text:
\section{Introduction}
We investigate in this paper a class of problems that arise in
the dynamical triangulation approach to 3-dimensional gravity restricted to the case of so-called causal triangulations. For an introduction to the dynamical triangulation approach to discrete quantum gravity we refer to \cite{book} and an account of causal dynamical triangulations can be found in \cite{al4}. For the case of 3-dimensional gravity in particular, one may consult \cite{adj1} and \cite{al2}.

In \cite{DJ} we gave a proof that the number $N(V)$ of causal 3-dimensional triangulations homeomorphic to a three-sphere and consisting of $V$ tetrahedra is exponentially bounded,
$$
N(V)\leq C^V\,,
$$
where $C$ is some positive constant. Validity of this bound is crucial in order for the relevant correlation functions
to exist, and thereby defining the discretised model (see \cite{book,al4}). The first step of the argument leading to this bound was to decompose the triangulations into slices (which is possible because of the causal structure defined below) and to show that it is sufficient to establish the bound for such causal slices. The number of causal slices
was then shown to be bounded by the number of certain coloured two-dimensional cell complexes homeomorphic to the 2-sphere.  These cell complexes arise
as mid-sections of the slices. Finally, bounding the number of those two-dimensional cell complexes can be done using well known techniques.

The method of associating a coloured 2-\-dimensional cell complex with a causal triangulation has been applied earlier by other workers in the field \cite{al2} and also used in numerical simulations and in combination with matrix model techniques to extract properties of the model. It has, however, not been established exactly what class of 2-dimensional complexes can occur as mid-sections of a causal slice. As noted in \cite{DJ}, at least some non-trivial constraints have to be imposed on top of its homeomorphism class. It follows that the connection between the model defined in terms of triangulations and the one realised in terms of a specific class of coloured two-dimensional cell complexes is obscure.  It might be the case that the precise class of cell complexes is unimportant in a possible scaling limit of the model, but this remains to be investigated.

In this paper we provide a complete characterisation of the coloured 2-dimensional cell complexes that correspond to
3-dimensional causal slices. In fact, we shall consider not only the standard notion of causal triangulations homeomorphic to $S^2\times [0,1]$, where $S^2$ denotes the 2-sphere, which in this paper will be referred to as causal sphere-triangulations, but find it useful to generalise the notion of causal triangulation to manifolds homeomorphic to $D^2\times [0,1]$, where $D^2$ is the 2-dimensional disc, and $\partial D^2\times [0,1]$ may be viewed as the time-like part of the boundary while the two discs $D^2\times \{0\}$ and $D^2\times \{1\}$ form the spatial parts of the boundary. Such triangulations will be called causal disc-triangulations and the corresponding slices will be called causal disc-slices. The coloured cell complexes are in this case homeomorphic to $D^2$. For both cases we will give a full characterisation of the mid-sections and show that they are
in bijective correspondence with the appropriate causal slices.

This result might serve as a starting point for obtaining an exact enumeration of causal slices by applying well known techniques for planar surfaces or maps to the coloured cell complexes homeomorphic to the disc. This interesting combinatorial problem is more complicated than those previously considered because of the colouring and the constraints identified in this paper. The result might also be instrumental in identifying an appropriate matrix model generating exactly the desired causal slices or as aid in designing effective numerical algorithms for simulations. These issues are, however, beyond the scope of the present paper.

\section{Preliminaries and notation}

 We will use notation consistent with that of \cite{DJ}. For the reader's convenience we briefly recall the main conventions, restricting the discussion to the 3-dimensional case. The basic building blocks of our triangulations are tetrahedra or
 3-simplices whose vertices have one of two colours: red or blue. Generally we will denote an unoriented simplex with vertices $x_1,\dots, x_n$ by $(x_1\dots x_n)$.  If all the vertices in a simplex have the same colour we say that the simplex is monocoloured and has that colour.
This means that if $x$ and $y$ are red vertices and $e=(xy)$ is a 1-simplex then we say that $e$ is red, and a triangle $\Delta=(xyz)$ is red if its vertices (or edges) are red, etc.  If a simplex is not monocoloured we say it is two-coloured. It is assumed that all tetrahedra are two-coloured. Thus the tetrahedra come in three types: type (3,1) with three red vertices, type (2,2) with two red vertices and type (1,3) with one red vertex.

A triangulation is by definition an abstract simplicial complex whose formal definition can be found in \cite{DJ} or \cite{plt}. For the purposes of this paper one may think of a triangulation as a collection of tetrahedra together with identifications of
some pairs of triangles in the boundaries of the tetrahedra, respecting the colouring, such that any triangle is identified with at most one other triangle.  When we identify triangles we also identify all their subsimplicies.
This point of view will be used and explained in more detail in Section 4 below. Any pair of identified triangles is called an interior triangle of the triangulation while the other triangles are referred to as
boundary triangles. It should be noted that it is implicit in the notion of a simplicial complex that
\begin{itemize}
\item[(i)] two triangles contained in the same tetrahedron cannot be identified,
\item[(ii)] two different triangles in a
given tetrahedron cannot be identified with two triangles contained in another tetrahedron.
\end{itemize}
\noindent It is common to speak about these two conditions as regularity conditions and about simplicial complexes as regular triangulations as opposed to singular triangulations, when one or both of these conditions are left out. In this paper we focus on regular triangulations.

One may think of a triangulation either as a purely combinatoric object or as a topological space
embedded in a Euclidean space. In the former case two triangulations are identified if there is a bijective correspondence between their vertices respecting the colouring and the pairwise identifications of triangles. In the latter case two triangulations are identified if there exists a homeomorphism between them mapping simplices to simplices and thus inducing a combinatorial identification. It is a fact, explained in
e.g. \cite{DJ}, that the two points of view are equivalent.

We now introduce the basic objects of study in this paper, the two types of causal slices that form the building blocks of the general causal triangulations.

\noindent
\begin{definition}\label{def1}
A \emph{causal sphere-slice} $K$ is a triangulation fulfilling the following conditions:

(i) $K$ is homeomorphic to the cylinder $S^2\times [0,1]$

(ii) all mono-coloured simplices of $K$ belong to the boundary $\partial K$, such that the red ones belong to one boundary component $\partial K_{\rm red}$ and the blue ones belong to the other component $\partial K_{\rm blue}$.

The set of all causal sphere-slices is denoted by ${\cal C}{\cal S}$.

\end{definition}

\noindent
\begin{definition}\label{def2}
  A {\it causal disc-slice} is a triangulation $K$ fulfilling the following conditions:

(i) $K$ is homeomorphic to the 3-dimensional ball $B^3$

(ii) all mono-coloured simplices of $K$ belong to the boundary $\partial K$, such that the red ones form a disc $D_{\rm red}$ and the blue ones form a disc $D_{\rm blue}$, which will be called the boundary discs of $K$.

The set of all causal disc-slices is denoted by ${\cal C}{\cal D}$.
\end{definition}

We note that the above definitions imply that all the vertices of a causal slice of either type lie on the boundary. Two-coloured edges are sometimes referred to as \emph{timelike edges}. Two-coloured triangles can be of two types, with two red vertices and one blue or vice versa, and are called \emph{forward directed} and \emph{backwards directed} triangles, respectively.

Relaxing the condition in Definition \ref{def2} that $D_{\rm red}$ and $D_{\rm blue}$ are
(homeomorphic to) discs to, say, the requirement that they are deformation retracts of discs
yields a larger class of causal triangulations, that will not be discussed in detail in this paper (see, however, Section~\ref{sec:6} for some further remarks).
In the course of the proof of Theorem \ref{thm2} we shall encounter particular examples of such triangulations and we use the same notation as for causal slices without further comment.

Although the main focus of this paper is on causal slices we introduce for the sake of completeness general causal triangulations in the following definition as a layered union of causal slices.   In general these triangulations have interior vertices.

\noindent
\begin{definition}\label{def3} A \emph{causal sphere-triangulation} is a triangulation of the form
$$
M= \bigcup_{i=1}^{N} K^i\,,
$$
where each $K^i$ is a causal sphere-slice such that $K^i$ is disjoint from $K^j$ if $i\neq j$ except that $\partial K_{\rm blue}^i =\partial K_{\rm red}^{i+1}$ for $i=1,\dots,N-1$ as uncoloured 2-dimensional triangulations.
The boundary components of $M$ are then $\partial K_{\rm red}^1$ and $\partial K_{\rm blue}^N$.

A \emph{causal disc-triangulation} is a triangulation of the form
$$
M= \bigcup_{i=1}^{N} K^i\,,
$$
where each $K^i$ is a causal disc-slice (with boundary discs $D^i_{\rm red}$ and $D^i_{\rm blue}$) such that $K^i$ is disjoint from $K^j$ if $i\neq j$ except that $D^i_{\rm blue} =D_{\rm red}^{i+1}$ for $i=1,\dots,N-1$ as uncoloured 2-dimensional triangulations. The two discs $D^1_{\rm red}$ and $D^N_{\rm blue}$ are called the red and blue boundary disc of $M$, respectively.
\end{definition}

The following lemma states that, apart from the two boundary discs, the boundary of a causal disc-triangulation consists of
a "timelike cylinder", that is a 2-dimensional causal sphere-triangulation, which is defined as in Definitions \ref{def1} and \ref{def3} with $S^2$ replaced by $S^1$.

\noindent
\begin{lemma}\label{le1}
Let $K$ be a causal disc-slice and denote by $C$ the subcomplex of $\partial K$ consisting of two-coloured triangles. Then $C$ is a 2-dimensional causal sphere-slice with boundary
$\partial D_{\rm red}\cup \partial D_{\rm blue}$, which will be called the {\it side} of $K$.

More generally, if $M$ is a causal disc-triangulation then the part of the boundary made up of triangles not in the red or blue boundary discs is a 2-dimensional causal sphere triangulation.

\end{lemma}
\noindent
{\it Proof.} It is clearly enough to prove the first statement. Choose an orientation of the circle $\partial D_{\rm red}$ and consider a vertex $v_1\in\partial D_{\rm red}$ and its two nearest neighbours $v_0$ and $v_2$ in $\partial D_{\rm red}$ with
$v_0$ preceding $v_1$ and $v_2$ succeeding $v_1$.  The edges $(v_0v_1)$ and $(v_1v_2)$ are each contained in exactly two triangles, one of which is red while the other one contains a blue vertex $v_1'$, resp.\  $v_2'$, both of which are in $\partial D_{\rm blue}$.
Note that, by the Jordan Curve Theorem, $\partial D_{\rm red}$ divides $\partial K$ into two discs.
It follows that if we denote by $u_1,\ldots ,u_k$ the neighbours of $v_1$ in $D_{\rm red}$ ordered cyclically around $v_1$ such that $u_1=v_0$
and $u_k=v_2$, then all other neighbours of $v_1$ in $\partial K$ are blue and constitute a segment $(w_1^1,w_2^1,\ldots , w_{\ell_1}^1)$ of
$\partial D_{\rm blue}$ with $w^1_1=v_1'$ and $w_{\ell_1}^1=v_2'$, see Fig.\ \ref{Fig1}.  Hence, the triangles
$$
(v_0w_1^1v_1), (w_1^1 v_1 w_2^1), \ldots ,(w_{\ell_1 -1}^1 v_1w_{\ell_1}^1), (w_{\ell_1}^1v_1v_2)
$$
make up a segment of a two-dimensional causal slice.

\begin{figure}[h]
  \begin{center}
\includegraphics[width=10truecm]{Fig1_SOS.pdf}
    \caption{The sequence of triangles on the side of the causal slice $K$ connecting the red and the blue boundary components.}
    \label{Fig1}
      \end{center}
      \end{figure}

Repeating this construction with $v_0,v_1,v_2$ replaced by $v_1,v_2,v_3$, where $v_3$ is the successor
of $v_2$ in $\partial D_{\rm red}$, we obtain a unique continuation of the segment constructed above by
$$
(w_1^2v_2w_2^2), (w_2^2 v_2 w_3^2), \ldots ,(w_{\ell_2 -1}^2 v_2w_{\ell_2}^2), (w_{\ell_2}^2v_2v_3)
$$
where $w_1^2=w_{\ell_1}^1$ and $(w_1^2, \ldots , w_{\ell_2}^2)$ is a segment of $\partial D_{\rm blue}$.

