text:
\section{Introduction}

Let $S_{g}$ denote the closed orientable surface of genus $g$. A finite collection of curves $\Gamma= \left\{\gamma_{1}, ... \gamma_{n} \right\}$ in pairwise-minimal position on $S_{g}$ is said to \textit{fill} if the complement $S_{g} \setminus \Gamma$ is a disjoint union of topological disks. When each $\gamma_{i}$ is simple and $n= 2$, we call $\Gamma$ a \textit{filling pair}.

Thurston used filling pairs to construct pseudo-Anosov mapping classes by composing Dehn twists in the appropriate way around the curves in the pair \cite{TS}. The dilatation of these pseudo-Anosovs directly relates to the geometric intersection number of the pair. Penner later generalized Thurston's construction to pairs of filling multicurves \cite{P}. Any pseudo-Anosov obtainable by such a construction is known as a \textit{Penner-Thurston map}.

Filling multicurve pairs with small intersection number yield maps with relatively small dilatation amongst all Penner-Thurston maps. Motivated by this and other connections to the mapping class group-- which will henceforth be denoted by $\mbox{Mod}(S_{g})$-- in joint work with Huang \cite{AH}, the first author constructed exponentially-many (in $g$) $ \mbox{Mod}(S_g)$-orbits of filling pairs whose geometric intersection number is the minimum possible among all filling pairs on $S_{g}$.

In this paper, using a new construction, we build factorially-many minimally-intersecting filling pairs, all of which satisfy the additional condition that the absolute value of the algebraic intersection is equal to the geometric intersection number.

\begin{theorem} \label{main} For odd $g \geq 3$, there exist at least $(g-2)!$ distinct $ Mod^{\pm}(S_g) $-orbits of filling pairs $\left\{\gamma_{1}, \gamma_{2} \right\}$ so that the absolute value of the algebraic intersection number is $2g-1$. For even $g >2$, there are at least $(g-5)\cdot (g-3)!$ such orbits.
\end{theorem}

\subsection{Square-tiled surfaces}

A \textit{square-tiling} of a surface $S= S_{g}$ is a decomposition of $S$ into quadrilaterals with at least four regions around each vertex. If each region is identified with a square whose area is scaled such that the total area is $1$, this induces a unit-area singular flat metric on $S$. The horizontal foliation on each individual square can be extended to a measured singular foliation $\mathcal{F}$ on $S$. A fundamental theorem of Hubbard and Masur then associates a unique quadratic differential to this horizontal foliation and, therefore, to the square-tiling \cite{HM}. In this way, one can interpret a square-tiling of $S$ as a point in the space of unit-area quadratic differentials. If $\mathcal{F}$ is orientable, the square-tiling supports the structure of a \textit{translation surface}, and the associated quadratic differential will be the square of an abelian differential. In this case, we refer to the surface equipped with its square-tiling as a \textit{square-tiled surface}.

An abelian differential on a surface of genus $ g>1 $ must have $2g-2$ zeros, counted with multiplicity.  The moduli space of abelian differentials is naturally stratified by the degrees of these zeros:  given $m_{1},..., m_{n} \in \mathbb{N}$ satisfying $m_{1}+...+m_{n}= 2g-2$, the \textit{stratum} $\mathcal{H}(m_{1},..., m_{n})$ of the moduli space consists of those abelian differentials which have $n$ zeros of degrees $m_{1},..., m_{n}$. Each stratum admits a volume element -- the \textit{Masur-Veech volume}-- and an invariant $SL(2,\mathbb{R})$-action.  Masur and Veech proved that each stratum has finite volume (\cite{M}, \cite{V1}, \cite{V2}).

To compute the volume of a stratum of abelian differentials, Eskin-Okounkov (\cite{EO}) use the following approach suggested by Zorich (\cite{Zo}): each stratum can be locally modeled on the relative cohomology $H^{1}(S, P; \mathbb{C})$, where $P$ is the finite set of singularities. The square-tiled surfaces in $ S $ are analogous to ``integer points,'' as their period coordinates lie in $\mathbb{Z} \oplus i \mathbb{Z}$. The asymptotics for the number of square-tiled surfaces constructed from at most $N$ squares in a given stratum can then be used to compute the volume.

For this reason, counting square-tiled surfaces has deep connections to Teichm{\"u}ller dynamics and to the study of flat surfaces. In the literature, a square-tiled surface is also called an \textit{origami} (alternatively defined as a Riemann surface obtained as a finite cover of the square torus branched over a single point). The $SL(2,\mathbb{R})$-action on any stratum restricts to an $SL(2,\mathbb{Z})$-action on the origamis in that stratum, and the study of the corresponding $SL(2,\mathbb{Z})$-orbits is a highly active area of research (\cite{HS} \cite{S}, \cite{HL}, \cite{Mc}).

A filling pair $\lbrace \alpha, \beta \rbrace$ on $S$ in minimal position naturally gives rise to a decomposition of $S$ into quadilaterals.  We can interpret $\alpha \cup \beta$ as a graph on $S$ whose vertices are elements of $ \alpha \cap \beta $ and whose edges are components of the symmetric difference $ \alpha \bigtriangleup \beta $; the dual of this graph will be the $1$-skeleton of a quadilateral decomposition. If the absolute value of the algebraic intersection number equals the geometric intersection number ($|\hat{\iota}(\alpha, \beta)|=\iota(\alpha, \beta)$), the decomposition yields an origami. In the language of origamis and flat geometry, when $g$ is odd, \autoref{main} can be reframed as:

\vspace{2 mm}

\textit{In the minimal stratum $\mathcal{H}(2g-2)$, there exist at least $(g-2)!$ square-tiled surfaces with $2g-1$ squares whose vertical and horizontal cylinder decompositions each consist of one cylinder.}

\vspace{2 mm}

Since these origamis have the simplest possible horizontal and vertical cylinder decompositions and the minimal number of squares possible, We will refer to such a surface as a \textit{minimal origami}. Using representation-theoretic tools and following the techniques of Zagier (\cite{Za}), Delecroix, Goujard, Zograf, and Zorich compute the exact number of square-tiled surfaces in $\mathcal{H}(2g-2)$ that have $2g-1$ squares and one maximal horizontal cylinder (see Equation (2.21) in \cite{DGZZ}). By contrast, both the vertical and horizontal cylinder decompositions of the origamis from \autoref{main} consist of a single cylinder, and they arise from a completely constructive recipe.

We advertise the constructive aspect of \autoref{main} as our key contribution---the recipe can be easily implemented on a computer to yield a large repository of origamis in $\mathcal{H}(2g-2)$ with the simplest cylinder decompositions and the smallest number of squares. During the 2018 Summer@ICERM program, a rudimentary version of the recipe was used to identify several distinct $SL(2, \mathbb{Z})$-orbits of these simple origamis; we summarize these findings in Section $5$.

We conjecture that there should be many distinct $SL(2,\mathbb{Z})$-orbits as genus grows:

\begin{conjecture} \label{many orbits} Let $N_{g}$ denote the number of distinct $SL(2, \mathbb{Z})$-orbits of origamis in $\mathcal{H}(2g-2)$ that have $2g-1$ squares and whose horizontal and vertical cylinder decompositions each consist of one cylinder. Then $N_{g}$ grows super-exponentially in $g$.
\end{conjecture}

\subsection{Outline}  In section $2$, we recall some preliminaries for both origamis and curves on surfaces.  In section $3$, we explain how to encode an origami with a minimally-intersecting filling pair as a permutation. In section $4$, we outline the main construction. Finally, in Section $5$, we summarize some results obtained in low genera after computationally implementing the main construction and investigating the orbits under the $SL(2, \mathbb{Z})$-action.

\subsection{Acknowledgements.} The first author thanks Jayadev Athreya and Samuel Taylor for many helpful conversations. The first author was supported by NSF Grant DMS 1939936 while this work was completed. The authors thank Zichen Cui, Ajeet Gary, Paige Helms, Ionnis Iakovoglu, Tasha Kim, and J.T. Rustad for their work during the 2018 Summer@ICERM program. The authors thank the Institute for Computational and Experimental Research in Mathematics (ICERM) for their hosting of the Summer@ICERM program during which the experimental data in Section $5$ was obtained. Finally, the authors thank Luke Jeffreys for extremely helpful discussions and for noticing an error in the original version of the even genus construction.

\section{preliminaries}

Let $S= S_{g}$ be oriented. If $\alpha, \alpha'$ are simple closed curves on $S$ which are isotopic to one another, we write $\alpha \sim \alpha'$. Then, given essential simple closed curves $\alpha, \beta$ on $S$, the \textit{geometric intersection number}, $\iota(\alpha, \beta)$, is defined by

\[ \iota(\alpha, \beta) = \min( |\alpha' \cap \beta'| : \alpha' \sim \alpha, \beta' \sim \beta). \]

The curves $\alpha$ and $\beta$ are said to be in \textit{minimal position} if $|\alpha \cap \beta|= \iota(\alpha, \beta)$.

Supposing that $\alpha, \beta$ are essential simple closed curves, equip each with a choice of orientation. Then, for $p \in \alpha \cap \beta$, define its \textit{index} $\epsilon(p) \in \left\{1,-1 \right\}$ to be $1$ if the orientation of $p$ from $\alpha$ to $\beta$ given by the right hand rule agrees with the orientation of $S$, and $-1$ otherwise. Then the \textit{algebraic intersection number}, $\hat{\iota}(\alpha, \beta)$, is defined by

\[ \hat{\iota}(\alpha, \beta) = \sum_{p \in \alpha \cap \beta} \epsilon(p). \]

Note that $\hat{\iota}(\alpha, \beta)= -\hat{\iota}(\beta, \alpha)$, and that the value of $\hat{\iota}(\alpha, \beta)$ does not change when we replace either $\alpha$ (or $\beta$) with another curve in its isotopy class. It follows that the absolute value of $\hat{\iota}$ is well-defined over pairs of isotopy classes $\left\{[\alpha], [\beta] \right\}$.

The \textit{mapping class group} of $S$, denoted $\mbox{Mod}(S)$, is the group of isotopy classes of orientation-preserving self-homeomorphisms of $S$. The \textit{extended mapping class group} of $S$, denoted $\mbox{Mod}^{\pm}(S)$, is the group of isotopy classes of self-homeomorphisms of $S$, including the orientation-reversing ones. Given two collections of isotopy classes $\Gamma= \left\{\gamma_{1},..., \gamma_{n} \right\}$, $\Omega= \left\{\omega_{1},..., \omega_{n} \right\}$, we say that $\Gamma$ and $\Omega$ are in the same $\mbox{Mod}(S)$ (resp. $\mbox{Mod}^{\pm}(S)$) orbit if there exists some $g \in \mbox{Mod}(S)$ (resp. in $\mbox{Mod}^{\pm}(S)$) so that $g \cdot \gamma_{i} = \omega_{i}$ for $i=1,.., n$.

