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\dateline{Jul 13, 2017}{Jan 3, 2018}{Aug 10, 2018}

\MSC{51E24, 52B11, 20F05}

\Copyright{  The authors. Released under the CC BY-ND license (International 4.0).}

\title{Hypertopes with tetrahedral diagram}


\author{ Domenico Catalano\qquad M. Elisa Fernandes  \\
\small Center for Research and Development in Mathematics and Applications\\[-0.8ex]
\small Department of Mathematics\\[-0.8ex] 
\small University of Aveiro, Portugal\\
\small\tt \{domenico, maria.elisa\}@ua.pt\\
\and
Isabel Hubard\\
\small Instituto de Matem\'aticas\\[-0.8ex]
\small Universidad Nacional Aut\'onoma de M\'exico\\[-0.8ex] 
\small Ciudad Universitaria,
\small M\'exico D.F.  04510\\
\small\tt isahubard@im.unam.mx\\
\and
Dimitri Leemans\\
\small  D\'epartement de Math\'ematique\\[-0.8ex]
\small C.P. 216 - Alg\`ebre et Combinatoire, Universit\'e Libre de Bruxelles\\[-0.8ex]
\small Boulevard du Triomphe, 1050 Brussels, Belgium\\
\small\tt dleemans@ulb.ac.be}


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\maketitle

\begin{abstract}
%A hypertope is a geometry generalizing a polytope.
In this paper we construct an infinite family of hypertopes of rank four having the complete graph $K_4$ as diagram. Their group of rotational symmetries is isomorphic to $PSL(2,q)$. It turns out some elements of this family are regular hypertopes and some are chiral.
Moreover, we show that the chiral ones have both improper and proper correlations simultaneously.
\end{abstract}



\section{Introduction}

Abstract polytopes generalize (the face lattice of) convex polytopes to combinatorial structures. The main interest of the theory of abstract polytopes has been the study of their symmetries. Hence, highly symmetric polytopes (in particular those regular and chiral), together with their automorphism groups, are the most studied ones.
A polytope can be regarded as a thin residually connected geometry with linear diagram. The concept of hypertope was introduced in \cite{FLW1} and generalizes the concept of a polytope by dropping the linear condition on the diagram. This generalization was made in such a way that the concept of chirality can be extended to hypertopes.
As it is the case with maps and polytopes, we are interested in understanding chiral hypertopes, and finding examples of them can be the first step to achieve this goal.

In the 1970's Branko Gr\"unbaum considered rank 4 polytopes that are locally toroidal, meaning that all their facets and vertex figures are either spherical or toroidal, and not all of them are spherical. In \cite{ARP} an almost complete answer to Gr\"unbaum's problem, the classification of  rank 4 regular locally toroidal polytopes, is given. In this paper we follow \cite{FLW1} and generalize the concept of a polytope to that of a hypertope, and study some interesting locally toroidal ones.

In \cite{FLW2} some examples of chiral hypertopes of rank 4 with certain diagrams are given, satisfying that their residues of rank 3 are either spherical or toroidal.
%Having in mind this kind of hypertopes, particularly those having as residues of rank 3 toroidal hypermaps of type $(3,3,3)$,
In this paper we continue with the study of locally toroidal 4-hypertopes, in the sense that their residues of rank $3$ are toroidal.
Using $PSL(2,q)$ as their rotational subgroups,
we found an infinite family of hypertopes that  contains both regular and chiral hypertopes.
Moreover, the chiral ones have the very interesting property that they admit both proper and improper correlations, a feature that is known to be impossible for  polytopes (see~\cite[Lemma 3.1]{HW2005}).

We construct these hypertopes as coset geometries $\Gamma=\left(G; \{G_0,G_1,G_2,G_3\}\right)$  where $G$ is $PSL(2,q)$ with $q=p$ or $q=p^2$ for certain primes $p$. In these geometries two maximal parabolic subgroups are alternating groups $A_4$ and the other two are $E_q:C_3$.  Therefore  $3$ must divide $q-1$ for our construction to work, and there is a third root of unity $e$ in the field of order $q$. In the case $q=p^2$ we require that $e$ is not in the subfield of order $p$, which is equivalent to requiring that $p\equiv 2\mod 3$. We get an infinite family of regular hypertopes when $q=p^2$ and an infinite family of chiral hypertopes when $q=p$.
More precisely  in the case $q=p^2$ two residues are hypermaps of type $(3,3,3)_{(2,0)}$ and the other two are  hypermaps of type $(3,3,3)_{(p,0)}$;  in the case $q=p$ two residues are hypermaps $(3,3,3)_{(2,0)}$ and the other two are both chiral hypermaps of types $(3,3,3)_{(1,e)}$ and $(3,3,3)_{(e,1)}$, respectively.

The paper is organised as follows.
In Section~\ref{prelim}, we give the definitions and notation needed to understand this paper.
In Section~\ref{SC+}, we construct $PSL(2,q)$, with $q\in \{p, p^2\}$ and $p$ a prime, as a $C^+$-group.
In Section~\ref{section4}, we show that the $C^+$-groups obtained in the previous section give hypertopes.
In Section~\ref{sec:symm} we show that the hypertopes obtained are either regular or chiral.
%, and that whenever they are chiral, they have both proper and improper correlations simultaneously.
In Section~\ref{section5}, we give a geometric description of the hypertopes we constructed, in terms of objects of the projective line $PG(1,q)$.
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\section{Preliminaries}\label{prelim}
\subsection{Incidence geometries}
Following~\cite{BuekCohen}, an {\it incidence system} $\Gamma := (X, *, t, I)$ is a 4-tuple such that
\begin{itemize}
\item $X$ is a set whose elements are called the {\it elements} of $\Gamma$;
\item $I$ is a set whose elements are called the {\it types} of $\Gamma$;
\item $t:X\rightarrow I$ is a {\it type function}, associating to each element $x\in X$ a type $t(x)\in I$;
\item $*$ is a binary relation on $X$ called {\em incidence}, that is reflexive, symmetric and such that for all $x,y\in X$, if $x*y$ and $t(x) = t(y)$ then $x=y$.
\end{itemize}
The {\it incidence graph} of $\Gamma$ is the graph whose vertex set is $X$ and where two vertices are joined provided the corresponding elements of $\Gamma$ are incident.
A {\it flag} is a clique of the incidence graph of $\Gamma$.
The {\it type} of a flag $F$ is $\{t(x) : x \in F\}$.
A {\it chamber} is a flag of type~$I$.

An element $x$ is {\em incident} to a flag $F$ and we write $x*F$ for that, when $x$ is incident to all elements of $F$.
An incidence system $\Gamma$ is a {\it geometry} or {\it incidence geometry} if  every flag of $\Gamma$ is contained in a chamber.
The {\it rank} of $\Gamma$ is the number of types of $\Gamma$, namely the cardinality of $I$.

Let $\Gamma:= (X, *, t, I)$ be an incidence geometry and $F$ a flag of $\Gamma$.
The {\em residue} of $F$ in $\Gamma$ is the incidence geometry $\Gamma_F := (X_F, *_F, t_F, I_F)$ where
\begin{itemize}
\item $X_F := \{ x \in X : x * F, x \not\in F\}$;
\item $I_F := I \setminus t(F)$;
\item $t_F$ and $*_F$ are the restrictions of $t$ and $*$ to $X_F$ and $I_F$.
\end{itemize}
An incidence system $\Gamma$ is {\em connected} if its incidence graph is connected; $\Gamma$ is {\em residually connected} when each residue of rank at least two of $\Gamma$ (including itself) has a connected incidence graph; $\Gamma$ is called {\it thin} (resp. {\it firm}) when every residue of rank one of $\Gamma$ contains exactly two (resp. at least two) elements.
As in~\cite{FLW1}, we say that a {\em hypertope} is a thin incidence geometry which is residually connected.

Let $\Gamma:=(X,*, t,I)$ be an incidence system.
An {\em automorphism} of $\Gamma$ is a mapping
$\alpha:(X,I)\rightarrow (X,I):(x,t(x)) \mapsto (\alpha(x),t(\alpha(x)))$
where
\begin{itemize}
\item $\alpha$ is a bijection on $X$;
\item for each $x$, $y\in X$, $x*y$ if and only if $\alpha(x)*\alpha(y)$;
\item for each $x$, $y\in X$, $t(x)=t(y)$ if and only if $t(\alpha(x))=t(\alpha(y))$.
\end{itemize}
Note that $\alpha$ induces a bijection on $I$. When $t(x)=i$ we say that $x$ is an element of \emph{type} $i$, or equivalently, that $x$ is an \emph{$i$-element}.
The set of automorphisms of $\Gamma$ is a group denoted by $Aut(\Gamma)$.

An automorphism $\alpha$ of $\Gamma$ is called {\it type preserving} when for each $x\in X$, $t(\alpha(x))=t(x)$ (i.e. $\alpha$ maps each element on an element of the same type).
The set of type-preserving automorphisms of $\Gamma$ is a group denoted by $Aut_I(\Gamma)$ and obviously $Aut_I(\Gamma)\leq Aut(\Gamma)$.

A {\em correlation} is a non-type-preserving automorphism, that is an element of $Aut(\Gamma)\setminus Aut_I(\Gamma)$.
A {\em duality} is correlation that induces an involutory permutation on $I$.

An incidence geometry $\Gamma$ is {\em chamber-transitive} if $Aut_I(\Gamma)$ is transitive on all chambers of $\Gamma$.
Finally, an incidence geometry $\Gamma$ is {\em regular} if $Aut_I(\Gamma)$ acts regularly on the chambers, that is, the action is semi-regular (free) and transitive.

Observe that chamber-transitivity implies flag-transitivity, that is, for each $J\subseteq I$, there is a unique orbits on the flags of type $J$ under the action of $Aut(\Gamma)$.

Sometimes, when $Aut(\Gamma)$ is not transitive on the chambers of $\Gamma$, it has two orbits.
We are also interested in those hypertopes having two orbits with an extra condition.
Two chambers $C$ and $C'$ of a thin incidence geometry of rank $r$ are called {\it $i$-adjacent} if $C$ and $C'$ differ only in their $i$-elements. We then denote $C'$ by $C^i$.
Let $\Gamma(X,*,t,I)$ be a thin incidence geometry.
We say that $\Gamma$ is {\it chiral} if
$Aut_I(\Gamma)$ has two orbits on the chambers of $\Gamma$ such that any two adjacent chambers lie in distinct orbits.
Moreover, if $\Gamma$ is residually connected, we call $\Gamma$ a {\it chiral hypertope}.

When $\Gamma$ is a chiral hypertope, if $Aut(\Gamma) \neq Aut_I(\Gamma)$, correlations may either interchange the two orbits or preserve them. A correlation that interchanges the two orbits is said to be {\em improper} and a correlation that preserves them is said to be {\em proper}.

The following proposition shows how to construct an incident geometry starting from a group.
\begin{proposition}(Tits Algorithm, 1956)~\cite{Tits61}\label{tits}
Let $n$ be a positive integer
and $I:= \{1,\ldots ,n\}$.
Let $G$ be a group together with a family of subgroups ($G_i$)$_{i \in I}$, $X$ the set consisting of all cosets $G_ig$ with $g \in G$ and $i \in I$, and $t : X \rightarrow I$ defined by $t(G_ig) = i$.
Define an incidence relation $*$ on $X\times X$ by :
\begin{center}
$G_ig_1 * G_jg_2$ iff $G_ig_1 \cap G_jg_2 \neq \varnothing$.
\end{center}
Then the 4-tuple $\Gamma := (X, *, t, I)$ is an incidence system having a chamber.
Moreover, the group $G$ acts by right multiplication as an automorphism group on $\Gamma$.
Finally, the group $G$ is transitive on the flags of rank less than 3.
\end{proposition}

When a geometry $\Gamma$ is constructed using the proposition above, we denote it by $\Gamma(G; (G_i)_{i\in I})$ and call it a {\em coset geometry}.
The subgroups $(G_i)_{i\in I}$ are called the {\it maximal parabolic subgroups}.

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\subsection{Coset geometries from C$^+$-groups}
As in~\cite{FLW1}, consider a pair $(G^+,R)$ with $G^+$ being a group and $R:=\{\alpha_1, \ldots, \alpha_{r-1}\}$ a set of generators of $G^+$.
Define $\alpha_0:=1_{G^+}$ and $\alpha_{ij} := \alpha_i^{-1}\alpha_j$ for all $i,j\in I:=\{0,\ldots,r-1\}$.
Let $G^+_J:= \langle \alpha_{ij} \mid i,j \in J\rangle$ for $J\subseteq I$.

If the pair $(G^+,R)$ satisfies the following condition called the {\em intersection property} IP$^+$, we say that $(G^+,R)$ is a {\em C$^+$-group}.
$$\mbox{$G^+_J \cap G^+_K = G^+_{J\cap K},$}$$
for all $J,K \subseteq I$, with $|J|, |K| \geq 2$.
The following construction produces an incidence system from a $C^+$-group.

\begin{construction}\label{hyper}\cite{FLW1}
Let $R=\{\alpha_1, \ldots, \alpha_{r-1}\}$ be an independent generating set of the group $G^+$.
Define $G_i := \langle \alpha_j | j \neq i \rangle$ for $i=1, \ldots, r-1$ and $G_0 := \langle \alpha_1^{-1}\alpha_j | j \geq 2 \rangle$.
The coset geometry $\Gamma(G^+,R) := \Gamma(G^+; (G_i)_{i\in \{0,\ldots,r-1\}})$ constructed using Tits' algorithm (see Proposition~\ref{tits}) is the incidence system associated to the pair $(G^+,R)$.
\end{construction}
We denote $\Gamma(G^+,R)$ simply by $\Gamma$ whenever $G^+$ and $R$ are clear from the context. We set $I:=\{0,\ldots,r-1\}$, where $r:=|R|+1$.