If the vertices of $\partial D_{\rm red}$ are $v_0,\ldots ,v_n$
then after $n$ steps we obtain a segment of a causal slice with horizontal edges in
$\partial D_{\rm red}\cup \partial D_{\rm blue}$ and whose first and last triangle share the vertex $v_0$.   Finally this segment can be completed to a two-dimensional causal slice $C$
by adjoining the triangles which contain $v_0$ and have one blue edge
(in $\partial D_{\rm blue}$) and two non-coloured edges.  By construction $C$ has boundary
$\partial D_{\rm red}\cup \partial D_{\rm blue}$ and evidently constitutes all of $\partial K\setminus {\rm int} (D_{\rm red}\cup D_{\rm blue})$.
\hfill $\Box $

\medskip

The next two propositions are elementary and ensure the existence of causal disc-slices and sphere-slices with prescribed boundary discs, respectively boundary components.

\noindent
\begin{proposition}\label{prop1}
Given two triangulated discs $D_1$ and $D_2$ there exists a causal disc-slice $K$ such that
$D_{\rm red}=D_1$ and $D_{\rm blue}=D_2$.

\end{proposition}
\noindent
{\it Proof.}  We give an inductive argument. Suppose first that $D_1$ and $D_2$ are triangles $\Delta_{\rm red}$ and $\Delta_{\rm blue}$, respectively. In this case, we can choose $K$ to be the prism, depicted in Fig.\ \ref{Fig2}, made up of one $(1,3)$ tetrahedron,
one $(3,1)$ tetrahedron and one $(2,2)$ tetrahedron.

\begin{figure}[h]
  \begin{center}
\includegraphics[width=6truecm]{Fig2_SOS.pdf}
    \caption{A prism made up of 3 tetrahedra $(xyzu)$,
    $(uvwy)$, and $(xyuw)$.}
    \label{Fig2}
      \end{center}
      \end{figure}

Now assume $K$ exists for some given $D_1$ and $D_2$. If $e$ is an arbitrary edge in
$\partial D_1=\partial D_{\rm red}\subset \partial K$ one can glue a
tetrahedron of type $(3,1)$ to $K$ along the unique triangle in the side of $K$ which contains $e$.  Thus we obtain a causal
disc-slice $K'$ whose red boundary disc has an extra triangle compared to that of $K$ but the blue boundary disc is the same.

Similarly, given two neighbouring edges $e=(x_1x_2)$ and $e'=(x_2x_3)$ in
$\partial D_1$, see Fig.\ \ref{Fig3}, one can first identify $e$
and $e'$ and then glue a sequence of $(2,2)$ tetrahedra sharing the identified $e$ and $e'$ and whose blue edges are
$$
(y_1y_2), (y_2y_3), \ldots , (y_{s-1}y_s)
$$
with the notation of Fig.\ \ref{Fig3}. The resulting triangulation
$K''$ is a causal disc-slice with red boundary disc $D'_{\rm red}$ obtained from $D_{\rm red}$ by identifying $e$ and $e'$
while $D_{\rm blue}$ is unchanged.
\begin{figure}[h]
  \begin{center}
\includegraphics[width=11truecm]{Fig3_SOSx2.pdf}
    \caption{Triangles in the side of a causal disc slice.}
    \label{Fig3}
      \end{center}
      \end{figure}

Similar constructions can of course be made with $D_{\rm blue}$ replacing $D_{\rm red}$.
Starting with an arbitrary triangle, it is well known (see e.g.\ \cite{dur} or \cite{bz}) that $D_1$ (and similarly $D_2$) can be constructed by repeated application of the process of either gluing on a triangle or by identifying two neighbouring boundary edges as described. We say that $D_1$ and $D_2$ are \emph{locally constructible}. Hence, the existence of $K$ follows from the preceding discussion by induction.
 Note that if $D_1$ and $D_2$ are regular triangulations then so is the causal disc-slice $K$.
\hfill$\Box$

\noindent
\begin{proposition}\label{prop2}
 Given two triangulated 2-spheres, $S_1$ and $S_2$, there exists a causal
sphere-slice $K$ such that $S_1$ is
the red boundary of $K$ and the blue boundary is $S_2$.

\end{proposition}

\noindent
{\it Proof.}  We remove one triangle from $S_1$ and another one from $S_2$.  Then we obtain two triangulated discs $D_1$ and
$D_2$.  By Proposition \ref{prop1} there exists a causal disc-slice $K'$ with boundary discs $D_{\rm red}=D_1$ and $D_{\rm blue}=D_2$.
The side of $K'$, $C$, is a causal two-dimensional disc-slice whose boundary components are triangles.  There are 3 different
possibilities for $C$, up to combinatorial equivalence, see Fig.\ \ref{Fig4}.

\begin{figure}[h]
  \begin{center}
 \includegraphics[width=10truecm]{Fig4_SOS.pdf}
    \caption{The 3 possible triangulations of the side of a disc-slice whose red and blue boundary components are both a single triangle.}
    \label{Fig4}
      \end{center}
      \end{figure}

Note that we may assume that no pair of neighbouring triangles in $C$ belongs to the same tetrahedron.  Indeed, if
the two triangles are both forward (or both backward) directed, this follows from the fact that
otherwise the removed triangle in either $S_1$ or $S_2$ would be glued to another triangle along two edges contradicting the regularity of $S_1$ and $S_2$.
If one backward triangle and one forward triangle in $C$ belong to the same tetrahedron in $K'$ it must be a
(2,2)-tetrahedron which can be removed from $K'$ without changing the red and blue boundary discs (only an edge in $C$
gets flipped).  Since there are only finitely many (2,2)-tetrahedra in $K'$ the claim follows.

If now $C$ is of type (a) in Fig.\ \ref{Fig4}, then we can glue to $K'$ a prism of the form indicated
in Fig.\ \ref{Fig2} to obtain $K$ as desired.
If $C$ is of type (b) we can glue onto the prism of Fig.\ \ref{Fig2} a
(2,2)-tetrahedron
to obtain
a regular prism whose side $C'$ is of the same type as $C$
and hence can be glued onto $K'$ to obtain $K$ as desired.  The argument when $C$ is of type (c) is similar.
\hfill $\Box$

\section{Disc-slices and midsections}
Given a causal disc-slice $K$ we define its midsection $S_K$ as in \cite{DJ} for a causal sphere-slice. More explicitly, we view $K$ as being embedded in a Euclidean space and consider a tetrahedron
$t=(v_1v_2v_3v_4)$ in $K$ with vertices $v_1, v_2, v_3, v_4$, i.e. $t$ consists of the points of the form
\begin{equation}\label{height}
x= s_1 v_1 + s_2 v_2 +s_3 v_3 + s_4 v_4\,,
\end{equation}
where $s_1, s_2, s_3, s_4 \geq 0$ and $s_1 + s_2 + s_3 + s_4 =1$. Letting the height $h(x)$ be defined as the sum of the coefficients
of the red vertices in \eqref{height} the 2-cell $F$ corresponding to $t$ is defined as the set of points
$x$ in $t$ of height $h(x)=\frac{1}{2}$. If $t$ is of type (3,1) then $F$ is a triangle whose edges by convention are coloured red; if $t$ is of type (1,3) then $F$ is likewise a triangle whose edges are coloured blue; if $t$ is of type (2,2) then $F$ is a quadrangle
each of whose boundary edges are contained in exactly one boundary triangle of $t$ containing a monocoloured edge of $t$;
by convention each edge of $F$ inherits the colour of the corresponding monocoloured edge in $t$.
It is easy to show (see \cite{DJ} for details) that the 2-cells so obtained define a 2-dimensional cell complex homeomorphic to $D^2$ with edges coloured red or blue in such a way that triangles are monocoloured while quadrangles are two-coloured with opposite edges of the same colour. Moreover, this cell complex is uniquely defined up to combinatorial isomorphism and is called the \emph{midsection} of $K$, denoted by
$S_K$.

By construction each vertex $a$ of $S_K$ is contained in a unique edge $e_a$ of $K$ whose endpoints have different colours and vice versa. Similarly, any red (blue) edge of $S_K$ is contained in a unique two-coloured triangle of $K$ which contains a red (blue) edge in $\partial K$, and vice versa. Finally, each 2-cell of $S_K$ is contained in a unique tetrahedron of $K$, the tetrahedron being of type (3,1), (2,2) or
(1,3) depending on whether the cell is a red triangle, a quadrangle, or a blue triangle, respectively.

In a two-dimensional coloured cell complex as described we shall use the notation $\langle ab\rangle$ and $\langle abc\rangle$ for edges and triangles with vertices $a,b$ and $a,b,c$, respectively, whereas a 2-cell with cyclically ordered vertices $a,b,c,d$, such that $\langle ab\rangle$ and $\langle cd\rangle$ are red edges, will be denoted by $\langle abcd\rangle$. Note that with this convention we have, e.g., $\langle abcd\rangle$ =$\langle dcba\rangle$ =$\langle cdab\rangle$.

By a red, resp. blue, path in $S_K$ we mean a sequence $e_1,\dots,e_k$ of red, resp. blue, edges such that
$e_i=\langle a_ia_{i+1}\rangle$ for each $i=1,\dots, k-1$, and some vertices $a_1,\dots,a_k$. In this case we say that the path connects $a_1$ and $a_k$. The path is called simple if either the vertices
$a_1,\dots,a_{k}$ are all different or if $a_1,\dots,a_{k-1}$ are different while $a_1=a_k$, in which case the path is said to be closed.

\noindent
\begin{remark} \label{rem1}
{\rm With the notation just introduced it follows from the definition of $S_K$
that if $a$ and $b$ are two vertices of $S_K$ then the red (blue) endpoints of $e_a$ and $e_b$ are identical if $a$ and $b$
are connected by a blue (red) path. Indeed, if $e=\langle ab\rangle$ is a red edge of $S_K$, then
$e$ is contained in a
triangle in $K$ two of whose edges are $e_a$ and $e_b$ sharing a blue vertex.
Evidently, the claim follows from this, and similarly if $e=\langle ab\rangle$ is blue.

 The converse statement that
$a$ and $b$ are connected by a red (blue) path if $e_a$ and $e_b$ share a blue (red) endpoint also holds as a consequence of the
proof of Theorem \ref{thm1} below. }

\end{remark}

Let us now note that since the monocoloured
boundary discs of $K$ are non-empty, it follows that $S_K$ contains at least one triangle of each colour. For the same reason the boundary of $S_K$ must contain edges of both colours. As a consequence, the boundary
of $S_K$ consists of a (cyclically ordered) alternating sequence of monocoloured paths which we shall call \emph{boundary arcs}.
The following few lemmas describe
properties satisfied by any midsection $S_K$.

\begin{lemma}
\label{le2}
 Two different vertices in a midsection $S_K$ cannot be connected by both a red and a blue path.

\end{lemma}

\noindent {\it Proof.} If two vertices $a$ and $b$ are connected by both a red and a blue path, then the two edges $e_a$ and $e_b$ in $K$ have identical endpoints and hence are identical, which implies $a=b$.
\hfill $\Box$

\medskip

Since $\partial D_{\rm red}$ is homeomorphic to $S^1$, there does not exist a sequence  of quadrangles $q_1,q_2,\ldots ,q_k$ in
$S_K$ such that $q_i$ and $q_{i+1}$ share a red edge for $i=1,2,\ldots ,k-1$ and the red edges in $q_1$ and $q_k$, not shared with $q_2$ or $q_{k-1}$, belong to $\partial D_{\rm red}$.   A sequence of quadrangles as we have described will be called a
{\it red path of quadrangles} connecting edges in $\partial D_{\rm red}$.  The corresponding statement with red replaced by blue
is of course also true.   We define a closed path of quadrangles (red or blue) analogously.  Obviously there cannot exist a closed path
of quadrangles in $S_K$ because then $K$ would contain a blue or a red edge in its interior.
The absence of paths of quadrangles just described follows from the following more restrictive conditions.

\begin{lemma}\label{le3}
Let $S_K$ be a midsection, let $\rho$ denote a closed simple red (blue)
path in $S_K$ and let $\mu$ be a simple red (blue) path connecting two vertices belonging to different
blue (red) arcs in the boundary $\partial S_K$. Then the following hold:

i) The interior of $\rho$, i.e. the component of $S_K\setminus\rho$ not containing any boundary edges, contains solely red (blue) edges.

ii) The two endpoints of $\mu$ are the two endpoints of a red (blue) boundary arc.