We say that a pair of minimal position curves $\left\{\alpha, \beta \right\}$ is a \textit{filling pair} if their complement is a disjoint union of disks. Given a filling pair $\left\{\alpha, \beta \right\}$, let $R_{1},..., R_{n}$ be the complementary disk regions. Since $\alpha \cup \beta$ is a $4$-valent graph on $S$, the graph has twice as many edges as vertices. Using Euler's formula, it follows that
\[ 2- 2g= \chi(S_{g}) = \iota(\alpha, \beta) - 2\iota(\alpha, \beta) + n \]
\[ \Rightarrow \iota(\alpha, \beta) \geq 2g-1. \]

Thus, when a filling pair $\left\{\alpha, \beta \right\}$ satisfies $\iota(\alpha, \beta) = 2g-1$, we call it a \textit{minimally-intersecting filling pair}. In \cite{AH}, the first author and Huang directly construct exponentially-many $\mbox{Mod}^{\pm}(S)$-orbits of minimally intersecting filling pairs. However, the construction \textit{never} produced filling pairs satisfying $|\hat{\iota}(\alpha, \beta)|= 2g-1$. Thus, the filling pairs constructed below in Section $4$ are all distinct from those constructed in \cite{AH}.

 An \textit{origami} is a Riemann surface obtained as a finite cover of the torus branched over a single point. The union of the core curves of each cylinder in the horizontal cylinder decomposition of an origami is called the \textit{horizontal multicurve}, denoted $\alpha$; likewise, the union of the vertical core curves is the \textit{vertical multicurve} and is denoted $\beta$.  Since the number of squares of the origami is equal to $ \iota(\alpha, \beta)$, the horizontal and vertical curves of a minimal origami form a minimally-intersecting filling pair.  If $ \lbrace \alpha, \beta \rbrace $ is a minimally-intersecting filling pair with $ \iota(\alpha, \beta)=|\hat{\iota}(\alpha, \beta)|$ (so that the underlying surface is a minimal origami), we refer to it as a \textit{minimal origami pair}.  If we orient $\alpha$ and label the squares from $1$ to $N$ (the total number of squares), we obtain an associated \textit{horizontal permutation} in the following manner:  each cycle of the permutation corresponds to a component of $ \alpha $ and its entries are given by the sequence of squares one obtains by traversing that component in the direction of orientation.  Similarly, we obtain a \textit{vertical permutation} associated with $ \beta $. A relabeling of the squares will produce horizontal and vertical permutations that are simultaneously conjugate to the first pair of permutations.

\section{Filling pairs as permutations}

Let $\left\{\alpha, \beta \right\}$ be a minimal origami pair.
Then we can represent the surface as a horizontal row of $2g-1$ squares with $ \alpha $ and $ \beta $ as its horizontal and vertical curves.  The horizontal mid-segments of each square will correspond to $\alpha$ (the left- and right-most vertical edges are identified so that these segments form a closed curve). The intersection pattern of $\alpha$ and $\beta$ provides gluing instructions for the remaining edges of the squares, ensuring that each top edge is identified to one bottom edge and that the vertical mid-segments form a single simple closed curve which corresponds to $ \beta $.

If we number the squares in the row from $1$ to $2g-1$ in a cyclic left-to-right order and orient $ \alpha $ to the right, then its associated horizontal permutation is simply $\sigma_{2g-1}=(1,2, \ldots, 2g-1) $.  We will denote by $\tau_{\alpha, \beta}$ the vertical permutation obtained by orienting $\beta$ so that each vertical mid-segment points upward.  We will see below that this permutation encodes all the relevant combinatorial information about the surface and its filling pair.  Since $ \beta $ is a single curve, $\tau_{\alpha, \beta}$ will necessarily be a $(2g-1)$-cycle.

\autoref{Fig:G3 example} illustrates a genus-$3$ example that is configured to be consistent with our construction in \autoref{main construction}.  At left is an origami in $\mathcal{H}(5)$  with the squares labeled in cyclic left-to-right order (starting with ``$3$'').  The numbering of the upper and lower edges corresponds to their gluing. At right is the filling pair drawn on $S_3$; the labeling of the curve segments corresponds to the labeling of the upper/lower edges. The permutation $\tau_{\alpha, \beta}$ is the one that maps the label of each upper edge to the label of its square.

\vspace{2 mm}

\begin{figure}[h]
\centering
\includegraphics[width=.8\linewidth]{G3-origami.pdf}
\caption{\ The left illustrates a filling pair and the associated origami.  The right illustrates its geometric realization on $S_3$. For this example,  $\tau_{\alpha , \beta} = (1,3,5,4,2)$.  Please compare with the $S_7$-example in \autoref{g=7 example}.}
\label{Fig:G3 example}
\end{figure}

\vspace{2 mm}

\begin{remark}[\textbf{A Note about Left and Right}]
In what follows, we adopt the convention of reading multiplication in the symmetric group from left to right:  if $ \sigma, \tau $ are permutations, then the permutation $ \sigma \tau $ is the one which performs $\sigma$ first and then $\tau$.  This choice affects certain other algebraic expressions that will be relevant to our arguments, namely conjugates and commutators.

When we refer to the operation ``conjugation by $ \tau $," we mean that, if a permutation is expressed in cycle notation, then the conjugate of that permutation is obtained by replacing each entry in that expression with its image under $\tau$.  Accordingly, ``the conjugate of $ \sigma $ by $ \tau $" is written in the form $ \tau^{-1} \sigma \tau $, rather than the form $ \tau \sigma \tau^{-1}$.

Finally, we define the commutator of $ \tau $ and $ \sigma $ by the formula  $ [\tau, \sigma]= \tau \sigma \tau^{-1} \sigma^{-1} $.
\end{remark}

We can trace around the boundary of a regular neighborhood of $\alpha \cup \beta$ as follows. Each square of $S$ contains one element of $ \alpha \cap \beta$, and $\alpha \cup \beta$ subdivides the square into $4$ quadrants. We then choose a starting intersection point and, beginning in the upper-right quadrant, we travel upwards along $\beta$ to the next intersection (which is contained in the square whose lower edge is glued to the upper edge of the starting square).  Because every intersection has the same index, we arrive in its lower-right quadrant; we then travel along $\alpha$ to the right until we arrive at the next intersection point
(in the lower-left quadrant of the square).  Next, we travel downwards along $\beta$ (opposite its orientation) to the upper-left quadrant of the next square.  Finally, we travel to the left along $\alpha$, arriving again in the upper-right quadrant of the next square. Repeating this pattern---up along $\beta$, right along $\alpha$, down along $\beta$, and then left along $\alpha$, never crossing $\alpha \cup \beta$---will trace around the boundary of the regular neighborhood.  Since $S \smallsetminus (\alpha \cup \beta)$ is a single disk, there is only one boundary component, so we will travel along both sides of every edge of $\alpha \cup \beta$ before arriving back at the starting place, after $2g-1$ iterations.

This process can be described with permutations: if we travel along $\beta$ in the direction of orientation beginning at the square labeled $ x $, then our destination will be the square labeled $\tau_{\alpha, \beta}(x)$; if we travel in the direction opposite the orientation, we arrive at $\tau_{\alpha, \beta}^{-1}(x)$.  Similarly, the destination along $ \alpha $ is given by $\sigma_{g}$ or $ \sigma_g^{-1}$ (depending on the direction of travel).  Thus, the boundary components of $\alpha \cup \beta$ correspond to the cyclic factors of $ [\tau_{\alpha, \beta}, \sigma_g ]$.  A regular neighborhood of $ \alpha \cup \beta $ has one boundary component precisely when $ [\tau_{\alpha, \beta}, \sigma_g ] $ is a $(2g-1)$-cycle.

These observations show that if $\tau_{\alpha, \beta}$ is obtained as the vertical permutation of a minimal origami, it must have the following properties:

\begin{enumerate}
\item $\tau_{\alpha, \beta}$ is a $(2g-1)$-cycle
\item $[\tau_{\alpha, \beta}, \sigma_{g}]$ is a $(2g-1)$-cycle.
\end{enumerate}

The following result establishes that these properties completely characterize such permutations.

\begin{lemma} \label{characterizing perm} Let $\tau$ be a permutation satisfying the two properties listed above. Then there corresponds a filling pair $\left\{\alpha, \beta \right\}$ on $S_{g}$ satisfying
\[ \iota(\alpha, \beta) = |\hat{\iota}(\alpha, \beta)|= 2g-1\]
such that $\tau$ is its vertical permutation. Furthermore, two such filling pairs are in the same $\mbox{Mod}^{\pm}(S_{g})$-orbit if and only if their vertical permutations are conjugate to one another by a permutation that commutes with $\sigma_{g}$.
\end{lemma}

\begin{proof} Starting with a horizontal row of $2g-1$ squares, we can use $\tau$ to specify a gluing pattern for the top and bottom edges, as described above. The fact that $\tau$ is a $(2g-1)$-cycle will guarantee that the vertical mid-segments will join up under this gluing to yield a single simple closed curve which we refer to as $\beta_{\tau}$. Gluing the left- and right-most edges together turns the concatenation of the horizontal mid-segments into a second simple closed curve, denoted by $\alpha_{\tau}$.  It's clear that, if we number the squares from left to right in consecutive order, the horizontal permutation of this origami is $ \sigma_g $ and the vertical permutation is $ \tau $; therefore, the second property of $\tau$ implies that the boundary of a regular neighborhood of $\alpha_{\tau} \cup \beta_{\tau}$ has a single connected component.

Now, suppose that $\left\{\alpha_{1}, \beta_{1} \right\}$ and $\left\{\alpha_{2}, \beta_{2} \right\}$ are two filling pairs which satisfy the required properties for geometric and algebraic intersection numbers and suppose there is a self-homeomorphism $ f $ of $ S_g $ mapping $\left\{\alpha_{1}, \beta_{1} \right\}$ to $\left\{\alpha_{2}, \beta_{2} \right\}$.  Since $\left\{\alpha_{1}, \beta_{1} \right\}$ is a minimal origami pair, there is a square-tiling $ R_1 $ of the surface which consists of a horizontal strip of $ 2g-1 $ squares whose horizontal curve is $\alpha_{1}$ and whose vertical curve is $\beta_{1}$.  Label the squares such that the horizontal permutation of the origami starting from the leftmost square is $ \sigma_g $ and let $ \tau $ be the corresponding vertical permutation.