It is convenient to represent $(G^+, R)$ by a  graph $\mathcal{B}$ with $r$ vertices which we call the {\it B-diagram} of $(G^+,R)$. %and denote by $\mathcal{B}(G^+,R)$.
The vertex set of $\mathcal{B}$ is the set $\{\alpha_0, \ldots, \alpha_{r-1}\}$.
The edges $\{\alpha_i,\alpha_j\}$ of this graph are labelled by $o(\alpha_i^{-1}\alpha_j)=o(\alpha_j^{-1}\alpha_i)$.
We take the convention of dropping an edge if its label is 2 and of not writing the label if it is 3.
Vertices of $\mathcal{B}$ are represented by small circles.
Finally, we sometimes attach to each vertex $\alpha_i$ the corresponding subgroup $G_i$ defined in Construction~\ref{hyper}.

Observe that, thanks to Proposition~\ref{tits} that ensures that $G^+$ is transitive on the flags of rank less than 3,  we not only know the number of elements of type $i$ for every $i$, that is the index of $G_i$ in $G^+$, but also the number of elements of type $j$ incident to a given element of type $i$, that is the index of $G_i \cap G_j$ in $G_i$.

The residue of the element $G_i$ is the coset geometry $\Gamma(G_i, (G_i \cap G_j)_{j \in I\backslash\{i\}})$.
Its diagram is obtained by removing the vertex $\alpha_i$ from the diagram of $\Gamma(G^+; (G_i)_{i\in I})$.

The coset geometry $\Gamma(G^+,R)$ gives an incidence system using Proposition~\ref{tits}.
In what follows we prove that any such coset geometry has a connected incidence graph if its rank is at least 3.
%coset geometry, of rank at least 3, constructed from a C$^+$-group has a connected incident graph.

\begin{proposition}\label{connected}
%For $r\geq 3$,
If $|R|\geq 2$, then $\Gamma(G^+,R)$ has a connected incidence graph.
\end{proposition}
\begin{proof}
%By construction, $\Gamma(G^+,R) := \Gamma(G^+; (G_i)_{i\in \red{I}})$.
%If $(G^+,R)$ is a $C^+$-group, we have that $G^+ = \langle G_i, G_j\rangle$. This implies that the subgraphs of the incidence graph with edges of type $i$ and $j$ are connected. Hence the whole incidence graph must be connected.
%
%Old proof of Proposition 2.2
As for any $g\in G^+$, $\{G_ig\,|\, i\in I\}$ is a set of mutually incident elements of $\Gamma$, it is sufficient to prove that $G_i$ and $G_ig$ are in the same connected component of the incidence graph (of $\Gamma$) for every $g\in G^+$ and every $i\in I$. As $G^+=\langle \alpha_i\,|\, i\in I\rangle$, we can assume $g\in\{\alpha_i:i\in I\}$.
%prove that  $G_i$ and $G_i\alpha_j$ are in the same component of the incidence graph.
If $i\neq 0$, then $G_i\cap G_i\alpha_j=G_i$ for any $j\in I\setminus\{i\}$. Moreover, $G_i*G_k$ and $G_k*G_i\alpha_i$ for any $k\in I\setminus\{0,i\}$, which is a non-empty set since $|I|-2=|R|-1\geq 1$. If $i=0$, then $G_0*G_k$ and $G_k*G_0\alpha_j$ for any $k\in I\setminus\{1,j\}$.
%When $j\neq i$ and $i\neq 0$, $\alpha_k \in G_i \cap G_i\alpha_j$ for every $k\neq i, j$.
%When  $i=0$ and $j\neq 0$, consider $k\notin\{0,j\}$. We have that $G_0\alpha_j*G_k$ and $G_k*G_0$, thus $G_0$ and $G_0\alpha_j$ are in the same component.
%Finally, let $i=j\neq 0$. We have that  $G_i\alpha_i*G_k$ for $k\notin\{0,i\}$ and $G_k*G_i$, which proves that $G_i$ and $G_i\alpha_i$ are in the same component.
\end{proof}

Although the incidence geometry $\Gamma$ has a connected incidence graph, it need not be residually connected. Moreover, $\Gamma$ might not be a thin geometry, and hence $\Gamma$ need not be a hypertope. Furthermore, in general $\Gamma$ might not be transitive on flags of rank $3$, and it might have many orbits of chambers. In the construction we shall give in the next section, the geometry that we obtain will in fact be a hypertope and have only two orbits of chambers.
The following theorem will help us to decide if a hypertope is regular of chiral.
%shows how to test whether a coset geometry obtained using  Construction~\ref{hyper} is regular or chiral.

\begin{theorem}\label{chiraltest}\cite{FLW1}
Let $(G^+,R)$ be a $C^+$-group.
Let $\Gamma := \Gamma(G^+,R)$ be the coset geometry associated to $(G^+,R)$ using Construction~\ref{hyper}.
If $\,\Gamma$ is a hypertope and $G^+$ has two orbits on the set of chambers of $\,\Gamma$, then
$\Gamma$ is chiral if and only if there is no automorphism of $G^+$ that inverts all the elements of $R$.
Otherwise, there exists an automorphism $\sigma\in Aut(G^+)$ that inverts all the elements of $R$ and the group $G^+$ extended by $\sigma$ is regular on $\Gamma$.
\end{theorem}

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\subsection{Toroidal hypermaps of type $(3,3,3)_{(a,b)}$}
The hypertopes we shall construct in Section~\ref{section4} have residues of rank $3$ with a known structure, namely, they are toroidal hypermaps of type $(3,3,3)$.

If $\Gamma$ is a hypertope of rank three, then it is also a hypermap.
A toroidal hypertope (or toroidal hypermap) is either a map or a hypermap embeddable on the torus. The toroidal (regular or chiral) hypertopes of rank 3 are divided into the following families: the toroidal maps $\{3,6\}_{(a,b)}$, $\{6,3\}_{(a,b)}$, $\{4,4\}_{(a,b)}$, and the hypermaps $(3,3,3)_{(a,b)}$
with $(a,b)\neq (1,1)$.
Note that the hypermap  $(3,3,3)_{(a,b)}$ is obtained from the toroidal map $\{6,3\}_{(a,b)}$ by doubling the fundamental region. Indeed as  $\{6,3\}_{(a,b)}$ is bipartite it is possible  to take one monochromatric set of vertices to be the hyperedges of the hypermap $(3,3,3)_{(a,b)}$  (see \cite{Walsh}).

The rotation subgroup $G$ of the automorphism group of a rank three toroidal hypermap is as follows for some integers $a$ and $b$.
\begin{equation}\label{eq:toroidal}
\langle x,y\,|\, x^3, y^3, (x^{-1}y)^3, (xy^{-1}x)^a(xy)^b\}
\end{equation}
This hypermap is denoted by $(3,3,3)_{(a,b)}$.
The above presentation readily shows that such a hypermap has a B-diagram that is a triangle with no numbers on the edges.
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\subsection{$PSL(2,q)$ acting on the projective line}\label{sub:action}
For a prime power $q=p^n$ let $GF(q)^\ast$ be the multiplicative group of the Galois field $GF(q)$ on $q$ elements.

A {\it primitive element} of $GF(q)$ is a generator
of the multiplicative group $GF(q)^\ast$.

For any positive integer $k<q$, a solution of $x^k=1$ which is not a solution of $x^j=1$ for any $j\in\{1,\dots,k-1\}$ is called a \emph{$k$th primitive root of unity}.
If $i$ is a $k$th primitive root of unity, then $k$ divides $q-1$ and $\,\sum\limits_{j=0}^{k-1}i^j=0$.


Let $V$ be the 2-dimensional vector space $GF(q)^2$ over $GF(q)$. Consider the relation $\sim$ in $V\setminus\{(0,0)\}$ defined as follows.
\[
(x_0,x_1)\sim(y_0,y_1)\mbox{ if and only if }(y_0,y_1)=\lambda(x_0,x_1),\,\mbox{for some }\lambda\in GF(q)^\ast.
\]
The \emph{projective line} $PG(1,q)$ is the set of equivalence classes  $V\setminus\{(0,0)\}/\sim$. The elements of the projective line $[x_0,x_1]$ can be identified with their non-homogeneous coordinates by the following bijection where $i$ is a $(q-1)$th root of unity.
\begin{eqnarray*}
PG(1,q)&\rightarrow&\{0,1, i, i^2,\dots,i^{q-2},\infty\}\,,\\
 \left[x_0,x_1\right]&\mapsto&\left\{\begin{array}{ll}\frac{x_0}{x_1} & \mbox{ if }x_1\neq 0,\\ \infty & \mbox{ if }x_1=0\,.
 \end{array}\right.
\end{eqnarray*}

Consider the special linear group $SL(2,q)=\{A\in GL(2,q)\mid\det(A)=1\}$ and denote by $Id$ its identity matrix. As $PSL(2,q)=SL(2,q)/Z$
where $Z=\{\pm Id\}$, the elements of $PSL(2,q)$ can be seen as unordered pairs $\pm A$ with $A\in SL(2,q)$. Observe that in characteristic 2, $|Z| = 1$.
For convenience, we shall denote $PSL(2,q)$ by $G^+$. An element of $G^+$ will be given by one of its two representative elements in $SL(2,q)$. Consequently equalities are to be taken modulo $Z=\{\pm Id\}$.
For $A=\m{a}{b}{c}{d}\in SL(2,q)$ and $z\in PG(1,q)$, consider the correspondence $$\varphi:\,G^+\times PG(1,q)\rightarrow PG(1,q),\ (\pm A,z)\mapsto Az$$defined as follows.
\[
Az=\frac{az+b}{cz+d}.
\]
As $\ -Az=Az\ $, $\varphi$ is well-defined and gives an action of $G^+$ on $PG(1,q)$.


\begin{lemma}\cite[Lemma 5.3]{SW1994}\label{sw94}
Let $q$ be a power of a prime. Then $SL(2,q)$ is generated by the elementary matrices of the form
\[
\m{1}{0}{c}{1}\mbox{ and }\m{1}{b}{0}{1}\mbox{ with }b, c\in GF(q).
\]
\end{lemma}


\section{A C$^+$-group for $PSL(2,q)$}\label{SC+}

Let $q\in\{p, p^2\}$ for a prime number $p$ with $3$ being a divisor of $q-1$, and let $e\in GF(q)$ be a third primitive root of unity. In the case $q=p^2$ we also assume that 3 is not a divisor of $p-1$ so that
$$GF(p^2)\cong\{a+eb:a,b\in GF(p)\}\,.$$
In this section we prove that $PSL(2,q)$ is a C$^+$-group with the B-diagram of Figure~\ref{tetra} (where the labels of the corners are the groups corresponding to each rank 3 residue).
\begin{figure}
$$\xymatrix@-1.5pc{
E_q:C_3 & *{\circ} \ar@{-}[rr] \ar@{-}[rrdd] \ar@{-}[dd]&& *{\circ} \ar@{-}[dd]\ar@{-}[lldd]& A_4\\
&&&&&&\\
E_q:C_3 &*{\circ}  \ar@{-}[rr] && *{\circ}&A_4}$$
\caption{Tetrahedral B-diagram for $PSL(2,q)$.}\label{tetra}
\end{figure}

In order to construct this C$^+$-group we consider the following elements of $PSL(2,q)$:
$$\alpha_1:=\m{0}{1}{-1}{1},\,\alpha_2:=\m{e}{e^2}{0}{e^2}\mbox{ and  } \alpha_3:=\m{e^2}{e}{0}{e},$$
and we let $G^+=\langle \alpha_1,\alpha_2, \alpha_3\rangle$, $G_i := \langle \alpha_j | j \neq i \rangle$ for $i=1, 2, 3$ and $G_0 := \langle \alpha_1^{-1}\alpha_j | j \geq 2 \rangle$.
In what follows we
%prove that $G^+$ is isomorphic to $PSL(2,q)$ and is a C$^+$-group with B-diagram as in Figure~\ref{tetra}.
%In fact, we shall
show that  $G_2 :=\langle \alpha_1,\alpha_3\rangle$ and $G_3:=\langle \alpha_1,\alpha_2\rangle$ are both isomorphic to  $A_4$, while
$G_0:=\langle \alpha_1^{-1}\alpha_2,\alpha_1^{-1}\alpha_3\rangle$ and $G_1:=\langle \alpha_2,\alpha_3\rangle$ are both isomorphic to  $E_q:C_3$.


\begin{lemma}\label{A4}
The groups $G_2$ and $G_3$ are both isomorphic to $A_4$ and their intersection is $\langle \alpha_1\rangle$. Moreover $(G_2,  \{\alpha_1,\alpha_3\})$ and $(G_3, \{\alpha_1,\alpha_2\})$ are C$^+$-groups with triangular B-diagram of type $(3,3,3)$.
\end{lemma}
\begin{proof}
We first observe that, for each $i\in\{2,3\}$,  $\{\alpha_1,\alpha_i\}$ is an independent set, which is sufficient to guarantee that it gives a set of generators of a C$^+$-group.
Now let  us compute the order of the elements that give the B-diagram.
It is straightforward to see that $\alpha_1$ has order 3.
%: $\alpha_1^2=\m{-1}{1}{-1}{0}$ and $\alpha_1^3=I$.
To compute  the order of the other elements we use the fact that $e$ is a third root of unity and hence $e^2+e+1=0$. Therefore the following equalities hold for any $c\in GF(q)$.
\[\begin{array}{c}
\m{e}{0}{c}{e^2}^2=\m{e^2}{0}{-c}{e}, \ \ \ \;\m{e}{0}{c}{e^2}^3=Id\, ;\\[10pt]
\m{e}{c}{0}{e^2}^2=\m{e^2}{-c}{0}{e}, \ \ \ \;\m{e}{c}{0}{e^2}^3=Id\, . \\[5pt]
\end{array}\]
In particular, this shows that both $\alpha_2$ and $\alpha_3$ have order 3. In addition,
$$\alpha_1^{-1}\alpha_2=\m{e}{0}{e}{e^2}\mbox{\quad and\quad}\alpha_1^{-1}\alpha_3=\m{e^2}{0}{e}{e},$$
hence these elements have also order 3.