\end{lemma}
\noindent {\it Proof.} i) It is sufficient to consider the case when $\rho$ is red. If $\rho$ encloses a blue edge it clearly also encloses a blue triangle. Since the boundary of $S_K$ contains blue edges the exterior of $\rho$ likewise contains a blue triangle. Evidently, all red paths connecting a vertex in an interior blue triangle with a vertex in an exterior blue triangle must intersect $\rho$. For $K$ this
means that $D_{\rm blue}$ consists of two nontrivial subcomplexes sharing a single vertex (namely the common blue endpoint of the edges $e_a$ for $a\in\rho$) which contradicts the fact that $D_{\rm blue}$ is a disc.

ii)  Assume $\mu$ is red and connects two vertices $a$ and $b$ belonging to two different blue boundary arcs $\alpha_1$ and $\alpha_2$. If the conclusion of ii) does not hold one sees that the common blue endpoint of the edges $e_a$ and $e_b$ in $K$ would separate $\partial D_{\rm blue}$ into two non-trivial parts sharing only this vertex. This contradicts the fact that $\partial D_{\rm blue}$ is a simple closed curve. Obviously, a similar argument applies when $\mu$ is blue.
\hfill $\Box$

\begin{lemma}\label{le4}
Let $e=\langle ab\rangle$ and $f=\langle a'b'\rangle$ be two disjoint blue (red) edges in the midsection $S_K$.  Suppose $a$ and $a'$ as well as $b$ and $b'$ are connected by a red (blue) path.  Then there exists a blue (red) path of quadrangles connecting $e$ and $f$.

\end{lemma}
\noindent
{\it Proof.} It suffices to prove the Lemma for blue edges.
Let $\Delta_e$ and $\Delta_f$ be the two-coloured triangles of $K$ that contain the blue edges $e$ and $f$, respectively.
Then $\Delta_e$ and $\Delta_f$ share a blue edge
$(xy)$ in the blue boundary of $K$ and also have red vertices $v_e$ and $v_f$ in $\partial K$.
Since $e$ and $f$ are disjoint $v_e\neq v_f$.  If $(xy)$ belongs to $\partial D_{\rm blue}$, the star of $(xy)$ consists of a sequence
$(xyv_1v_2), (xyv_2v_3),\ldots ,(xyv_{k-1}v_k)$ of (2,2)-tetrahedra sharing $(xy)$ and an additional
(1,3)-tetrahedron $(xyzv_k)$.  We have $v_e,v_f\in\{ v_1,\ldots ,v_k\}$ and the claim follows since each (2,2)-tetrahedron corresponds to a quadrangle in the midsection.  If $(xy)$ is an interior edge in $D_{\rm blue}$ then the first (2,2)-tetrahedron above has to be preceded by a (1,3)-tetrahedron $(xyz'v_1)$ and then the rest of the argument is unchanged.
\hfill $\Box$

\medskip
\noindent
\begin{proposition}\label{prop3}
The midsection $S_K$ of a causal disc-slice $K$ determines $K$ uniquely up to
combinatorial equivalence.

\end{proposition}

\noindent
{\it Proof.} This follows by arguments identical to those in \cite{DJ} for causal sphere-slices.
\hfill $\Box$

\noindent
\begin{definition}
We let ${\cal S}{\cal D}$ denote the set of all coloured cell complexes $S$ homeomorphic to a disc (with cells as described previously) which have at least one triangle of each colour and satisfy the following conditions:
\begin{itemize}

\item[($\alpha$)] No pair of different vertices in $S$ are connected by both a red and a blue path.

\item[($\beta_1$)] Each closed simple red (blue) path encloses solely red (blue) triangles in its interior.

\item[($\beta_2$)] Considering the division of $\partial S$ into red and blue arcs, there is no red (blue) path connecting
two vertices belonging to different blue (red) arcs unless they are the two endpoints of a red (blue) arc.

\item[($\gamma$)]  If $e=\langle ab\rangle$ and $f=\langle cd\rangle$ are two disjoint blue (red) edges in $S$ such that $a$ and $c$ as well as $b$ and $d$ are connected by a red (blue) path, then there exists a blue (red) path of quadrangles connecting $e$ and $f$.

\end{itemize}
\end{definition}

\begin{remark}\label{rem2} {\rm
As mentioned previously, condition $(\beta_1)$ implies the absence of closed red (blue) paths of quadrangles. Indeed, the outer blue (red) boundary component of such a path would violate $(\beta_1)$.  Similarly, condition $(\beta_2)$ implies that no two different edges in $\partial S$ can be connected by a red or blue path of quadrangles. Likewise, it follows that two different edges in the same triangle cannot be connected by a path of quadrangles, since it would contradict ($\alpha$).

One can demonstrate by explicit examples that property ($\gamma$) does not follow from the first three properties. }

\end{remark}

\begin{definition}\label{delta}
We say that a two-dimensional coloured cell complex satisfies condition $(\delta)$ if for any pair of distinct
red, resp.\ blue, triangles it is not possible to join their vertices pairwise by blue, resp.\ red, paths.
\end{definition}

\noindent
\begin{lemma}\label{le6}
Condition ($\delta$) holds for all $S\in{\cal S}{\cal D}$.

\end{lemma}

\noindent
{\it Proof.}
Given the two triangles, let us assume they are red and that three blue paths exist connecting $a$ to $a'$,
 $b$ to $b'$, and $c$ to $c'$. It follows easily from property ($\alpha$) that the two triangles must be disjoint. Using ($\gamma$) there exists a (non-trivial) path $\rho_{ab}$ of quadrangles connecting $\langle ab\rangle$ to $\langle a'b'\rangle$. Similarly, a path $\rho_{ac}$ of quadrangles exists connecting
$\langle ac\rangle$ to $\langle a'c'\rangle$. Each of these two paths contains a blue path connecting $a$ to $a'$ which do not intersect each other (although they may touch at some vertices or edges). Hence they define a closed curve, whose interior consists of blue triangles by ($\beta_1$). In particular, all edges of those blue triangles as well as all red edges of $\rho_{ab}$ and $\rho_{ac}$ are interior edges of $S$, and the exterior of the curve contains the two original triangles and also the path $\rho_{bc}$ of quadrangles connecting $\langle bc\rangle$ to $\langle b'c'\rangle$, whose existence again follows from ($\gamma$). Similarly, considering the closed blue curves determined by $\rho_{ab}$ and $\rho_{bc}$, respectively $\rho_{bc}$ and $\rho_{ac}$, we conclude that all edges are interior in $S$ which contradicts the fact that $S$ is a disc.
\hfill $\Box$

\medskip

Lemmas \ref{le2} - \ref{le6} together with Proposition \ref{prop3} show that the mapping
$\psi : K \mapsto S_K$ is a well-defined injective map from the set ${\cal C}{\cal D}$ of causal disc-slices
into ${\cal S}{\cal D}$.  We now aim to prove the following main result of the present paper.

\noindent
\begin{theorem}\label{thm1}
The mapping $\psi :K\mapsto S_K$ is bijective from the set ${\cal C}{\cal D}$
of causal disc-slices onto the set ${\cal S}{\cal D}$ of coloured two-dimensional cell complexes.

\end{theorem}

\section{Proof of the main result}

In this section we prove Theorem \ref{thm1}.
The strategy is to show first that from any $S\in {\cal S}{\cal D}$ we can construct a unique simplicial complex.
We then proceed to show that this simplicial complex has the topology of a ball and is actually a causal disc-slice.
The midsection of this causal slice is the coloured cell complex we started with.

Let $S\in{\cal S}{\cal D}$ be given.  In order to construct the corresponding $K\equiv K_S\in{\cal C}{\cal D}$ we start by
associating to each vertex $a\in S$ a pair of new vertices $r_a, b_a$ which will form the vertex set $K_S^0$
of $K_S$ with the following identifications:
\begin{itemize}
\item[($\star$)]
$r_a=r_b$, resp. $b_a=b_b$, if $a$ and $b$ are connected by a blue, resp. red, path,
\end{itemize}
\noindent
where $a,b$ are arbitrary vertices in $S$.
Thus $K_S^0$ consists of all the vertices $r_a,b_a$ with $a\in S$ subject to the identifications ($\star$). By definition we attach the colour red to the vertex $r_a$ while $b_a$ is coloured blue.

The set $K_S^3$ of coloured tetrahedra is obtained from the collection of
2-cells of $S$ as follows:
for each red triangle $\Delta =\langle abc\rangle$ let $t_\Delta$ be the (3,1)-tetrahedron $(r_ar_br_cb_a)$
where we notice that all 4 vertices are different by ($\alpha$) and $b_a=b_b=b_c$;
similarly, if $\Delta$ is a blue triangle, let $t_\Delta =(b_ab_bb_cr_a)$; finally, for the quadrangle $q=\langle abcd\rangle$, we let $t_q$ be the (2,2)-tetrahedron $(r_ar_bb_ab_c)$;  again it follows from
($\alpha$) that the four vertices are different and $t_q$ depends only on the quadrangle $q$.

Thus to each 2-cell $F$ of $S$ there corresponds a tetrahedron $t_F$ with vertices in $K_S^0$.  This defines an abstract
3-dimensional coloured simplicial complex $K_S$ whose edges and triangles are obtained as sub-simplices of the tetrahedra.

Let us first verify that $F\mapsto t_F$ is bijective between 2-cells in $S$ and tetrahedra in $K_S$.  By definition the mapping is
surjective.  Consider two 2-cells $F$ and $F'$ such that $t_F=t_{F'}$.  Clearly $F$ and $F'$ are both triangles with the same colour or they are both quadrangles.

Suppose $F=\langle abc\rangle$ and $F'=\langle a'b'c'\rangle$ are, say, red triangles.  Then $t_F=t_{F'}$ means that
$b_a=b_b=b_c=b_{a'}=b_{b'}=b_{c'}$ and $\{r_a,r_b,r_c\}=\{r_{a'},r_{b'},r_{c'}\}$.  By ($\alpha$) this implies that $\{a,b,c\}=\{a',b',c'\}$ and hence $\langle abc\rangle =\langle a'b'c'\rangle$.

If $F=\langle abcd\rangle$ and $F'=\langle a'b'c'd'\rangle$ are quadrangles then
$t_F=t_{F'}$ implies that $\{r_a,r_b\} = \{r_{a'}, r_{b'}\}$ and $\{b_a, b_c\}=\{b_{a'},b_{c'}\}$.
 Using ($\alpha$) it is then straightforward to check that
$\langle abcd\rangle$ is equal to $\langle a'b'c'd'\rangle$.  We have thus established that the tetrahedra of $K_S$ are labelled by the 2-cells of $S$.

Next let us consider the triangles in $K_S$.  These fall into four disjoint classes:
\begin{itemize}

\item[(i)] Red triangles $(r_a r_b r_c)$ where $\langle abc\rangle$ is a red triangle in $S$.
\item[(ii)] Blue triangles $(b_a b_b b_c)$ where $\langle abc\rangle$ is a blue triangle in $S$.
\item[(iii)] Triangles $(r_a r_b b_a)$ where $\langle ab\rangle$ is a red edge in $S$.
\item[(iv)]  Triangles $(b_a b_b r_a)$ where $\langle ab\rangle$ is a blue edge in $S$.

\end{itemize}
Using property ($\delta$), which holds by Lemma \ref{le6}, we
see that red triangles in $K_S$ corresponding to different red triangles in $S$ are different.
Such triangles are not shared by different tetrahedra and therefore lie in the boundary of $K_S$.  The corresponding
statement about blue triangles is also clearly true.

Next consider a triangle $(r_a r_b b_a)$ where $\langle ab\rangle$ is a red edge in $S$. By definition of $t_F$ it holds that, if $\langle ab\rangle$ belongs to a 2-cell $F$ in $S$, then $(r_ar_b b_a)$ belongs to the boundary of $t_F$.