The map $ f $ lifts to a map $ \tilde{f} $ mapping $ R_1 $ to a square-tiling $ R_2 $ for $\left\{\alpha_{2}, \beta_{2} \right\}$.  This $ \tilde{f} $ also acts as a permutation of the labels of the squares.  Label the squares of $ R_2 $ in consecutive order from left to right and let $ k+1=\tilde{f}(1) $.  Since $ \tilde{f} $ is a homeomorphism of $\alpha_{1}$ onto $\alpha_{2}$, it must map consecutive squares to consecutive squares, and so $ \tilde{f}(i)=k+i $ for each $ i \in \lbrace 1, 2, \ldots, 2g-1 \rbrace $.  Now let $ \tau' $ be the vertical permutation of $\left\{\alpha_{2}, \beta_{2} \right\}$.  Since $ \tilde{f} $ is a homeomorphism of $\beta_{1} $ onto $ \beta_{2} $, the observation above implies that, if $ j=\tau(i) $, then $ j+k=\tau'(i+k) $ for any $ i,j \in \lbrace 1, 2, \ldots, 2g-1 \rbrace $.  This means that, if we express both permutations in cycle notation, then the expression for $ \tau' $ is obtained from the expression for $ \tau $ by adding $ k $ to each entry.  In other words: \[ \tau' = \sigma_{g}^{-k} \tau \sigma_{g}^{k} \ . \]So $ \tau $ and $ \tau' $ are conjugate by a permutation that commutes with $ \sigma_g $.

Conversely, suppose there exist square-tilings $ R_1 $ and $ R_2 $ for $\left\{\alpha_{1}, \beta_{1} \right\}$ and $\left\{\alpha_{2}, \beta_{2} \right\}$ (respectively) with vertical permutations $ \tau $ and  $ \tau' $ (respectively) and that $ \tau $ and $ \tau' $ are conjugate by a permutation that commutes with $ \sigma_g $ (i.e., an element of the centralizer of $ \sigma_g $).  This centralizer is simply the subgroup $\langle \sigma_g \rangle $ generated by $\sigma_{g}$, so the two vertical permutations must be conjugate by a power of $ \sigma_g $.  Then it's possible to define a map from $ R_1 $ to $ R_2 $ that maps consecutive squares to consecutive squares and, therefore, maps $ \alpha_1 $ onto $ \alpha_2 $ and $ \beta_1 $ onto $ \beta_2 $.  This map descends to a self-homeomorphism of $ S_g $ mapping $\left\{\alpha_{1}, \beta_{1} \right\}$ to $\left\{\alpha_{2}, \beta_{2} \right\}$, as claimed.

Conversely, note that any homeomorphism sending $\alpha_{i}$ to $\beta_{i}$ will lift to a square-preserving homeomorphism from $R_{1}$ to $R_{2}$, \textit{after} potentially replacing $R_{2}$ with an equivalent diagram $R'_{2}$ obtained by cyclically rotating the squares so that the first square of $R_{1}$ is sent to the first square of $R'_{2}$. This homeomorphism will necessarily preserve the combinatorics of the gluing pattern. It follows that the permutations $\tau_{\alpha_{1}, \beta_{1}}, \tau_{\alpha_{2}, \beta_{2}}$ must coincide, up to conjugation by the centralizer of $\sigma_{g}$.

\end{proof}

\autoref{characterizing perm} implies that the problem of building minimal origami pairs can be replaced by the problem of finding $(2g-1)$-cycles $\tau$ whose commutator with $\sigma_{g} $ is a $(2g-1)$-cycle. We will call such permutations \textit{minimal origami permutations}. Since the centralizer of $\sigma_{g}$ has order $2g-1$, we have the following direct consequence of \autoref{characterizing perm}:

\begin{proposition} \label{counting} Let $O_{g}$ denote the number of $\mbox{Mod}^{\pm}(S_{g})$-orbits of minimal origami pairs and let $P_{g}$ be the number of distinct minimal origami permutations. Then

\[ O_{g} \geq \frac{P_{g}}{2g-1} \ . \]

\end{proposition}

We conclude this section by again reframing the problem at hand in more convenient terms. Let $\rho$ be a $(2g-1)$-cycle so that: \medskip

\begin{enumerate}
\item $\rho \sigma_{g}^{-1}$ is a $(2g-1)$-cycle;
\item $\rho$ is conjugate to $\sigma_{g}$ by a $(2g-1)$-cycle.
\end{enumerate} \medskip

Given such a $\rho$, let $\tau$ be the $(2g-1)$-cycle conjugating $\rho$ to $\sigma_{g}$. Thus
\[ \sigma_{g} = \tau^{-1} \rho \tau \Rightarrow \rho = \tau \sigma_{g} \tau^{-1}. \]

But then
\[ [\tau, \sigma_g]=\tau \sigma_g \tau^{-1} \sigma_g^{-1}=\rho \sigma_g^{-1} \]
By the first property above, this commutator is a $(2g-1)$-cycle and by second property above, $\tau$ is a $(2g-1)$-cycle as well. Thus, by \autoref{characterizing perm}, $\tau$ is a minimal origami permutation. We replace the problem of finding the permutation $ \tau $ with the problem of finding the permutation $ \rho $.

\begin{lemma} \label{rephrase} Suppose $\rho_{1}, \rho_{2}$ satisfy the above two properties, and let $\tau_{1}, \tau_{2}$ be the $(2g-1)$-cycles that conjugate $\rho_{1}, \rho_{2}$ (respectively) to $\sigma_{g}$. Then if $\tau_1$ and $\tau_2$ are conjugate by an element of $\langle \sigma_{g} \rangle$, then the same holds for $\rho_1$ and $\rho_2$.
\end{lemma}

\begin{proof}

If there is some $k$ so that
\[ \tau_2 = \sigma^{-k} \tau_1 \sigma^{k}, \]
then
\[ \rho_2 = \tau_{2}\sigma_{g} \tau_{2}^{-1} \]
\[= (\sigma_{g}^{-k} \tau_{1} \sigma_{g}^{k}) \sigma_{g} (\sigma_{g}^{-k} \tau_{1}^{-1} \sigma_{g}^{k}) \]
\[ = \sigma_{g}^{-k}(\tau_{1}\sigma_{g} \tau_{1}^{-1})\sigma_{g}^{k} = \sigma_{g}^{-k}\rho_{1} \sigma_{g}^{k}.\]

\end{proof}

Thus, distinct solutions to the new permutation problem correspond to distinct solutions to the old one. We now go on to build solutions to the new problem.

\section{The main construction}
\label{main construction}

Our goal is to construct $(2g-1)$-cycles $\rho$ satisfying \medskip

\begin{enumerate}
\item $\rho \sigma_{g}^{-1}$ is a $(2g-1)$-cycle;
\item $\rho$ is conjugate to $\sigma_{g}$ by a $(2g-1)$-cycle
\end{enumerate} \medskip

We begin by outlining the construction in the case where $g$ is odd. Begin by placing the numbers $3,4,5,..., 2g-1,1,2$ (in that order) in a horizontal row. This will represent the permutation $\sigma_{g}$: placing parentheses around this list gives an expression of $\sigma_{g}$ in cycle notation. Placing the $1$ and $2$ at the end is a notational convenience whose purpose will be made clear later in the section.

We will build our permutation $\rho$ by placing the numbers $\left\{1,2,..., 2g-1 \right\}$ in a second row above the $ \sigma_g $-row.  The permutation $\rho$ will be represented in cycle notation by placing parentheses around the numbers in this top row. The stacking of the two rows defines a third permutation $ \tau $ which maps each entry in the top row to the entry below it.  This is the permutation that conjugates $\rho$ to $\sigma_{g}$.

We proceed by first placing the number $ 1 $ in the top row above the number $ 3 $.  We then place each of the pairs $ (i, i+1) $ ($ i $ odd) in turn such that: \medskip

\begin{enumerate}
    \item The odd entry of the pair is placed over an even entry $ j $ of the bottom row (until there is no other choice but to place it over an odd entry, which will necessarily be $1$).
    \item The next pair to be placed in the top row is determined by the number $ j $.
\end{enumerate} \medskip

Since we placed $ 1 $ over $ 3 $, this determines that $ \tau(1)=3 $, so we begin by placing the pair $(3,2)$ over an arbitrary pair of numbers $(j, j+1)$ where $j$ is even. This dictates that $ \tau(1)=3 $, $\tau(3)= j$, and $\tau(2)= j+1$. Thus, the cycle expression for $\tau$ will contain a cycle that begins with $(1,3,j,...)$.  We do not yet know where the $2$ or the $j+1$ will appear in cycle notation, only that $2$ must appear directly before $j+1$ in some cycle.

Because $ 3 $ was placed over $ j $, the next pair to be placed in the top row is $ (j+1, j) $, which we place over some other pair of consecutive entries in the bottom row (making sure that $ j+1 $ is placed above an even number).  We continue in this manner until we have only one pair left to place (over the entries $ (1,2) $ in the bottom row; this was the reason for beginning the bottom row with $ 3 $ instead of $ 1 $).  We claim that this results in a cycle expression for $\tau$ which consists of a single $(2g-1)$-cycle, thus verifying the second of the two desired properties of $ \rho $.