To prove that the groups are isomorphic to $A_4$ we first observe that for $i\in\{2,3\}$
$$\alpha_1^3 = \alpha_i^3 = (\alpha_1^{-1}\alpha_i)^3=(\alpha_1\alpha_i^{-1}\alpha_1)^2 = Id.$$
As $Id, \, \alpha_1, \,\alpha_i, \,\alpha_1^2, \,\alpha_i\alpha_1,\, \alpha_1\alpha_i,\,\alpha_i^2,\,\alpha_i\alpha_1^2, \,\alpha_i^2\alpha_1, \,\alpha_1^2\alpha_i,\, \alpha_1\alpha_i^2,\, \alpha_1\alpha_i^2\alpha_1$ are distinct elements of $\langle\alpha_1,\alpha_i\rangle$, from \ref{eq:toroidal} we conclude that $G_i$ is isomorphic to $A_4$, the rotational subgroup of the hypermap of type $(3,3,3)_{(2,0)}$. From the given enumeration of the elements of these two groups we conclude that their intersection is $\langle \alpha_1\rangle$.
\end{proof}

Figures~\ref{ActionG2} and \ref{ActionG3} show that $G_2$ is in the stabilizer of $\{0,1,\infty, -e^2\}$, while $G_3$ is in the stabilizer of $\{0,1,\infty, -e\}$ (and both groups are isomorphic to $A_4$).

\begin{figure}
\[\begin{array}{l}
\mbox{}\hspace{19ex}(0,1,\infty,-e^2)\\
\mbox{}\hspace{14ex}\mbox{}^{\alpha_1}\hspace{-1ex}\swarrow\hspace{15ex}\searrow\hspace{-1ex}\mbox{}^{\alpha_3}\\[2pt]
\mbox{}\hspace{4ex}(1,\infty,0,-e^2)\hspace{18ex}(1,-e^2,\infty,0)\\[3pt]
\mbox{}\hspace{7ex}\mbox{}^{\alpha_1}\hspace{-1ex}\swarrow\hspace{3ex}\searrow\hspace{-1ex}\mbox{}^{\alpha_3}\hspace{20ex}\mbox{}^{\alpha_1}\hspace{-1ex}\swarrow\hspace{3ex}\searrow\hspace{-1ex}\mbox{}^{\alpha_3}\\[2pt]
\mbox{}\hspace{0ex}(\infty,0,1,-e^2)\hspace{2ex}(-e^2,\infty,1,0)\hspace{2ex}(\infty,-e^2,0,1)\hspace{2ex}(-e^2,0,\infty,1)\\[3pt]
\mbox{}\hspace{4ex}{\mbox{}^{\alpha_3}\hspace{-1ex}\downarrow}\hspace{14ex}{\mbox{}^{\alpha_3}\hspace{-1ex}\downarrow}\hspace{13ex}{\downarrow\hspace{-1ex}\mbox{}^{\alpha_1}}\hspace{11ex}{\downarrow\hspace{-1ex}\mbox{}^{\alpha_1}}\\[2pt]
\mbox{}\hspace{0ex}(\infty,1,-e^2,0)\hspace{3ex}(0,\infty,-e^2,1)\hspace{2ex}(0,-e^2,1,\infty)\hspace{2ex}(-e^2,1,0,\infty)\\[2pt]
\mbox{}\hspace{22ex}\mbox{}_{\alpha_1}\hspace{-1ex}\searrow\hspace{6ex}\swarrow\hspace{-1ex}\mbox{}_{\alpha_3}\\[2pt]
\mbox{}\hspace{23ex}(1,0,-e^2,\infty)
\end{array}
\]
\caption{Action of $G_2$ on the 4-tuple $(0,1,\infty,-e^2)$.}\label{ActionG2}
\end{figure}


\begin{figure}
\[\begin{array}{l}
\mbox{}\hspace{19ex}(0,1,\infty,-e)\\
\mbox{}\hspace{14ex}\mbox{}^{\alpha_1}\hspace{-1ex}\swarrow\hspace{14ex}\searrow\hspace{-1ex}\mbox{}^{\alpha_2}\\[2pt]
\mbox{}\hspace{8ex}(1,\infty,0,-e)\hspace{13ex}(1,-e, \infty,0)\\[3pt]
\mbox{}\hspace{7ex}\mbox{}^{\alpha_1}\hspace{-1ex}\swarrow\hspace{3ex}\searrow\hspace{-1ex}\mbox{}^{\alpha_2}\hspace{16ex}\mbox{}^{\alpha_1}\hspace{-1ex}\swarrow\hspace{3ex}\searrow\hspace{-1ex}\mbox{}^{\alpha_2}\\[2pt]
\mbox{}\hspace{0ex}(\infty,0,1,-e)\hspace{2ex}(-e,\infty,1,0)\hspace{2ex}(\infty,-e,0,1)\hspace{2ex}(-e,0,\infty,1)\\[3pt]
\mbox{}\hspace{4ex}{\mbox{}^{\alpha_2}\hspace{-1ex}\downarrow}\hspace{11ex}{\mbox{}^{\alpha_2}\hspace{-1ex}\downarrow}\hspace{15ex}{\downarrow\hspace{-1ex}\mbox{}^{\alpha_1}}\hspace{9ex}{\downarrow\hspace{-1ex}\mbox{}^{\alpha_1}}\\[2pt]
\mbox{}\hspace{0ex}(\infty,1,-e,0)\hspace{2ex}(0,\infty,-e,1)\hspace{2ex}(0,-e,1,\infty)\hspace{2ex}(-e,1,0,\infty)\\[2pt]
\mbox{}\hspace{20ex}\mbox{}_{\alpha_1}\hspace{-1ex}\searrow\hspace{6ex}\swarrow\hspace{-1ex}\mbox{}_{\alpha_2}\\[2pt]
\mbox{}\hspace{21ex}(1,0,-e, \infty)
\end{array}
\]
\caption{Action of $G_3$ on the 4-tuple $(0,1,\infty,-e)$}.\label{ActionG3}
\end{figure}


\begin{lemma}\label{E:3}
The groups $G_0$ and $G_1$ are both isomorphic to $E_q:C_3$, more precisely, $G_0=H_0:\langle \alpha_2^{-1}\alpha_3\rangle$ and $G_1=H_1:\langle \alpha_2^{-1}\alpha_3\rangle$ where $H_0$ and $H_1$ are the following elementary abelian $p$-groups of order $q$.
\[
H_0=\left\{\m{1}{0}{c}{1}\mid c\in GF(q)\right\}\mbox{ and }H_1=\left\{\m{1}{b}{0}{1}\mid b\in GF(q)\right\}.
\]
Moreover $G_0\cap G_1= \langle \alpha_2^{-1}\alpha_3\rangle$. Finally, $(G_0,  \{\alpha_1^{-1}\alpha_2,\alpha_1^{-1}\alpha_3\})$ and $(G_1, \{\alpha_2,\alpha_3\})$ are C$^+$-groups with triangular B-diagram of type $(3,3,3)$.
\end{lemma}

\begin{proof}
First observe that the sets $\{ \alpha_2,\alpha_3\}$ and $\{ \alpha_1^{-1}\alpha_2,\alpha_1^{-1}\alpha_3\}$ are independent, thus $G_0$ and $G_1$ are C$^+$ -groups.
Now let  us compute the order of the elements that give the B-diagram.
As noted above, the elements  $\alpha_2,\, \alpha_3, \, \alpha_1^{-1}\alpha_2$ and $\alpha_1^{-1}\alpha_3$ have order 3.
Now $E:=\alpha_2^{-1}\alpha_3=\m{e}{0}{0}{e^2}$
%, by arguments similar to those used in the proof of Lemma~\ref{A4}, we conclude that
has also order 3 and belongs to $G_0\cap G_1$.

Consider the following subgroups of $PSL(2,q)$:
\[
H_0=\left\{\m{1}{0}{c}{1}\mid c\in GF(q)\right\}\mbox{ and }H_1=\left\{\m{1}{b}{0}{1}\mid b\in GF(q)\right\}
\]
%It is obvious that $H_0$ and $H_1$ are (elementary) abelian groups of order $q$.
%The following equalities show that $H_0$ is an elementary abelian $p$-group of order $q$.
%Indeed as $\m{1}{0}{c}{1}\m{1}{0}{d}{1}=\m{1}{0}{c+d}{1}$, $H_0$ is abelian and $\m{1}{0}{c}{1}^i=\m{1}{0}{ic}{1}$, hence $\m{1}{0}{c}{1}^p=I$.
%Analogously $H_1$ is also an elementary abelian $p$-group of order $q$.

When $q=p$, $H_0$ and $H_1$ are cyclic groups of order $p$. More precisely,
$$H_0=\langle \m{1}{0}{1}{1}\rangle\mbox{ and }H_1=\langle \m{1}{1}{0}{1}\rangle.$$

When $q=p^2$, $H_0$ and $H_1$ are elementary abelian groups of order $p^2$. More precisely,
$$H_0=\langle \m{1}{0}{1}{1},\m{1}{0}{e}{1}\rangle\mbox{ and }H_1=\langle \m{1}{1}{0}{1},\m{1}{e}{0}{1}\rangle,$$
since we assumed that 3 does not divide $p-1$ and therefore $GF(q)=\{a+eb:a,b\in GF(p)\}$.

We claim that
$G_0\cong H_0:\langle E\rangle\mbox{ and }G_1\cong H_1:\langle E\rangle$.
The following equalities prove that $H_0$ and $H_1$ are subgroups of $G_0$ and  $G_1$, respectively.
%\[
%\begin{array}{ll}
%\m{1}{0}{1}{1}=(\alpha_1^{-1}\alpha_3)(\alpha_1^{-1}\alpha_2)^{-1}(\alpha_1^{-1}\alpha_3)&
%\m{1}{0}{e}{1}=(\alpha_1^{-1}\alpha_2)^{-1}(\alpha_1^{-1}\alpha_3)^2\\[10pt]
%\m{1}{1}{0}{1}=\alpha_2\alpha_3^{-1}\alpha_2&
%\m{1}{e}{0}{1}=\alpha_3^{\red{-1}}\alpha_2^{-1}
%\end{array}
%\]
\begin{equation}\label{eq:H0H1}
\begin{array}{ll}
\m{1}{0}{1}{1}=(\alpha_1^{-1}\alpha_3)(\alpha_1^{-1}\alpha_2)^{-1}(\alpha_1^{-1}\alpha_3)&
\m{1}{1}{0}{1}=\alpha_2\alpha_3^{-1}\alpha_2\\[10pt]
\m{1}{0}{e}{1}=(\alpha_1^{-1}\alpha_2)^{-1}(\alpha_1^{-1}\alpha_3)^2&
\m{1}{e}{0}{1}=\alpha_3^{-1}\alpha_2^{-1}
\end{array}
\end{equation}

On the other hand, as
$$\alpha_1^{-1}\alpha_2=\m{1}{0}{1}{1}\m{e}{0}{0}{e^2}\mbox{\quad and\quad}\alpha_1^{-1}\alpha_3=\m{1}{0}{e^2}{1}\m{e^2}{0}{0}{e}$$
we have that $\alpha_1^{-1}\alpha_2,\alpha_1^{-1}\alpha_3\in H_0:\langle E\rangle$, while
$$\alpha_2=\m{1}{1}{0}{1}\m{e}{0}{0}{e^2}\mbox{\quad and\quad}\alpha_3=\m{1}{1}{0}{1}\m{e^2}{0}{0}{e}$$
show that $\alpha_2,\alpha_3\in  H_1:\langle E\rangle$.

With this we have that the groups $G_1$ and $G_0$ are both isomorphic to $E_q:\langle E\rangle$.

