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\begin{document}
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\begin{center}
\bf $c$ - extensions of  $P$- and $T$-geometries 
\end{center}
\medskip
\begin{center}{\sc
Gernot Stroth, Corinna Wiedorn\\
Martin-Luther-Universit\"at Halle-Wittenberg\\
Institut f\"ur Algebra und Geometrie\\
06099 Halle, Germany}
\end{center}
\bigskip
  
We consider affine extensions of geometries of Petersen  and of tilde type,
i.e., of  flag-transitive geometries $\Gamma $ belonging to 
a diagram 
$$\hbox{\Diagn{1}{2}{n-2}{n-1}{n}{$X$},}$$
where  either
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$=$
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i.e., the geometry of edges and vertices of the Petersen graph,
or
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$=$
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i.e., the 3-fold cover of the generalized $Sp_4(2)$-quadrangle.
We call such geometries
 $c$-$P$- resp. $c$-$T$-geometries because the residue of an element
of type 1 is a $P$- resp. $T$-geometry. Such geometries have been
classified by different authors and an overview over the results is
given in \cite{IS1}. In particular, there exist only two
$P$-geometries of rank 3 with automorphism groups $Aut(M_{22})$
resp. $3Aut(M_{22})$ and three $T$-geometries of rank 3 with automorphism
groups 
$M_{24}$, $He$, and $3^7Sp_6(2)$. In this paper we will assume:
\begin{enumerate}
\item[($*$)] If $F$ is a flag of $\Gamma $ of type $\{ 1,\ldots ,n-3\}$ then
the residue   $res(F)$ of $F$ in $\Gamma $ is the $P$-geometry for
$M_{22}$ or the   $T$-geometry for $M_{24}$. 
\end{enumerate}
Since there are no $P$-geometries of rank $\ge 5$ and no
$T$-geometries of rank $\ge 5$ satisfying ($*$), this forces
$n\le 6$. On the other hand, it is known that for $n=3$ the universal
cover of $\Gamma $ is infinite, and          
the case that  $\Gamma $ is a $c$-$P$-geometry of rank 4 satisfying ($*$)
has been considered in \cite{AC}. There it is shown  that
$Aut(\Gamma )$ is isomorphic to a factorgroup  of  one of  
the groups $2^{11}\!\!n :\! Aut(M_{22})$, $M_{24}$ or $2\!\cdot\!
U_6(2)\! :\! 2$ (where by $H\! :\! K$ and $H\!\cdot\! K$ we denote as
usually a split- resp. non-split extension of a group $K$ by a group $H$).
In 
the first case, the normal 2-subgroup is the universal representation group
(see below) of the $P$-geometry for $M_{22}$. It is isomorphic to a
submodule of index two (the even half) of the Golay co-code.
Models for the other two geometries are described in \cite{AC}.

\newtheorem{defi}{Definition}
\begin{defi}[see \cite{IPS}] Let $\Gamma $ be any geometry which
  contains some sets of 
objects ${\cal P}$ and ${\cal L}$ (``points'' and ``lines'') such that
any line $l\in {\cal L}$ is incident to exactly three points of ${\cal P}$.
Let $U(\Gamma )$ be the abstract group defined by
$$
\begin{array}{rl}
U(\Gamma )=&\langle u_x|x\in {\cal P}, u_x^2=1=u_xu_yu_z \hbox{ whenever
$x,y,z$ are the }\\
&\hbox{ three points incident to a line } l\in{\cal L}\rangle .
\end{array}
$$
Then $U(\Gamma )$ is called the {\bf universal representation group} of the
point-line system $({\cal P},{\cal L})$.\par
If $\Gamma $ is a $P$- or $T$-geometry with diagram 
\begin{picture}(45,5)
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\put(1,2.5){\scriptsize {1}}
\put(11,2.5){\scriptsize{2}}
\put(21,2.5){\scriptsize{n-2}}
\put(31,2.5){\scriptsize{n-1}}
\put(41,2.5){\scriptsize{n}}
\end{picture} 
we define the group $U(\Gamma )$ to be the universal representation group of
the point-line system $({\cal P},{\cal L})$ where $\cal P$ and $\cal L$ are the objects of $\Gamma $ of type 1 and 2.
\end{defi}