\subsection{Example of the construction.}
\label{g=7 example}
We outline the above construction in an example for $g= 7$ (so $2g-1 = 13$). The construction begins with the following diagram:

\[ \begin{array} {ccccccccccccc}
1 & . & . & . & . & . & . & . & . & . & . & . & . \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 1 & 2
\end{array}\] \medskip

As described above, the first choice we make in the construction is to decide the placement of the pair $(3,2)$; we can choose arbitrarily between $(4,5), (6,7), (8,9), (10,11)$, and $(12,13)$ as numbers over which to place this pair. In this example, we will chose to place it over $(8,9)$, producing the following diagram:

\[ \begin{array} {ccccccccccccc}
1 & . & . & . & . & 3 & 2 & . & . & . & . & . & . \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 1 & 2
\end{array}\] \medskip

This dictates that $\tau(3)$ will be $8$, so the next choice will be the placement of $ (9,8) $; the possible options are $(4,5), (6,7), (10,11)$, and $(12,13)$. We will select $(12, 13)$, producing:

\[ \begin{array} {ccccccccccccc}
1 & . & . & . & . & 3 & 2 & . & . & 9 & 8 & . & . \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 1 & 2
\end{array}\] \medskip

We must next choose the position of the pair $(13,12)$, and we choose to place it over $(4,5)$:

\[ \begin{array} {ccccccccccccc}
1 & 13 & 12 & . & . & 3 & 2 & . & . & 9 & 8 & . & . \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 1 & 2
\end{array}\] \medskip

Now we can choose to place $ (5,4) $ over $(6,7)$ or $(10,11)$; we choose $(6,7)$:

\[ \begin{array} {ccccccccccccc}
1 & 13 & 12 & 5 & 4 & 3 & 2 & . & . & 9 & 8 & . & . \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 1 & 2
\end{array}\] \medskip

We now have only the pair $ (7,6) $ left to place and no choice but to place it over $(10,11)$, and finally to place $(11,10)$ over $(1,2)$:

\[ \begin{array} {ccccccccccccc}
1 & 13 & 12 & 5 & 4 & 3 & 2 & 7 & 6 & 9 & 8 & 11 & 10 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 1 & 2
\end{array}\] \medskip

The top row is the cylce expression of our permutation $\rho=(1,13,12,5,4,3,2,7,6,9,8,11,10)$. We now verify the desired properties. The permutation $\tau$ is the map sending each number in the top row to the number directly below it in the bottom:

\[ \tau= (1,3,8,13,4,7,10,2,9,12,5,6,11), \]
which is indeed a $13$-cycle, as desired. To check the first property, we verify that $\rho \sigma_{7}^{-1}$ is a $13$-cycle:

\[ \rho \sigma_{7}^{-1} = (1,12,4,2,6,8,10,13,11,9,7,5,3). \]

\subsection{Proof of the construction in odd genus.}

\begin{theorem} \label{the construction} The permutation $ \rho $ produced by the construction described above always satisfies the desired two properties.
\end{theorem}

\begin{proof}[Proof. Odd genus] The reader is encouraged to refer back to the example above during the course of the argument.  To show that $ \tau $ is a $ (2g-1) $-cycle, we will argue that every entry of $ \tau $ is either in the cycle containing $ 1 $ (denoted $ \kappa_1 $) or in the cycle containing $ 2 $ (denoted $ \kappa_2 $); we conclude by showing that $ 2 \in \kappa_1 $ and, therefore, $ \kappa_1=\kappa_2 $, so the entire permutation consists of a single cycle.

The permutation is determined by the placement of $ (g-1) $ pairs of the form $ (i+1, i) $, where $ i $ is even.  The last such pair is placed over the entries $ 1 $ and $ 2 $ so that, if $ (k+1, k) $ is that pair, then $ 1 $ is the image of $ k+1 $ and $ 2 $ is the image of $ k $; thus $ k+1 \in \kappa_1 $ and $ k \in \kappa_2 $.  The entries of this pair, though, are the images of the entries of the $ (g-2)^{nd} $ pair to be placed; since $ k $ is the image of its odd entry, that odd entry belongs to $ \kappa_2 $ and, since $ k+1 $ is the image of the even entry, the even entry belongs to $ \kappa_1 $.

We can proceed in this manner, ``walking back" to the first pair to be placed; with each step, we alternate whether the odd entry or the even entry belongs to $ \kappa_1 $:  after an odd number of steps, the even entry belongs to $ \kappa_1 $ and, after an even number of steps, the odd entry belongs to $ \kappa_1 $.  Since $ g-1 $ is even, it will take an odd number of steps to reach the first pair, which means that the even entry of this pair belongs to $ \kappa_1 $.  But the first pair to be placed was $ (3,2) $, which implies that $ \kappa_1=\kappa_2 $ as we claimed.  Thus $ \tau $ is a $ (2g-1) $-cycle.

To verify that $\rho \sigma_{g}^{-1}$ is a $(2g-1)$-cycle, we first remark that, from the second entry onward, the entries of $\rho$ alternate in parity, starting with odd. Moreover, for $j$ even, the $j^{th}$ entry is always an odd number that is one larger than the $(j+1)^{st}$ entry. It follows that $\rho \sigma_{g}^{-1}$ has a cycle expression that starts with

\[ (1, 3^{rd} \hspace{1 mm} \mbox{entry of} \hspace{1 mm} \rho, 5^{th} \hspace{1 mm} \mbox{entry of} \hspace{1 mm} \rho,..., (2g-1)^{st} \hspace{1 mm} \mbox{entry of} \hspace{1 mm} \rho...) \].

Note that this sequence contains all the even numbers in $ \lbrace 1, 2, \ldots, 2g-1 \rbrace $.  The $(2g-1)^{st}$ entry of $\rho$ is, of course, sent to $1$ by $\rho$, which is then sent to $2g-1$ by $\sigma_{g}^{-1}$. By construction, every odd entry $ i>1 $ of $ \rho $ is followed by the number $ i-1 $; thus $ \rho \sigma_g^{-1}(2g-1)=2g-3, \rho \sigma_g^{-1}(2g-3)=2g-5 $, and so on through all the odd numbers in $ \lbrace 1, 2, \ldots, 2g-1 \rbrace $ (ending with $ 3 $, which then maps back to $ 1 $).  Since it contains all possible entries, $ \rho \sigma_g^{-1} $ is a $ (2g-1) $-cycle, completing the proof.

\end{proof}

\begin{corollary} \label{odd theorem} For odd $g$,

\[ P_{g} \geq (g-2)! \]
\end{corollary}

\begin{proof} There are $g-2$ pairs of the form $(j,j+1)$ ($j$ even) in the bottom row.  For each such pair, we place above it a pair $(i+1, i)$ ($i$ even).  Thus there are $ g-2 $ places for the first $ i $-pair, $ g-3 $ for the second, and so on, until the last $ j $-pair has been taken and we are forced to place the last $ i $-pair above $ (1,2) $.  Thus, the number of $ \rho $'s that can be constructed is $ (g-2)! $.
\end{proof}

\begin{remark} \label{no conjugates} Lemma \ref{rephrase} implies that if solutions $\rho_1, \rho_2$ are not conjugate by an element of $\langle \sigma_{g} \rangle$, the corresponding minimal origami permutations will correspond to distinct $\mbox{Mod}^{\pm}(S)$ orbits of minimally intersecting filling pairs. Proposition \ref{counting} implies that at most $2g-1$ of the solutions constructed above can correspond to the same $\mbox{Mod}^{\pm}(S)$ orbit, but in fact, the orbits of our solutions under conjugation by $\langle \sigma_{g} \rangle$ are pairwise disjoint, as we now demonstrate.

Fix a solution $\rho$; a conjugate of $\rho$ by an element of $\langle \sigma_{g} \rangle$ can be obtained by adding some $k$ (modulo $2g-1$) to each entry of the cycle notation for $\rho$. Assume that one of these conjugates produces another one of our solutions.

If $k$ is odd, the resulting permutation will not be built from pairs of the form $(i+1, i)$ with $i$ odd, as all of our solutions are, and thus we can assume that $k$ is even. Since $2g-1$ is odd, the unique solution to $k + x  = 1$ (mod $2g-1$) must be even-- denote this number by $j$. By construction, $j+1$ appears directly before $j$ in the cycle notation for $\rho$, and thus the conjugate $\sigma_{g}^{-k} \rho \sigma^{k}$ must have $2$ appearing directly before $1$. For this permutation to be one of our other solutions, $3$ must then occur immediately before $2$, as $(3,2)$ is one of the pairs from which our solutions are built.

It follows that $(3,2)$ is the last pair placed in the construction of $\rho$; indeed, the last pair is always the one which cyclically appears directly in front of $1$. However, by design, $(3,2)$ is always the first pair placed.

\end{remark}

\autoref{odd theorem} together with \autoref{no conjugates} immediately implies \autoref{main} for odd $g$.

\subsection{The even case.} We next consider the case of even $g \geq 4$ (as shown in \cite{AH}, there are no filling pairs intersecting $3$ times on a genus $2$ surface). We begin with a permutation $ \xi $ on the numbers $ \lbrace 1, 2, \ldots, 2g-3 \rbrace $ obtained by the construction described above (since $ 2g-3=2(g-1)-1 $, this fits into the odd-genus case), and let $\eta$ denote the permutation conjugating $\xi$ to $\sigma_{g-1}$. Arbitrarily choose any odd $k \in \left\{3,5,7,..., 2g-3 \right\}$ not equal to $\eta^{-1}(1)$. In a horizontal row, we place the entries of $ \sigma_g $ in ascending order. As in the odd-genus case, we will build our permutation $ \rho $ by arranging the numbers $\left\{1,2,..., 2g-1 \right\}$ above this $ \sigma_g$-row.  The top row will then be a cycle expression for $ \rho $ and $ \tau $ will be the permutation that conjugates $ \rho $ to $ \sigma_g $.

We place each entry of $ \eta $ except for $ k $ above its image (so $ \tau(x)=\eta(x) $ for all $ x \notin \lbrace k, 2g-2, 2g-1 \rbrace $); we complete the top row by placing $k$ over $2g-1$, $2g-1$ over $2g-2$, and $2g-2$ over $\eta(k)$.

\subsection{Example in the even case} We give an example of the construction in the even case where $g= 10$, starting with the following $17$-cycle, which is a minimal origami permutation for $g=9$:

\[ \xi= (1,11,10,15,14,3,2,7,6,17,16,9,8,5,4,13,12). \] \medskip

The permutation $ \eta $ conjugating $ \xi $ to $ \sigma_{9} $ is
\[ \eta= (1,3,8,15,6,11,4,17,12,2,9,14,7,10,5,16,13). \] \medskip

We arbitrarily choose the odd number $k=11$ and carry out the construction described above. We begin by placing the numbers from $1$ to $19$ in ascending order from left to right:

\[ \begin{array} {ccccccccccccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19
\end{array}\]

We then place each number other than $ k=11,\ 2g-2=18 $, and $ 2g-1=19 $ above its image under $ \eta $. This leads to the diagram

\[ \begin{array} {ccccccccccccccccccc}
13 & 12 & 1 & & 10 & 15 & 14 & 3 & 2 & 7 & 6 & 17 & 16 & 9 & 8 & 5 & 4 & &  \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19
\end{array}\] \medskip

Finally, we place $11$ over $19$, $19$ over $18$, and $18$ over $ \eta(11)= 4 $, yielding

\[ \begin{array} {ccccccccccccccccccc}
13 & 12 & 1 & 18 & 10 & 15 & 14 & 3 & 2 & 7 & 6 & 17 & 16 & 9 & 8 & 5 & 4 & 19  & 11 \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19
\end{array}\] \medskip

Denote by $\rho$ the permutation obtained by reading the top row from left to right and interpreting it as a $19$-cycle. We compute:
\[ \rho \sigma_{10}^{-1} = (1,17,15,13,11,12,19,10,14,2,6,16,8,4,18,9,7,5,3), \] \medskip
verifying the first property of a minimal origami permutation. Letting $\tau$ denote the permutation conjugating $\rho$ to $\sigma_{10}$, we see that
\[ \tau = (1,3,8,15,6,11,19,18,4,17,12,2,9,14,7,10,5,16,13), \]
verifying the second property.

