n$ is a partial cyclic order on $\Lag{2n}$, enabling us to apply the constructions and results from section $\ref{Sec:pingpong}$.\\
We also remark that the definition makes sense for any isotropic subspace $Q$, not only the maximal isotropic ones.

The following property of the Maslov index is well-known.

\begin{Prop}
    The Maslov index classifies orbits of triples of pairwise transverse Lagrangians, i.e. the map
    \[ (P,Q,R) \mapsto \mathsf{M}(P,Q,R) \] induces a bijection from orbits of pairwise transverse Lagrangians under $\Sp{2n}$ to the set $\{ -n,-n+2,\ldots,n\}$.
\end{Prop}
The Maslov index and the reflection in a pair of Lagrangians are related in the following way:
\begin{Prop}
\[\mathsf{M}(P,\Inv{P}{Q}(V),Q)=-\mathsf{M}(P,V,Q).\]
\begin{proof}
\[\B{P}{Q}\big(\Inv{P}{Q}(u),\Inv{P}{Q}(v)\big)=\omega(\Inv{P}{Q}(u),v)=-\omega(u,\Inv{P}{Q}(v))=-\B{P}{Q}(u,v). \qedhere  \]
\end{proof}
\end{Prop}
 The proposition above means that reflections reverse the partial cyclic order.
\subsection{Fundamental domains}
\label{sec:FundamentalDomains}
In the special case of $\Sp{2n}$, the Schottky groups we obtain admit nice fundamental domains for their action on $\RP{2n-1}$. The domain of discontinuity which is the orbit of this fundamental domain is in general hard to describe, but it simplifies in some cases.\\
We will proceed as follows: First, we associate a ``halfspace'' in $\RP{2n-1}$ to each interval in $\Lag{2n}$ and explain how to construct the fundamental domain. Then we cover some preliminaries which will allow us to explicitly identify the domain of discontinuity for generalized Schottky groups modeled on an infinite volume hyperbolization. More specifically, we explain how to identify an interval with the symmetric space associated with $\GL{n}$ and how to use a contraction property from \cite{Contraction} for maps sending one interval into another.

\subsubsection{Positive halfspaces and fundamental domains}\label{sec:pos_halfspaces}
\begin{Def}
Let $P,Q$ be an ordered pair of transverse Lagrangians. We define the $\emph{positive halfspace}$ $\PH{P}{Q}$ as the subset
\[\PH{P}{Q}:=\{ \ell \in \RP{2n-1} ~|~ \B{P}{Q}|_{\ell\times\ell} > 0 \}.\]
It is the set of positive lines for the form $\B{P}{Q}$.
\end{Def}
The positive halfspace $\PH{P}{Q}$ is bounded by the conic defined by $\B{P}{Q}=0$. This type of bounding hypersurface was introduced by Guichard and Wienhard in order to describe Anosov representations of closed surfaces into $\Sp{2n}$. They are also the boundaries of $\mathbb{R}$-tubes defined in \cite{BurgerPozzetti}. A symplectic linear transformation $T\in\Sp{2n}$ acts on positive halfspaces in the following way : $T\PH{P}{Q} = \PH{TP}{TQ}$.

\begin{Prop}
Let $P,Q$ be an ordered pair of Lagrangians. Then,
\[\PH{Q}{P}=\overline{\PH{P}{Q}}^C = \Inv{P}{Q}(\PH{P}{Q})\]
\begin{proof}
For the first equality,
\[\B{Q}{P}(v,w)=\omega(v,\Inv{Q}{P}(w))=\omega(v,-\Inv{P}{Q}(w))=-\B{P}{Q}(v,w).\]
For the second equality, notice that $\B{P}{Q}(\Inv{P}{Q}(v),\Inv{P}{Q}(w)) = -\B{P}{Q}(v,w)$.
\end{proof}
\end{Prop}
\begin{Prop}
\label{Prop:proj}
A positive halfspace is the projectivisation of an interval, that is,
\[\PH{P}{Q}=\bigcup_{L\in (P,Q)} \mathbb{P}(L)\]
\begin{proof}
If $\ell \subset L$ for some $L\in(P,Q)$, then
\[\B{P}{Q}|_{\ell\times\ell} > 0\]
and so $\ell \in \PH{P}{Q}$.

Conversely, if $\ell \in \PH{P}{Q}$, then we wish to find a Lagrangian $L\supset \ell$ with $\mathsf{M}(P,L,Q)=n$. Consider the subspace $V=\langle\ell,\Inv{P}{Q}(\ell)\rangle$. The form $\B{P}{Q}$ has signature $(1,1)$ on that subspace, and so its orthogonal has signature $(n-1,n-1)$. Moreover, the form $\omega$ is nondegenerate on $V$ so $V^{\perp_\omega}$ is a symplectic subspace. Notice that
\[V^{\perp_\mathcal{B}}=\langle\ell,\Inv{P}{Q}(\ell)\rangle^{\perp_\mathcal{B}} = \ell^{\perp_\mathcal{B}} \cap (\Inv{P}{Q}(\ell))^{\perp_\mathcal{B}}= \ell^{\perp_\mathcal{B}} \cap \ell^{\perp_\omega}=V^{\perp_\omega}.\]
So we can pick a positive definite Lagrangian $L'\subset V^\perp$, which will be orthogonal to $\ell$ for both $\omega$ and $\B{P}{Q}$, so $L=\langle L',\ell \rangle$ is a positive definite Lagrangian containing $\ell$.
\end{proof}
\end{Prop}
\begin{Lem}
\label{Lem:Disjointness}
If $(P,Q,R,S)$ is a cycle in $\Lag{2n}$ and $V\in\Lag{2n}$ such that $\mathsf{M}(P,V,Q)=n$, then $\mathsf{M}(R,V,S)=-n$.
\begin{proof}
Using the cocycle relation,
\[\mathsf{M}(V,Q,R)-\mathsf{M}(P,Q,R)+\mathsf{M}(P,V,R)-\mathsf{M}(P,V,Q)=0\]
so \[\mathsf{M}(V,Q,R)+\mathsf{M}(P,V,R)=2n\]
which implies that $\mathsf{M}(V,Q,R)=\mathsf{M}(P,V,R)=n$.