Since an element of $G_0$ is equal to $\m{e^i}{0}{c}{e^{2i}}$ for some $c\in GF(q)$ and $i\in\{0,1,2\}$, while an element of $G_1$ is  equal to $\m{e^j}{b}{0}{e^{2j}}$ for some $b\in GF(q)$ and $j\in\{0,1,2\}$, we have that $G_0\cap G_1=\langle E\rangle$.
\end{proof}

\begin{lemma}\label{C+}
$(G^+,  \{\alpha_1,\alpha_2,\alpha_3\})$ is a C$^+$-group with tetrahedral B-diagram as in Figure~\ref{tetra}.
\end{lemma}

\begin{proof}
By Lemmas~\ref{A4} and \ref{E:3} we have that $G_i$, $i=0,1,2,3$ are C$^+$-groups, $G_2 \cap G_3=\langle \alpha_1\rangle$ and $G_1\cap G_0=\langle\alpha_2^{-1}\alpha_3\rangle$. We have that $G_0$ is a subgroup of the stabilizer of $0$ in $G^+$ and $G_1$ is a subgroup of the stabilizer of $\infty$ in $G^+$.
Figure~\ref{ActionG2} shows that the only elements of $G_2$ fixing $0$ are the identity, $\alpha_3^2\alpha_1$ and $\alpha_1^2\alpha_3$, hence $G_0\cap G_2=\langle \alpha_1^{-1}\alpha_3\rangle$.
The same  figure shows that the only elements of $G_2$ fixing $\infty$ are the identity, $\alpha_3$ and $\alpha_3^2$, thus $G_1\cap G_2=\langle \alpha_3\rangle$.
Figure~\ref{ActionG3} shows that the only elements of $G_3$ fixing $0$ are the identity, $\alpha_2^2\alpha_1$ and $\alpha_1^2\alpha_2$, hence $G_0\cap G_3=\langle \alpha_1^{-1}\alpha_2\rangle$.
The same figure shows that the only elements of $G_3$ fixing $\infty$ are the identity, $\alpha_2$ and $\alpha_2^2$, thus $G_1\cap G_3=\langle \alpha_2\rangle$. From this we conclude that $(G^+,  \{\alpha_1,\alpha_2,\alpha_3\})$ is a C$^+$-group. Moreover, as $G_i\cap G_j$ are cyclic groups of order three, for $i,j\in\{0,1,2,3\}$ with $(i\neq j)$, the B-diagram is a complete graph $K_4$ with edges labelled $3$.
\end{proof}


\begin{figure}
$$\xymatrix@C=60pt@R=30pt{
& G_1 \ar@{-}[d]|{\langle\alpha_2^{-1}\alpha_3\rangle}\ar@{-}[ddl]|{\langle\alpha_3\rangle} \ar@{-}[ddr]|{\langle\alpha_2\rangle}                              &\\
& G_0 \ar@{-}[dl]|{\langle\alpha_1^{-1}\alpha_3\rangle}\ar@{-}[dr]|{\langle\alpha_1^{-1}\alpha_2\rangle}                      &\\
G_2\ar@{-}[rr]|{\langle\alpha_1\rangle} &  & G_3
}$$
\caption{The intersections $G_i\cap G_j$.}\label{intersections}
\end{figure}

\begin{lemma}The group $G^+$  is isomorphic to $PSL(2,q)$.
\end{lemma}
\begin{proof}
This follows from the description of the subgroups $H_0$ and $H_1$ in Lemma~\ref{E:3} and Lemma~\ref{sw94}.
\end{proof}
%===
\begin{comment}
\begin{proof}
Let $A=\m{a}{b}{c}{d}\in GL(2,p)$ (with $ad-bc=1$).
If $d\neq 0$,
$$A=\m{d^{-1}}{0}{0}{d}\m{1+bc}{bd}{cd^{-1}}{1}$$
otherwise,  $b\neq 0$ and
$$A=\m{a}{b}{-b^{-1}}{0}.$$

Consider the following matrices {\color{red}$X,Y \in H_0$ and $Z, W \in H_1$}.
\[X:=\m{1}{0}{1}{1},\,Y:=\m{1}{0}{e}{1},\,Z:=\m{1}{1}{0}{1},\,W:=\m{1}{e}{0}{1}\]
 In what follows we prove that $A$ can be written in terms of $X$, $Y$, $Z$ and $W$, and as these matrices belong to $G^+$, by Lemma~\ref{E:3}, we conclude that any element of ${\color{red}PSL}(2,q)$ is in $G^+$.

For $q=p$ and $d= 0$, $A=X^{-b^{-1}}Z^bX^{(a-1)b^{-1}}$. For $q=p$ and $d\neq 0$:
$$\m{d^{-1}}{0}{0}{d}=X^dZ^{1-d^{-1}}X^{-1}Z^{1-d}\mbox{ and }\m{1+bc}{bd}{cd^{-1}}{1}=Z^{bd}X^{cd^{-1}}.$$

Now consider the case $q=p^2$. In this case any element of $GF(q)$ can be written as
$a_0+a_1e$ for  $a_0,a_1\in GF(p)$.
Thus we  have the equalities:
$$\m{d^{-1}}{0}{0}{d}= X^{d_0}Y^{d_1}Z^{1-f_0}W^{-f_1}X^{-1}Z^{1-d_0}W^{-d_1}$$
and
$$\m{1+bc}{bd}{cd^{-1}}{1}=X^{\beta_0}W^{\beta_1}X^{\gamma_0}Y^{\gamma_1},$$
where $d=d_0+d_1e\in GF(q)^\ast$; $f_0,f_1\in GF(p)$ are such that $(d_0+d_1e)(f_0+f_1e)=1$; $\beta_0+\beta_1e=bd$ and $\gamma_0+\gamma_1e=cd^{-1}$.
If $d=0$,  $A=X^{c_0}Y^{c_1}X^{b_0}W^{b_1}X^{a_0}Y^{a_1}$ where $\,c_0+c_1e=-b^{-1}\,$ and $\,a_0+a_1e=(a-1)b^{-1}$ .

\end{proof}
\end{comment}
%====
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

\section{A hypertope for $PSL(2,q)$}\label{section4}

As pointed out in \cite{FLW1}, not every coset geometry is a hypertope.
In this section we shall see that the coset geometry $\Gamma:=\Gamma(G^+,R) := \Gamma(G^+; (G_i)_{i\in I})$ where $I=\{0,1,2,3\}$, $G^+$ is isomorphic to $PSL(2,q)$ and the $G_i$'s are the groups defined in Section~\ref{SC+}, is a hypertope.
Since $\Gamma$ is a coset geometry, then we only have to show that it is thin and residually connected.

We observe that $\Gamma$ has a lot of symmetry, and we shall use such symmetry to show that it is a hypertope.
Note that since $G^+$ is a $C^+$-group, then $G_1\cap G_2\cap G_3$ is trivial. This implies that the action of $G^+$ on the flags by right multiplication is free (semi-regular) on the chambers.
Moreover, as $\Gamma$ is a coset geometry,  it is transitive on the
 flags of rank less than $3$. We shall show that in fact $\Gamma$ is transitive on the flags of rank $3$.
 To show this, in the following two lemmas we  analyze some incidences among  elements of $\Gamma$, and some incidences among  elements and flags of $\Gamma$.

%Consider the coset geometry $\Gamma:=\Gamma(G^+,R) := \Gamma(G^+; (G_i)_{i\in I})$ where $I=\{0,1,2,3\}$, $G^+$ is isomorphic to $PSL(2,q)$ and the $G_i$'s are the groups defined in Section~\ref{SC+}. An element of $\Gamma$ of type $i$ is a right coset $G_ig$ with $g\in G^+$.
From the B-diagram of $G^+$, since $G_i$ with $i\in\{2,3\}$ is isomorphic to $A_4$ and $G_i \cap G_j \cong C_3$, we know that each element of type $i\in\{2,3\}$ is incident to exactly four elements of type $j\neq i$. Similarly, since $G_i$ with $i\in \{0,1\}$ is isomorphic to $E_q:C_3$, we have that each element of type $i\in\{0,1\}$ is incident to exactly $q$ elements of type $j\neq i$.
% In the following two lemmas we describe the elements that are incident to $G_i$ for each $i\in I$.
Remark also that, if $G_ig*G_j$ then $G_ig=G_ih$ for some $h\in G_j$. Moreover, if $T$ is a transversal for $G_i\cap G_j$ in $G_j$, then $G_ig=G_ih=G_it$ for some $t\in T$. More precisely we have the following lemma.

\begin{lemma}\label{01}
Let $i,j \in I$ with $i\neq j$ and let $g \in G^+$. Then\\[3pt]
\footnotesize{
\begin{math}
G_ig*G_j\Leftrightarrow\left\{
\begin{array}{ll}
G_ig \in \{G_ih \mid h\in H_j\} & \mbox{ if } j\in\{0,\,1\};\\[3pt]
G_ig \in \{G_i, G_i\alpha_{i}, G_i\alpha_i\alpha_k, G_i\alpha_i\alpha_k^{-1} \} &\mbox{ if }\, j\in\{2,\,3\} \mbox{ and } \{i,\,j,\,k\}=\{1,\,2,\,3\};\\[3pt]
G_ig \in \{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_k\} & \mbox{ if }\, j,\,k\in\{2,\,3\} \mbox{ and }\{i,\,j,\,k\}=\{0,\,2,\,3\}.
\end{array}
\right.
\end{math}}
\end{lemma}

\begin{proof}
It can be easily checked that all of the cosets in the set on the right side of the equivalences are incident to the respective $G_j$.

Suppose that $j\in\{0,1\}$. By Lemma~\ref{E:3}, $H_j$ is a subgroup of $G_j$ of order $q$. Thus $g\in H_j$ implies that $G_ig\cap G_j\neq \varnothing$. As $G_i\cap G_j$ has order $3$, if $G_ig*G_j$, then $|G_ig\cap G_j|=3$. It remains to prove that the cardinality of $\{G_ig\,|\, g\in H_j\}$ is $q$.
Suppose that $g,h\in H_j$ and $G_ig=G_ih$. Then $gh^{-1}\in G_i\cap H_j=(G_i\cap G_j)\cap H_j$. In all the cases $G_i\cap G_j$ is generated by an element of order 3 which does not belong to $H_j$, therefore
%Thus $gh^{-1}\in\langle \alpha_1^{-1}\alpha_2\rangle\cap H_i$ or $gh^{-1}\in\langle \alpha_1^{-1}\alpha_2\rangle\cap H_i$  or $gh^{-1}\in\langle \alpha_1^{-1}\alpha_3\rangle\cap H_i$. As
all the intersections $(G_i\cap G_j)\cap H_j$ are trivial, and so $g=h$.
%Thus any two distinct elements of $H_i$ will give two distinct cosets of $G_j$ that are incident to $G_i$ and therefore, since $|H_i| = q$, the number of elements of type $j$ incident to an element of type $i$ is $q$ for every $j\neq i$.

For the cases when $j\in\{2,3\}$, by Lemma~\ref{C+} (see also Figure \ref{intersections} we have the following:
\begin{itemize}
\item If $i\neq 0$, then $G_i\cap G_j=\langle \alpha_k\rangle$  where $\{i,j,k\}=\{1,2,3\}$. Hence $$T=\{Id,\ \alpha_i,\ \alpha_i\alpha_k,\ \alpha_i\alpha_k^{-1}\}$$ is a transversal for $G_i\cap G_j$ in $G_j$ and $G_ig=G_it$ for some $t\in T$.
\item If $i=0$, then $G_0\cap G_j=\langle \alpha_1^{-1}\alpha_k\rangle$ where $\{j,k\}=\{2,3\}$. Hence $$T=\{Id,\ \alpha_1,\ \alpha_1^{-1},\ \alpha_k\}$$
is a transversal for $G_0\cap G_j$ in $G_j$ and $G_0g=G_0t$ for some $t\in T$.\qedhere
\end{itemize}
\end{proof}


%\begin{lemma}\label{2ou3}
%Let $i\in\{2,3\}$ and $j\neq i$.
%For  $j\neq 0$,
%$$G_i*G_jg\Leftrightarrow G_jg\in\{G_j, G_j\alpha_j, G_j\alpha_j\alpha_k, G_j\alpha_j\alpha_k^{-1}\}$$
%where $k\in \{1,2,3\}\setminus \{i,j\}$,
%$$G_i* G_0g\Leftrightarrow G_0g\in\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_k\}$$
%where $k\in\{2,3\}\setminus\{i\}$.
%\end{lemma}
%\begin{proof}
%It can be checked easily that each coset in the set on the left side of the equivalences is incident to $G_i$. The direct implication follows from Lemma~\ref{C+}, particularly from the equalities: $G_i\cap G_j=\langle \alpha_k\rangle $ and $G_i\cap G_0=\langle \alpha_1^{-1}\alpha_k\rangle$ where $i,j$ and $k$ are as in statement of this lemma. Indeed these equalities ensure that $|\{G_j, G_j\alpha_j, G_j\alpha_j\alpha_k, G_j\alpha_j\alpha_k^{-1}\}|=|\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_k\}|=4$.
%\end{proof}

%\begin{lemma}\label{01}
%Let $i\in\{0,1\}$ and $H_i$ be the groups of Lemma~\ref{E:3}. For any $j\neq i$,
%$$G_i*G_jg\Leftrightarrow G_jg\in\{ G_jh\,|\, h\in H_i\}.$$
%\end{lemma}
%\begin{proof}
%By Lemma~\ref{E:3}, the subgroup $H_i$ is a subgroup of $G_i$ thus $g\in H_i$ implies that $G_i\cap G_jg\neq \varnothing$. As $G_i\cap G_j$ has order $3$, if $G_i*G_jg$, then $|G_i\cap G_jg|=3$. It remains to prove that the cardinality of $\{G_jg\,|\, g\in H_i\}$ is $q$.
%Suppose that $g,h\in H_i$ and $G_jg=G_jh$. Then $gh^{-1}\in G_j\cap G_i$. Thus $gh^{-1}\in\langle \alpha_1^{-1}\alpha_2\rangle\cap H_i$ or $gh^{-1}\in\langle \alpha_1^{-1}\alpha_2\rangle\cap H_i$  or $gh^{-1}\in\langle \alpha_1^{-1}\alpha_3\rangle\cap H_i$. As all these intersections are trivial, $h=g$.
%\end{proof}
%
%\subsection{Action of $G^+$}
%Consider the action of $G^+$ on the flags by right multiplication. As $G_1\cap G_2\cap G_3$ is trivial, $G^+$ acts freely on the chambers.
%In what follows we prove that $G^+$ is transitive on the flags of rank less than $4$ of the same type.
%For convenience we first  describe some incidences to flags of rank 2.
%


\begin{proposition}\label{2+3}
Let $i,j,k\in I$ be distinct and let $g\in G^+$. Then
\[
G_ig*\{G_j,G_k\}\Leftrightarrow G_ig=G_ih\,\mbox{ for some }\,h\in G_j\cap G_k\,.
\]
\end{proposition}
\begin{proof}
Obviously the right hand side of the equivalence implies the left one. The proof of the converse will be divided in cases covering all the possibilities for the set $\{j,k\}$. Assume that $G_ig*\{G_j,G_k\}$.