To demonstrate the necessity of requiring that $k \neq \eta^{-1}(1)$, we include the following example shown to the authors by Luke Jeffreys.

Beginning with $\xi = (1,3,2,5,4)$ in genus $3$, we can take $\eta$ to be $(1,3,4,2,5)$. Then suppose we choose $k= 5= \eta^{-1}(1)$ and proceed as above. This will yield the permutation $\rho= (1,3,2,7,5,6,4)$. One then computes that $\rho \sigma_{4}^{-1}$ is \textit{not} a $7$-cycle, as required:
\[ \rho \sigma_{4}^{-1} = (1,2,6,3)(4,7)(5) \]

\begin{proof}[Proof. Even genus]
We now prove that the construction demonstrated above always produces a minimal origami permutation. For $g$ even, let $\xi$ denote the $(2g-3)$-cycle used to construct the $(2g-1)$-cycle $\rho$ and note that, beginning with $ 1 $, the entries of $ \rho, \xi $ in cycle notation coincide \textit{until} the odd number $k$ appears in $ \xi $.  We will describe the cycle expression of $ \rho \sigma_g^{-1} $ one entry at a time; this description is diagrammed in \autoref{evencase} and we encourage the reader to refer to it as we go.

We begin the first (and, we claim, only) cycle of $ \rho \sigma_g^{-1} $ with 1 and assume, for the moment, that $ k \neq \xi(1) $.  To find the next entry, we evaluate $ \rho(1) $, which is equal to $ \xi(1) $, and is therefore an odd number greater than 1.  By construction, almost every odd entry of $ \xi $ is followed by its predecessor:  if $ i+1>1 $ is odd, then $ \xi(i+1)=i $.  This means that the $ 2^{nd} $ entry of $ \rho \sigma_g^{-1} $ will be the $ 3^{rd} $ entry of $ \xi $.  This entry will be even and the behavior described above will repeat: $ \rho \sigma_g^{-1} $ will land on each even entry of $\xi$ until we reach the position just before $ k $, since $k$ is to the right of $1$ in the top row owing to the fact that $\eta^{-1}(1) \neq k$. What we have just described accounts for the red portion of \autoref{evencase} at the top left.  Note that, if $ \xi(1)=k $, then what we describe next happens at the beginning of the cycle.

In the permutation $ \rho $, the position occupied by $ k $ in the permutation $ \xi $ is now occupied by the entry $ 2g-2 $, which is then sent to $ 2g-3 $ by $ \sigma_g^{-1} $.  This is the largest odd entry of $ \xi $, which means that the next entry of $ \xi $ (and also of $ \rho $) is its predecessor $ 2g-4 $, which $ \sigma_g^{-1} $ then sends to $ 2g-5 $.  From this point onward, the entries of $ \rho \sigma_g^{-1} $ will continue counting downwards through the odd entries of $ \xi $ (which agree again with the entries of $ \rho $) until we reach $ k $.  In the construction of $ \rho $, $ k $ was placed above $ 2g-1 $ as the last entry in the top row; this means that, if $ \rho(k)=j $, then $ j $ will be the last odd entry placed in $\xi $ (since $ j $ would have been placed above 1 in that construction and 1 has now been cycled to the front of the bottom row).  The permutation $ \sigma_g^{-1} $ sends $ j $ to $ j-1 $; but, since $ j $ is an odd entry of $ \xi $ greater than 1, $ j-1 $ is also the entry following $ j $ in the cycle notation of $ \xi $.  The entry $ j-1 $ was last in $ \xi $, which means that $ \rho(j-1)=1 $, so $ (\rho \sigma_g^{-1})(j-1)=2g-1 $. This paragraph has accounted for the black portion of \autoref{evencase} in the top right.

The entry $ 2g-1 $ was placed above $ 2g-2 $ in the construction of $ \rho $ and $ k $ was placed above $ 2g-1 $, so $ \rho(2g-1)=k $, which $ \sigma_g^{-1} $ then sends to $ k-1 $.  But $ k-1 $ follows $ k $ in the cycle expression of $ \xi $, so now we return to the pattern at the beginning of the cycle:  the next entry of $ \rho $ is an odd entry of $ \xi $ greater than 1, which means that the next entry in both $ \rho $ and $ \xi $ is its predecessor. Thus the entries of $ \rho \sigma_g^{-1} $ now run through the even entries of $ \xi $ appearing after $k$ in $\xi$.  This continues until we reach the even entry $ \tau^{-1}(2g-3) $ (again, see \autoref{evencase}), which is just before 2g-1 near the end of the cycle.  From this position, $ \rho \sigma_g^{-1} $ sends us to $ 2g-2 $. The above accounts for the lower red portion of \autoref{evencase}.

The entry $ 2g-2 $ in $ \rho $ occupies the position that $ k $ occupied in the cycle expression of $ \xi $, which means that the next entry in $ \rho $ will be $ k-1 $, which $ \sigma_g^{-1} $ sends to $ k-2 $.  And now we once again resume a previously-established pattern:  since $ k-2 $ is odd, the next entry of $ \rho $ is its predecessor $ k-3 $, which $ \sigma_g^{-1} $ sends to $ k-4 $.  The remaining entries of $ \rho \sigma_g^{-1} $ will be the odd numbers less than $ k $ in descending order, until we finally reach 1, completing the cycle.  The cycle we have just described consists of all the numbers in $ \lbrace 1, 2, \ldots, 2g-1 \rbrace $; since we have exhausted all possible entries of $ \rho \sigma_g^{-1} $, this permutation is a $ (2g-1) $-cycle, as claimed.  \medskip

\begin{figure}
\centering
\includegraphics[width= \textwidth, height= 80mm]{EvenCaseOrigami}{}
\caption{}
\label{evencase}
\end{figure}

It remains to show that $\tau$, a permutation conjugating $\rho$ to $\sigma_{g}$, is a $(2g-1)$-cycle. A representative for the permutation $\tau$ in cycle notation beginning with the cycle containing $1$ will agree with the representative for $\eta$ that begins with $1$, until the occurrence of $k$ within $\eta$. At that point, the next entries of $\tau$ will be $2g-1, 2g-2, \eta(k)$, and after this, the subsequent entries will coincide exactly with what occurs in $\eta$ after $\eta(k)$. It follows that $\tau$ is a $(2g-1)$-cycle. \end{proof}

\vspace{3 mm}

 Since there are $g-3$ odd numbers between $3$ and $2g-3$ (inclusive) not equal to $\eta^{-1}(1)$, we obtain:

\begin{corollary}  For $g$ even,
\[ P_{g} \geq (g-3)! \cdot (g-3) = (g-3)(g-3)! \]
\end{corollary}

\begin{remark} \label{no conjugates even} Analogous to Remark \ref{no conjugates}, we now show that the solutions described above can be conjugate to one another by elements of $\langle \sigma_{g} \rangle$ only under very specific circumstances. To this end, assume there are solutions $\rho_1, \rho_2$ constructed above and some $k$ so that $\sigma_{g}^{-k} \rho_1 \sigma_{g}^{k} = \rho_2$.

Let $t_{1}$ (resp. $t_{2}$) denote the odd number in $\left\{3,..., 2g-3 \right\}$ used to construct $\rho_{1}$ (resp. $\rho_{2}$) from solutions in genus $g-1$. In cycle notation for $\rho_{i}$, each odd number $n$ except for $1$ and $2g-1$ appears directly before $n-1$. Moreover, $2g-1$ appears directly before $t_{i}$ and $2g-2$ appears directly before $t_{i}-1$. It follows that
\[ \rho_{i}(2g-1)- 1 = \rho_{i}(2g-2). \]

Adding $k$ (modulo $2g-1$) to each entry of $\rho_{1}$ produces $\rho_{2}$. Let $j \in \left\{1,..., 2g-1 \right\}$ satisfy $j+k = 2g-1$; since $k$ is (wlog) not a multiple of $2g-1$, $j \neq 2g-1$.

If $j=1$, we can take $k= 2g-2$ and thus $2g-1 + k = 2g-2$ (mod $2g-1$), so the position of $2g-2$ in $\rho_{2}$ will coincide with the  position of $2g-1$ in $\rho_{1}$. Hence, $t_{2} - 1 = t_{1} + k$ (mod $2g-1$), which implies that $t_{1} = t_{2}$. The position of $t_{2}$ in $\rho_{2}$ coincides with the position of $\rho_{1}(1)$ in $\rho_{1}$, and thus $\rho_{1}(1)= t_{2}+1= t_{1}+1$, which is even. By construction, this is only possible if $\rho_{1}(1) = 2g-2$, and this would imply that $t_{1} = 2g-3$.

Next, assume that $j= t_{1}$. In this case we can take $k= 2g-1 -t_{1}$ (which is even), and this implies that the position of $2g-1$ in $\rho_{1}$ coincides with the position of $2g-1 - t_{1}$ in $\rho_{2}$. Any even number $n$ in $\rho_{2}$ either has the property that it is directly preceded by $n+1$, or it is one of $t_{2}-1, 2g-2$. If $2g-1 - t_{1}$ is preceded by $2g-t_{1}$ in $\rho_2$, then $1$ directly precedes $2g-1$ in $\rho_{1}$, contradicting the construction of $\rho_{1}$.

If $2g-1-t_{1} = 2g-2$, then $\rho_{2}(2g-1-t_{1})= t_{2}-1$, but by assumption this is also $t_{1}+k$ since the position of $2g-1$ in $\rho_1$ coincides with the position of $2g-1-t_{1}$ in $\rho_{2}$. Thus, (mod $2g-1$) we obtain
\[ t_{2}- 1 = t_{1}+k = t_{1} + 2g-1 - t_{1} = 2g-1, \]
which implies that $t_{2} = 1$, a contradiction.

If $2g-1 -t_{1} = t_{2}- 1$, then $2g-2$ must immediately precede it in $\rho_{2}$. It then follows that $t_{1}- 1$ immediately precedes $2g-1$ in $\rho_{1}$. This in turn implies that at the end of the cycle for $\rho_1$, one sees $(...2g-2, t_{1}-1, 2g-1, t_{1})$. This specifies $\rho_1$ amongst all solutions obtained from the same permutation in odd genus: $\rho_1$ is the one for which the associated odd number $t_{1}$ is $\eta^{-1}(2g-3)$.

Next, assume $j$ is odd and not equal to either $1$ or $t_{1}$. Then $\rho_{1}(j)= j-1$, and so $j-1+k = t_{2}$. However, $k$ is even in this case because both $2g-1$ and $j$ are odd, so $j-1+k$ is even, but $t_{2}$ is odd.