Similarly,
\[\mathsf{M}(V,R,S)-\mathsf{M}(P,R,S)+\mathsf{M}(P,V,S)-\mathsf{M}(P,V,R)=0\]
so
\[\mathsf{M}(V,R,S)+\mathsf{M}(P,V,S)=2n\]
which means that $\mathsf{M}(V,R,S)=n$ and so $\mathsf{M}(R,V,S)=-n$.
\end{proof}
\end{Lem}
Now we can prove the disjointness criterion for positive halfspaces.
\begin{Prop}
\label{Prop:Disjointness}
If $(P,Q,R,S)$ is a cycle in $\Lag{2n}$, then $\PH{P}{Q}$ is disjoint from $\PH{R}{S}$.
\begin{proof}
Let $\ell\in \PH{P}{Q}$. By Proposition \ref{Prop:proj}, $\ell \subset L$ for some Lagrangian $L$ with $\mathsf{M}(P,L,Q)=n$. By Lemma \ref{Lem:Disjointness}, $\mathsf{M}(R,L,S)=-n$ which means that $\B{R}{S}|_\ell<0$ and so $\ell \notin \PH{R}{S}$.
\end{proof}
\end{Prop}
\begin{figure}
    \centering
    \includegraphics[width=.4\textwidth]{disjointhalfspaces.png}
    \caption{A pair of disjoint positive halfspaces in $\RP{3}$}
    \label{fig:halfspaces}
\end{figure}
For any generalized Schottky group, we can use this previous proposition to construct a fundamental domain. If the defining intervals for the Schottky group are $(a_1^\pm,b_1^\pm),\dots,(a_g^\pm,b_g^\pm)\subset \Lag{2n}$, let
\[D = \bigcap_{j=1}^g \left(\PH{a_j^+}{b_j^+}\cup\PH{a_j^-}{b_j^-}\right)^C.\]
That is, $D$ is the subset of $\RP{2n-1}$ which is the complement of the positive halfspaces defined by each interval. It is a closed subset since each positive halfspace is open. The interiors of the translates of $D$ are all disjoint by the two previous propositions and the boundary components are identified pairwise, so $D$ is a fundamental domain for its orbit (Fig. \ref{fig:sp4halfspaces}). This orbit is in general hard to describe, but in some cases we can identify it precisely.

\begin{figure}
    \centering
    \includegraphics[width=\textwidth]{sp4schottky.png}
    \caption{The first two generations of positive halfspaces for a two-generator Schottky group in $\Sp{4}$.}
    \label{fig:sp4halfspaces}
\end{figure}

In the definition of generalized Schottky subgroups, we required that the model be a finite volume hyperbolization. This is an artificial requirement which made the analysis of maximal representations simpler. In what follows, we will assume that the model Schottky group acting on $\RP{1}$ is defined by intervals with disjoint closures, so it corresponds to an infinite volume hyperbolization. The advantage of using intervals with disjoint closures lies in the contraction property proven in \cite{Contraction} which we will exploit later.
\subsubsection{Intervals as symmetric spaces}
We will now describe how to identify an interval in $\Lag{2n}$ with the symmetric space associated with $\GL{n}$, endowing any interval with a Riemannian metric.\\
Let $P,Q \in \Lag{2n}$ be two transverse Lagrangians. As we saw earlier in Corollary \ref{Lem:PosTriplesTransverse}, all Lagrangians in the interval $(P,Q)$ have to be transverse to $Q$, so they are graphs of linear maps $f:P\to Q$. The isotropy condition on $f$ is given by
\[ \omega(v+f(v),v'+f(v')) = \omega(v,f(v')) + \omega(f(v),v') = 0 \quad \forall v,v' \in P. \]
Now we recall from our discussion of the Maslov index that we can associate the bilinear form
\begin{align*}
    \B{P}{Q}: P\oplus Q & \to \mathbb{R} \\
    (v,w) & \mapsto \omega(v,\Inv{P}{Q}(w))
\end{align*}
to this splitting, and the index of its restriction to $\mathrm{graph}(f)$ is the Maslov index $\mathsf{M}(P,\mathrm{graph}(f),Q)$. We observe that this restriction is given by
\[ \B{P}{Q}(v+f(v),v'+f(v')) = \omega(v,f(v')) - \omega(f(v),v') = 2\omega(v,f(v')), \]
where the last equation follows from the isotropy condition on $f$. This bilinear form on $\mathrm{graph}(f)$ can also be seen as a symmetric bilinear form on $P$. Maximality of the Maslov index then translates to this form being positive definite.\\
Conversely, given a symmetric bilinear form $b$ on $P$, we obtain, for any $v'\in P$, a linear functional
\[ \left(v \mapsto \frac 12 b(v,v')\right) \in P^*. \]
Using the isomorphism
\begin{align*}
    Q & \to P^* \\
    w & \mapsto \omega(\cdot,w),
\end{align*}
we see that there is a unique vector $f(v') \in Q$ such that $b(v,v') = 2\omega(v,f(v')) \ \forall v\in P$. This uniquely defines a linear map $f:P\to Q$, and we have
\[ 2\left(\omega(v,f(v')) + \omega(f(v),v')\right) = b(v,v') - b(v',v) = 0, \]
so $\mathrm{graph}(f)$ is a Lagrangian. The Maslov index $\mathsf{M}(P,\mathrm{graph}(f),Q)$ is maximal if and only if $b$ is positive definite. This gives an identification of $(P,Q)$ with the space of positive definite symmetric bilinear forms on $P$, which is the symmetric space of $\mathrm{GL}(P)$.\\
The stabilizer in $\Sp{2n}$ of the pair $(P,Q)$ can be identified with $\mathrm{GL}(P)$ since any element $A\in\mathrm{GL}(P)$ uniquely extends to a symplectomorphism of $\mathbb{R}^{2n}$ fixing $Q$: The linear forms $v \mapsto \omega(A(v),w)$ on $P$, for $w\in Q$, give rise to a unique automorphism $A^*:Q \to Q$ such that
\[ \omega(A(v),w) = \omega(v,A^*(w)). \]
Then $A\oplus (B^*)^{-1}$ is the unique symplectic extension of $A$ fixing $Q$; we abuse notation slightly and denote it by $A$ as well. It acts on graphs $f:P\to Q$ by
\[ f \mapsto AfA^{-1}, \]
on bilinear forms on $P$ by
\[ (A\cdot b)(v,v') = b(A^{-1}v,A^{-1}v'), \]
and the identification of graphs and bilinear forms is equivariant with respect to these actions. In particular, $\mathrm{Stab}_{\Sp{2n}}(P,Q)$ is identified with the isometry group of the symmetric space $(P,Q)$.