Conversely, suppose the triangle $(r_a r_b b_a)$ belongs to the boundary of a tetrahedron $t_F$.
If $F$ is a red triangle $\langle a'b'c'\rangle$, then $(r_a r_b b_a)$
equals one of the triangles $(r_{a'} r_{b'} b_{a'})$, $(r_{a'} r_{c'} b_{a'})$, $(r_{b'} r_{c'} b_{a'})$.
By ($\alpha$) this implies that $\{a,b\}$ equals one of $\{ a',b'\}$,
 $\{ a',c'\}$, $\{ b',c'\}$ and hence the edge $\langle ab\rangle$ equals one of the edges $\langle a'b'\rangle$,
 $\langle a'c'\rangle$, $\langle b'c'\rangle$.
Thus, $\langle ab\rangle$ belongs to $F$.
Similarly, if $F$ is a quadrangle $\langle a'b'c'd'\rangle$, it follows that
$(r_a r_b b_a)$ equals $(r_{a'}r_{b'}b_{a'})$ or   $(r_{a'}r_{b'}b_{c'})$.
By ($\alpha$) this implies that $\langle ab\rangle=\langle a'b'\rangle$ or $\langle ab\rangle=\langle c'd'\rangle$ and hence $\langle ab\rangle$ is
an edge in $F$.

From these observations follows that two tetrahedra $t_F$ and $t_{F'}$ share a triangle
$\Delta =(r_ar_bb_a)$ (or $\Delta =(b_ab_br_a)$) if and only if $F$ and $F'$ share the
edge $\langle ab\rangle$.  In particular, the interior triangles in $K_S$
are labelled by the interior edges in $S$.  The boundary triangles in $K_S$ are labelled by the boundary edges of $S$ together with the monocoloured triangles in $S$ which label the monocoloured boundary triangles in $K_S$.

We next consider the edges in $K_S$. In particular, we want to show that monocoloured edges lie in $\partial K_S$.  Let
$(r_ar_b)$ be a red edge in $K_S$.  This means that $a,b$ can be assumed to belong to some 2-cell $F$ in $S$ such that $\langle ab\rangle$ is  a red edge in $F$.  If $F=\langle abc\rangle$ is a red triangle it follows from the preceding paragraph that $(r_ar_b)$ belongs to $\partial K_S$ since it belongs to the red triangle $(r_ar_br_c)$.
Alternatively, if $\langle ab\rangle$ belongs to a quadrangle $q$, then by ($\beta_1$) and ($\beta_2$)
we have that $q$ belongs to a red path of quadrangles connecting either two red triangles
or a red triangle and a red boundary edge in $S$.  This shows that $(r_ar_b)$ is an edge in a red triangle in $\partial K_S$.
Applying similar arguments to blue edges in $K_S$ shows that all monocoloured edges lie in the boundary.

The correspondence $a\to (r_ab_a)$ between vertices in $S$ and the two-coloured edges of $K_S$ is bijective by property ($\alpha$). Moreover, $(r_ab_a)$ belongs to $\partial K_S$ if and only if $a$ belongs to $\partial S$, since it belongs to the two boundary triangles corresponding to the edges in $\partial S$ incident on $a$.

Finally, consider a red vertex $x$ in $K_S$.  It belongs to some tetrahedron $t_F$ where $F$ is a 2-cell in $S$.  If $F$ is a quadrangle or a red triangle, then $x$ belongs to a red edge and hence, by the preceding paragraph, it belongs to $\partial K_S$.  If $F$ is a blue triangle
$\langle abc\rangle$ then $x=r_a=r_b=r_c$.
Since we assume that there is a least one triangle of each colour in $S$ we can pick
a path in $S$ starting at $a$ and ending at vertex $a'$ in a red triangle $\langle a'b'c'\rangle$.  There is a first vertex $d$ in the path which is contained in a red edge and we have $x=r_d$.  Hence, $x$ belongs to $\partial K_S$.  An identical argument shows that a blue vertex in $K_S$
is necessarily contained in the boundary.   This completes the argument that all
monocoloured simplices in $K_S$ belong to the boundary.

The next thing to consider is the structure of $\partial K_S$.  We have seen above that there is a one-to-one correspondence between the boundary edges in $S$ and the two-coloured triangles in $\partial K_S$.  It follows that these triangles form a sequence
$\Delta_1,\ldots ,\Delta_k$ corresponding to the boundary edges $e_1,\ldots ,e_k$ in $\partial S$ ordered cyclically.
Setting $e_i=\langle a_ia_{i+1}\rangle$, where $a_1, \ldots ,a_k$ are the cyclically ordered vertices of
$\partial S$ (with $a_{k+1}=a_1$), we see that
$\Delta_i$ and $\Delta_{i+1}$ share the edge $(r_{a_{i+1}}b_{a_{i+1}})$.
By ($\beta_2$) the triangles
$\Delta_1,\ldots ,\Delta_k$ make up a two-dimensional causal slice $C$, homeomorphic to the cylinder $S^1\times [0,1]$, whose red and blue
boundary circles are denoted $\partial C_{\rm red}$ and $\partial C_{\rm blue}$.
At this stage we need the following lemma whose proof we will postpone a little.

\noindent
\begin{lemma}\label{le7}
The simplicial complex $K_S$ is homeomorphic to the 3-ball so its boundary is homeomorphic to
the 2-sphere.
\end{lemma}

By the above lemma $\partial K_S\setminus C$ consists of two discs $D_{\rm red}$ and $D_{\rm blue}$ whose boundaries are
$\partial C_{\rm red}$ and $\partial C_{\rm blue}$.  We claim that $D_{\rm red}$ is made up exactly of the red triangles in $\partial K_S$ and similarly for $D_{\rm blue}$.  Indeed, since $D_{\rm red}$ is a two-dimensional pseudomanifold any triangle in $D_{\rm red}$ can be edge-connected to $\partial D_{\rm red}=\partial C_{\rm red}$ by a sequence of triangles in $D_{\rm red}$.  The triangles in $D_{\rm red}$
are mono-coloured and are therefore either red or blue.  But a
blue triangle cannot be glued to a red triangle so $D_{\rm red}$
consists of red triangles.  These must be all the red triangles in $\partial K_S$ since $D_{\rm blue}$ consists solely of blue triangles.
We have therefore shown that $K_S$ is a causal disc slice.

Now it is not hard to verify that the midsection of $K_S$ is combinatorially identical to the original midsection $S$ used to construct $K_S$. We already noted above that the vertices of $S$ are in bijective correspondence with the two-coloured edges in $K_S$. By the definition of the midsection $S'$ of $K_S$ its vertices are the midpoints of the two-coloured edges of $K_S$, and hence can be be labeled by the vertices of $S$. In this way the midsection of $t_F$ clearly gets identified with $F$ and the edge in $S'$ corresponding to an interior triangle $(r_ar_bb_a)$, respectively $(b_ab_br_a)$, of $K_S$ is the red, respectively blue, edge $\langle ab\rangle$ in $S$. Hence, the correspondence between vertices induces bijective correspondences between 2-cells and 1-cells as well, and so $S$ and $S'$ are combinatorially identical. This completes the proof of Theorem \ref{thm1}.      \hfill $\Box$

 \medskip

It remains to prove Lemma \ref{le7}. For this purpose an alternative
construction of $K_S$ by a gluing procedure is a useful tool.  We begin by explaining this construction.

For each 2-cell $F$ in $S$ and each vertex $a$ of $F$ we define a red vertex $r_a^F$ and a blue
vertex $b_a^F$ such that $r_a^F=r_b^F$, resp. $b_a^F=b_b^F$, if $\langle ab\rangle$ is a blue, resp. red, edge in F. In this way four different vertices are defined for each 2-cell $F$ and they in turn define a coloured (abstract) tetrahedron $\tau_F$ considered as a simplicial complex including its sub-simplices. As before $\tau_F$ is of type $(3,1),(2,2)$, or $(1,3)$ depending on whether $F$ is a red triangle, a quadrangle, or a blue triangle, respectively. Without further identifications the tetrahedra so defined are pairwise disjoint and we note that the definition of $K_S$ can be reformulated by stating that $K_S$ is obtained from the collection of tetrahedra $\tau_F$ labeled by the 2-cells of $S$ by imposing the identifications of simplices implied by the relations

\begin{itemize}
\item[($\star\star$)]
$r_a^{F_1}=r_b^{F_2}$, resp. $b_a^{F_1}=b_b^{F_2}$, if either $a=b$ or $a$ is connected to
$b$ by a blue, resp. red, path,
\end{itemize}
\noindent
where $a,b$ are arbitrary vertices in $S$ and $F_1, F_2$ are arbitrary 2-cells in $S$ containing $a$ and $b$, respectively.

 We next show that $K_S$ can equivalently be obtained by applying a suitable gluing procedure to the collection of
 tetrahedra defined above.
Given two 2-cells $F_1$ and $F_2$ sharing a red edge $\langle ab\rangle$ we have that $\tau_{F_1}$ contains the triangle $(r_a^{F_1}r_b^{F_1}b_a^{F_1})$ and $\tau_{F_2}$ contains the triangle
$(r_a^{F_2}r_b^{F_2}b_a^{F_2})$. We say that $\tau_{F_1}$ is glued to $\tau_{F_2}$ along $\langle ab\rangle$ if
$r_a^{F_1}$ is identified with $r_a^{F_2}$, $r_b^{F_1}$ is identified with $r_b^{F_2}$, and $b_a^{F_1}$ is identified with $b_a^{F_2}$ in $\tau_{F_1}$ and $\tau_{F_2}$  and their sub-simplices. Similarly, gluing along a blue interior edge in $S$ is defined. In this way, given an interior edge $\langle ab\rangle$
of $S$, the tetrahedra $\tau_{F_1}$ and $\tau_{F_2}$ corresponding to the 2-cells $F_1$ and $F_2$
sharing $\langle ab\rangle$ can be glued along $\langle ab\rangle$.

We now define the simplicial complex $K'_S$ by imposing the identifications of the simplices in the collection $\{\tau_F\}$ implied by gluing pairs of tetrahedra as described along all interior
edges of $S$. We claim that $K_S=K'_S$. In order to prove this we need to verify that the identifications of vertices implied by the gluing conditions are identical to those given
by $(\star\star)$.

First, consider $r_a^{F'}$ and $r_a^{F''}$ where $a$ is a vertex in two different 2-cells $F'$ and $F''$. Since $S$ is a manifold
there exist 2-cells $F_1, F_2,\dots,F_n$ such that $F_1=F'$ and $F_n=F''$ and $F_i$ and $F_{i+1}$ share an edge $e_i$
containing $a$ for each $i=1,\dots,n$. From the gluing of $\tau_{F_i}$ and $\tau_{F_{i+1}}$ along $e_i$
it follows that $r_a^{F_i}= r_a^{F_{i+1}}$ and $b_a^{F_i}= b_a^{F_{i+1}}$ for $i=1,\dots,n-1$, and hence we conclude that $r_a^{F'}= r_a^{F''}$ and $b_a^{F'}= b_a^{F''}$.

Next consider two different vertices $a$ and $b$ in $S$ and assume they are connected by a blue path with edges $\langle a_1a_2\rangle, \langle a_2a_3\rangle,\dots, \langle a_ma_{m+1}\rangle$, where $a_1=a$ and $a_{m+1}=b$. Choosing arbitrary 2-cells $F_1,\dots, F_m$ such that $\langle a_ia_{i+1}\rangle$ belongs to $F_i, i=1,\dots,m$, we have that $r_{a_i}^{F_i}= r_{a_{i+1}}^{F_i}$ and by the preceding paragraph $r_{a_{i+1}}^{F_i}= r_{a_{i+1}}^{F_{i+1}}$ for all $i$. It follows that $r_a^{F_1}= r_b^{F_m}$. Of course, the corresponding result for blue vertices
holds if $a$ and $b$ are connected by a red path.

Conversely, it is clear that if $a$ and $b$ are different vertices belonging to 2-cells $F'$ and $F''$, respectively, and $r_a^{F'}= r_b^{F''}$, then $a$ and $b$ are connected by a blue path. Indeed, there exists a sequence of vertices $b_1,\dots, b_m$ and corresponding 2-cells $F_1,\dots, F_m$ such that $a=b_1, F'=F_1$ and $b=b_m, F''=F_m$ and $r_{b_i}^{F_i}$ is identified with $r_{b_{i+1}}^{F_{i+1}}$
either through a gluing of $\tau_{F_i}$ to $\tau_{F_{i+1}}$ along an edge containing $b_i$, in which case $b_i=b_{i+1}$, or else $F_i=F_{i+1}$ and $b_i$ is connected to $b_{i+1}$ by a blue edge in $F_i$. Similarly, if $b_a^{F'}= b_b^{F''}$ then $a$ and $b$ are connected by a red path.