On the other hand, if $j$ is even, $k$ is odd, and this will imply that one sees $g-3$ occurrences of $(... i, i-1,...)$ with $i$ even in the cycle for $\rho_{2}$, which is impossible by construction.

To summarize: given $\rho_1$, there exists $\rho_2$ conjugate to it by an element of $\langle \sigma_{g} \rangle$ only if it is constructed from a genus $g-1$ solution by choosing either $\eta^{-1}(2g-3)$ or $2g-3$ as the odd number centered in the construction of an even genus solution from an odd one.

Thus, given a solution in odd genus, there are $g-5$ ways of extending it to a solution in even genus such that none of the resulting permutations are conjugate by $\langle \sigma_{g} \rangle$.

\end{remark}

\section{Exploring $SL(2, \mathbb{Z})$-orbits}

As discussed in the Introduction, the $SL(2, \mathbb{R})$-action on $\mathcal{H}(2g-2)$ restricts to an $SL(2, \mathbb{Z})$-action on the origamis contained in the minimal stratum.

The $SL(2, \mathbb{Z})$-action can be easily described in terms of the generators

\[ A = \left( \begin{array} {cc}
			1 & 1 \\
			1 & 0
			\end{array} \right),  B = \left( \begin{array} {cc}
									0 & -1 \\
									1 & 0
									\end{array} \right).  \]
The generator $A$ has the effect of a rightward shear, transforming the unit square into a parallelogram by rotating its vertical edges $45$ degrees clockwise; the generator $B$ rotates the plane by $90$ degrees counter-clockwise. We define the action of either $A$ or $B$ on an origami by cutting the origami into its constituent squares, applying the appropriate generator to each square individually (after identifying it with the standard unit square in $\mathbb{R}^{2}$), and then gluing the images back together. The $B$ generator thus sends origamis to origamis. The shear generator $A$ yields a surface tiled by congruent parallelograms, each comprised of two isosceles right triangles; by cutting and re-gluing right triangles, one obtains a new square-tiling of the surface.

Since there are only finitely many ways to glue $2g-1$ squares together, the orbit of any origami constructed in Section $4$ is finite. The cardinality of the orbit can be related to the index $ |SL(2, \mathbb{Z}) : V| $, where $ V $ is the Veech group of the origami.

The number of squares is an obvious invariant of the $SL(2, \mathbb{Z})$-action. A slightly subtler invariant is the \textit{monodromy}, which is the (conjugacy class of the) subgroup of the symmetric group $ \Sigma_N $ generated by the vertical and horizontal permutations (where $N$ is the number of squares).

In collaboration with Zichen Cui, Ajeet Gary, Paige Helms, Ionnis Iakovoglou, Tasha Kim, and J.T. Rustad, we implemented the construction outlined in the previous section in order to form a computational library of origamis in $\mathcal{H}(2g-2)$ with $2g-1$ squares and one cylinder in each of the vertical and horizontal cylinder decompositions, for $g=3,4,5,6,7$. We summarize some of the findings in \autoref{table}.

\begin{figure}[htp]
\centering
\includegraphics[width= .8\textwidth, height= 80mm]{OrbitTable3}{}
\caption{}
\label{table}
\end{figure}

In fact, for $g \leq 5$, the search spaces are sufficiently small that it is possible to find all minimal origami permutations. Thus, for these genera, we determine the exact number of orbits of origamis in $\mathcal{H}(2g-2)$ with $2g-1$ squares and one cylinder for each of the vertical and horizontal cylinder decompositions.

The group $M_{11}$ is the sporadic Mathieu group on $11$ symbols, while $L_{2}(8)$ is the group that admits the following presentation:
\[ \langle a,b \ | \  a^{2} = b^{3}= ((ababab^{-1})^{2}ab^{-1})^{2}= 1 \rangle \]
While we conjecture that the number of orbits grows super-exponentially as a function of $g$ (and, furthermore, that many orbits can be found amongst the origamis constructed in Section $4$), we note that we do not even know if the number of orbits is monotonically increasing in $g$.

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


We made it! Broadway Direct brought you their 100th issue this week and #TGIFTunes continues the celebration with the second half of the playlist celebrating the longest-running shows in Broadway history! From Funny Girl to The Wiz to The Phantom of the Opera, Broadway's most famous shows are all here.
How high does your favorite show rank? Will anyone ever top the Phantom? Let us know what you think in the comments!


A personal quest to re-learn how to walk through cherished, northern landscapes introduces a gifted new voice. Gathering fragments from memory, myth, archaeology and geology, Kate Davis's debut is a nimble exploration of what it means not only to exist, but to persist.
The Girl Who Forgets How to Walk feels to me incredibly timely. Coming on the back of a surge of so-called 'new nature' writing, Davis's perspective – female, middle-aged, disabled – is refreshing. In what has become de rigueur, a (usually white, male, able-bodied) narrator will be inspired to set off following some river or shoreline, tracing Situationist ambiance onto the terrain in order to offer mild critique of late capitalist ennui. Thankfully, Davis's book shuns what we might term 'common denominator psychogeography' to present the reader with something altogether more fulsome.
Split across three sections, The Girl Who Forgets How to Walk is a sustained rumination on how we come to terms with disability and loss. It is also a book charting the wildness of Cumbria and Lancashire and how we, as human actors, figure in the landscape irrespective of our flaws. 'When they say geology, she sees anatomy', Davis writes in 'She examines her fear that the ground is not to be trusted…', one of a series of poems which has a self-reflexive narrator ruminate on the circumstances that led to her forgetting how to walk.
Not to know that 'the virus in her spine would bring a terrible dream of heat', such a lack of omniscience is immediately countered in the following poem, 'The hot, terrible dream', in which a fragmented narrative uses the run-on line to deft effect, channelling the refracted memories of care and convalescence.
Interested in scale and the playing-off of everyday objects of modernity against glacial time, the poem, with its clever ear for internal rhyme, seems to be measuring the personal against the prehistoric (the 'Costa […] mocha' against the 'Permo-Triasic rocks'). The last poem in this introductory section, 'Disaster', sees Davis toying with line lengths, foregoing a conventional account of a train crash which 'everyone remembered' to instead focus on the surreal image of the 'whole train falling slowly' into a sinkhole. I have not copied verbatim Davis's use of line and even word breaks here, but it should be noted that their cumulative effect makes for a kind of stalling: the blunt terror of a one-hundred foot chasm opening in the earth and swallowing a carriage full of people is apprehended. Are we – is our narrator – certain that this disaster actually happened?
Two preceding poems would have us be sceptical. We begin to wonder if it is animism – the 'troll teeth' in the 'mouth' of the pit in 'Limekiln' – or myth and folklore – the 'stories of farmers' in 'Haematite' – which is at fault here, but then Davis seems more intent on the bigger picture: might both be different sides of the same coin? The fallibility of memory and the impermanence of apparently-solid elemental structures are routinely clashed together.
By the time we reach its closing section, images and motifs have begun reappearing. While this forms a pleasing narrative arc – this is one of few collections which I've felt obliged to read in order – it also makes for thematic unity; some order amid the chaos of circumstance. 'The Taxidermist's Apprentice Discovers Palaeontology' sees the return of the pink dress in 'Peninsula', while 'gigantic bones' and 'jelly-fish in a warm viscous sea,/tree-ferns decaying to sediment' remind the reader of the cornucopia of 'certain creatures' met elsewhere in the collection ('Bodies').
geologists gave it — Boulder Drift.
You can find out more about The Girl Who Forgets How To Walk from Penned in the Margins.


In 2008, the IRS paid out over $70 billion in refundable credits to people who had no income tax liability. Congress's Joint Committee on Taxation recently produced some estimates on the number of tax filers who receive refundable credits larger than what they pay in payroll taxes. This is an important contribution to the debate over the use of credits because in any discussion about the record number of non-payers, we frequently hear the refrain that "well, they do at least pay FICA taxes."
In a May 28, 2010 letter to Representative Dave Camp and Senator Kent Conrad, JCT reports that between 2000 and 2006 the number of returns with refundable credits in excess of the employee's share of payroll taxes increased from 11.8 million to 16.1 million. In 2009 and 2010, those figures jumped to 23 million because of such things as the making work pay credit and the lowering of the income threshold for determining the refundable portion of the child credit to $3,000.
JCT projects that the number of returns with refundable credits exceeding the employee's share of payroll taxes will hover between 14 million and 15 million for the next ten years.
Of course, employers pay the other half of an employee's payroll taxes so JCT estimated how many returns receive refundable credits in excess of both portions of the payroll tax. In 2000, 8.7 million returns got more in refundable credits than they paid in total payroll taxes. The number jumped to nearly 12 million in 2006 and then up to 15.5 million in 2009 and 2010. It will hover between 10.6 million and 11.3 million for the next decade.
There are two elements to this story. The obvious first one is the increased role of the IRS in delivering what are essentially welfare benefits to people who don't pay income taxes. Whatever we think of the IRS, this is generally not a function it should play. As I've said repeatedly, too many people see April 15th as payday, not Tax Day.
Lastly, not only are these people not paying any income taxes to fund the general cost of government from which they benefit, but they are effectively not paying into the Social Security or Medicare systems for which they will benefit in retirement. That's a free ride in any book.