\subsubsection{The Riemannian distance on intervals}
There is a simple formula for the Riemannian distance between two points in the interval $(P,Q)$:
\begin{Def}
Let $f,g$ be linear maps from $P$ to $Q$ whose graphs are elements of $(P,Q)$. Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of the automorphism $fg^{-1}$. Then, define
\[d_{PQ}(f,g) = \sqrt{\sum_{i=1}^n \log(\lambda_i)^2}.\]
\end{Def}
The following useful proposition is proved in \cite{Contraction}.
\begin{Prop}
\label{prop:contr}
Let $T\in\Sp{2n}$ such that $\overline{T(P,Q)} \subset (P,Q)$. Then, $T$ is a Lipschitz contraction for the distance $d_{PQ}$.
\end{Prop}
\begin{Cor}
\label{cor:contr}
Let $T\in \Sp{2n}$ such that $\overline{T(P,Q)} \subset (R,S)$. Then, for any $X,Y\in(P,Q)$,
\[d_{RS}(TX,TY) \leq C d_{PQ}(X,Y)\]
for some constant $0<C<1$.
\end{Cor}
Now we are ready to prove the main lemma that lets us describe the domains of discontinuity. Let $\rho:\Gamma\rightarrow\Sp{2n}$ define a generalized Schottky group in $\Sp{2n}$. We assume that the model $\Gamma$ is defined by intervals with distinct endpoints, so that the intervals in $\Lag{2n}$ have disjoint closures.
\begin{Lem} \label{lem:contr}
Let $\gamma \in \rho(\Gamma)$ be a word of reduced length $\ell$ in the generators $T_i$ and their inverses, with first letter $T$ and last letter $S$. We denote their attracting and repelling intervals by $I^\pm$ and $J^\pm$. Then, for any Schottky interval $K \neq J^-$ and $X,Y\in K$, we have
\[d_{I^+}(\gamma(X),\gamma(Y)) < C^{\ell} d_{K}(X,Y)\]
for some $0<C<1$ depending only on the set of generators.
\begin{proof}
Since a generator $T_k$ maps the interval $-I_k^-$ into $I_k^+$, we can consider it as a map from any Schottky interval $L\neq I_k^-$ into $I_k^+$. All of these maps are Lipschitz contractions by Corollary \ref{cor:contr}.\\
Now let $C$ be the maximum Lipschitz constant of all such maps, for $1\leq k\leq 2g$. We have
\[d_{J^+}(SX,SY) < C d_{K}(X,Y).\]
Composing contractions, we obtain
\[d_{I^+}(\gamma(X),\gamma(Y)) < C^\ell d_{K}(X,Y).\]
\end{proof}
\end{Lem}
\subsubsection{The domain of discontinuity}
We now have the necessary ingredients to analyze the orbit $\rho(\Gamma)\cdot D$ of the fundamental domain $D\subset\RP{2n-1}$ we defined in section \ref{sec:pos_halfspaces}. Using Lemma \ref{lem:contr}, we first define a map from the boundary of $\Gamma$, which is a Cantor set, into $\Lag{2n}$.
\begin{Prop}
Let $L\in \bigcap_{j=1}^{g}(-I_j^+\cap -I_j^-)$ where $I_j^\pm$ are the defining intervals of the generalized Schottky group. The evaluation map $\eta_0(\gamma) = \rho(\gamma)(L)$ induces an injective map $\eta:\partial \Gamma \rightarrow \Lag{2n}$ independent of the choice of $L$. Moreover, $\eta$ is continuous and increasing.
\begin{proof}
Let $x\in\partial\Gamma$ be a boundary point. Then $x$ corresponds to a unique infinite sequence in the generators $T_i$ and their inverses, where this sequence is reduced in the sense that no letter is followed by its inverse. We denote by $x^{(k)}\in\Gamma$ the word consisting of the first $k$ letters of $x$. Then the map $\eta$ will be defined by taking the limit
\[ \eta(x) = \lim\limits_{k\to\infty} \rho(x^{(k)})(L). \]
Let us first check that this limit does in fact exist. Recall that we introduced $k$-th order intervals and a bijection between words of length $k$ and $k$-th order intervals in section \ref{Sec:pingpong}. By the specific choice of $L$, its image $\rho(x^{(k)})(L)$ has to lie in the interval $I_{x^{(k)}}$ corresponding to the word $x^{(k)}$. Since the first $k$ letters of any word $x^{(m)}, \ m>k$ agree with $x^{(k)}$, the intervals $I_{x^{(k)}}$ form a nested sequence. Now we want to make use of the contraction property from the previous subsection. We first observe that since our model uses Schottky intervals with disjoint closures, second order intervals are relatively compact subsets of first order intervals. Let $I^{(2)} \subset I^{(1)}$ be such a configuration. Since the number of second order intervals is finite, there is a uniform bound $M$ such that
\[ \mathrm{diam}_{I^{(1)}}(I^{(2)}) < M, \]
where we used the metric on the symmetric space $I^{(1)}$. Then, denoting the first letter of $x$ by $T$, Lemma \ref{lem:contr} tells us that
\[ \mathrm{diam}_{I_T}(I_{x_k}) < MC^{k-2}. \]
This contracting sequence of nested subsets of the symmetric space $I_T$ thus has a unique limit, and $\eta$ is well-defined. By the same argument, we see that this limit does not depend on the choice of $L$.

We now show continuity of $\eta$. Let $y_n \to x$ be a sequence in $\partial\Gamma$ converging to $x$. This implies that for any $N \in \mathbb{N}$, we can find $n_0$ such that for all $n\geq n_0$, the first $N$ letters of $y_n$ and $x$ agree. In this situation, $\eta(y_n)$ and $\eta(x)$ lie in the same interval $I_{x^{(N)}}$ and so we conclude, if the first letter of $x$ is $T$, that \[d_{I_T}(\eta(x),\eta(y_n)) < M C^{N-2}.\]