This shows that the identifications pertaining to $K_S$ and $K'_S$ are the same and hence $K_S=K'_S$ as claimed. We are now ready to prove Lemma \ref{le7}.

\medskip
 \noindent
{\it Proof of Lemma \ref{le7}.}
Let $S_1,S_2,\ldots  ,S_N =S $ be a local construction (see \cite{dj,bz}) of the midsection $S$.  This means that $S_1$ is a coloured
2-cell and $S_{n+1}$ is obtained from $S_n$ by either (i) gluing a coloured 2-cell to $S_n$ along an edge $e$ in $\partial S_n$ or (ii)
by identifying two edges $e_1$ and $e_2$
in $\partial S_n$ which have the same colour and share a vertex. The existence of such a construction is well known and easy to establish
in the 2-dimensional case. For further details we refer to \cite{bz,dur}.  Evidently, all the $S_n$'s have the topology of a disc and the 2-cells as well as interior edges of $S_n$ can be identified with corresponding 2-cells and interior edges in $S$.

The correspondence between tetrahedra $\tau_F$ in $K_S$ and 2-cells $F$ of $S$ and the local construction of $S$ gives rise to a sequence of coloured simplicial complexes $K_n$, $n=1,\ldots ,N$, where $K_n$ is defined by gluing the tetrahedra $\tau_F$ assigned to the 2-cells in $S_n$ along the interior edges of $S_n$. In particular, $K_1$ is a single tetrahedron and $K_N=K_S$, since $S_N=S$ and $K_S=K'_S$ as shown above. Moreover, $K_{n+1}$ is obtained from $K_n$ by gluing a tetrahedron $\tau_F$ to a tetrahedron $\tau_{F'}$ in $K_n$ along an edge $\langle ab\rangle$. In case $\tau_F$ is not in $K_n$ already it is clear that the topological class of $K_n$ equals that of $K_{n+1}$. We shall now argue that the same holds if $\tau_F$ and $\tau_{F'}$ both belong to $K_n$. Since $K_1$ is homeomorphic to the 3-ball the same will consequently hold for $K_S$, and hence the proof of the lemma will be completed.

If $\tau_F$ and $\tau_{F'}$ belong to $K_n$ and are glued along the edge $e$
in $K_n$
then $F$ and $F'$ belong to $S_n$, and $S_{n+1}$ is obtained from $S_n$ by identifying
two edges $e_1$ and $e_2$ in $\partial S_n$ sharing a vertex $v$, such that the identified edges equal $e$.

\begin{figure}[t]
  \begin{center}
 \includegraphics[width=9truecm]{Fig5_SOSx2.pdf}
    \caption{The local construction of the midsection and the corresponding 3-dimensional simplical complex.
    After identifying the edges $e_1$ and $e_2$ we identify the triangles $\Delta_1$ and $\Delta_2$.}
    \label{figY}
      \end{center}
      \end{figure}

Let us set $e_1=\langle vv_1\rangle$ and $e_2=\langle vv_2\rangle$ and assume $e$ is red.
Then $K_{n+1}$ is obtained from $K_n$ by identifying the
triangles $\Delta_1=(r_{v_1}^{F_1}r_v^{F_1}b_v^{F_1})$ and $\Delta_2=(r_{v_2}^{F_2}r_v^{F_2}b_v^{F_2})$ which share the edge $(r_v^{F_1}b_v^{F_1}) = (r_v^{F_2}b_v^{F_2})$ in $K_n$.
This is illustrated on Fig.~\ref{figY}.
We need to show that no further identifications of simplices are implied.

Additional identifications can only arise if there is a vertex $x\in\partial K_n$ which is a neighbour
of both $r_{v_1}^{F_1}$ and $r_{v_2}^{F_2}$
in which case the edges $(xr_{v_1}^{F_1})$ and $(xr_{v_2}^{F_2})$ are identified in the step from $K_n$ to $K_{n+1}$.
We claim that no such $x$ exists except $b_v^{F_1}$ and $r_v^{F_1}$.

First, suppose $x$ is a blue vertex, $x\neq b_v^{F_1}$.  Then
there are two vertices $w_1$ and $w_2$ in the midsection such that $b_{w_1}^{F'_1}=x=b_{w_2}^{F'_2}$
and $r_{w_i}^{F'_i}=r_{v_i}^{F_i}, i=1,2$, in $K_n$ for some 2-cells $F'_1, F'_2$ in $S_n$. This implies that $w_1$ and $w_2$ are connected by a red path in $S$ while $w_i$ is connected to $v_i$ by a blue path
in $S$ for $i=1,2$.  When we now take the step
from $S_n$ to $S_{n+1}$ and identify $e_1$ and $e_2$ the vertices $v_1$ and $v_2$ get identified and
hence $w_1$ and $w_2$ are connected both by a blue path in $S$
and by a red path, see Fig.\  \ref{Z0}.
Hence, $w_1=w_2$, and we get a closed blue path enclosing  the red edge $e$ which is impossible by
$(\beta_1)$.

\begin{figure}[h]
  \begin{center}
 \includegraphics[width=8truecm]{Fig6_SOS.pdf}
    \caption{Paths arising in the local construction of the midsection when $x$ is blue.}
    \label{Z0}
      \end{center}
      \end{figure}

Now suppose $x$ is a red vertex.  Then there are four vertices $w_1,w_2,u_1,u_2$ in the midsection such that
$\langle w_1w_2\rangle$ and $\langle u_1u_2\rangle$ are red edges and
$x=r_{w_2}^{F'}=r_{u_2}^{F''}$, $r_{w_1}^{F'}=r_{v_1}^{F_1}$, $r_{u_1}^{F''}=r_{v_2}^{F_2}$ for some 2-cells $F', F''$ in $S_n$. It follows that $w_2$ and $u_2$ are connected by a blue path $\rho_1$ and similarly that
$w_1$ is connected to $v_1$ and $u_1$ is connected to $v_2$ by a blue path in $S$, see Fig.\ \ref{Za}.
\begin{figure}[h]
  \begin{center}
 \includegraphics[width=6truecm]{Fig7_SOS.pdf}
    \caption{Paths arising in the local construction of the midsection when $x$ is red.}
    \label{Za}
      \end{center}
      \end{figure}

Upon merging $e_1$ and $e_2$ we get as above a blue path $\rho_2$ from $w_1$ to  $u_1$.
It follows by condition $(\gamma)$
that the edges $\langle w_1w_2\rangle$ and $\langle u_1u_2\rangle$ are connected by a blue path of quadrangles.
This path of quadrangles either lies inside the closed loop made up of $\rho_1, \rho_2$ and the edges $\langle w_1w_2\rangle$ and $\langle u_1u_2\rangle$ or outside it, see Fig.\ \ref{figZ}.
In both cases we obtain a closed blue loop containing a red edge in its interior which violates $(\beta_1)$.
This completes the proof of Lemma~\ref{le7} and concequently that $K_S$ is a 3-ball.
\begin{figure}[h]
  \begin{center}
 \includegraphics[width=10truecm]{Fig8_SOS.pdf}
    \caption{Paths of quadrangles arising in the local construction of the midsection when $x$ is red.}
    \label{figZ}
      \end{center}
      \end{figure}
\hfill $\Box $

\section{Sphere-slices}

In this section we generalise Theorem \ref{thm1} to the case of sphere-slices.
It is clear from the proofs of Lemmas \ref{le2} and \ref{le4} that the midsection of any sphere-slice has
properties $(\alpha)$ and $(\gamma)$, while ($\beta_2$) is irrelevant as it refers to the boundary. Furthermore, the proof of
Lemma \ref{le3}  shows that an analogue of ($\beta_1$) still holds in the following form:
\begin{itemize}
\item[($\beta$)] Each simple closed red (or blue) path divides the midsection into two components,
one of which is red (blue).
\end{itemize}
Since a simplicial decomposition of the 2-sphere requires at least 4 triangles it follows that $S$ must contain at least 4 triangles of each colour.
Let $\phi$ denote the mapping that takes any sphere slice $K$ to its midsection $S$.

\begin{definition}\label{def5}
We let ${\cal S}$ denote the set of all coloured cell complexes $S$ homeomorphic to the 2-sphere (with cells as described previously) which have at least four triangles of each colour and satisfy the conditions
($\alpha$), ($\beta$), and ($\gamma$).

\end{definition}
\noindent
\begin{lemma}  \label{le8}
Property ($\delta$) holds for all $S\in {\cal S}$.

\end{lemma}
\noindent
{\it Proof.}
Applying the same argument as in the proof of Lemma \ref{le6} we conclude that if ($\alpha$), ($\beta$), and
$(\gamma)$ hold for a coloured spherical cell complex $S$ and there are two different red triangles in
 $S$ whose vertices are pairwise connected by blue paths, then there exist three paths of quadrangles connecting the edges of the triangles pairwise and the complement of those paths consists entirely of blue triangles together with the two given red triangles in $S$. In particular, there can be only two red triangles, which contradicts $S\in{\cal S}$. Clearly, a similar argument holds for blue triangles.
\hfill $\Box $

We now state and prove the main result in this section.

\noindent
\begin{theorem} \label{thm2}
  The map $\phi: K\to S$ is a bijection from the set ${\cal C}{\cal S}$ of causal sphere-slices to the
  set ${\cal S}$ of coloured 2-dimensional cell complexes.

\end{theorem}

\noindent
{\it Proof.}
Given $S_0\in{\cal S}$ it is enough to construct a sphere-slice $K$ such that $\phi (K)=S_0$.  The idea of the proof
is to cut a piece out of $S_0$ in such a way that we obtain a cell complex $S$ in ${\cal C}{\cal D}$.  Then we use
Theorem \ref{thm1} to obtain a disc-slice with the given midsection $S$.  Finally, we fill
in the "hole`` of the disc-slice to obtain a sphere-slice with midsection $S_0$.   We divide the argument into 5 parts.
\medskip

\noindent
(1) Let $R$ be a maximal edge-connected cluster of red triangles in $S_0$ and let $S=S_0\setminus {\rm int}(R)$.
Condition $(\beta)$ and the Jordan Curve Theorem imply that $R$ and $S$ are
(closed) discs.   Moreover,
since $R$ is maximal, all edges in $\partial S$ are red and contained in quadrangles in $S$.  Pick one of these
quadrangles $q$ and let $\Gamma =(q_1,q_2,\ldots ,q_n)$ be the maximal blue path of quadrangles containing $q$.
Let $\rho=(v_1,v_2,\ldots ,v_n)$ be the red path in $\Gamma$ that shares at least one red edge with $\partial S$,
see Fig.\ \ref{Fig18}.
\begin{figure}[h]
  \begin{center}
 \includegraphics[width=8truecm]{Fig9_SOS.pdf}
    \caption{A maximal path of quadrangles in $S$ touching the boundary.}
    \label{Fig18}
      \end{center}
      \end{figure}
      The other red path in $\Gamma$, $\rho'=(w_1,w_2,\ldots ,w_n)$,
  shares no vertex with $\partial S$ as a consequence of condition $(\alpha)$. Likewise, $\rho$ and $\rho'$ do not intersect and neither $\rho$ nor $\rho'$ can have multiple vertices, since it would contradict
	$(\alpha)$. Moreover, if $1\leq i<j\leq n$ are such that $v_i,v_j\in \partial S$, then $v_k\in\partial S$ for all $k$ between $i$ and $j$ since otherwise $R$ is not maximal due to $(\beta)$.  Hence, $\Gamma$ intersects $\partial S$ along a curve segment $\mu$, see Fig.\ \ref{Fig18}.
Note that $\mu$ does not contain all of $\partial S$ because this would again contradict condition
$(\alpha)$, see Fig.\ \ref{Fig20}.
\begin{figure}[h]
  \begin{center}
 \includegraphics[width=10truecm]{Fig10_SOS.pdf}
    \caption{If $\partial S=\mu$ then $w'$ and $w''$ are connected both by a red and by a blue path.}
    \label{Fig20}
      \end{center}
      \end{figure}