\smallskip

\mbox{\bf{Case 1.} }If $\{j,k\}=\{0,1\}$, then $i\in\{2,3\}$ and
\[ G_ig\in\{G_ih_0\mid h_0\in H_0\}\cap\{G_ih_1\mid h_1\in H_1\} \]
by means of Lemma~\ref{01}. Looking at the 12 elements of $G_i$, we see that for $h_0=\m{1}{0}{c}{1}\in H_0$ and $h_1=\m{1}{b}{0}{1}\in H_1$,
\[
h_0h_1^{-1}=\m{1}{-b}{c}{1-bc}\in G_i\Leftrightarrow (b,c)\in\{(0,0),(1,1),(-e^{1-i},-e^{i-1})\}\,.
\]
As before, we set $E:=\alpha_2^{-1}\alpha_3$. We have to prove that $G_ig=G_ih$ for some $h\in\langle E\rangle=G_0\cap G_1$ (Figure \ref{intersections}).
If $(b,c)=(0,0)$, then $G_ig=G_ih_0=G_ih_1=G_i$.
If $(b,c)=(1,1)$, then $h_1=\alpha_2\alpha_3^{-1}\alpha_2$ by \ref{eq:H0H1}. Hence
\[
G_ig=G_ih_0=G_ih_1=G_i\alpha_2\alpha_3^{-1}\alpha_2=G_iE^{i-1}
\]
\begin{tabbing}
since \= $G_2\alpha_2\alpha_3^{-1}\alpha_2=G_2(\alpha_3^{-1}\alpha_2)^2=G_2E^{-2}=G_2E$\\
and \> $G_3\alpha_2\alpha_3^{-1}\alpha_2=G_3\alpha_3^{-1}\alpha_2=G_2E^{-1}=G_3E^2$.
\end{tabbing}
If $(b,c)=(-e^{1-i},-e^{i-1})$, then
\[ G_ig=G_ih_0=G_ih_1=G_ih_1E^{i-1}E^{1-i}=G_iE^{1-i} \]
since $h_1E^{i-1}=\m{e^{1-i}}{e^{i-1}}{0}{e^{i-1}}^{-1}\in G_i$.
Thus $G_ig=G_ih$ for some $h\in\langle E\rangle=G_0\cap G_1$.

\smallskip

\mbox{\bf{Case 2}}. If $\{j,k\}=\{0,2\}$ or $\{0,3\}$, then $i\neq 0$ and we can assume $k=0$, $j\in\{2,3\}$. By means of Lemma~\ref{01} there is $h_0\in H_0$ such that
\[
G_ig=G_ih_0\in\{G_i,G_i\alpha_i,G_i\alpha_i\alpha_l,G_i\alpha_i\alpha_l^{-1}\}
\]
where $l\in I$ is such that $\{i,j,k,l\}=I$. In other words, $G_ig$ must belong to the set
\[ S:=\{G_it\mid t\in\{Id,\alpha_i,\alpha_i\alpha_l,\alpha_i\alpha_l^{-1}\}\,,\ H_0\cap G_it\neq\varnothing\}\,.\]
Obviously $G_i\in S$. To see that $G_i\alpha_i\in S$, consider the following elements:
\[ \beta_1:=\m{1}{-1}{0}{1}\in G_1\,,\quad\beta_2:=\alpha_1\alpha_3^{-1}\alpha_1\in G_2\quad\mbox{and}\quad\beta_3:=\alpha_1\alpha_2^{-1}\alpha_1\in G_3\,.\]
Then $\beta_i\alpha_i\in H_0\cap G_i\alpha_i$ for any $i\in\{1,2,3\}$. To show that $G_i\alpha_i\alpha_l\not\in S$ we remark that
\[
\alpha_i\alpha_l(0)=\alpha_i(1)=\left\{\begin{array}{ll} \infty & \mbox{ if }\ i=1\,,\\ -e & \mbox{ if }\ i=2\,,\\ -e^2 & \mbox{ if }\ i=3\,,\end{array}\right.
\]
from which it follows $H_0\cap G_i\alpha_i\alpha_l=\varnothing$, since there is no element in $G_i$ sending $\alpha_i(1)$ to 0. Finally, considering the following elements
\[
\gamma_1:=E^{j-1}\in G_1\,,\quad\gamma_2:=\alpha_1\alpha_3\in G_2\quad\mbox{and}\quad\gamma_3:=\alpha_1\alpha_2\in G_3\,,
\]
we have that $\gamma_i\alpha_i\alpha_l^{-1}\in H_0$ for any $i\in\{1,2,3\}$. Hence $G_i\alpha_i\alpha_l^{-1}\in S$ and so
\[
S=\{G_i,G_i\alpha_i,G_i\alpha_i\alpha_l^{-1}\}\,.
\]
It is now straightforward to check that $S=\{G_ih\mid h\in\langle\alpha_1^{-1}\alpha_l\rangle\}$, from which we conclude that $G_ig=G_ih$ for some $h\in\langle\alpha_1^{-1}\alpha_l\rangle=G_0\cap G_j$
(Figure~\ref{intersections}).

\smallskip

\mbox{\bf{Case 3}}. If $\{j,k\}=\{1,2\}$ or $\{1,3\}$ we assume $k=1$ and $j\in\{2,3\}$. Let $l\in I$ such that $\{j,l\}=\{2,3\}$. According to Lemma~\ref{01} we have the following:
\begin{itemize}
\item If $i=0$ then $G_ig=G_0g$ belongs to
\[
\{G_0t\mid t\in\{Id,\alpha_1,\alpha_1^{-1},\alpha_l\},\ H_1\cap G_0t\neq\varnothing\}
\]
which is equal to $\{G_0,G_0\alpha_1^{-1},G_0\alpha_l\}$. Namely, $E$ ,$E^2$ and $\m{1}{0}{-1}{1}$ are elements of $G_0$ such that
\[ E^2\alpha_2, E\alpha_3,\m{1}{0}{-1}{1}\alpha_1^{-1}\in H_1\,,\]
while $\alpha_1(\infty)=0$ gives $H_1\cap G_0\alpha_1=\varnothing$. As $G_0\alpha_1^{-1}=G_0\alpha_l^{-1}$, we conclude that $G_0g=G_0h$ for some $h\in\langle\alpha_l\rangle=G_j\cap G_1=G_j\cap G_k$ (Figure~\ref{intersections}).
\item If $i\neq 0$ then $G_ig$ belongs to
\[
\{G_it\mid t\in\{Id,\alpha_i,\alpha_i\alpha_1,\alpha_i\alpha_1^{-1}\},\ H_1\cap G_it\neq\varnothing\}
\]
which is equal to $\{G_i,G_i\alpha_i,G_i\alpha_i\alpha_1\}$. Namely, $\alpha_j\alpha_i,\alpha_i\alpha_j\in H_1$, $G_i\alpha_i\alpha_1=G_i\alpha_j^{-1}\alpha_i^{-1}=G_i\alpha_i^{-1}$, while $\alpha_i\alpha_1^{-1}(\infty)=\alpha_i(1)=-e^{i-1}$ gives $H_1\cap G_i\alpha_i\alpha_1^{-1}=\varnothing$. We conclude so that $G_ig=G_ih$ for some $h\in\langle\alpha_i\rangle=G_j\cap G_1=G_j\cap G_k$ (Figure~\ref{intersections}).
\end{itemize}

\smallskip

\mbox{\bf{Case 4}}. If $\{j,k\}=\{2,3\}$, then by Lemma~\ref{01} we have the following:
\begin{itemize}
\item If $i=0$, then $G_ig=G_0g$ belongs to
\[\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_2\}\cap\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_3\}\]
which is equal to $\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}\}$ since $\alpha_2\alpha_3^{-1}\not\in G_0$.
\item If $i=1$, then $G_ig=G_1g$ belongs to
\[
\{G_1, G_1\alpha_1, G_1\alpha_1\alpha_2, G_1\alpha_1\alpha_2^{-1}\}\cap\{G_1, G_1\alpha_1, G_1\alpha_1\alpha_3, G_1\alpha_1\alpha_3^{-1}\}\,.
\]
Remark that $G_1\alpha_1\alpha_k=G_1\alpha_k^{-1}\alpha_1^{-1}=G_1\alpha_1^{-1}$ by means of the fact that $\alpha_1\alpha_k$ is an involution for $k=2,3$. Moreover $(\alpha_1\alpha_2^{-1})(\alpha_1\alpha_3^{-1})^{-1}\not\in G_1$ because it does not fix $\infty$. Hence $G_1g$ belongs to $\{G_1, G_1\alpha_1, G_1\alpha_1^{-1}\}$.\qedhere
\end{itemize}
\end{proof}

%
%
%\begin{lemma}\label{2+3}
%\red{Let $i\in\{0,1\}$ and $\{j,k\}=\{2,3\}$. Then
%\begin{itemize}
%\item[(a)] $G_ig*\{G_2,G_3\}\Leftrightarrow G_ig=G_ih$ for some $h\in\langle\alpha_1\rangle=G_2\cap G_3$.
%\item[(b)] $G_0g*\{G_1,G_k\}\Leftrightarrow G_0g=G_0h$ for some $h\in\langle\alpha_j\rangle=G_1\cap G_k$.
%%\item[(c)] If $\{i,k\}=\{2,3\}$ then $G_ig*\{G_1,G_k\}\Leftrightarrow G_ig=G_ih$ for some $h\in\langle\alpha_i\rangle$.
%\end{itemize}}
% \end{lemma}
%
%\begin{proof}
%\red{Obviously the right hand side of both equivalences (a) and (b) implies the left hand side.}
%
%\red{(a) If $G_ig*\{G_2,G_3\}$, then by Lemma~\ref{01} we have the following:
%\begin{itemize}
%\item If $i=0$, then $G_ig=G_0g$ belongs to
%\[\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_2\}\cap\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_3\}\]
%which is equal to $\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}\}$ since $\alpha_2\alpha_3^{-1}\not\in G_0$.
%\item If $i=1$, then $G_ig=G_1g$ belongs to
%\begin{equation}\label{intersection(a)}
%\{G_1, G_1\alpha_1, G_1\alpha_1\alpha_2, G_1\alpha_1\alpha_2^{-1}\}\cap\{G_1, G_1\alpha_1, G_1\alpha_1\alpha_3, G_1\alpha_1\alpha_3^{-1}\}\,.
%\end{equation}
%Remark that $G_1\alpha_1\alpha_k=G_1\alpha_k^{-1}\alpha_1^{-1}=G_1\alpha_1^{-1}$ by means of the fact that $\alpha_1\alpha_k$ is an involution for $k=2,3$. Moreover $(\alpha_1\alpha_2^{-1})(\alpha_1\alpha_3^{-1})^{-1}\not\in G_1$ because it does not fix $\infty$. Hence \ref{intersection(a)} is equal to $\{G_1, G_1\alpha_1, G_1\alpha_1^{-1}\}$.
%\end{itemize}}
%
%\red{(b) If $G_0g*\{G_1,G_k\}$, then by Lemma~\ref{01} $G_0g$ belongs to
%\begin{equation}\label{intersection(b)}
%\{G_0h:h\in H_1\}\cap\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_j\}\,.
%\end{equation}
%Remark that $G_0\alpha_1\cap H_1=\varnothing$ since $\alpha_1(\infty)=0$.
%As $\alpha_1^{-1}\alpha_j\in G_0$ we have that $G_0\alpha_1^{-1}=G_0\alpha_j^{-1}$.  From $\alpha_j,\alpha_j^{-1}\in G_1=H_1:\langle\alpha_2^{-1}\alpha_3\rangle$ and $\alpha_2^{-1}\alpha_3\in G_0$ we conclude that
%$G_0\alpha_j,G_0\alpha_j^{-1}\in\{G_0h:h\in H_1\}$.
%Hence \ref{intersection(b)} is equal to $\{G_0,G_0\alpha_j,G_0\alpha_j^{-1}\}$.
%}
%Straightforward from the definition of the subgroups $G_i$'s and the fact that $G^+$ is a C$^+$-group.
%Similarly as in the previous lemma, it can be checked easily that if $g$ belongs to the groups on the right, then the incidences occur.