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Our newly restored barn is the perfect location for an incredible farm wedding venue. Accommodating your guests in rustic elegance. With space for 300 guests we cater to weddings large and small.
Looking for a great location for showers, parties, and other special events? We offer unique features and spaces to make your event stand out. With the freedom to utilize multiple indoor and outdoor spaces, your event is sure to be unforgettable.
We love to host area business and non-profit events. Our peaceful farm setting is a great place for groups and has plenty of space for your team or guests to work, play and roam.
As a registered historical landmark, Meadow Brook has been hosting group tours for many years. Join countless others who have made their way through our unique and informative historical tours of this classical Midwestern family farm.
Three incredible ceremony locations to choose from.
The beautifully restored barn is rustic yet elegant with its hand hewn timbers and vaulted gambrel roof. The eighteen chandeliers provide romantic lighting for the ceremony or reception. Antique tables and hand built barrel tables can be used for gifts, cake, pictures, and a guest book while white tables and chairs are provided for guests.
Our rectangular chapel gazebo is located north of the large original barn doors to provide a grand entrance for the wedding party as they promenade to the service. The lovely setting is surrounded by farm fields and a rustic granary. Archways, flowers and antique farm equipment invite photo opportunities.
Our beautiful garden along the east lawn features award-winning scarecrows with "flower hair" who are dressed for the occasion. A round white gazebo can be decorated using your wedding colors and white chairs are provided for your guests. The landscape is dotted with arches, a wooden bridge, a whimsical playhouse, and antique farm equipment.
The legacy of Meadow Brook Barn began many years ago. The farmstead is well known for hosting the local community fairs in the early 1900's, making its mark in Indiana's listing of historical places. Not only does the property hold cherished memories of the past but also is a current destination where family and friends can reconnect and enjoy good food and fellowship. Filled with unique rustic spaces, magnificent views, and whimsical flowers guests will enjoy the comforts of the modern day while taking a step back in time. We, the owners and family of Meadow Brook Barn, take great pride in making your experience a story that will be passed on for generations to come.
"The beauty of the barn at Meadow Brook Farm will not disappoint. The amazing craftsmanship, beautiful beams, down to the details of re purposed wood on the walls, captures the charm of the past. There are little nooks and crannies for you to unwind and enjoy the beauty of the barn. You will find this country setting warm and welcoming."
We, at Meadow Brook Barn, invite you to choose your own caterer. We would be glad to give you suggestions about what caterers have worked well with us in the past, but we feel your special day should be filled with foods and treats of your choosing!
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Joe Carnahan on 'Nemesis' And the Crooked Hollywood Business Model [/Film Interview Part 2]
Posted on Wednesday, May 7th, 2014 by Germain Lussier
Read part one of our Joe Carnahan interview here, in which we primarily discussed Stretch and Death Wish.
Below is the second part of our interview with director Joe Carnahan. The director recently won the the Mendez Award at the Maryland International Film Festival-Hagerstown and took the occasion to offer both updates on his career and thoughts about the current state of Hollywood.
In this section, we talk about potential alternative distribution models for Stretch, what's going on with his Mark Millar comic book adaptation Nemesis, his admiration for Steven Soderbergh, his hatred of the foreign sales model and the dark side of the Hollywood business model.
So are you exploring different ways to distribute Stretch? Are VOD models something you are looking at in general moving forward?
Yeah dude, but there was this discussion, because it's so much fun, "Do we just make it like 10 episodes?' Because it's such a cool little story. But then I thought it has its kind of place and time and it's representative of this moment and we should just kind of stick to that. And yeah, I think there's absolutely [VOD]. There's also this old-fashioned part of me that's like 'Let's take it to Toronto and have some fun, man.' Let's go on the back to the festival circuit, which I always had a blast doing. And I always loved going to film festivals. So why not, man? It'll play like gangbusters, you know? Like that would be a lot of fun. And it would be ultimately kind of worthwhile. But again, these are off concepts. These are exercises. Listen, the guy whose career I love is like Steven Soderbergh. Soderbergh will go off and do, like The Girlfriend Experience, he'll do Che, a four hour version of Che Guevara. And then he'll do an Ocean's film, you know what I mean?
And then Haywire with Gina Carano. Who saw that coming? You know, there's a total kind of contrarian spirit in that guy as an artist, as a filmmaker, that I just love. I mean, who has the balls to re-cut Heaven's Gate? Like you just lop off the Harvard section and just start in Wyoming on the plains. I love that guy's spirit. And I think that's kind of where Stretch came from. I wanna do a Soderbergh. Just do a little script I've got laying around and go have some fun. But again, when there's an agreement in place, then it's like 'Well wait, we didn't think it was gonna be this, we thought it would be that.' And there's this weird unraveling. My first thought is 'Well, guys, we all read the script, man. We knew what we were getting into.'
They know what the movie is.
You know, I thought it was an extremely wider market. But again, I think the landscape of contemporary big cinema is constantly [changing]. Somebody said recently, I don't know who it was, but I heard a quote when somebody at Warner Brothers said, 'We would not make Argo today.' Which I thought was like 'Wow,' you know? If that's true, I mean, that's crazy. You wouldn't have made the film that won Best Picture last year.
But at the same time, I get it, man. It's there's a lot on the line now, dude. And they'd rather spend 150 million dollars marketing something than take a runner at something. Also we were trying to do an action comedy. Forget about the horror genre, can you do an action comedy on the cheap? And have it be of really great quality? And it is, dude. That's the thing about Stretch. You'd never know it was a five million dollar flick. You just wouldn't know it, you know. But again, dude, listen, I appreciate everything that's going on in the business right now. And oftentimes you have to also decide, 'Right now it ain't for me.' You know what I mean?
Like I'm having fun doing this other thing.
We talk about $100-million dollar movies, and those being the ones that get made. And you're a part of Nemesis, which you think would that kind of movie. What's going on with that? It seems like it's an R-rated movie about a villain.
Yeah, [Nemesis] is an extremely controversial script, man. To encapsulate, it's a very thorny, very dark paean to anarchy in this country. You know what I mean? Listen, I know how much Mark [Millar] loves it. I think the script's great. But again, you come back to that same thing. It's like we can continue to generate these things. And if they ain't getting made, for whatever reason, are you gonna keep lamenting the fact they're not getting made or are you gonna go do something else? I feel like there's a time and a place for that script. I would like it to be sooner rather than later 'cause I think there's a lot of extremely topical, extremely contemporary, extremely hot button things within that script. But you know what? If we're having this conversation in 1975 and I wanna make Nemesis, it's a greenlit film.
You know what I mean? Post Watergate, we're not having that conversation. We're having that conversation in 2014. It's a different time and a different place. Radically so. Is my hope that we will see the way through and get everybody on the same page about Nemesis? Absolutely. Am I gonna hold my breath? Absolutely not. To be candid, I can't tell the studio what's gonna work for them. That's what they're there for. They know what works for them.
My brother feels the same way. We can only write the version that…the thing that we pitched, the thing we laid out is exactly what we put down. And it's scary, dude. It's dystopian. It's kind of bleak. It's also very hopeful in an odd way, you know? And then what Nemesis ultimately represents is really I find fascinating. And I know my brother does. But again, that doesn't mean 500 million dollars worldwide. You know what I mean?
Someone has to make the leap. If another studio made an R-rated super-villain movie and it made 250 million dollars, your movie would be made instantly.
Absolutely. I think John Landis said recently, and I'm paraphrasing, so forgive me, but he said Hollywood is not interested in the idea so much as the previously executed idea. Which is so spot on, man. It's like everyone has to show some metric of prior success with a certain genre or certain type of film for them to even, you know what I mean?
But that also goes to like, nothing's more destructive or ridiculous or absurd or complete horseshit than the foreign sales model. I'm sorry. You know what I mean? It's a valuation based on….like my brother tried to put this movie together recently with Garrett Hedlund who I just think is phenomenal. That kid's talented. A couple years ago coming off of Tron, Garrett gets [anything.] And nothing's changed between then, you know what I mean? He was still the guy that starred in Tron. But for whatever reason, you're not worth whatever this idiotic, measureless, abacus they have says he's worth. They just plug into the most formulaic, nonsensical system and yet, and I think it's systematically destroyed the way that we finance films and make films. Gone is the idea of 'Hey, this is a great script. And I wanna go make this film.' It's gotta be, 'We have to hedge that bet and then we have to hedge that bet on top of that.' You know what I mean?
It's this never-ending, circular saw of bullshit. It just is. And I don't wanna participate in that, man. It's like I get that there are, you know, the Avi Lerner's of the world, man, and I love it. Did you watch that James Toback, Alec Baldwin documentary? He comes right out and says it, man. 'I'm here to make money.' Again, I appreciate that. Of course, yes, I don't wanna eat soup the rest of my life. We want nice things. But I think it's we've done it now. It's like we cut off our nose to spite our face and we've also taken the ears and gouged the eyes out.I's just this kind of Grand Canyon-like gulf opening up between the Philomena and August: Osage Countys of the world and The Avengers. You know what I mean?
It's like, nobody's making 35, 40 million dollar films anymore. They're just not making them. It just ain't happening. You know? You start to lead that, they go well, 'We should go to 100.' Or it's like three million. It's one of the two. And that's cause for concern.
The Spider-Man reboot was once going to cost like 50 million dollars. They really wanted to make a grimy Spider-Man. And then it ended up being 250. Because they had to make it big.
So do you find that your point of view here about the foreign sales is the prevalent feeling amongst filmmakers you know? Or do you think people follow the 'It's a business' mindset?
Well the directors I know, my friends, are always more interested in the craft than they should be, obviously. We all want the experience and we all wanna… You know, dude, I want another experience like The Grey. Because that was an adventure and we just happened to be making a movie at the same time. It's a very rare occurrence. That was something that is s a very, very fond memory for me, that whole experience. But I think the frustration comes because mostly everybody on the foreign sales side, they're just crooks. Literally they're crooks. There's no modicum of honesty or decency or polite business acumen. They're just cutthroats. I mean, I don't think they even disguise the fact that if they can fuck you, they're gonna fuck you.
It's like you know how much money I lost on The Grey because of these clowns that I was dealing with? And by the way, a guy like [producer] Mickey Liddell comes in and saves my ass because these other bozos were complete vipers. And Mickey made a really aggressive deal as he should have. He's not a bad guy, you know. But the rest of them to the man, they're just scum. And it's like, if that's how I have to go make a movie and go deal with these guys? I'll just skip it, man. You know, I'll do something else. Like that's the fear.
You know, I got The Grey made because Liam Neeson wanted to make that movie. And Liam is worth X, which is great. I owe that film being made to Liam. If you don't have that, it's the whole Johnny Depp thing. "He's done this movie, it hasn't worked." It doesn't matter, Johnny Depp's phenomenal. You know what I mean? It's like it's in a market, they're gonna knock on Johnny. Who gives a shit? But as soon as Johnny Depp puts on the Captain Jack Sparrow stuff again, they turn the money faucets back on. So it doesn't matter. Johnny Depp probably doesn't wanna play Captain Jack every fucking movie, you know? And what happens, he tries to do something like Transcendence and they punish him. He's punished because of wanting to do something different. I just find it really restrictive and confining in a way that you wanna be comfortable when you work sometimes. You know what I mean? I don't wanna put myself in those situations. I've done it too many times.
Have you thought about crowd-sourcing funds for any of these scripts you have?
No. I don't want some poor bastard to give me 100 bucks. I feel bad. I mean, dude, I might get to the point where I just like… part of the reason you do television, too, is a successful show is so wonderful and can be so potentially lucrative. It'd be nice to be able to put enough money aside to just go pay to do the shit myself. So six people see my movie, I don't care. You know what I mean?
People always say, whenever you use your own money, you never…I don't know. But if you get kind of fed up with the process, why not? If you believe in something strongly enough, why not? And thank God for Kickstarter 'cause it gives me hope that, when people get these things financed, they're really decent, fans and benefactors out there. Supporters that'll actually dig in their own pocket. I have a very warm feeling about Kickstarter 'cause I think it's the best of what we can be. It's people who actually help out our fellow artists. We actually kind of go into our pocket for something. It's very rare. I just don't want…it's Catholic guilt, man, or lapsed Catholic guilt. I can't shake it.
Check back tomorrow for the final part of our interview where we discuss the upcoming Daredevil TV show and more.
'The Magic Order': James Wan to Direct Mark Miller's Netflix Pilot
'El Chicano' Co-Writer and Producer Joe Carnahan Explains What It Takes to Direct a First Film [Interview]
'The Raid' Remake is Still Happening and Joe Carnahan Shares New Plot Details
Joe Carnahan Left 'Bad Boys III' Because He Clashed With Will Smith
Comic Book/Superhero, Gossip/Controversy, Interviews, Joe-Carnahan, Mark-Millar, Nemesis, Stretch, The Grey