Finally, we prove positivity in a similar way to theorem \ref{Thm:bdry_map_gen_Schottky}. For any $x,y,z\in \partial\Gamma$ such that $\cycle{x}{y}{z}$ (where we use the natural embedding of $\partial \Gamma$ in $S^1$ to get the cyclic order) we can find a large enough $K$ so that $I_{x_K}$,$I_{y_K}$ and $I_{z_K}$ have disjoint closures. But since the cyclic relations on $k$-th order intervals are the same in $S^1$ as in $\Lag{2n}$, for any $P \in I_{x_K},Q\in I_{y_K},R\in I_{z_K}$ we have $\cycle{P}{Q}{R}$. In particular, $\cycle{\eta(x)}{\eta(y)}{\eta(z)}$.
\end{proof}
\end{Prop}
\begin{Rem}
The map $\eta$ that we define is related to the map $\xi$ of theorem \ref{Thm:bdry_map_gen_Schottky}. In this case, the endpoints of $k$-th order intervals are not dense in $S^1$, so we cannot get a map on the whole circle. However, because the intervals have disjoint closures, we get continuity on both sides rather than just left-continuity.
\end{Rem}
The next lemma relates the construction of the limit map $\eta$ with the positive halfspaces that intervals define in $\RP{2n-1}$.
\begin{Lem} \label{lem:conv_intervals}
Let $L_1^k,L_2^k$ be sequences of Lagrangians such that $L_1^k \rightarrow L$ and $L_2^k \rightarrow L$ with $\cycle{L_1^k}{L}{L_2^k}$ for all $k$. Then,
\[\bigcap_{k=1}^\infty \overline{\PH{L_1^k}{L_2^k}} = \bigcap_{k=1}^\infty \PH{L_1^k}{L_2^k} = \mathbb{P}(L).\]
\begin{proof}
Assume $\B{L_1^k}{L_2^k}(v,v)\geq 0$ for all $k$. Then we can find $v_k\xrightarrow{k\to\infty}v$ such that $\B{L_1^k}{L_2^k}(v_k,v_k)> 0$ for all $k$. Now, by proposition \ref{Prop:proj}, $v_k$ can be completed to a Lagrangian $L^k$ with $\mathsf{M}(L_1^k,L^k,L_2^k)= n$, so $L^k \subset (L_1^k,L_2^k)$ for all $k$, which implies $L^k \rightarrow L$, and so $v\in L$.
\end{proof}
\end{Lem}
Now we are ready to describe the orbit $\rho(\Gamma) D$.

The union of $D$ with the positive halfspaces defining the Schottky group is all of $\RP{2n-1}$, by definition of $D$. Denote by $\Gamma_\ell$ the set of words in $\Gamma$ of length up to $\ell$. Then, the union of $\rho(\Gamma_\ell) D$ with the projectivizations (positive halfspaces) of all $\ell$-th order intervals again covers all of $\RP{2n-1}$. Thus, when taking words of arbitrary length in $\Gamma$, these two pieces become respectively the full orbit $\rho(\Gamma) D$ and limits of nested positive halfspaces, which by lemma $\ref{lem:conv_intervals}$ collapse to the projectivization of a single Lagrangian. We conclude:
\begin{Thm}
\label{Thm:dod}
The orbit $\rho(\Gamma) D$ is the complement of a Cantor set of projectivized Lagrangian $n$-planes in $\RP{2n-1}$. This Cantor set is exactly the projectivization of the increasing set of Lagrangians defined by the boundary map $\eta$.
\end{Thm}
\begin{Rem}
The symplectic structure on $\mathbb{R}^{2n}$ induces a contact structure on $\RP{2n-1}$ preserved by the symplectic group. The projectivizations of Lagrangian subspaces correspond to \emph{Legendrian} $(n-1)$-dimensional planes in $\RP{2n-1}$.
\end{Rem}


Feng Shui is more than just an interior design style. It's a 3.000 year old Chinese dogma, meant to positively affect important aspects of life, like love, happiness and good fortune. Even if you don't believe in this, Feng Shui tips can help you turn your home into a more harmonious and relaxing place. To which extend you want to follow the Feng Shui principles, is all up to you.
Different rooms serve different goals, so they should be arranged in a different way. But in general, it is important to keep enough space available for new energy. Keep the doors free, so Chi can run through your entire home.
The living room serves family life. It's the place where you host your friends and family, and where you relax together. According to Feng Shui principles, this room is ideally situated in the front part of the house. Sofas and chairs should be directed to each other in a square or rectangular form. Make sure you leave enough circulation space so Chi can run through it with the least resistance. The shape of the room should also be square or rectangular. You can give an L-shaped room a rectangular character using curtains, sideboards or room dividers.
The bedroom is a private room where you spend a lot of time. It is a multifunctional space to sleep, read and relax. Feng Shui advises to position your bed with a view on the door and to make sure it does not reflect in any mirrors. Try to ban electrical devices from your bedroom, because they can obstruct Chi. Soft wall painting colors are important for positive energy in your bedroom. Use white, pink, light green or soft blue. For more inspiration, check out our wall painting tips & tricks.
A Feng Shui kitchen creates a happy feeling while cooking. Try to arrange your kitchen in such a way that you have a good sight on the entire room and the door and that there is enough space to move around. Make sure your stuff can be put away in sideboards easily.
Keeping the 5 nature elements water, fire, earth, metal and wood, in balance is an important principle of Feng Shui. In the bathroom this is challenging, because there is a lot of water energy there. To keep the balance with the other elements, compensate with wooden furnishings and plants. A large mirror can also serve positive energy in your bathroom. Ideally the orientation of the bathroom should be south-west. Read more bathroom design tips here.


Bridal magazines and jewelry companies say that 40 percent of all proposals of marriage occur in the month of February. These proposals are centered around the celebration of St. Valentine whose Feast Day is February 14th. This day of love usually includes flowers, candy, and a special meal. When you add a proposal to the mix, the festivities also include a ring.
The engagement ring traces back to the 1400s. Archeological digs have revealed women, and occasionally men, wore special rings to signify they were promised or engaged to someone. In the United States, jewelers experienced an upswing in the sale of diamond engagement rings right as the Depression was ending. The increasing sales were credited to the advertising campaign of the DeBeers jewelry company who made a diamond engagement ring de rigueur for couples.
Engagement rings range from the traditional to custom one-of-a-kind pieces that reflect the tastes and personalities of the couple. The traditional diamond engagement ring is a band of platinum, gold, or silver with a single stone, in a prong setting, in the center of the band. Popular diamond shapes are the round, square, pear, marquise, heart, rectangular, and oval. Cushion is the same as square and emerald, princess, and radiant are all cuts of the rectangular shape.
The shape is the appearance of the stone. The cut of the diamond is the way that it reflects light. This reflection is what makes the diamond sparkle. Certain shapes can be cut to allow greater reflection than others. These cuts obtain names of their own.
Pure diamonds are perfectly colorless and transparent. When grading a diamond, you will hear about clarity or how transparent or pure the stone appears. While the clear diamond is the most popular engagement ring stone choice, many betrothed sport colored diamonds on their ring finger.
Yellow diamonds, brown or chocolate diamonds, black diamonds, pink diamonds, and blue diamonds have all set fashions trends in the past 50 years. Color is many times made a popular choice when a celebrity or fashionista betrothed gifts them with a colored diamond. In the last decade, canary yellow diamonds were on the hand of Jennifer Lopez, Carrie Underwood, Nicki Minaj, and Heidi Klum.
Diamonds are not the only popular stone in engagement rings. By far, one of the most famous engagement rings in the world is a blue sapphire surrounded by diamonds. That ring was the one selected by a nineteen-year-old Lady Diana Spencer when Prince Charles asked her to marry him in the early '80s. The ring presently is worn by Her Royal Highness the Duchess of Cambridge since it was Prince Williams choice to present to her when she was asked to marry him.
Iconic engagement rings, in addition to Princess Diana's heirloom ring that Duchess Kate wears, is the emerald and diamond ring of Jacqueline Bouvier Kennedy, the emerald cut, platinum set, 18 karats on Beyoncé's ring finger, and Lady Gaga's heart shaped diamond presented on Valentine's Day a few years ago.
Regardless of the day, type of stone, or if the ring is an heirloom or not, your engagement ring is special to you. Once engaged, you want your wedding venue to be just as outstandingly special. Le Pavilion in Parc Lafayette invites you to consult with them and their many experts to plan your perfect day.
In addition to being an exquisite wedding venue, Le Pavilion can host your special events like rehearsal dinners, engagement teas, and private parties. Come tour this engaging space today as a wedding and reception venue.