Of course, the group $U(\Gamma )$ might become trivial, as it does
e.g. in  case of the
rank 4 $P$-geometry for $M_{23}$. But if it is nontrivial, then
$Aut(\Gamma )$ acts on $U(\Gamma )$ in a natural way and the
semidirect product of $U(\Gamma )$ with $Aut(\Gamma )$ acts
flag-transitively on  a $c$-extension of $\Gamma $.\par
The universal representation groups of the  $P$- and $T$-geometries
occurring as residues in this paper are all known (see \cite{IPS}). For the
$P$-geometry for $Co_2$ it is
an elementary abelian 2-group of order $2^{23}$, and it is  isomorphic to
the submodule 
of index 2  stabilized by $Co_2$ in the Leech lattice mod 2. This
module contains a one-dimensional subspace invariant under the action
of $Co_2$ and the quotient over this subspace is also a representation
group (\cite[(5.7)]{IPS}). The representation groups of the $T$-geometries
for $M_{24}$ 
and $Co_1$ are elementary abelian of order $2^{11}$ resp. $2^{24}$ and
isomorphic to the 11-dimensional irreducible submodule
of the Golay co-code resp. to the Leech lattice mod 2
(\cite[(3.2),(5.8)]{IPS}). Also by \cite{IPS} the  universal 
representation  group of the $P$-geometry for the Babymonster
sporadic simple group $BM$ is a non-split
extension $2\!\cdot\! BM$ and the universal representation group of the
$T$-geometry for the Monster $M$ is isomorphic to $M$. By an unpublished
result of A. Ivanov and S. Shpectorov there exist examples of
$c$-extensions with automorphism groups $(2\!\cdot\!
BM\! *\! 2\!\cdot\! BM)\! :\! 2$ and $(M\!\times\! M)\! :\! 2$, which
are related to the representation group in these cases as well.\par
The aim of the present paper is to show that under the assumption of ($*$) all
$c$-$T$-geometries  arise in the context of the representation group and
that there are only three more
examples of $c$-$P$-geometries , which we will describe in Section
\ref{sec3}. More precisely we prove

\begin{satz}\label{satz1}
Let $G\le Aut(\Gamma )$ be a flag-transitive automorphism-group of a
$c$-$P$-geometry 
$\Gamma$ of rank $n\ge 5$. Assume that the residue of 
each flag of type $\{1,\ldots ,n-3\}$ is the $P$-geometry
for $M_{22}$.
\begin{enumerate}  
\item[(a)] If $n=5$ and the residue of an element of type 1 is the $P$-geometry for $M_{23}$ then  $G\cong M_{24}$.
\item[(b)] If $n=5$ and the residue of an element of type 1 is the
  $P$-geometry for $Co_2$ then either  $G\cong 2^{22}\! :\!Co_2$ resp. $2^{23}\! :\!Co_2$ and $O_2(G)$ considered as module for $G/O_2(G)$ is isomorphic to a section of  the Leech lattice mod 2;  or $G\cong Co_1$. 
\item[(c)] If $n=6$ then either $G$ is isomorphic to one of the groups $BM\!\times\! BM$, $BM\wr {\bf Z}_2$,
$(2\!\cdot\! BM\! *\! 2\!\cdot\! BM)$, $(2\!\cdot\! BM)\wr {\bf Z}_2$,
or $G\cong M$. 
\end{enumerate}
\end{satz}

\begin{satz}\label{satz2}
Let $G=Aut(\Gamma )$ be the automorphism-group of a $c$-$T$-geometry
$\Gamma$ of rank $n\ge 4$. Assume that the residue of 
each flag of type $\{1,\ldots ,n-3\}$ is the tilde geometry
for $M_{24}$. 
\begin{enumerate}
\item[(a)] If $n=4$ then 
$G$ is the semidirect product of  $M_{24}$ with the
(11-dimensional) Golay code module.
\item[(b)] If $n=5$ then $G$ is the semidirect product of  $Co_1$ with
  the Leech lattice mod 2.
\item[(c)] If $n=6$ then $G\cong M\times M$, or $M\wr {\bf Z}_2$.  
\end{enumerate}
\end{satz}

We prove Theorems 1 and 2 in a series of lemmas in Section 4. In Section 2 we derive some general results about subgeometries of $\Gamma $, which apart from being useful for the determination of $Aut(\Gamma )$ also might be of their own interest. Section 3 contains some further preliminary results.\par

For the rest of the paper we fix the following notation:
$\Gamma $ is a $c$-$P$- or a $c$-$T$-geometry, $\Gamma ^{(i)}$ is the set of
objects  
of $\Gamma $ of type $i$, $\{ \alpha _1,\ldots ,\alpha _n\} $ is a maximal
flag in $\Gamma $ with $\alpha _i\in \Gamma ^{(i)}$, $G\le Aut(\Gamma )$ is
a  flag-transitive automorphism group of $\Gamma $,
$G_i=G_{\alpha _i}$ is the stabilizer of $\alpha _i$ in $G$, $K_i$ is the
kernel of the action of $G_i$ on the residue $res(\alpha _i)$ of $\alpha _i$
in $\Gamma $, $P_i=\bigcap_{j\neq
i}G_j$ are the so-called ``minimal parabolics'' and $B=P_1\cap \ldots \cap
P_n$ is the ``Borel subgroup''. Elements of type 1 will sometimes be
called ''points'', elements of type 2 ''lines'', and elements of type 3
''planes''. Further, for $x\in\Gamma ^{(i)}$, by
$res(x)^-$ we denote the geometry induced on the objects of types
$1,\ldots ,i-1$ in $res(x)$ and by $res(x)^+$ the geometry induced on the objects of types $i+1,\ldots ,n$.\par
By $2^n$ we usually denote an elementary abelian 2-group of order $2^n$, by $[2^n]$ any 2-group of order $2^n$, and by $2^{n_1+n_2+\ldots +n_k}H$ we mean the extension of a 2-group $P$ by a group $H$ such that $P$ has an $H$-invariant composition series $1=P_0\unlhd P_1\unlhd \ldots\unlhd P_k=P$ with $P_i/P_{i-1}\cong  2^{n_i}$. The rest of the notation is standard.