redpajama_set_name: RedPajamaCommonCrawl


Corner Bed sofa Enhance First Impression » Today sofas come in virtually hundreds of styles with unlimited options of upholstery. You are able to order your couch email order through companies like Pottery Barn and Dog crate and Barrel. Most people will tell you that after style the most important thing they look for in a sofa is comfort. Deciding on the best style of couch to enhance your property is key. There are various styles of couch to choose from, but most manufacturers break them into two distinct styles, traditional and fashionable. Corner Bed Sofa is For example, more traditional styles of couch would include barrel back again, British arm, tufted back again, tight back again and tuxedo.
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Corner Bed sofa Enhance First Impression » However, depending on fabric alternatives a Tuxedo couch could be used in a modern day home. Contemporary styles include sock arm, route back and sectional. Some sofas include pillows attached to the sofa, some come with loose cushions that you can free style any way you choose. Corner Bed Sofa Another thing that provides your sofa a definite style is cloth choice. When you visit most furniture stores you should have the capability to first choose the style sofa you want and then choose the selection of textile. Some manufacturers have 30 or even more different fabrics you can choose from if you are designing a couch, with the one limit being how much cash you want to invest. The decision of fabric can add thousands of dollars to the ultimate cost of a sofa. Bold bright colors and geometric designs would be utilized on modern-day sofas and softer muted colors and prints would be used on traditional sofas.
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Corner Bed sofa Enhance First Impression Much like any major purchase, shop around. Shop around online at different sofa selections so as to get a concept of what you would like before you visit store after store after store. If you see a couch doing a web based search you should be in a position to find a store that offers that one brand.
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PARAGON INITIATIVE
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The Quick Guide to Simple and Secure Automatic Updates
August 30, 2017 6:38 pm by Scott Arciszewski
Should your software update itself automatically? YES!
If you aren't convinced, we've previously made the case for automatic software updates as a means of preventing yesterday's software vulnerabilities from being exploited today.
However, as our previous article on the subject notes, implementing automatic software updates requires a nontrivial amount of engineering effort in order to be secure.
Our company has been hard at work for the past few years to diminish the effort required to achieve secure automatic software updates in the PHP community. Most of our efforts are reproducible and/or relevant to any other programming stack, although PHP remains the first major programming language to decide to adopt modern cryptography in its standard library.
Let's explore how to use our existing work to build a secure automatic software update system, without having to do any of the heavy lifting.
Simple and Secure Automatic Software Updates
How to Implement Secure Automatic Updates, the Easy Way
First, make sure your deliverables are reproducible from the source code. If you're working with scripting languages that are never compiled into binary code, this merely requires your software be open source.
Second, use an update framework (i.e The Update Framework) that enforces code signing. This means that your update files must be signed by a private key controlled only by you, but can be verified by anyone with your public key.
If you don't understand what "private key" or "public key" means, this page is an approachable introduction to cryptography terms and concepts and will shed some light on the matter.
Finally, run a Chronicle instance. Every time you release an update, publish the new release information to your Chronicle. Make your code that interfaces with The Update Framework verify that the release you're seeing is also published in the Chronicle (or, especially for enterprise customers, their own replica of your Chronicle that resides on the corporate network).
That's it. The Update Framework (or a similar implementation relevant to your stack) and Chronicle are all you need (as far as tooling goes). Make your software open source, and your builds reproducible, and you'll drastically reduce your customer's attack surface in terms of both space and time.
Secure Automatic Software Updates in PHP
There isn't currently a PHP implementation of The Update Framework. If there's enough community interest, we may commit to building one in the future. However, that might not be necessary.
If you're developing modern PHP, you're almost certainly using Composer and Packagist. If not, it's highly recommended that you learn it ASAP.
Earlier this year, I opened a proposal to the Packagist team to run their own Chronicle instance, which would be used to publish information about software releases in real time. We're working on other proposals to enforce signature validation and solve the Public Key Infrastructure (PKI) problems.
In other words: If you're using Composer, then in the near future this may already be a solved problem for you.
If you cannot wait for our work to be accepted and deployed in Composer, you'll have to either roll your own implementation or hire Paragon Initiative Enterprises to build it for you. We've previously implemented secure automatic updates in two different products (one of them is Free Software). Refer to the implementation in CMS Airship if you need a starting point.
If your users won't have the Sodium extension available, look into sodium_compat. This probably won't be a problem once everyone is running PHP 7.2 and newer.
Secure Automatic Updates for Embedded Devices and the Internet of Things (IoT)
Embedded development faces unique challenges and there hasn't been a lot of guidance on implementing secure automatic update protocols, especially for so-called "smart devices". Due to low memory or power usage requirements, it's often not feasible to just staple cryptography onto your product design without using up your entire power or memory budget.
For extremely constrained devices, libhydrogen is an attractive option. It's very lightweight, and the current implementation uses only two primitives to provide a full-featured cryptography library: the Gimli permutation and Curve25519.
Commercial support for libhydrogen is available from Primulinus.
The protocol designs that went into The Update Framework and Chronicle can be easily re-implemented using libhydrogen, but the Hydrogen version will not be compatible with the Sodium version. If a Hydrogen variant of Chronicle is desired, get in touch with our team.
License: Creative Commons Attribution-ShareAlike 4.0 International CC-BY-SA 4.0 Intl
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Public Key Cryptography
Scott Arciszewski
With 15 years of software development, application security, and system administration experience, Scott aspires to help others attain a happier work-life balance by solving difficult problems and automating trivial tasks. He is mostly known in the community for his open source software security research and strong progressive positions on providing tools and frameworks that are secure by default. @CiPHPerCoder
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Elica received two IDA Design Awards for NikolaTesla Libra and NikolaTesla Flame. Both products were designed by Fabrizio Crisà for Elica and won in the subcategory "Home Interior Products – Kitchen Appliances".
NikolaTesla Libra had an Honorable Mention, while NikolaTesla Flame a Silver Award. In the previous edition, Elica's NikolaTesla Switch was a Gold winner as well as Lullaby which was also Product of the Year, and Haiku was a Bronze winner.


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Amii Stewart is an American R&B/disco/dance-pop singer, dancer and actress most famous for her hit disco record "Knock on Wood". Stewart is the stepsister of actress-singer Miquel Brown and aunt to Brown's actress-singer daughter Sinitta.
Amy Stewart is the youngest of six children.Her father, Joseph Stewart II, signed her up for singing and dancing lessons in 1960, at the age of four. There already was an Amy Stewart registered with Actor's Equity, so she changed the spelling of her name to Amii. She began attending Howard University in Washington but soon left to work with the D.C. Repertory Dance Company, studying ballet and modern dance. Before being signed to Ariola Records, Stewart was in the touring company of the stage production Bubbling Brown Sugar in 1975, firstly in Miami, then Broadway, and eventually London's West End, where she met Barry Leng, a record producer for Hansa records. The heavy drum sounds heard on "Knock On Wood", and many other of Amii's tracks, were provided by session drummer supremo Peter Boita ( also credited as the "Drum Doctor" on many tracks ). Using a combination of Linn Drum Computers and Simmons SDS7 electronic drums, Peter developed drum sounds and synching techniques that were way ahead of anything else at that time.
At the end of 1977, "You Really Touched My Heart" a Leng/Simon May composition, produced by Leng, was Stewart's first recording. An album followed, which contained five Leng/May songs, one Leng/Morris song and three cover versions.
Her first single, a disco cover version of the 1966 Eddie Floyd hit "Knock on Wood" (Floyd/Cropper), reached number one in the U.S. in April 1979, and earned her a platinum record and a Grammy Award nomination. It also reached number 6 in the UK, and number 2 in Australia. In the same year, Amii Stewart followed up with another cover of The Doors' classic, "Light My Fire", which reached #5 in the UK, although only managed a lowly #69 in the US. She recorded it as a medley, incorporating "Light My Fire/137 Disco Heaven".
Among her other successes in music recordings is the song entitled "Friends" from 1985. This song was a collaboration with the Italian composer/musician Mike Francis and was a big hit in the UK and the Philippines.


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Private Lessons can be purchased individually or in Packages to Save Money and will be schedule based on the Customers and our instructors availability.
Individual Lessons are $75 per hour plus Floor Fee. This option is recommended for individuals that want to simply take 1 Lesson every so often to enhance their Group Lessons.
4 Lesson Package: This is our most common Package for Private Lessons as it establishes commitment and yet provides flexibility of only committing to 4 Lessons. Each Lesson will be progressive and we will determine the best path to take with each customer.
Other Large Packages are offered and will SAVE you money accordingly. Packages of 6 or 8+ are some of the options we have.
Bahhari also offers Specific Services for different occasions such as Weddings, Corporate Events, Sweet 16's etc. Contact Us for more details.


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Use WASD for movement. And left mouse button for actions.


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21st Century Fox Makes Senior HR Appointments
By Mark K. Miller | July 20, 2017 | 8:26 a.m. ET.
21st Century Fox today announced two new roles in its global human resources group with the promotion of Julie Franke to SVP of employer brand and strategic talent acquisition, and the appointment of Carsten von der Linden to SVP of talent management.
Von der Linden, who joins 21CF's HR group from Boston Consulting Group and Franke from Fox Entertainment Group, both report to Thomas Gaissmaier, EVP, chief human resources officer for 21CF, and will be based in Los Angeles.
During a 10-year tenure at BCG, Von der Linden served as a principal based in Munich and Boston and helped establish BCG's HR practice globally. He has a background in large HR transformations, HR operating models, talent management, leadership development, digital HR and many other HR topics. In addition, Von der Linden has authored numerous reports focusing on HR.
Franke has been VP, talent acquisition operations for Fox Entertainment Group since 2011. In this role, she led recruiting and onboarding for 21CF's Los Angeles-based film, television and sports businesses. Prior to that, she held a number of senior recruiting roles for clients in the biotech and banking sectors.
Tags: 21st Century Fox, Carsten von der Linden, Human resources, Julie Franke