What's the "more" in Custom Floors & More?
The "more" goes beyond the array of backsplashes, custom showers and special tile projects (in addition to flooring of course) that Custom Floors provides customers in the Belleville, Illinois area.
When Julie and Doug Nitz first thought of opening a flooring store, they had a vision of a company that would be there for customers every step of the way. The Nitzes wanted to help with the selection of not only the products that CFM provides but also the other important items that come along with designing a home. The CFM sales team helps customers take their chosen cabinets and countertops and then pairs them with paint colors and flooring materials that will complete the desired look.
Helping customers in every way is what makes a CFM experience so special. Custom Floors' goal is to take away the frustration that many feel in tying so many pieces of their home together. After all decorating your home should be fun! With the combined decades of experience and knack for color, the Nitzes knew their business could give each customer a personalized experience and look. Custom Floors proudly assists customers from the selection process to after the final installation no matter how unique the project may be.
Want to experience the "more" in Custom Floors & More, Inc.? Visit our testimonials page and see what our customers have to say: https://customfloors.biz/floor-installation-testimonials/. Then give us a call at 618-277-7233 to set up your free one-on-one consultation. You won't regret it!
Why Custom Floors & More, Inc?
We take the time to meet with you one-on-one to look at your floor plan, give advice and answer any questions. We provide all types of flooring: wood, vinyl, carpet, laminate, ceramic tile and marble.
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Are you a homebuilder or remodeler?
We partner with metro east Illinois area home builders and remodelers to provide professional flooring installation, plus we also assist in making color choices of the products based on each individual home.
Click here to find out more about partnering up!
© Custom Floors & More, Inc.


text:
Bokurano
Bokurano Season 1 Episode 22
Jun takes Kana to her father's house, having Machi traveling with them. On the way, he reminisces about his infancy days, and later, he asks Machi's help to protect Kana from Dung Beetle's intentions.
Serie: Bokurano
Episode Title: Distance
UFO Ultramaiden Valkyrie
UFO Ultramaiden Valkyrie, aka UFO Princess Valkyrie, is an anime series that follows the story of Kazuto Tokino and the alien Princess, Valkyrie, who gives half her soul to Kazuto…
Yu-Gi-Oh! 5DS is an anime series and the third installment in the franchise of Yu-Gi-Oh! anime series. It aired in Japan between April 2, 2008 and March 30, 2011, following…
Inu X Boku Secret Service
Protected by the highest security, the Maison de Ayakashi is rumored to be haunted, where only eccentrics could live. In reality, it is a place for humans with non-human ancestors…
Genre: Animation, Comedy, Drama, Sci-Fi & Fantasy
Hatsukoi Limited.
Ayumi and her friends are all just entering the eighth grade, as, one by one, they discover that love is infectious and you can catch it from anyone! Now Ayumi's…
A group of friends have customized their microwave so that it can send text messages to the past. As they perform different experiments, an organization named SERN who has been…
Genre: Action & Adventure, Animation, Comedy, Drama, Mystery, Sci-Fi & Fantasy
Speed Racer: The Next Generation
Speed Racer: The Next Generation is an American animated television series based on the classic Japanese Speed Racer franchise, in which the internal events take place decades after those in…
Marvel's Guardians of the Galaxy
Peter Quill is Star-Lord, the brash adventurer who, to save the universe from its greatest threats, joins forces with a quartet of disparate misfits — fan-favorite Rocket Raccoon, a tree-like…
Genre: Action & Adventure, Animation
Umineko: When They Cry
As customary per year, the entire family is gathering on the island for a conference that discusses the current financial situations of each respective person. Because of the family head's…
Genre: Action & Adventure, Animation, Comedy, Drama, Sci-Fi & Fantasy
The protagonist, Baki Hanma, trains with an intense focus to become strong enough to surpass his father, Yujiro Hanma, the strongest fighter in the world. Five of the world's most…
Kobayashi lives alone in an apartment, until one day, Tooru appeared and they ended up living together. Tooru looks down on humans as inferior and foolish, but having been saved…
Domestic Girlfriend
Natsuo is a high school boy who is experiencing the crushing despair of unrequited love. To make matters worse, the person he is in love with is his teacher, Hina….
Genre: Animation, Drama
Tsutsui Hikari (a.k.a "Tsuttsun") is a high school student who is content with his virtual life of anime and video games. One day, he gets stuck cleaning the pool with…

redpajama_set_name: RedPajamaCommonCrawl


text:
Il distretto di Adıyaman (in turco Adıyaman ilçesi) è uno dei distretti della provincia di Adıyaman, in Turchia.

Adıyaman

redpajama_set_name: RedPajamaWikipedia


Source: In: Stanturf, John A., ed. Proceedings of the 14th biennial southern silviculture research conference; 2007 February 26-March 1; Athens, GA. Gen. Tech. Rep. SRS–121. Asheville, NC: U.S. Department of Agriculture, Forest Service, Southern Research Station: 229-234.
Jackson, D. Paul; Dumroese, R. Kasten; Barnett, James P.; Patterson, William B. 2010. Effects of liquid fertilizer application on the morphology and outplanting success of container longleaf pine seedlings. In: Stanturf, John A., ed. Proceedings of the 14th biennial southern silviculture research conference; 2007 February 26-March 1; Athens, GA. Gen. Tech. Rep. SRS–121. Asheville, NC: U.S. Department of Agriculture, Forest Service, Southern Research Station: 229-234.


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The added responsibility is gained by the management side of career training. For people who find themselves quite frequent in air travel and have been loyal to a limited variety of carriers are conscious of the frequent flier applications run by the airways.