The above observations show that the boundary of the strip covered by the path of quadrangles
$\Gamma$, which consists
of $\rho$ and $\rho'$ and the two blue edges $\langle v_1w_1\rangle$ and $\langle v_nw_n\rangle$, is a simple curve.
This curve shares exactly a segment $\mu=(v_i,v_{i+1},\ldots ,v_j)$ with $\partial S$ where
$1\leq i<j\leq n$.  Hence, removing the strip covered by $\Gamma$ from $S$ we obtain a new disc $S'$ whose boundary consists of the blue edges $\langle v_1w_1\rangle$ and $\langle v_nw_n\rangle$ together with the red
paths $\rho'$, $\partial S\setminus \mu$, $(v_1,\ldots ,v_i)$ and $(v_j,\ldots ,v_n)$, see Fig.\ \ref{Fig21}.
\begin{figure}[h]
  \begin{center}
\includegraphics[width=10truecm]{Fig11_SOS.pdf}
    \caption{The cell complex $S'$.}
    \label{Fig21}
      \end{center}
      \end{figure}
Now consider the blue triangles $\Delta_1$ and $\Delta_2$ in $S'$ containing $\langle v_1w_1\rangle$
and $\langle v_nw_n\rangle$, respectively.
Denote the third vertex in these triangles by $u_1$ and $u_2$, respectively, see Fig.\ \ref{Fig21}.
It follows from condition $(\delta)$ (see Lemma \ref{le8}) that $u_1\neq u_2$ and they cannot be connected by a red path.  Moreover, $u_1$ and $u_2$ are not in $\partial S'$ since this would violate $(\alpha)$.  Now, removing $\Delta_1$ and $\Delta_2$ (including edges $\langle v_1w_1\rangle$ and $\langle v_nw_n\rangle$) from $S'$
we obtain another 2-dimensional cell complex $S''$
with the topology of a disc and a boundary consisting of the blue paths
$(v_1,u_1,w_1)$ and $(v_n,u_2,w_n)$ and the same red paths as $\partial S'$.

\medskip
\noindent
(2) We now claim that $S''$ belongs to ${\cal S}{\cal D}$.  Since paths in $S''$ are also paths in $S_0$, conditions $(\alpha)$ and $(\beta_1)$ are obviously fulfilled.  Furthermore, $S''$ contains at least two blue triangles, and since there exist red edges in $\partial S''$ the existence of at least one red triangle in $S''$ will follow once $(\beta_2)$ has been established for $S''$.

In order to verify $(\beta_2)$ we recall that $u_1$ and $u_2$ cannot be connected by a red path.  Connecting $u_1$ to $w_n$ or $v_n$ or connecting $u_2$ to $w_1$ or $v_1$ by a red path evidently contradicts $(\alpha)$.
Similarly, a red path from $v_1$ to $w_n$ or from $w_1$ to $v_n$ contradicts $(\alpha)$.
 This proves $(\beta_2)$ for $S''$ as far as red paths are concerned.

Now let $\eta$ be a blue path in $S''$ connecting the vertex $w_k$ to a vertex $v$ in the other red arc of $\partial S''$.  Viewing  $\eta$ as a path in $S_0$ and extending it by the edge $\langle w_kv_k\rangle$ we obtain a blue path $\eta'$ in $S_0$ connecting $v_k$ to $v$.
If $v\neq v_k$ then condition $(\alpha)$ is violated.   If $v=v_k$ then $\eta'$ is a closed blue path in $S_0$ with edges of both colours inside and outside unless $k=1$ or $k=n$.  This proves condition $(\beta_2)$.

In order to verify condition $(\gamma)$ for $S''$ let $e_1=\langle x_1y_1\rangle$ and $e_2=\langle x_2y_2\rangle$ be two blue edges in $S''$ such that $x_1$ and $x_2$ as well as $y_1$ and $y_2$ are connected by red paths.  Then, since $(\gamma)$ holds for $S_0$, there is a blue path of quadrangles $\Lambda$ in $S_0$ connecting $e_1$
and $e_2$.  Since the blue boundary edges of $S''$ are not contained in quadrangles outside $S''$ the path $\Lambda$ cannot leave $S''$.  This establishes
condition $(\gamma)$ for the blue edges.

Let now $e_1=\langle x_1y_1\rangle$ and $e_2=\langle x_2y_2\rangle$ be two red edges in $S''$ such that $x_1$ and $x_2$ as well as $y_1$ and $y_2$ are connected by blue paths $\lambda'$ and $\lambda''$.  By condition
$(\alpha)$ the paths $\lambda'$ and $\lambda''$ do not intersect and we can assume without loss of generality that they have no multiple vertices.
Hence, the closed curve consisting of $\lambda'$, $\lambda''$ and the two edges $e_1$ and $e_2$ is simple and encloses a disc $D$ in $S''$.   Since condition $(\gamma)$ holds for $S_0$ there exists a red path $\Gamma'$ of quadrangles in $S_0$
connecting $e_1$ and $e_2$ and evidently $\Gamma'$ is either contained in $D$ or in its exterior.  However, the latter case can be excluded as follows. Observe that neither $D$ nor $\Gamma'$ can intersect $R$. It follows that one of the closed curves made up of either $\lambda'$ or
$\lambda''$ and one of the blue paths in $\Gamma'$ must separate $R$ (in $S_0$) from the union of $D$ and $\Gamma'$ in contradiction with condition $(\beta)$, see Fig.\ \ref{Fig22}. Note, however, that the curve in question is not necessarily simple, since $\lambda'$ may touch the blue path in $\Gamma'$ with the same endpoints, and similarly for $\lambda''$. Using property ($\beta$) one can in this case remove closed parts of the curve so as to obtain a simple closed blue curve with the same separating property.
Thus $\Gamma'$ is contained in $D$ (and hence in $S''$) and this proves that condition $(\gamma)$ holds for $S''$.

\begin{figure}[h]
  \begin{center}
\includegraphics[width=10truecm]{Fig12_SOS.pdf}
    \caption{A path of quadrangles outside $D$ connecting the edges $e_1$ and $e_2$
    and the blue paths $\lambda'$ and $\lambda''$ connecting their endpoints. A priori, parts of the path may be outside $S''$.}
    \label{Fig22}
      \end{center}
      \end{figure}

\medskip
\noindent
(3) We can now apply Theorem \ref{thm1} to conclude that there is a unique disc-slice $K''\in{\cal C} {\cal D}$
whose midsection is $S''$.
In view of the structure of $\partial S''$, the side of $K''$
consists of two pairs of adjacent backwards directed triangles and two arrays of forward directed triangles
corresponding to the two red arcs in $\partial S''$.

We now first construct from $K''$ a new simplicial complex $C$ by gluing to $K''$ the tetrahedra corresponding to the
quadrangles $q_1,\ldots ,q_n$ in $\Gamma$ and those corresponding to the blue triangles at the ends of $\Gamma$.
One way to accomplish this is to consider the simplicial complex $K_0$ whose midsection is $\Gamma$ with the two
blue triangles attached to its ends.  Clearly $K_0$ is a ball and we may think of it as a causal slice whose side consists of two pairs of adjacent backwards directed triangles and two arrays of forward directed triangles of size $n$ whose red edges are pairwise
 identified and constitute $D_{\rm red}$. On the other hand, $D_{\rm blue}$ consists of two triangles with one common edge, see Fig.\ \ref{Fig23}.
 \begin{figure}[h]
  \begin{center}
\includegraphics[width=8truecm]{Fig13.pdf}
    \caption{The cell complex $K_0$ in the case $n=3$.}
    \label{Fig23}
      \end{center}
      \end{figure}
 Now consider one array of forward directed triangles in $\partial K_0$ together with the two adjacent backwards directed triangles.  These triangles
 form a disc in $\partial K_0$ and we can glue $K_0$ to $K''$ along this disc and the corresponding disc in $\partial K''$ or, more precisely, the gluing is performed along the edges $\langle u_1w_1\rangle, \langle w_1w_2\rangle,\dots,\langle w_nu_2\rangle$,
see Fig.\ \ref{Fig21}.   Since both $K_0$ and
 $K''$ are simplicial balls, so is the resulting complex $K''\sharp K_0$.  Its Euler characteristic is therefore $-1$.

 Note also that $K''\sharp K_0$ has a side consisting of two pairs of adjacent backwards directed triangles
 and two arrays of of forward directed triangles, one of which has length $n$.  The blue edges in the backwards directed triangles form a closed curve.  We now identify the two backwards directed triangles in each pair, i.e.\ we glue tetrahedra in $K''\sharp K_0$ along the edges $\langle u_1v_1\rangle$ and $\langle u_2v_n\rangle$.  Doing this, the number of triangles in the simplicial complex decreases by 2, the number of edges decreases by 4 and the number of vertices decreases by 1 and we obtain a new simplicial complex $C'$ with Euler characteristic $-2$.

 The side of $C'$ is a closed circuit of forward directed triangles sharing a single blue vertex.  We now obtain
 the simplicial complex $C$ by gluing tetrahedra in $C'$ along the egdes $\langle v_kv_{k+1}\rangle$
 in the midsection, for $k=1,\ldots ,i-1$ and
 $k=j,\ldots ,n-1$.  It is easy to check that the Euler characteristic does not change with these identifications so
 $$
 \chi (C)=-2.
 $$
 By construction, the midsection of $C$ is $S_0\setminus R$.

 \medskip
 \noindent
 (4)  Now return to the disc $R$ that was removed from $S_0$ in the first step.  Let $\overline{R}$ be the cone over $R$, i.e.\
 $\overline{R}$ consists of the tetrahedra obtained by adding a common blue
 vertex to all the triangles in $R$ or alternatively, by gluing the tetrahedra corresponding to
triangles in $R$ along interior edges in $R$.  We can now
 glue $\overline{R}$ to $C$ along their sides, i.e.\ along the edges in $\partial S =\partial R$.  The
 resulting simplicial complex $K$ has Euler characteristic
 $$
 \chi (K)=\chi (C)+\chi(\overline{R})+\chi_2 (\overline{\partial R})=-2\,,
 $$
 since $\chi (\overline{R})=-1$, and $\chi_2(\overline{\partial R}) = 1$ is the Euler characteristic of the cone $\overline{\partial R}$ over $\partial R$, which is in fact a disc.

 \medskip
 \noindent
 (5) Since $K''$ is a disc-slice it is evident from the preceding construction that $\partial K$ consists of two triangulated
 2-spheres, one red and one blue.  The midsection of $K$ is by construction $S_0$.
By gluing on cones over the two boundary components of $K$ we obtain a simplicial complex $\tilde{K}$
with Euler characteristic 0.  Since $\tilde{K}$ is clearly a pseudomanifold we conclude
(see \cite{seifert}) that $\tilde{K}$ is also a manifold.  In fact, $\tilde{K}$ is simply
connected since any closed curve can be deformed to a closed curve on the midsection which is simply connected.
Hence, $\tilde{K}$ is a 3-sphere by \cite{perel} and $K$ is a cylinder as desired.
This completes the proof of Theorem \ref{thm2}.
\hfill $\Box $

\section{Discussion}\label{sec:6}

A few remarks pertaining to extensions or variations of the present work are in order. As already indicated, the requirement in Definition~\ref{def2} that $D_{\rm red}$ and $D_{\rm blue}$ be discs is unnecessarily restrictive.
Thus, replacing $(ii)$ of Definition~\ref{def2} by

\medskip

{\it (ii') all mono-coloured simplices of $K$ belong to the boundary $\partial K$, such that the red (resp. blue) ones form a connected and simply connected subsimplex $D_{\rm red}$ (resp. $D_{\rm blue}$) of $\partial K$,}

\medskip

\noindent leads to a convenient class of causal triangulations. Let us call this class ${\cal C}{\cal D}'$. In particular, any coloured tetrahedron belongs to ${\cal C}{\cal D}'$ (but not to ${\cal C}{\cal D}$) and so does the complex depicted in Fig.\ \ref{Fig23}.

By inspection of the proofs given above it is seen that condition ($\beta_2$) is used in the construction of a causal disc slice from a cell complex in ${\cal C}{\cal D}$ solely to ensure that $\partial D_{\rm red}$ and $\partial D_{\rm blue}$ are simple curves. It follows that dropping ($\beta_2$) one obtains a one-to-one correspondence between ${\cal C}{\cal D}'$ and the set ${\cal S}{\cal D}'$ of coloured cell complexes homeomorphic to a disc and fulfilling conditions ($\alpha$), ($\beta_1$), and ($\gamma$).