%Start by considering the case when $i=1$.
%\begin{lemma}\label{2+3}
%For $i\in\{0,1\}$, $G_ig*\{G_2,G_3\}\Leftrightarrow g\in\langle \alpha_1\rangle$.
%\end{lemma}
%\begin{proof}
%Let $i=1$.
%By Lemma~\ref{01},
%$$G_1g\in \{G_1, G_1\alpha_1, G_1\alpha_1\alpha_2, G_1\alpha_1\alpha_2^{-1}\}\cap \{G_1, G_1\alpha_1, G_j\alpha_1\alpha_3, G_1\alpha_1\alpha_3^{-1}\}.$$
%We have that $G_1\alpha_1\alpha_3=G_1\alpha_2^{-1}\alpha_3\alpha_1\alpha_3=G_1\alpha_2^{-1}\alpha_3^{-1}=G_1\alpha_1\alpha_2=G_1\alpha_1^{-1}$.  In addition $G_1\alpha_1\alpha_3^{-1}\neq G_1\alpha_1\alpha_2^{-1}$ as $\alpha_1\alpha_3^{-1}\alpha_2\alpha_1^{-1}$ does not fix $\infty$. Thus $G_1g\in\{G_1, G_1\alpha_1, G_1\alpha_1^{-1}\}$.
%For $i=0$ similar arguments can be used.
%\end{proof}
%Now let $i=0$ and $\{j,k\} = \{2,3\}$.
%Again, by Lemma~\ref{01},
%$$G_0g\in\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_2\} \cap \{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_3\}.$$
%As $\alpha_3\alpha_2^{-1}$ does not fix $0$,  $G_0\alpha_2\neq G_0\alpha_3$, thus $G_0g\in\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}\}$.
%
%The implication from the right follow from  the  equality $G_2\cap G_3=\langle \alpha_1\rangle$.
%\end{proof}
%
%
%\begin{lemma}\label{not2+3}
%For $\{i,j\}=\{2,3\}$ and $i\neq j$,  $G_0g*\{G_1,G_i\} \Leftrightarrow g\in \langle \alpha_j\rangle$.
%\end{lemma}
%\begin{proof}
%Suppose that $ G_0g*\{G_1,G_i\}$
%By Lemma~\ref{01}, $G_0g\in\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}, G_0\alpha_j\}$.
%As $\alpha_1(\infty)=0$ and $G_0$ stabilizes $0$, $G_1\alpha_1$ is not  incident to $G_1$.
%Hence  $G_0g\in\{G_0, G_0\alpha_1^{-1}, G_0\alpha_j\}$, or equivalently   $G_0g\in\{G_0, G_0\alpha_j^{-1}, G_0\alpha_j\}$.
%
%The implication from the right follows from  the  equality $G_0\cap G_i=\langle\alpha_j\rangle$.
%\end{proof}

We are now ready to show that $G^+$ is transitive on the flags of rank $3$.

\begin{proposition}\label{action}
Let $J=\{i,j,k\}\subset \{0,1,2,3\}$ with $i,j$ and $k$ all distinct.
The action of $G^+$ on the flags of type $J$ is transitive.
\end{proposition}
\begin{proof}
Let $F=\{G_i, G_j, G_k\}$ and $F'=\{G_ia_i, G_ja_j, G_ka_k\}$. By Proposition~\ref{tits}, $G^+$ is transitive on the flags of rank $2$. Hence, there exists a flag $F''$ and $a\in G^+$ such that  $F'=F''a$ with $F''=\{G_ig, G_j,G_k\}$.
Without loss of generality we may assume that $(i,j,k)\in \{(0,2,3), (1,2,3), (0,1,3), (0,1,2)\}$.
Then by Proposition~\ref{2+3}, % and \ref{not2+3},
  there exists $h\in G_ig\cap G_j\cap G_k$.
  Hence $F''=Fh$ and therefore $F'=Fha$.
\end{proof}

%\subsection{Thinness}

To show that $\Gamma$ is a thin geometry, we need to show that every residue of rank $1$ has exactly two elements.
A residue of rank $1$ is the set of elements of $\Gamma$ that are incident to a flag of type $J\subset \{0,1,2,3\}$, with $| J | = 3$. By Proposition~\ref{action}, all residues of rank $1$ corresponding to flags of a given type $J$ are isomorphic. Hence, it is enough to show that, for $\{i,j,k,l\}=\{0,1,2,3\}$, the sets $\{G_jg \mid G_jg * \{G_i,G_k,G_l\}\}$ all have cardinality two.

\begin{proposition}\label{thin}
The coset geometry $\Gamma$ is a thin geometry.
\end{proposition}

\begin{proof}
Following the above discussion, we need to show that four sets have cardinality two. To do this, we shall show that:
\begin{equation}\label{thin1}%\label{thin2}
\begin{array}{lcl}
G_0g * \{G_1,G_2, G_3\} &\Leftrightarrow& G_0g \in \{G_0,G_0\alpha_1^{-1}\}; \\
G_1g * \{G_0,G_2, G_3\} &\Leftrightarrow& G_1g \in \{G_1,G_1\alpha_1\}; \\
G_2g * \{G_0,G_1, G_3\} &\Leftrightarrow& G_2g \in \{G_2,G_2\alpha_2\}; \\
G_3g * \{G_0,G_1, G_2\} &\Leftrightarrow& G_3g \in \{G_3,G_3\alpha_3\}.
\end{array}
\end{equation}

It is straightforward to see that the elements on the sets of the right are incident to the flags on the left.

We start by assuming that $G_ig * \{G_j,G_2, G_3\}$ where $\{i,j\}=\{0,1\}$.
As $G_ig*\{G_2,G_3\}$, by Proposition~\ref{2+3}, $G_ig\in\{G_i, G_i\alpha_1, G_i\alpha_1^{-1}\}$.
As we also have $G_ig*G_j$ we observe that
\[
\m{1}{0}{-1}{1}\alpha_1^{-1}\in G_0\alpha_1^{-1}\cap G_1
\]
while $G_0\alpha_1\cap G_1=\varnothing$ since any element of $G_0\alpha_1$ sends $\infty$ to $0$ and so it does not belong to $G_1$ which is in the stabilizer of $\infty$. As $G_0g\ast G_1\Leftrightarrow G_0\ast G_1g^{-1}$,
this proves the first two equivalences in \ref{thin1}.

Now let $G_ig*\{G_0,G_1,G_j\}$ where $\{i,j\}=\{2,3\}$.
As $G_ig*G_0$ and $G_ig*G_1$, there exists an element of $G_ig$ fixing $0$ and an element fixing $\infty$.
In addition, as $G_ig*G_j$, by Lemma~\ref{01}, $G_ig\in \{G_i, G_i\alpha_i, G_i\alpha_i\alpha_1, G_i\alpha_i\alpha_1^{-1}\}$.

Let $i=2$. Looking at Figure~\ref{ActionG2} and Figure~\ref{ActionG3}, we have that $\alpha_2\alpha_1(0)=-e$ and  $\alpha_2\alpha_1^{-1}(\infty) =-e$. As $G_2$ stabilizes $\{0,1,\infty,-e^2\}$, there is no element in $G_2\alpha_2\alpha_1$ fixing $0$ and there is no element in $G_2\alpha_2\alpha_1^{-1}$ fixing $\infty$. Thus $G_2g\in\{G_2,G_2\alpha_2\}$, as $\alpha_3^{-1}\alpha_2\in G_2\alpha_2$ fixes both $0$ and $\infty$.

Now consider $i=3$. We have that $\alpha_3\alpha_1(0)=-e^2$ and  $\alpha_3\alpha_1^{-1}(\infty) =-e^2$. As $G_2$ stabilizes $\{0,1,\infty,-e\}$, there is no element in $G_3\alpha_3\alpha_1$ fixing $0$ and there is no element in $G_3\alpha_3\alpha_1^{-1}$ fixing $\infty$. Thus $G_3g\in\{G_3,G_3\alpha_3\}$, as $\alpha_2^{-1}\alpha_3\in G_3\alpha_3$ fixes both $0$ and $\infty$.

Therefore all the equivalences in (\ref{thin1}) are satisfied and so $\Gamma$ is a thin geometry.
\end{proof}


%\begin{lemma}\label{thin1}
%For $i,j\in\{2,3\}$ and $i\notin\{0,1, j\}$,
%$$G_jg*\{G_0,G_1,G_i\}\Leftrightarrow G_jg\in\{G_j,G_j\alpha_j\}.$$
%\end{lemma}
%\begin{proof}
%Clearly $G_j*\{G_0,G_1,G_i\}$ and $G_j\alpha_j *\{G_0,G_1,G_i\}$, which gives the implication from the right to the left.
%
%Now suppose that $G_jg*\{G_0,G_1,G_i\}$.
%As $G_jg*G_0$ and $G_jg*G_1$, there exists an element of $G_jg$ fixing $0$ and an element fixing $\infty$.
%In addition, as $G_jg*G_i$, by Lemma~\ref{01}, $G_jg\in \{G_j, G_j\alpha_j, G_j\alpha_j\alpha_1, G_j\alpha_j\alpha_1^{-1}\}$.
%
%Let $j=2$. We have that $\alpha_2\alpha_1(0)=-e$ and  $\alpha_2\alpha_1^{-1}(\infty) =-e$. As $G_2$ stabilizes $\{0,1,\infty,-e^2\}$, there is no element in $G_j\alpha_2\alpha_1$ fixing $0$ and there is no element in $G_j\alpha_2\alpha_1^{-1}$ fixing $\infty$. Thus $G_2g\in\{G_2,G_2\alpha_2\}$
%
%Now consider $j=3$. We have that $\alpha_3\alpha_1(0)=-e^2$ and  $\alpha_3\alpha_1^{-1}(\infty) =-e^2$. As $G_2$ stabilizes $\{0,1,\infty,-e\}$, there is no element in $G_j\alpha_3\alpha_1$ fixing $0$ and there is no element in $G_j\alpha_3\alpha_1^{-1}$ fixing $\infty$. Thus $G_3g\in\{G_3,G_3\alpha_3\}$
%\end{proof}
%
%\begin{lemma}\label{thin2}
%For $i,j\in\{0,1\}$ and $i\notin\{ j,2,3\}$,
%$$G_jg*\{G_i,G_2,G_3\}\Leftrightarrow \left\{\begin{array}{c}
%G_jg\in\{G_j,G_j\alpha_1\}\,\mbox{ if }j=1\\
%G_jg\in\{G_j,G_j\alpha_1^{-1}\}\,\mbox{ if }j=0.\end{array}\right.$$
%\end{lemma}
%\begin{proof}
%We have that $G_j*\{G_i,G_2,G_3\}$, $G_0\alpha_1*\{G_1,G_2,G_3\}$ and $G_1\alpha_1^{-1}*\{G_0,G_2,G_3\}$, which gives the implication from the right to the left.
%Suppose that $G_jg*\{G_i,G_2,G_3\}$.
%
%Consider first $j=1$.
%As $G_1*\{G_2,G_3\}$, by Lemma~\ref{2+3}, we have $G_1g \in \{G_1, G_1\alpha_1,$ $G_1\alpha_1\alpha_2\}$.
%In addition $G_1g*G_0$. As $\alpha_1\alpha_2(0)=\infty$ and $G_1$ fixes $\infty$, $G_1\alpha_1\alpha_2\cap G_0=\varnothing$. Therefore
%$G_1g*\{G_0,G_2,G_3\}\Leftrightarrow G_1g\in\{G_1,G_1\alpha_1\}.$
%
%Now let $j=0$.  As $G_0g*\{G_2,G_3\}$, by Lemma~\ref{2+3}, $G_0g\in\{G_0, G_0\alpha_1, G_0\alpha_1^{-1}\}$.
%As we also have $G_0g*G_1$ and $G_0\alpha_1$ does not contain an element fixing $\infty$,
%$G_0g\in\{G_0,G_0\alpha_1^{-1}\}.$
%\end{proof}
%
%\begin{lemma}\label{thin}
%$\Gamma$ is thin.
%\end{lemma}
%\begin{proof}
%This is a consequence of Lemmas~\ref{action}, \ref{thin1} and \ref{thin2}.
%\end{proof}
%\subsection{Residual connectedness}

In order to show that the geometry $\Gamma$ is a hypertope, we need to show that it is residually connected. That is, we need to show that each residue of rank at least two has a connected incidence graph. We already showed that all the residues of rank $3$ are hypermaps of type $(3,3,3)$ (Lemmas~~\ref{A4} and \ref{E:3}). Since we know that all such hypermaps are residually connected, as observed in  \cite{{FLW1}}, that will conclude the proof. More precisely, we have the following theorem.
%For this, we shall need the following lemma.

%In what follows, let $\Gamma_i$ denote a residue of a flag of type $i$, for each $i\in I$.
%
%\begin{lemma}\label{residues1}
%$\Gamma_i$ is isomorphic to the coset geometry $\Gamma(G_i, (G_{i,j})_{j\in I\setminus\{i\}})$ where $G_{i,j}=G_i\cap G_j$.
%\end{lemma}
%\begin{proof}
%As all residues of type $i$ are isomorphic, we may consider that $\Gamma_i$ is the residue of $\{G_i\}$. \red{An element of $\Gamma_i$ is a coset $G_jg$ for some $j\in I\setminus\{i\}$ and some $g$ that we may assume lying in $G_i$ as $G_jg\ast G_i$. For $g,h\in G_i$ and $j,k\in I\setminus\{i\}$ we have that
%\[
%G_jg\ast G_kh\Leftrightarrow G_jgh^{-1}\ast G_k\Leftrightarrow G_jgh^{-1}\ast\{G_i,G_k\}\,.
%\]
%According to Proposition~\ref{2+3}, this implies the existence of an element $\gamma\in G_i\cap G_k$ such that $G_jgh^{-1}=G_j\gamma$. Then $G_jg=G_j\gamma h$ and therefore $\beta g=\gamma h$ for some $\beta\in G_i\cap G_j$ since $\gamma hg^{-1}\in G_i\cap G_j$. Thus $G_jg\ast G_kh$ implies
%$G_{i,j}g\ast G_{i,k}h$.
%Obviously $G_{i,j}g\ast G_{i,k}h$ implies $G_jg\ast G_kh$ proving that $\Gamma_i$ is isomorphic to $\Gamma(G_i, (G_{i,j})_{j\in I\setminus\{i\}})$.}
%
%Consider first $i\in\{0,1\}$.
%By Lemma~\ref{01}, we have that $G_jg*G_i$ if and only if $g\in H_i$.
%Thus  $\Gamma_i$ has $q$ elements of each type. By Lemma~\ref{E:3}, the subgroup $G_i\cong E_q: C_3$ hence the number of elements of $\Gamma_i$ is precisely the number of elements of $G_i$.
%Moreover for $g,h\in H_i$, $G_jg*G_kh\Leftrightarrow G_{i,j}g*G_{i,k}h $.
%
%Now consider $i\in\{2,3\}$. By Lemma~\ref{01} the elements of type $j\neq 0$ that are incident to $\{G_i\}$ are
%$$\{G_j, G_j\alpha_j, G_j\alpha_j\alpha_k, G_j\alpha_j\alpha_k^{-1}\}$$
%with $k\in \{1,2,3\}\setminus \{i,j\}$. The elements of type $0$ incident to $\{G_i\}$ are  $\{G_0, G_0\alpha_1,$ $G_0\alpha_1^{-1}, G_0\alpha_k\}$ where $k\in \{2,3\}\setminus \{i\}$.
%In total there are $12$ elements incident to $\{G_i\}$, precisely the same number of elements of $G_i\cong A_4$. Moreover if $G_jg*G_i$ then $g\in G_i$, hence $G_jg*G_i$.
%
%In both cases the geometry is isomorphic to $\Gamma(G_i, (G_{i,j})_{j\in I\setminus\{i\}})$.
%\end{proof}