When we think of Turf most of us think of Buffalo Turf. So what is it?
Buffalo grass is a species of turf grass. There are many different varieties of Buffalo Turf on the market like Kings Pride Buffalo and Sapphire Buffalo. Most are soft leafed and possess quality traits.
Buffalo varieties are the most popular selling turf grasses for the residential market. The reason for this are the excellent features a Buffalo lawn can offer for the everyday problems a suburban gardens face.
While some Buffalo varieties are better than others, you need to choose a Buffalo grass that is best suited to your type of area and the intended use of your area. Whether it be a highly used back yard that needs something durable yet soft for those sensitive little knees or it might be your weekender that requires a low maintenance low mow lawn.
If you like the sound of a new Buffalo grass lawn but you're unsure of what type would best suite your home; contact us and we can help you make the right decision. Then you too can enjoy the quiet splendour and comfort a new Buffalo lawn offers to the areas we love to live and play.


Brenda and I just returned from San Antonio Texas. We spent the last four days being dazzled by Toro. The Toro RLC National Dealer Conference was held at the Hyatt Regency Hill Country Resort & Spa. The weather was perfect, sunny and 70+ degrees. We enjoyed over the top hospitality, and an exceptional program loaded with useful information from expert speakers and instructors. A couple of highlights included a dinner at the Buckhorn Hall of Horns around the corner from the Alamo and Nolan Ryan as one of the keynote speakers. We are honored to have been chosen to attend and grateful for all that we got out of our time in Texas. THANK YOU TORO!


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Making Games for Amazon Appstore Part 5: Publishing and Marketing Your Game : Appstore Blogs
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Making Games for Amazon Appstore Part 5: Publishing and Marketing Your Game
Jesse Freeman
Fire phone
Welcome to the final part of this five part series on making games for the Amazon Appstore. Part one, two, three and four helped you build the foundation for your game's design and in this final post we will talk about how to actually market your game.
So you have finally finished your game and are ready to release it to the world. While there is a lot you can do after you have completed your game to help make it a success, you should always be thinking about marketing your game from the very beginning. The following section will help you with everything from naming your game to how to help it stick out in an overcrowded mobile store. While marketing a game is not an exact science, and its success has a lot to do with the quality of your game and pure luck, you can still take the necessary steps to ensure it gets the most attention possible out of the gate.
Naming Your Game
The name of your game is going to be the single most important decision you make. Companies spend huge amounts of money doing market research to come up with product names, and what you come up with for your game has lasting effects. The most basic thing you can do to help your game be more successful is to simply give it a descriptive name. Look at other games and how they came up with their names:
"Mario" – Named after the main character. While it doesn't describe what you will be doing in the game, it helps establish the lead and his name defines the brand.
"Legend of Zelda" – The word legend implies a deep story told a long time ago, and saving Princess Zelda is the main objective of the game.
"Grand Theft Auto" – The name alone implies stealing cars and breaking the law. It clearly describes the main gameplay mechanic the player is going to expect to be doing.
"Angry Birds" – Doesn't get any more basic than this. You play as a bunch of angry birds.
Naming a game isn't always an easy task, but make sure your game's name is clear, descriptive, and easy to remember. Also, make sure your game's name makes sense for sequels and continuations if you find yourself with a hit on your hands.
Getting People to Play Your Game
This topic really goes back to the beginning section on what platform to build your game in. The key to success is really getting your game in the hands of as many players as possible. To do that, you will need to pick a platform with the most reach possible. While HTML5 has the incredible advantage of being published to the Web, you will find that it is more difficult to publish to a native store. Frameworks like GameMaker and Unity will allow you to publish to multiple platforms, but they charge extra for that service. Some features are free, and while GameMaker allows you to export an HTML5 version for the Web and Unity has a Web Player for its game, you really need to do your research and pick the right tool from the beginning.
The ideal scenario is that you are able to release your game on all of the major platforms: Web, iOS, Android, FireOS and desktop. From there, you have some granular options to decide on, like what devices to support on each mobile platform, and even what version of the OS you can support. Don't forget Web stores, such as Chrome Market and Firefox's new Web-based Phone OS. Also, there are online game portals that do revenue sharing based on ads around your game. While mobile is all the rage, don't forget how many people still use Web browsers, and every modern mobile device also supports playing Web games, so you get even more bang for your buck. I am now working on having all of my games as Web-playable demos pointing to the app store versions to help broaden my audience, since being found on the Web is much easier than in an app store.
If the Web isn't your thing, also keep in mind cross-platform desktop publishing. A lot of indies have had success with Steam Greenlight, as well as bundle deals like Humble Bundle. Both of these distribution solutions leverage desktop compiling and usually want games that support Windows, Mac, and Linux. If you intend to ever get your game on Steam, make sure you are able to offer a desktop installer.
Charging for Your Game
With all of this talk about distribution, you are probably thinking through all of your monetization options. Let's take a look at the three most common options out there now:
Flat rate – Your first instinct is probably to charge a flat rate for your game, like 99 cents or more. Some games can command a very high premium on the mobile market, but gone are the days of charging $5, $10, or more, especially if you are just starting out. When it comes to pricing your game, you should do a lot of research on your game's genre, its competition, and what it costs. If you see a lot of high quality free games similar to yours, it's probably best to consider an alternative option.
Free with ads – This was popular for a long time before in-app purchases were introduced. Basically, you make the game free and use ads to help generate money. While there are clear advantages to making your game free, which allows anyone to download it, it is also incredibly hard to monetize from ads. As an example, my most successful game on Android with almost 50k downloads (roughly 1-2k players a day) only generates $1 or less each day. In order to really make money with ads, you will need at least 100k players a day or more, and show larger, more obtrusive ads that tend to slow down the game's flow or frustrate players.
IAP (in-app purchases) – The final approach, which has gained considerable popularity lately, is relying on IAP (or in-app purchases). Usually, the game is free and you try to upsell so users buy more levels, items in the game, or even ways to speed up their in-game character's development. What's good about this approach is that your game is completely free, so you get my people playing your game, and if they like it or you have compelling IAP, they end up paying more than they normally would had your game been sold at a flat rate. This model is extremely hard to pull off successfully, and it's a fine line between adding additional value to your game and trying to exploit the player's compulsion loop.
The reality is that, if this is your first game, chances are high that you will not make a lot of money off of it. Having a successful game is kind of like winning the lottery. I am not telling you this to discourage you; I want first-time game developers to have a clear idea of what they are in for. Even seasoned game developers have a difficult time monetizing their games without help from marketing firms. Even getting good reviews on game sites, which is hard to do, doesn't always translate into lots of sales. The best thing you can do is get featured on the app store itself, but once that promotion is over, unless you have a real hit on your hands, you will see a natural drop off in your game's sales or downloads.
The last thing I want to talk about, which leads into our next topic, is that making games is a learning experience. The more games you make the better you become as a game developer and grow your chances of making a hit game. While it's very frustrating for first-time game developers to work hard on something and not see people download it or buy it, you shouldn't give up. One of the best things your game can be is a promotional tool for you and even your next game. Always be sure to have links to your other games inside of your game to help promote everything you are doing. You may be surprised to realize that, over time, you actually end up building up a following of people more willing to pay for your next game if they continue to play and like the ones you are releasing now.
The More Games You Make the Better
I talk to a lot of game developers each day, and they all have the same story: the more games they have in the store the better their sales are. This happens for two reasons. The first is that, when you make more than one game, they are all featured on your game's download page under a section that highlights other games made by the same author. That means that each additional game you make basically gives you free advertising for your other games. The second thing that happens is that, when a person likes your game, they naturally want to play more games you have made. This means they will look at anything else you have created, which is the basis for building your own fan base. So each successive game you release will naturally help drive downloads and sales of your past games as well. This is another reason why you should not give up if your first game isn't an instant success and just keep making as many games as you can!
The good news is that publishing to the Amazon Appstore is free and easy, especially if you are already building Android games. Simply sign up for a developer account at the developer portal to get started today. Here are some additional links to help you gets started:
Register for a free developer account
Download our SDKs
- Jesse Freeman (@jessefreeman)
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History Research
The Centre for Scottish Studies at the University of Guelph is the oldest and most eminent academic unit in North America to carry out research, graduate training and teaching on Scottish history and Scottish culture, with special emphasis on the history of the Scottish Canadian community. Our graduate program was founded in 1968 and we are proud to host special events and public lectures that presents current research by graduate students and scholars from Scotland, Canada and elsewhere around the world.
We began publishing Scottish Tradition in 1971, which was relaunched as the International Review of Scottish Studies in 2003. The IRSS is the leading interdisciplinary journal for international scholarship on Scottish history and culture, with a mission to create a space for scholars at all career levels exploring Scotland's past and present. From 2023, the IRSS, which has been publishing innovative research for almost 50 years, will be published by Edinburgh University Press.
In 2004, the Scottish Studies Foundation endowed the position of Scottish Studies Foundation Chair. This was a singular achievement of fundraising and generosity by the Scottish Canadian community towards the advancement of Scottish scholarship. Both our past and our present international reputation give us pride and we look forward to placing Guelph at the centre of innovative research uncovering Scotland's past and exploring its future, in a global framework.
IRSS Journal
Digital Archive Initiative
Meet the Chair
Frank Watson Book Prize
Supporting Scottish Studies
Prof Kevin James - Scottish Studies Foundation Chair, Professor of History
Prof Elizabeth Ewan - Professor of History Emerita
Prof Linda Mahood - Professor of History
Prof James Fraser - Professor of History
Prof Cynthia J. Neville - Professor of History
Prof Susannah Ferreira - Professor of History
Prof Cathryn Spence - Professor of History
Melissa McAfee - Archival and Special Collections, McLaughlin Library
The Centre for Scottish Studies office is located in room 1008 of the MacKinnon Extension, in the Department of History. The Centre is staffed remotely on a part-time basis. Visitors are welcome, although it is important to e-mail in advance to confirm office hours on the day of your visit. A collection of Scottish Studies theses completed at the Centre is kept in the office and the Scottish Studies Archival Collection is held at the McLaughlin Library.
Office Staff:
Melissa Turner
scottish@uoguelph.ca