On a different note, one may observe that condition ($\gamma$) is used partly to ensure the validity of condition ($\delta$) (see Lemma~\ref{le6}) and otherwise only in the final part of the proof of Lemma~\ref{le7}
ensuring that the complex $K$ is a simplicial ball. It is natural to consider replacing property ($\gamma$) by property ($\delta$) or dropping both of them. This would require stepping outside the category of simplicial complexes and adopt a different setting encompassing singular triangulations.

Finally, one might also envisage characterising causal triangulations in higher dimensions, which evidently would lead to more involved higher dimensional coloured cell complexes of which little is known at present.   In \cite{DJ} it is explained how the midsections of
4-dimensional causal triangulations are made up of tetrahedra and prisms which replace the triangles and quadrangles which make
up the midsections considered in this paper.

\medskip
\noindent
{\bf Acknowledgements.}  BD acknowledges support from the Villum Foundation via the QMATH Centre of Excellence
(Grant no. 10059).

redpajama_set_name: RedPajamaArXiv


text:
8 Incredible People to Replace Racist Statues With
BY Georgia Dodd
IN Living
With all these statues coming down, many have asked who should replace them? Should we even have statues? What's the purpose of them? Many say it's for history, but museums are for history. Statues are not for history, they're for recognition. Most of the confederate statues were put up during the Jim Crow era and the Civil Rights Movement as a way to strike fear and "put people in their place". That's why they're so easy to take down, they were put up quickly with little to no regard about who they were really putting up.
As a way to confront some of the long-held institutions in America, many have also tried to change town names. Recently, Rhode Island proposed to change its name from the State of Rhode Island and Providence Plantations to just Rhode Island. There's even a petition for Columbus, Ohio to change its name to Flavortown as an ode to famous chef Guy Fieri. So, as we address our racist history, we should also recognize incredible Americans that advocated for change and deserve recognition. Here's a list of eight people who actually deserve a statue:
The statue of Major General Philip Schuyler in front of City Hall was ordered to be removed by Mayor Kathy Sheehan, who noted Schuyler was reportedly the largest slave owner in Albany. We recommend a statue of Sojourner Truth, famous abolitionist and women's rights activist who penned the famous piece, Ain't I A Woman, should be erected in its place. She was born in Swartekill, New York and later moved to New York City. She was born into slavery but was able to escape with her daughter and later went to court for her son, making her the first black woman to win a case against a white man. The statue of Theodore Roosevelt outside the American Museum of Natural History has also been ordered to be removed. We think that a statue of famous writer, feminist, and civil rights activist, Audre Lorde, should be put in its place. She was born in Harlem and describes herself as "black, lesbian, mother, warrior, poet." Both Truth and Lorde are perfect examples of amazing people who used their art to protest for change.
In Alexandria, Virginia, the Confederate statue, "Appomattox," which depicted a Civil War soldier, was removed from an intersection. A forgotten hero of the civil rights movement in Virginia, Samuel Tucker should get the recognition he deserves. He was born in segregated Alexandria and became an attorney who represented African Americans in civil and criminal cases. Tucker attended and led many peaceful protests, sit-ins. He also became the leading attorney for the NAACP in Virginia. Tucker was an incredible man who played a big role in the civil rights movement in Virginia.
A statue of Confederate officer Charles Linn was taken down by protestors on June 2. Mayor Woofin plans to remove a Confederate Sailors and Soldiers Monument. In the place of the statue of Charles Linn, civil rights activist Angela Davis should have a statue. Davis was born in Birmingham and is an activist, philosopher, Marxist feminist. She was a member of the Black Panther Party and has written many books on class, race, feminism and the US Prison System. Recently, Davis' philosophies and ideas have been taken much more seriously.
A Texas Ranger statue that went by the name "One Riot, One Ranger," was removed from Love Field Airport on June 11. A statue of Texas Native Juanita Craft, activist and politician, should be put in its place. She participated in sit-ins and peaceful protests. Craft also served as a member of the Dallas City Council and helped a black student enroll at North Texas State College. She played a big role in Civil Rights Movement in Texas and has been praised by many.
A work crew lifted the statue of John Sutter, a 19th century man who enslaved Indigenous people, off its pedestal in Sacramento, California. In his place, we think a state of actor LeVar Burton should be placed. Famous for his work on Roots, Star Trek and Reading Rainbow, Burton is both an incredible actor and director. He's donated to many organizations and is on the board of directors for the AIDS Research Alliance.
A statue honoring fallen Confederate soldiers was removed from Hemming Park in Jacksonville, Florida. The statue had been at the park since 1898. We think a statue of A. Phillip Randolph should be put there instead. He was labor unionist, activist and politician. Randolph led the Brotherhood of Sleeping Car Porters, the first predominately black labor union and was the head of the March on Washington. Born in Jacksonville, Randolph was an incredible leader and activist and deserves to be celebrated.
The statue of former Mayor of Philadelphia Frank Rizzo was removed June 2. Rizzo has been accused of mistreating members of the Black and LGBT communities during his two terms in the 1970s. We propose that Questlove, musician and music journalist, should replace Rizzo's statue. He is an activist for the issues and has definitely made an impact on culture.
A statue of politician Edward Carmack, a known critic of Tennessee civil rights journalist Ida B. Wells, was taken down outside the state capitol. In his place why don't we put Ida B Wells. Although there has been talks of giving Wells a statue in Chicago, she should also get a statue in Tennessee. She was an investigative journalist, teacher, activist and one of the founders of the NAACP. While many have said that Dolly Parton should have her own statue in Tennessee, and we agree, we should first give one to Ida B. Wells. She was an incredible woman who deserves a monument to her hard work and activism.
Header image courtesy of Cville dog via WikiCommons.
Kristen Stewart Set To Play Princess Diana In Upcoming Biopic "Spencer"
Apparently Americans Were Much Happier 50 Years Ago
5 Black Womxn Artists You Should Be Following On Instagram
Georgia is a journalism student at The New School in Manhattan who loves writing, watching cartoons and intersectional feminism. She is an avid napper and cat lover. Because she is behind on the times, follow her only recently made twitter @georgiagrdodd.
Tags: confederate statues ,	black lives matter ,	Christopher Columbus ,	Robert E. Lee

redpajama_set_name: RedPajamaCommonCrawl


Whether you want to channel Cleopatra or King Tut, dressing up as an Egyptian is a relatively easy Halloween costume to throw together, even at the last minute. With just a few simple materials, you can create a look that mimics the traditional garb of ancient Egypt.
Create a base for your costume. While a long white dress will do well for most women, many men and women will want more coverage from their costume. It is easy to create a unisex caftan. Purchase a piece of white fabric, about one yard long, depending on your height. Fold in half the long way, cut a hole in the fold at the top to stick your head through, and sew up the sides. Women may want to wear a slip underneath the base costume, and men may want to wear a pair of shorts. For more information, see "How to Make a Simple Caftan" in the Resources section.
Accessorize with gold bangles, bracelets, earrings and armbands. Both men and women can wear armbands or thick wristbands or bracelets to complete their look, as pharaohs often showcased their opulence with gold accessories.
Choose footwear. You may choose to go barefoot if you will not be leaving the house; otherwise, gold or white sandals or even flip flops will do in a pinch. Gladiator-style sandals with straps that crisscross up the leg work well for women, whereas men may choose to wear leather sandals, such as Birkenstocks.
Style your hair. Traditionally, Egyptian men/pharaohs would have shaved heads, paired with a chin beard and headdress. Men who do not wish to shave their heads may consider wearing a bald cap, or simply leaving their hair as is. Women should have long dark hair that can be worn straight, pulled into tiny braids or crimped. A long black wig is another option that may work for both sexes, and can be cut to just above the shoulder for men or women.
Apply your makeup. Men should outline their eyes with black kohl eye pencil. Women will do the same, further accentuating the eyes by adding mascara and blue or gold eyeshadow to finish off the dramatic look of the Egyptian eye. Lipstick should be crimson, and the whole face can get a light dusting of bronzing powder. Self-tanning spray, lotion, or bronzing powder should also be added to exposed skin, especially if you have a naturally fair skin tone.


text:
Die   ist eine Lagune auf der Insel  , die zur Region  von Neuseeland gehört.

Geographie
Die   befindet sich im westlichen Teil von  , die rund 14 km südlich der Südküste der Südinsel von Neuseeland zu finden ist. Die Lagune besitzt bei Überflutung einen Zugang im nördlichen Teil zum Meer. Dann nimmt die Lagune Besitz von einer Fläche von rund 13 Hektar ein. Der vom Meer abgeschottete Teil umfasst hingegen eine Fläche von rund 4,3 Hektar. Mit einer Südwest-Nordost-Ausdehnung erstreckt sich die Lagune über eine Länge von rund 880 m und besitzt ein maximale Breite von rund 170 m.

Siehe auch
 Liste der Seen in Neuseeland

Weblinks

Einzelnachweise

Lagune in Neuseeland
Lagune in Australien und Ozeanien

redpajama_set_name: RedPajamaWikipedia


Sink deep into this passionate romance. Kate Pleasance is on her best behavior. Matthew McLeod is certainly not. She really needs the job he's interviewing her for. He totally wants the unexpected candidate in his bed.
But is Kate spying for her famous father? Should Matthew trust her in the least? Join them in the beautiful New Zealand alpine resort of Queenstown, famous for its daredevil extreme sports.
Will Kate dare? And is Matthew the devil he seems to be?
WARNING: Contains one tall tortured man who's super-talented in bedrooms and boardrooms.


text:
David Haye vs Anthony Joshua – rival trainer calls for it
Action Images/Reuters/Suzanne Plunkett
Peter Fury would like to see a tantalising domestic showdown that may produce his nephew's future foe: David Haye vs Anthony Joshua
THERE is no love lost between the camps of Tyson Fury and David Haye. Equally, there is at least a rivalry with the Furys and Anthony Joshua – Hughie Fury would like to face "AJ" while the latter aspires to one day challenge unified world heavyweight king Fury. Peter Fury – Hughie's father and Tyson's uncle – is not bothered about the global champ meeting either Joshua or Haye anytime soon, but would like to see David Haye vs Anthony Joshua. He told Boxing News this after reflecting on Haye's comeback win over Mark de Mori.
"Haye did nothing wrong but fought a dummy," Fury explained. "That guy was not a boxer, I don't know where they found him, but I think a good fight is Haye and Joshua. They can both punch, Joshua is younger but Haye has that experience."
by Matt Christie
Matt Christie has been hooked on boxing since he picked up a copy of A Pictorial History of Boxing in 1985 as a seven-year-old. He worked as producer of KOTV from 2006 until he joined Boxing News in 2010. He has been honoured by the Sports Journalists' Association and the Boxing Writers Association of America. In 2017 he won the Sam Kellerman Media Award in New York and The Reg Gutteridge Annual Award in London two years later. He has also written about boxing for The Guardian and the Daily Mirror and has been used by the BBC, Sky Sports, BT Sport, HBO Boxing and Talk Sport as a boxing analyst. He has been the editor of Boxing News since 2015.