%\begin{lemma}\label{residues2}
%The residues of rank three are toroidal hypertopes.
%\end{lemma}
%\begin{proof}
%This is a consequence of Lemmas~\ref{A4}, \ref{E:3} and \ref{residues1}.
%\end{proof}


%\begin{corollary}\label{RC}
%$\Gamma$ is residually connected.
%\end{corollary}
%\begin{proof}
%This is a consequence of Lemmas~\ref{residues1} and \ref{residues2} and Proposition~\ref{connected}.
%\end{proof}

%Now we have all the material needed to show the following theorem.
\begin{theorem}
$\Gamma$ is a hypertope.
\end{theorem}
\begin{proof}
As pointed out before, $\Gamma$ is a geometry, and by Proposition~\ref{thin}, it is thin. By Lemmas~\ref{A4} and \ref{E:3}, its rank 3 redidues $\Gamma_i$ are toroidal hypertopes isomorphic to the coset geometries $\Gamma(G_i,(G_i\cap G_j)_{j\in I\setminus\{i\}})$. This, together with Proposition~\ref{connected}, implies that $\Gamma$ is residually connected. Therefore $\Gamma$ is a hypertope.
%This is a consequence of Lemmas~\ref{thin} and Corollary~\ref{RC}.
\end{proof}

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\subsection{The residues of rank 3}

The residues of rank three of $\Gamma$ are hypermaps of type $(3,3,3)_{(a,b)}$ for some $a$ and $b$ that depend on the type of the residue.
To determine the vector $(a,b)$ of the rank 3 residues of $\Gamma$ we now fix an order for the generators of each $G_i$. Let the first generator of $G_i$ be the one with minimal label when $i\neq 0$ and let the first generator of $G_0$ be $\alpha_1^{-1}\alpha_2$.

\begin{lemma}\label{residues3}
The residues $\Gamma_i$ with $i\in \{2,3\}$ are regular hypermaps of type $(3,3,3)_{(0,2)}$. When $q=p^2$ the residues  $\Gamma_i$ with $i\in \{0,1\}$ are regular hypermaps of type $(3,3,3)_{(0,p)}$. When $q=p$ the residue $\Gamma_0$ is the chiral hypermap of type $(3,3,3)_{(1,e)}$ while the residue  $\Gamma_1$ is the chiral hypermap of type $(3,3,3)_{(e,1)}$, where $e$ is a third root of unity (which is an integer, as $q=p$).
\end{lemma}
%Observe that, in the above lemma, when $q=p$, $e$ is an integer.
\begin{proof}
For the following see \ref{eq:toroidal}.  As $(\alpha_1\alpha_2)^2=(\alpha_1\alpha_3)^2=1_{G^+}$, the residues $\Gamma_2$ and $\Gamma_3$ are regular hypermaps of type $(3,3,3)_{(0,2)}$.

Consider first the case $q=p^2$. In this case the order of $(\alpha_1^{-1}\alpha_2)(\alpha_1^{-1}\alpha_3)$ is $p$. Hence $\Gamma_0$ and $\Gamma_1$ are regular hypermaps of type $(3,3,3)_{(0,p)}$.

Now let $q=p$.
We have that $\m{1}{e}{0}{1}=\m{1}{1}{0}{1}^e$ which gives the following relation in $G^+$,
$$(\alpha_2\alpha_3^{-1}\alpha_2)^e(\alpha_2\alpha_3)=1_{G^+}.$$
Hence $\Gamma_1$ is a chiral hypermap of type $(3,3,3)_{(e,1)}$.
We also have that $(\alpha_1^{-1}\alpha_2)(\alpha_1^{-1}\alpha_3)=\m{1}{0}{1+e}{1}$ and
$(\alpha_1^{-1}\alpha_2)(\alpha_1^{-1}\alpha_3)^{-1}(\alpha_1^{-1}\alpha_2)=\m{1}{0}{1}{1}$.
As $\m{1}{0}{1+e}{1}^e= \m{1}{0}{-1}{1}$ and $\m{1}{0}{-1}{1}$ is the inverse of $\m{1}{0}{1}{1}$,
$$(\alpha_1^{-1}\alpha_2)(\alpha_1^{-1}\alpha_3)^{-1}(\alpha_1^{-1}\alpha_2)[(\alpha_1^{-1}\alpha_2)(\alpha_1^{-1}\alpha_3)]^e=1_{G^+}.$$
Hence $\Gamma_0$ is a chiral hypermap of type $(3,3,3)_{(1,e)}$.
\end{proof}

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\section{The symmetries of $\Gamma$}\label{sec:symm}
As pointed out in the introduction, the hypertopes we have constructed in this paper are highly symmetric. Recall that we have two families: a family of hypertopes $\Gamma^{(p)}=\Gamma(G^+,R)$ with $G^+=PSL(2,p)$ for any prime number $p$ satisfying $p\equiv 1\!\!\mod 3$, and a family of hypertopes $\Gamma^{(p^2)}=\Gamma(G^+,R)$ with $G^+=PSL(2,p^2)$ for any prime $p$ satisfying $p\equiv 2\!\!\mod 3$, where $R=\{\alpha_1,\alpha_2,\alpha_3\}$ for both families. %constructed, a hypertope for each $PSL(2,q)$, with $q$  a prime, and another hypertope for each $PSL(2,q)$, with $q$ a square of a prime.
In this section we shall show that all hypertopes $\Gamma^{(p)}$ are chiral, while all hypertopes $\Gamma^{(p^2)}$ are regular. Furthermore, we study the correlations of the hypertopes and find that all the chiral hypertopes $\Gamma^{(p)}$ have both proper and improper correlations.
%This is an interesting characteristic of hypertopes, as in the case of polytopes this cannot happen.
For any $\Gamma$ in one of these two families we have the following lemma.

\begin{lemma}\label{2orbits}
$G^+$ acts with two orbits on chambers of $\,\Gamma$ with adjacent chambers in different orbits.
\end{lemma}
\begin{proof}
This is a consequence of Propositions~\ref{action} and \ref{thin}.
\end{proof}

\begin{lemma}\label{chirality}
Any hypertope $\Gamma^{(p)}$ is chiral while any hypertope $\Gamma^{(p^2)}$ is regular.
\end{lemma}
\begin{proof}
By Lemma~\ref{residues3}, one residue of $\Gamma^{(p)}$ is chiral. Hence $\Gamma^{(p)}$ is itself chiral, indeed by Theorem~\ref{chiraltest} the residues of a regular hypertope must all be regular.
For $\Gamma^{(p^2)}$ we have that $G^+=PSL(2,p^2)$. Then, the Frobenius automorphism of $GF(p^2)$ gives rise to the involutory automorphism
$$\varphi : G^+\to G^+,\,\m{a}{b}{c}{d}\mapsto \m{a^p}{b^p}{c^p}{d^p}$$
which fixes $\alpha_1$ and interchanges $\alpha_2$ and $\alpha_3$ (since $p\equiv 2\!\!\mod 3$ and therefore $e^p=e^2$ in that case).
Let $\psi$ be the automorphism of $G^+$ given by conjugation with
$$A :=\m{-1}{1}{0}{1}\in GL(2,p^2).$$
Then $\psi\varphi $ is an automorphism of $G^+$ inverting $\alpha_1$, $\alpha_2$ and $\alpha_3$.
Hence, by Theorem~\ref{chiraltest} $\Gamma^{(p^2)}$ is a regular hypertope.
\end{proof}

%\red{
%\begin{corollary}\label{cor:Aut}
%$Aut_I(\Gamma^{(p)})$ acts with two orbits on chambers and is isomorphic to $G^+=PSL(2,p)$.
%\end{corollary}
%\begin{proof}
%The statement follows from Lemma~\ref{2orbits} and Lemma~\ref{chirality}.
%\end{proof}}

%If $q=4$, the hypertope is universal, meaning that the relations given by the rank three residues are sufficient to define $G^+$. Indeed the size of the type preserving automorphism groups in this case is $120$, that is precisely the size of the geometric group $[112]^q$, that is a group with tetrahedral diagram (see Theorem 9E14 of~\cite{ARP}). %Hence $\alpha_i\mapsto \alpha_i^{-1}$ is an automorphism of $G^+$.

%\begin{lemma}\label{correlations1}
%The automorphism of $G^+$ sending
%%Consider the automorphism $\varphi$ of $G^+$ defined as follows.
%\begin{eqnarray*}
%%\varphi : G^+&\to& G^+,\\
%\m{a}{b}{c}{d}&\mapsto& \m{a^5}{b^5}{c^5}{d^5},
%\end{eqnarray*}
%induces an automorphism of $\Gamma$, that is a correlation.
%%If $q=p^2$, $\varphi$ is a correlation of $\Gamma$.
%When $q=p$ (and hence $\Gamma$ is chiral),  such correlation is an improper correlation of $\Gamma$.
%\end{lemma}
%\begin{proof}
%This automorphism interchanges $\alpha_2$ and $\alpha_3$ and fixes $\alpha_1$. All regular residues remain the same when this automorphism is applied, while chiral residues are mapped into their mirror images, that is, $(3,3,3)_{(1,e)}\mapsto (3,3,3)_{(e,1)}$ and vice-versa.
%\end{proof}

For the chiral hypertopes $\Gamma^{(p)}$ we will now exhibit proper and improper correlations.
In order to give a proper correlation of $\Gamma^{(p)}$, we remark the following general fact. Given a coset geometry $\Gamma:=\Gamma(G,(G_i)_{i\in I})$, any $\varphi\in Aut(G)$ satisfying $\{\varphi(G_i)\mid i\in I\}=\{G_i\mid i\in I\}$ gives rise to an automorphism of $\Gamma$ mapping $G_ig$ to $\varphi(G_ig)=\varphi(G_i)\varphi(g)$. In particular, we have the following lemma.
\begin{lemma}\label{correlations2}
The automorphism $\eta$ of $\,G^+=PSL(2,p)$ defined by conjugation with the involution $$B:=\m{0}{1}{1}{0}\in GL(2,p)$$
gives rise to a proper duality of $\,\Gamma^{(p)}$.
\end{lemma}
\begin{proof}
As $\eta$ is an involution satisfying
\[
\eta(\alpha_1)=\alpha_1^{-1}\,,\quad \eta(\alpha_2)=\alpha_1^{-1}\alpha_3\,,\quad \eta(\alpha_3)=\alpha_1^{-1}\alpha_2\,,
\]
it interchanges $G_0$ with $G_1$ and $G_2$ with $G_3$.
\end{proof}

To give an improper correlation of $\Gamma^{(p)}$, we extend the assignment
\[
\mu:G_0\mapsto G_0\alpha_1^{-1}\,,\ G_1\mapsto G_1\,,\ G_2\mapsto G_3\,,\ G_3\mapsto G_2
\]
to a bijection of the set of elements of $\Gamma^{(p)}$ by setting
\[
\mu(G_ig):=\mu(G_i)g^A\,,\qquad\mbox{where}\quad A:=\m{-1}{1}{0}{1}\in GL(2,p)\,.
\]
Note that $A$ is an involution and that $\alpha_1^A=\alpha_1^{-1}$, $\alpha_2^A=\alpha_3^{-1}$, $\alpha_3^A=\alpha_2^{-1}$. Therefore $\mu(G_0)^A=\alpha_1G_0$ and $\mu(G_i)^A=G_i$ for any $i\in\{1,2,3\}$. Then, for any $j\in I$ we have
\[
\mu(G_jg)=\mu(G_jh)\Leftrightarrow\mu(G_j)g^A=\mu(G_j)h^A\Leftrightarrow\mu(G_j)^Ag=\mu(G_j)^Ah\Leftrightarrow G_jg=G_jh\,,
\]
showing that $\mu$ is well-defined and a bijection. To prove that $\mu$ is a correlation we need the following lemma.