Fabric Shopping... a few places you should know about!
Over the last few years, I've gathered quite a few sources for fabric needs. Before I started Marifly, I rarely went to a fabric shop, beyond JOANNS, Michael's and Hobby Lobby - mainly because I lived in an area without boutique fabric shops. And though I still head to these shops for various needs, there are a lot of other fun places to snag premium cottons and specialty textiles.
There are shops in every nook along the streets featuring every dream of textile. You will find apparel wear, silks, faux furs, athletic wear/sportswear, home dec. fabric and so many more options. Here's a directory if you're headed down there. Feel free to bargain, and if your a business bring copies of your tax deductible information.
Fabric Bubb Bundles are crafted with style & ingenuity. It's an amazing spot for inspiration for your next quilt.
Hawthorne Supply Co. - This used to be named Hawthorne Threads and they have so many fabric options + they've started their own lines.
Cottoneer Fabric - A Modern Fabric shop This is also a husband and wife duo that photographs all of their fabrics and sells the most lovely linen fabrics.
Shop Cabin has started a collection with Hawthorne Supply Co.
Spoonflower: I know I know, it's expensive, but their knits are fun and great for specific projects, and you can pretty much get the exact print you want with all of its options.
Girl Charlee: I don't love the quality of all their knits, but if you want to make a baby swaddle that helps a baby sleep because it has the right amount of stretch and breathes well, this is your place. Also, it's super super reasonable.
Birch Organics: I love their organic knit fabric. It's a little thicker, but the colors and finish is amazing.
Most Textile Companies wholesale to retailers, and don't sell their goods through their company.
Cotton + Steel: Features some of the most beautiful wovens and canvas options. Rifle Paper Co. prints through C+S.
Art Gallery Fabrics: Amazing quality that focuses on quilting cotton, but also designs rayons, poplins, canvas, knit and denims…. top notch stuff.
Shannon Fabrics: Best cuddle fabrics on the block. I suggest going on ETSY to find this soft plush, faux furs, llama sherpas, etc.


We only purchase responsibly sourced specialty grade coffee beans. By working with our importers to source sustainably grown coffees, we're insuring you get the best value and farmers get the support they so greatly need. Great coffee begins with hard-working farmers and co-ops.
The Hot Air Roasting method is unique. Hot Air Roasting produces a very clean taste that is intensely aromatic, smooth and balanced, minus the acids and bitter tars that are produced by conventional roasters. The aroma alone will tell you this is very special coffee.
The Nguvu is a blend of coffees harvested from small-holder cooperatives the Murang'a district. The area is considered the traditional home of the Kikuyu tribe, which currently makes up 22% of the Kenyan population. Red, loamy soil and startlingly high altitudes contribute to a juicy, berry-flavored cup.
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Flavor Notes: Cane sugar, intense stone fruit flavors, lemony, herbaceous florals, clean, juicy, complex acidity. Molasses, rich body. Even sweeter as it cools.
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A complex coffee with an exotic and syrupy body with hints of chocolate, vanilla and spice.
Altitude: 3,900 to 5,250 feet.
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Taste the uniqueness that can only be found in Nariño Colombia. Coffee from these slopes is well-balanced, with high acidity, medium body and a lingering citric sweetness.
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text:
Le bailliage de Stade (Landdrostei Stade) est ancienne division administrative du royaume de Hanovre et puis du royaume de Prusse.