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Introductory Academic Program
The Introductory Academic Program (IAP) is a mandatory program held at all participating institutions across Australia. The IAP is designed to help you make the transition to a new educational institution and a new cultural setting, offering an ideal beginning to your studies and a way to establish contact with the people who can help you throughout your degree. You will be automatically registered once you accept your offer, and are required to attend. During the program, you will be introduced to your new environment, assisted with enrolment, given an explanation of academic expectations, and offered the opportunity to practise skills in academic writing, approaches to study, computing and the effective use of the library. For more information about the IAP at your university, please contact your Student Contact Officer.
Orientation Week (O-Week) is a university-wide celebration for international and domestic students who are beginning their studies at university. It offers you social and sporting opportunities, study and support sessions, and events to help familiarise you with the campus, meet people and prepare for study. O-Week is held in the week before the start of the first and second semesters, and attendance is compulsory for faculty-based programs. O-Week is a great opportunity to sign up to various university clubs and societies, meet other students, make new friends and settle into life at your university.
For a list of African community associations by state and African student associations at participating institutions, visit http://afsaap.org.au/resources/the-australia-awards-for-africa-scholarships.
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Thank you very much for using Roland products. We appreciate your support and remain committed to serving you with exceptional products that meet the highest standards for quality, performance and safety.
1.On a gradual basis, Roland will expand its labeling for products according to Chinese RoHS guidelines. New product labels with key safety information will be introduced on a global basis. These labels have no impact on product performance, quality or packaging dimensions.
2.Roland will record the presence or absence of the six substances specified by the Chinese RoHS directive in the user's manual of each product.
3.Roland will add a recycle mark to all product packaging.
For most products, this label will be attached directly to the product near the serial plate. If the product is not large enough to accommodate the label, it will appear in the user's manual.
The number in the center of the label represents the product's environmentally safe period.
The circular arrows indicate that the product is recyclable. In this example, the product is environmentally safe for 10 years.
*The "environmentally safe period" is defined as the time period in which the product will not seriously impact the environment when used under ordinary conditions. The environmentally safe period does not represent or guarantee the product lifespan.
The sample chart below indicates the presence or absence of these substances.
*An "o" mark is present wherever the levels of the substance are within the guidelines established by the Chinese RoHS Directive. An "x" mark indicates that the substance in question exceeds the limits stipulated by the directive.
**The sample chart above is printed in English; Actual safety charts in the user's manual will be printed in Chinese.
The following recycle mark will appear on external product packaging which is comprised of recyclable corrugated board.


Get ready to show off those arms and get some laughs, from these funny men's graphic t-shirts of course! Your arms are fine.
As if we couldn't love summer any more than we do, these snappy graphic tees just walked into our lives and we can't get enough of them. Personally, we love this kind of t-shirt because it really makes a statement and, well, has people think we're a bit funnier than we actually are. Plus, who doesn't love an easy ice-breaker when you're trying to spark up a conversation with someone you don't really know (hello people we don't know on our daily run!).
You can always sport these tees with your favorite pair of jeans, casual chinos and, of course, shorts. We typically also sleep in ours and wear them to the gym because we're that kind of dumpster fire. We don't judge. If you're ready to update your t-shirt collection this season because, well, your old ones are getting a little too dingy, then check out our picks for some of the coolest graphic t-shirts for men (so far) in 2019.
Because who doesn't love all things Bob Ross?! This t-shirt comes in sky blue, features the giant head of PBS painter Bob Ross, and makes us content because we're thinking of his happy little trees all over the place.
BOOM! That cute little hedgehog certainly just told you! This tee comes in cream and is made from ultra-soft cotton. Take that comfort!
Because you totally love running, but at the same time, you totally hate running. This t-shirt will get a few smiles and head turns as you run that path every other day (or, let's face it, once a week).
Because, let's face it, the best part of brunch is the drinks. Whether you go for the classic mimosa, the Bloody Mary, or you just crush beers, this is the tee for you. Pick up one for your brunching buddies too! Check it out in black, white, grey, and more.
Because at the end of the day it's all about 'merica! Enjoy this tee in a bunch of different colors, but the white really makes the red and blue pop, especially with those aviator sunglasses!
Because who doesn't need a daily reminder to stay zen? We love, love, love this grass green t-shirt, but it also comes in some other cool options like light blue, white, and grey, and light pink.
Innuendos all over the place! Go back to your tripy place with this 'Don't Trip' subtle tie-dye t-shirt in grey and blue hues.
Because we totally stand by our 'dad bod' that, thanks to recent pop-culture events, is totally on-trend. Finally, it's happened! This one comes in a bunch of other color options too like navy blue, black, slate grey, heather, and more.
Because who doesn't love a little extra attention at the gym? This tee has the font printed backwards so, you know, when you're looking in the mirror it actually says "Gym #Selfie" and will make your, you know, "Selfie" get extra likes on the 'Gram.
For everyone who ever remembers this being the absolute best part of summer and lunch. Who didn't try to con their way into getting a second or third blue drink each time?? Guilty as charged.
Because we love donuts and we don't really give AF about anything else. Check it out in other cool colors like blue, yellow, grey, and light pink.
We love all things 'sloth.' This slow riding sloth t-shirt comes in white and is made from super soft cotton so wear it to bed or just out for the day. Or both. We won't tell.
Because every now and then we need a little motivational reminder.
The pineapple is the staple icon of any summer graphic tee. Well, we made part of that up, but it surely dresses up this plain white t-shirt, don't ya think?
Because real guys show their support. This 'Feminist' t-shirt comes in light heather grey, blue, green, and dark grey. Make a statement STAT.
Because these are your real best friends and, well, they'll never turn their back on you. This one also comes in other colors like white, black, yellow, grey, and pink.


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Midwest District Review
It has been some time since we have sent out a district newsletter to remind you of the good things that have been happening around the Midwest District. As you read the following news items, please be praying that God will continue to accomplish His purposes in and through the people of the Midwest District of the Missionary Church!
Another wonderful summer of camping ministries!
Reports from all of the district's three summer camps have been exciting. The summer began with Junior Camp for kids age 9-12. The week's theme was "Wanted," and the activities focused on the old west. Speakers Sam and Mandy Petro encouraged the young campers with the truth that God truly loves them and wants them to be a part of His family. It was a week that the campers and the wonderful camp staff will not soon forget!
Youth camp was held in mid-July and once again the teens met at Timber Lakes. Throughout the week energy and enthusiasm were high! Along with the traditional times of fun and play, the services with worship leader Elias Martinez and speaker Anthony Adams challenged the teens to be RELENTLESS in their love for and service to the Lord.
To complete the 2019 schedule of district camps, 165 people gathered at Timber Lakes at the end of July for the annual Midwest District Family Camp. With the theme, "Us and Them," speaker Sean Azzaro gave helpful insights regarding seeing people like Jesus did. Inspiring times of worship and family devotions, great Bible studies and age-related ministries, reports from Bulgaria by the Hawkins family, along with wonderful food and fellowship, made this year's Family Camp one of the best!
2020 Midwest District Family Camp
Circle the dates for next summer's Family Camp (July 28-August 2) and plan to be a part of a wonderful family vacation with life-changing spiritual purpose! The speaker will be Gary Aupperle, Senior Pastor at Hope Missionary Church in Bluffon, IN. Registration information will be available by no later than January 1st -- register early to secure the housing of your choice.
50th Anniversary Celebration in Freeman
The following account was written with information taken from an article by Tim L Waltner, Contributing Editor of the FREEMAN COURIER regarding the 50th anniversary activities at the Freeman Missionary Church.
A GOLDEN SUNDAY
On September 22nd, close to 180 members, former members, and friends of the Freeman Missionary Church came together to celebrate 50 years of congregational life. It was, as Pastor Stephen Roussos described, "a family reunion" that focused on the faithfulness of God.
The six-plus hours of the event began with a fellowship time that allowed those attending to renew and strengthen acquaintances. At 10:00 AM a worship service filled the sanctuary with music, prayer, sharing and a message by District Superintendent, Andy Petro, taken from Joshua 24.
Following the worship service, a catered meal was served and then an afternoon service was held that included a 13-minute slide show featuring scores of historic photos ranging from the construction of the church to events and ministries throughout the years. That service concluded with a concert by an abbreviated version of the Tooley Family Singers (Jim and Dee Tooley, their son Dennis, and their granddaughter Jonie)
Several of the people who shared during the anniversary celebration were Jan Secor, wife of the late Walter Ortmann, who founded the church. She told the congregation "I've come home. I can't believe it's been 50 years since we started this church. It thrills my soul that God called us to build this church." Stephen Ortmann, who was in grade school when his family moved to Freeman, also shared and said "I have so many good memories of the people who helped shape my life. The influence from many people of the Freeman Missionary Church has been profound - you've made a huge impact on my life. Thank you for your faithfulness." Former Pastor Lloyd George, who now lives in Indiana, was unable to attend but shared his greetings via a recorded video.
Other speakers included Jim Horning, who has been a member since he moved to Freeman in 1977, and Rick Wuertz, who called himself the "new kid on the block" since he has only been attending the church for about 9 years. Both men testified to God's love, grace and faithfulness throughout the history of the church and to the ministry and impact that the church has had in their own personal lives.
At the conclusion of the anniversary celebration, Tena Preheim, one of the few remaining charter members of the Freeman Missionary Church, commented that "It was a good day." Pastor Roussos echoed those thoughts when he said "It truly was a great day. The positive comments I received, saying it was meaningful, really affirmed that." He also noted that the emphasis of the day was right on target - "God was number one; He received all the credit and glory."
The June 26, 1969 issue of the Freeman COURIER reported that "Rev. Walter L. Ortmann and family of Bedford, Iowa, will be moving to Freeman on July 1 to give leadership in the establishing of a fundamental, evangelical, holiness church for the Mid-West District of the Missionary Church." Services began July 6 at Freeman High School but by the end of the month, the new church was meeting in the vacated City Cafe building on Main Street. Subsequently it met in Pioneer Hall and then in a tool shed located on property along Sixth Street that would become home to the permanent structure today known as the Freeman Missionary Church. A ground breaking service was held April 30, 1972, and the dedication service for the completed structure was held in October 1978. Three pastors followed Ortmann, who died in 1984: Dudley Horning (1984-86), Lloyd George (1986-91) and Stephen Roussos, who arrived in 1991 and has the longest tenure, with 28 years of service as pastor.
Anthony Adams To Be Ordained
At the request of the Church Board of First Missionary Church in Newton, KS; and following an interview with a committee of area pastors, the District Executive Board, has approved Anthony Adams ordination within the Missionary Church. Pastor Anthony is serving as part of the pastoral team at First Missionary Church, assisting in preaching, leading worship, serving as a Life Group leader, and helping with youth ministry. He is active in discipleship within the church, at work, and in his neighborhood. Speaking on behalf of the church board, Pastor Ray Adams stated, "Pastor Anthony and his wife Whitney have been a blessing to the church family, and their lives demonstrate spiritual fruit and giftings that clearly show God's hand upon their ministry." The date of Anthony's ordination service has yet to be determined.
Bryan Conrad To Be Ordained
Following the recent Pastors' and Wives' Retreat in Omaha, NE, an appointed committee of pastors and their wives interviewed Bryan and Faith Conrad in preparation for Bryan's ordination. In that interview, Pastor Bryan reaffirmed God's call on his life and his commitment to pastoral ministry. He is currently serving as the Senior Pastor of Grace Missionary Church in Milford, NE. Prior to that he served alongside Pastor Brian Bontrager at Hillside Missionary Church in Clarinda, IA. All of this came after what Bryan had initially thought would be a life-long career in law enforcement, but obviously God had other plans! Following the interview, the committee wholeheartedly recommended that the District Executive Board grant Bryan's ordination which they did in a unanimous vote. His ordination service will be held at Grace Missionary Church in Milford. NE on Sunday, November 17th.
2019 Pastors' and Wives' Retreat
Omaha, NE was the setting for the 2019 Pastors' and Wives' Retreat which was held on September 24-26. Thirteen pastoral couples were able to attend either part or all of the retreat which featured times of laughter and fellowship, times of devotion and prayer, and time just to relax and renew relationships. Thanks to Bob and Dawn Thompson for planning and directing this wonderful retreat, and thanks to the churches who encouraged and enabled their pastor(s) to attend.
Eating together
Sharing communion
Work continues on the new sanctuary at First Missionary Church in Newton, KS. Although they've had some weather-related delays, construction is actually ahead of schedule and the prospect of the congregation being able to begin holding services in that facility is encouraging.
DISTRICT CONTACT INFORMATION
6920 Willow Ave., Lincoln, NE 68507 andypetro.mwd@gmail.com 402-840-5056 themidwestdistrict.com
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Just sitting around Vwaza's main camp will bring plenty of animal sightings, for it overlooks Lake Kazuni, which is inhabited by hundreds of hippos. On most days you'll see crocodiles lying out in the sun, hippos popping their heads out of the water, and a steady parade of animals coming down for a drink.
A good network of drivable tracks in the reserve is navigable in a 4WD or high-clearance vehicle; if you're in a 2WD, ask at the gate or camp for advice on the condition of the tracks. The best driving route is along the southern edge of the reserve, parallel to the river, heading to Zoro Pools.