\begin{lemma}\label{lem:=}
$\{G_ig^A\alpha_1\mid g\in H_0\}=\{G_ih\mid h\in H_0\}$ for any $i\in\{1,2,3\}$.
\end{lemma}
\begin{proof}
Let $g=\m{1}{0}{c}{1}\in H_0$. Then
\[
\begin{array}{ll}
g^A\alpha_1=\m{-c}{1}{-c-1}{1} & =\m{1}{1}{0}{1}\m{1}{0}{-c-1}{1}\in \{G_1h\mid h\in H_0\}\\
                               & =\alpha_1\m{1}{0}{-c}{1}\in \{G_2h\mid h\in H_0\}\cap\{G_3h\mid h\in H_0\}\,.
\end{array}
\]
Hence $G_ig^A\alpha_1\in\{G_ih\mid h\in H_0\}$.
\end{proof}



\begin{lemma}\label{correlations1}
The bijection $\mu$ is an improper duality of $\,\Gamma^{(p)}$.
\end{lemma}
\begin{proof}
%One can easy check that $\mu$ is an involution.
By definition $\mu$ sends the chamber $\{G_0,G_1,G_2,G_3\}$ to $\{G_0\alpha_1^{-1},G_1,G_2,G_3\}$, the adjacent chamber. 
Hence we have only to prove that $\mu$ is a correlation.
% Let $i,j\in I$ be distinct and let $g,h\in G^+$. Then $\mu(G_ig)*\mu(G_jh)\Leftrightarrow\mu(G_i)g^A*\mu(G_j)h^A\Leftrightarrow\mu(G_i)\left(gh^{-1}\right)^A*\mu(G_j)\Leftrightarrow\mu(G_igh^{-1})*\mu(G_j)$\,.
%This show that
It is enough to prove that for any $g\in G^+$ and any $i,j\in I$ with $i>j$
\[ \mu(G_ig)*\mu(G_j)\Leftrightarrow G_ig*G_j\,. \]
According to the definition of $\mu$ and Lemma~\ref{01},
%taking into account all the above mentioned properties of the conjugation in $G^+$ with the involution $A$,
we have
\[
\begin{array}{ll}
\mu(G_ig)*\mu(G_j) & \Leftrightarrow\mu(G_i)g^A*\mu(G_j)\\[6pt]
                   & \Leftrightarrow\left\{
\begin{array}{ll}
\mu(G_i)g^A\alpha_1\in\{\mu(G_i)h\mid h\in H_0\} & \mbox{ if }\, j=0\,,\\
\mu(G_i)g^A\in\{\mu(G_i)h\mid h\in H_1\} & \mbox{ if }\, j=1\,,\\
G_3g^A\in\{G_3,G_3\alpha_3,G_3\alpha_3\alpha_1,G_3\alpha_3\alpha_1^{-1}\} &  \mbox{ if }\, j=2\; (\Rightarrow i=3).
\end{array}
\right.\\[24pt]
                   & \Leftrightarrow\left\{
\begin{array}{ll}
G_ig\alpha_1^{-1}\in\{G_ih^A\mid h\in H_0\} & \mbox{ if }\, j=0\,,\\
G_ig\in\{G_ih^A\mid h\in H_1\} & \mbox{ if }\, j=1\,,\\
G_3g\in\{G_3,G_3\alpha_3^{-1},G_3\alpha_3^{-1}\alpha_1^{-1},G_3\alpha_3^{-1}\alpha_1\} &  \mbox{ if }\, j=2\,.
\end{array}
\right.\\[24pt]
                   & \Leftrightarrow\left\{
\begin{array}{ll}
G_ig\in\{G_ih^A\alpha_1\mid h\in H_0\} & \mbox{ if }\, j=0\,,\\
G_ig\in\{G_ih^A\mid h\in H_1\} & \mbox{ if }\, j=1\,,\\
G_3g\in\{G_3,G_3\alpha_3^{-1},G_3\alpha_3^{-1}\alpha_1^{-1},G_3\alpha_3^{-1}\alpha_1\} &  \mbox{ if }\, j=2\,.
\end{array}
\right.
\end{array}
\]
Using Lemma~\ref{lem:=} ($j=0$), checking that $A$ normalizes $H_1$ ($j=1$) and verifying that
\[
\{G_3,G_3\alpha_3^{-1},G_3\alpha_3^{-1}\alpha_1^{-1},G_3\alpha_3^{-1}\alpha_1\}=\{G_3,G_3\alpha_3\alpha_1,G_3\alpha_3,G_3\alpha_3\alpha_1^{-1}\}\quad (j=2)\,,
\]
we conclude that $\mu(G_ig)*\mu(G_j)\Leftrightarrow G_ig*G_j$ for any $i,j\in I$ with $i>j$ using again Lemma~\ref{01}.
\end{proof}

As it has been shown in Lemmas~\ref{correlations2} and~\ref{correlations1}, the chiral hypertopes $\Gamma^{(p)}$ have both proper and improper correlations simultaneously. We therefore have the following theorem.

\begin{theorem}
There exist chiral hypertopes of rank $4$ having both proper and improper correlations simultaneously.
\end{theorem}

We finish this section with the following two remarks.
\begin{enumerate}
\item[1.] All residues of rank 3 of the regular hypertope $\Gamma^{(4)}$ are regular hypermaps of type $(3,3,3)_{(0,2)}$ (Lemma~\ref{residues3}).  Consequently a correspondence fixing a generator of $G^+$ and interchanging the other two, can be extended to an automorphism of $G^+$. Thus $\Gamma^{(4)}$ has all possible kinds of correlations.
In other words, $Aut(\Gamma^{(4)} )\cong S_5\times S_4 \cong PSL(2,4)\times S_4$.
\item[2.] We observe that the tetrahedral groups $G^+$ arise naturally as subgroups of index 4 in the symmetry group $[6,3,6] =\langle r_0, r_1, r_2, r_3\rangle$ of the hyperbolic tessellation $\{6,3,6\}$ given in~\cite[Section 11]{SW1994} (generated by $r_1$, $r_2$, $(r_1)^{r_0}$ and $(r_2)^{r_3}$).
\end{enumerate}

\section{A geometric construction of $\Gamma$}\label{section5}

We now give a geometric description of the rank 4 hypertopes $\Gamma = (X, *, t, I)$  constructed in the previous sections.
The set $I=\{0,1,2,3\}$ of types gives a partition of the set of elements of $\Gamma$ into four sets $X_0, X_1, X_2$ and $X_3$, where $x\in X_j$ if and only if $t(x)=j$.
The elements of $X_0, X_1, X_2$ and $X_3$ can be described from a certain action in the projective line $PG(1,q)$, where $q=p$ if $\Gamma=\Gamma^{(p)}$ and $q=p^2$ if $\Gamma=\Gamma^{(p^2)}$.
In what follows we define each of the $X_i$ and the incidence between them.
%The proof that the defined $\Gamma$ is a hypertope using the geometry instead of the algebra is left to the interested reader.

%Let $q$ be either a prime or the square of a prime, such that $3$ divides $q-1$.
Let $G^+$ be the group $PSL(2,q)$ acting on the projective line $PG(1,q) = GF(q) \cup \{\infty\}$  as follows.
\[
PG(1,q)\times G^+\rightarrow PG(1,q),\ (z,g)\mapsto z^g:=g^{-1}(z),
\]
where $g^{-1}(z)=Az$ if $g^{-1}=\pm A$ with $A\in SL(2,q)$, as in Section \ref{sub:action}.

Consider a transversal $U$ for the stabilizer $G_\infty^+$ of $\infty$. The index of $G_\infty^+$ in $G^+$ is $q+1$, since $G_\infty^+$ is the semi-direct product of the elementary abelian $p$-group $H_1$ of order $q$ (given in Lemma~\ref{E:3}) with the cyclic group $D$ of diagonal matrices in $G^+$, which has order $\frac{q-1}{d}$ with $d=\gcd(2,q-1)$. As $q+1=|PG(1,q)|$, we can identify any $u\in U$ with the element $z:=\infty^u$ of $PG(1,q)$ and write $u=u_z$.

By Lemma~\ref{E:3} the group $G_1=H_1:(G_0\cap G_1)$ is a subgroup of $G_\infty^+=H_1:D$. Consider a transversal $T$ for $G_1$ in $G_\infty^+$.
Let $X_0$ and $X_1$ be two disjoint copies of $T\times PG(1,q)$. For $x_0=(t,z) \in X_0$ and $x_1=(s,w)\in X_1$ we say that
%Let $U$ be a transversal \red{for} $G_\infty$ in $G^+$.
%The stabilizer $G_{\infty,0}\red{:=G_\infty\cap G_0}$ in $G^+$ is a cyclic group $C$ of order $\frac{q-1}{d}$. It contains a unique cyclic subgroup $D$ of order $3$.
%Similarly, $G_\infty$ contains a unique subgroup $H_\infty := E_q:D$.
%Let $T$ be a transversal of $D$ in $C$ (or similarly a transversal of $H_\infty$ in $G_\infty$).
%Elements of $X_0$ and $X_1$ correspond to cosets of $H_\infty$ in $G^+$. Geometrically, a coset $H_\infty*g$ can be written as $H_\infty*t*u$ with $t\in T$ and $u\in U$. It corresponds to a pair $(x,u)$ where $\infty^u = x$.
%Let $X_0$ and $X_1$ be two copies of the cosets of $H_\infty$ in $G^+$.
%We shall say that, if $x_0= (x,t)$ and $x_1=(y,s)$, then
\begin{eqnarray}\label{Inci0-1}
x_0 * x_1\mbox{ whenever } t=s\mbox{ and } z \neq w.
\end{eqnarray}
It is straightforward to see that an element $x_0 \in X_0$ is incident with exactly $q$ elements of $X_1$ and vice-versa.

Let $X_2$ and $X_3$ be the orbits of the sets $\{1, e, e^{2}, \infty\}$ and $\{1,  e, e^{2}, 0\}$ under the action of $G^+$, respectively.
%Note that, since $G$ is sharply $3$-transitive on $PG(1,q)$, then $\{1,  k, k^{2}, \infty\}$ and $\{1, k, k^{2}, 0\}$ are never in the same orbit under $G^+$. THIS IS FALSE AS YOU CAN MAP ONE SET TO ANOTHER
Remarking that all possible cross ratio of four distinct points in $\{1, e, e^{2}, \infty\}$ or in $\{1,  e, e^{2}, 0\}$ take only two values, we conclude that both sets have a stabilizer isomorphic to $A_4$. Hence  $|X_2|=|X_3|=\frac{(q+1)q(q-1)}{12d}$.
%Sometimes $X_3$ is just a copy of $X_2$ and sometimes not. It actually depends on how many conjugacy classes of subgroups $A_4$ the group $G^+$ has. If $q \equiv\pm 3 \mod 8$ (meaning also that $q$ is a prime), there is only one and $X_3$ is a copy of $X_2$. If $q \equiv \pm 1 \mod 8$, then there are two conjugacy classes of subgroups $A_4$ and $X_2$ and $X_3$ are disjoint sets.

Given $x_2\in X_2$ and $x_3\in X_3$, we shall say that
\begin{eqnarray}\label{Inci2-3}x_2*x_3\mbox{ whenever }|x_2\cap x_3| = 3.\end{eqnarray}
Then, each element $x_2 \in X_2$ is incident with exactly  $4$ elements of $X_3$ and vice-versa.

Before defining the incidences between any other two elements, we shall first define the incidence between
$\{1, e, e^{2}, \infty\}\in X_2$ and an element $(t,z)$ of $X_0$ or $X_1$. We say that
\begin{eqnarray}\label{Inci2-j}
\{1, e, e^{2}, \infty\} * (t,z)\mbox{ if and only if }t=1_{G^+}\,\mbox{ and }z\in\{1, e, e^{2}, \infty\}.
\end{eqnarray}
Having \ref{Inci2-j} in mind, the transversal $U$ can be choosen in such a way that $u_z\in G^+_{\{1,e,e^2,\infty\}}$ for $z\in\{1,e,e^2,\infty\}$. This is possible and
motivated by the following. If the coset $G_1tu_z$ intersects $G^+_{\{1,e,e^2,\infty\}}$, then $z\in\{1,e,e^2,\infty\}$ and therefore
$u_z\in G^+_{\{1,e,e^2,\infty\}}$ by our choice of $U$. It follows then that $G_1t$ intersects $G^+_{\{1,e,e^2,\infty\}}$. As $G_1t$ is in $G^+_\infty$ and
$G^+_\infty\cap G^+_{\{1,e,e^2,\infty\}}<G_1$, we have that $G_1t\cap G_1\neq\varnothing$, that is $t=1_{G^+}$. Now we can say that an element
$\{1,e,e^2,\infty\}^g$ in $X_2$
will be incident to those $(t,z)$ (in $X_0$ or $X_1$) such that $G^+_{\{1,e,e^2,\infty\}}\cap G_1tu_zg^{-1}\neq\varnothing$. Note that, if
$g\in G^+_{\{1,e,e^2,\infty\}}$, then $\{1,e,e^2,\infty\}^g=\{1,e,e^2,\infty\}$ will be incident to those $(t,z)$ such that $G^+_{\{1,e,e^2,\infty\}}$
is incident with $G_1tu_zg^{-1}=G_1tu_{z^{g^{-1}}}$ (by the choice of $U$). As we saw, these are the pairs $(t,z)$ with $t=1_{G^+}$ and $z\in\{1,e,e^2,\infty\}$.
Thus incidence is well-defined.

Finally, an element $\{1,e,e^2,0\}^g\in X_3$ will be incident with an element $(t,z)$ in $X_0$ or $X_1$ if the corresponding element
$\{1,e,e^2,\infty\}^g\in X_2$ is incident with $(t,z)$.

By definition, $\Gamma= (X, *, t, I)$ is an incidence system, where $X=\cup_{j=0}^3X_j$, $I=\{0,1,2,3\}$,
$t:X\to I$ is such that $t(x)=j$ whenever $x\in X_j$ and the incidence $*$ is given by (\ref{Inci0-1}), (\ref{Inci2-3}), (\ref{Inci2-j}) and
the last paragraph above. Furthermore, it is straightforward to see that every flag is in a chamber, implying that $\Gamma$ is in fact
an incidence geometry. By construction, $\Gamma$ is isomorphic to the hypertope given in section \ref{section4}.



\subsection*{Acknowledgements}
This research was accomplished when Dimitri Leemans and Isabel Hubard were visiting the University of Aveiro. This visit was support by the Portuguese funds through the CIDMA - Center for Research and Development in Mathematics and Applications, and the Portuguese Foundation for Science and Technology (FCT- Funda\c c\~{a}o para a Ci\^encia e a Tecnologia), through CIDMA - Center for Research and Development in Mathematics and Applications, within project UID/MAT/04106/2013. This work was completed when Dimitri Leemans was visiting Mexico, thanks to the support of PAPIIT-DGAPA-UNAM under project IN101615.
Part of this research was also sponsored Marsden Grant UOA1218 of the Royal Society of New Zealand and by CONACyT (Mexico) under project 166951.

The authors also thank Asia Ivi\`c Weiss for useful comments on a preliminary version of this paper.


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