Description
Il était compris entre les bouches de l'Elbe au nord, l'Oldenbourg et la République de Brême à l'ouest, le  au sud, la République de Hambourg et le Holstein à l'est

Il mesurait 120 kilomètres sur 65 et renfermait une population de  habitants.

Baillis
 1823–1841:
 1841–1855: Ernst von Bülow
 1856–1858:
 1858–1862:
 1863–1872:
 1872–1885: Heinrich Küster

Notes et références

Bibliographie

Liens externes

Stade
Stade
Stade
Stade
Stade bailliage
Stade (Allemagne)


Goal Setting 101- How do you do it?
Goal setting. Do the words send shivers down your spine, or do they make you want to dig out your day planner and your prettiest pen, and dig in?
I've studied goal setting under the Franklin Planner people; I've been a devotee for over twenty years of their system. I've learned goal setting (and mission planning) from Bob Mayer. I've done a five main goals for five years, completely under Debbie Macomber's spell, during an RWA National Conference Keynote Luncheon.
And yet, I'm still figuring out how goal setting and I get along. I know that a big mission is propped up with smaller tasks that must get done in order for the big mission to be successful. In other words, "life chunks" – those big rock goals – should be put into your life bucket first, so the smaller goals – those pebbles, the sand – can fit into and around the bigger goals. And your bucket still has room for water (lesser goals).
But I'm looking for more. I'm looking for your opinions. How do you set goals? How often do you review them? Do you have daily, weekly, monthly goals? A one year plan, five year plan, ten years? Career goals, personal goals, fitness goals? Do you separate out your goals (financial, fitness, career, house), or are they all a part of a master plan?
I'd love to know. Please leave a comment and tell me how you work your goals list. The comments will be open all week – and at the end of the week I'll pick one lucky commenter to receive a $10 Amazon card as a thank you!
This entry was posted in Life and tagged Amazon, Caine Brothers, Christine Ashworth, Debbie Macomber, Franklin Covey, giveaway, goal setting, RWA National Conference. Bookmark the permalink.
14 Responses to Goal Setting 101- How do you do it?
Great question, Christine. I love goal setting! In my world, goals are flexible, measurable, and are like a personal assistant to keep me on track. Starting with the macrocosm, I do have broad, three, five, and ten year goals–a general vision of where I see myself heading over time. That said, God always seems to have bigger plans for me than I have for myself, LOL. On a smaller scale, I set yearly goals in Jan. And check in quarterly to see how I'm doing and if they need adjusting. I have a calendar with monthly and weekly tasks to keep me on point, but the real work is in the daily and weekly to do list that moves me forward step by step. The actual goals for my writing life require much more attention and micromanagement than my general life goals which are a part of everyday life for me. Things like working out, making time for fun, and trying to stay balanced are always in my head. I think folks who have trouble with setting goals have a hard time figuring out exactly what they want. If you know that, getting there is just one goal after the other.
PJ, I like the broad, overarching goals. I need to do that again – it's been a while. Thanks for dropping by!
I'm also a member of Jenny Crusie's Forums and we have a category titled monthly goals and submissions. We post there and do the rah rah, or commiserate on rejections, etc.
In my opinion, I think it doesn't matter where you record goals it just matters that you have a plan.
Roben, I like the monthly goals idea. That's marvelous!
You are going to have to share your methods with me. I don't seem to have any goals – only I do, but they are on vision boards. For whatever reason that works better for me. This reminds me that it is time for a new one.
I just don't work well with linear goals. Visual stimulation works better for me.
Vision boards are a great way to keep your goals in sight. Do you put dates beside the visual cues, Maria?
Sam, you are too funny! Often I write my goals on a calendar – and then never look at that calendar again. I'm trying to really wrap my head around goal-setting!
Goal setting, what a great post! I am just now considering making goals every first of the month. I write a daily list (that often turns out to be a weekly list of goals LOL) and of course the New Year is a great time to goal set, but I'm thinking once a month. I'll let you know how it goes.
Suzanne, monthly goals – what a terrific idea! I'm gonna have to steal that if you don't mind.
I think I've always admired people who set goals, but it seems that I'm more comfortable existing in the moment.
Laurie, if I just existed in the moment I would never get anything done – but I appreciate those who can!
Good post. Ever since I started my freelance writing business I've set one-, three-, and five-year business goals. In my one-year plan I also do mid-year reviews to make sure I'm still on-track, and the longer term goals get updated as life flexes to make for new opportunities and challenges. Because most of my pay comes from nonfiction, I also put money goals in place. I've found if I consistently budget future sales at at least 25% more than I "think" at first I can make (and what I need currently to make bills), I consistently reach that higher goal. I've only missed a couple of years, and even then I made at least half of the bonus percentage I was shooting toward. I think anytime you quantify goals, put a "true value" on your product, you're much more likely to reach the goal. Brownie points are nice, but extra zeros on the paycheck are even better.


After months of long, cold winter, who doesn't want to be outside soaking up the sun and fresh air? Our canine companions often long to be outdoors as well. With increased people, other dogs, birds and animals out and about, some of our furry dog friends may be prone to excessive barking. This can be quite annoying for you, and often times, your neighbors. How do you put an end to this challenging behavior?
The first step is to not yell at your dog. While it may seem counter to reason, yelling at your dog stimulates him more and he may think that you are joining in on the fun. Speak to your dog in a clear, even but firm voice.
Secondly, use a command word. Say your command word – it could be something like, "quiet" or "stop" and wait until your dog is quiet and then reward him with a treat. Do not give your dog attention while he is barking. Wait until he is quiet and settled down before you reward him with your attention. After all, that's why he is barking. You can also distract him by saying his name and then giving him a task to do such a trick or a command.
Limit your dog's exposure to stimulation by closing blinds, installing privacy fencing and or use noise reduction techniques such as softly playing classical music or a TV turned down low. Finally, make sure that your dog is exercised properly for her activity level. Some dogs bark out of boredom. Exercise and mental stimulation activities such as puzzle feeders will help keep the noise level down.
Most towns and municipalities have ordinances regarding howling or barking dogs. In the City of Rochester, it is violation for dogs to bark continuously for more than 15 minutes during the hours of 8:00 am to 10:00 pm or to bark continuously for more than five minutes between 10:00 pm and 8:00 am. If you are cited, you could be subject to fines and penalties. Please check with your town for the local ordinance in your area.
A quiet dog helps everyone in the neighborhood enjoy the outdoors.


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