\section{Examples of $c$-$P$-geometries}\label{sec3}

\subsection{A geometry for $M_{24}$ with diagram}\label{exm24}
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This example is probably well known but we include it for completeness.
Let ${\cal S}={\cal S}(24,8,5)$ be the Steiner system for $G=M_{24}$
with underlying set $\Omega $ (for some properties of $\cal S$ see
e.g. \cite[chapters 6,7]{Asch}). Let ${\cal O}$ denote the set of octads and
${\cal T}$ the set of trios.  We set
\begin{enumerate}
\item[--] $\Gamma^{(1)}=\Omega $,
\item[--] $\Gamma ^{(2)}=\{ \{ a_1,a_2\} |a_i\in\Omega , a_1\neq a_2\} $
\item[--] $\Gamma ^{(3)}=\{ \{ a_1,a_2,a_3,a_4\} |a_i\in \Omega $, $a_i\neq
a_j$ for $i\neq j\}$
\item[--] $\Gamma ^{(4)}=\{ (O_1,\{ O_2,O_3\})|\{ O_1,O_2,O_3\}\in{\cal T}\}$ 
\item[--] $\Gamma ^{(5)}={\cal O}$.
\end{enumerate}
Incidences between objects of types 1,2,3 are defined by inclusion. An
element  $a\in
\Gamma ^{(1)}$ is incident to $(O_1,\{ O_2,O_3\})\in \Gamma ^{(4)}$ if
$a\in O_1$, and to $O\in \Gamma ^{(4)}$ if  $a\not\in O$. Elements $x\in \Gamma
^{(2)}\cup \Gamma ^{(3)}$ and $y\in \Gamma ^{(4)}\cup \Gamma ^{(5)}$ are
incident if all elements of $x$ are incident to $y$. The elements of type 5
in $res(x)$ for $x=(O_1,\{ O_2,O_3\})\in \Gamma ^{(4)}$ are $O_2$ and $O_3$. 

\subsection{A geometry for $Co_1$ with diagram} 
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\put(7,2.5){c}
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\end{picture}}$$
In order to define the geometry for $Co_1$ first we briefly recall the
definition of the P-geometry for $Co_2$ as it is given in \cite{Sh1}.\par
Let $\Lambda $ denote the Leech lattice with inner
product ( , ) and let $\Lambda _2=\{ v\in \Lambda , (v,v)=32\} $. Consider the
group $H=Co_2$ as the stabilizer in $Aut(\Lambda )$ of a fixed vector $v_0\in 
\Lambda _2$. Let $\Sigma =\{ \{ v,-v\} |v\in \Lambda _2, (v_0,v)=0\} $ and
define a graph  on $\Sigma $ in the way that $\sigma $ and $\tau
\in \Sigma $ are adjacent if $H_{\sigma}\cap H_{\tau }\cong 2^9.2^5\Sigma
_5$. A clique ${\cal C}$ in this graph is called closed if for each 
$\{ v,-v\} , \{ w,-w\}\in {\cal C}$ there exists $\{ u,-u\}\in {\cal C}$ such
that $v_0+v+w+u\in 2\Lambda $. It is shown in \cite{Sh1} that each closed
clique is of size 1,3,7 or 15, and that, if we take as
objects of type $i$ the closed cliques of size $2^i-1$ and define incidence
by inclusion, then we get the P-geometry for $Co_2$. \par
Now we can define the $c$-$P$-geometry for $Co_1$: Let
\begin{enumerate}
\item[--]$\Gamma ^{(1)}=\{ \{ v,-v\} |v\in \Lambda _2\} .$
\end{enumerate}
For $x=\{ v,-v\}\in \Gamma ^{(1)}$ let $\Sigma _x=\{ \{ w,-w\}\in \Gamma
^{(1)}| (v,w)=0\} $ and consider $\Sigma _x$ as the graph just described. Let
\begin{enumerate}
\item[--]$\Gamma ^{(i)}=\{ \{ x\} \cup {\cal C}|x\in \Gamma ^{(1)}, {\cal
C}\subseteq \Sigma _x, |{\cal C}|=2^{i-1}-1$,  ${\cal C}$ is a closed
clique in the graph on $\Sigma _x\} $, for $2\le i\le 5$,
\end{enumerate}
and define incidence  by inclusion. Then it is 
straightforward to show that $\Gamma =\bigcup_{i=1}^5\Gamma ^{(i)}$ is a
geometry with the desired diagram. Since $Co_1$ is transitive on
$\{ \{ v,-v\} |v\in \Lambda _2\} $, it is clear that $Aut(\Gamma )\cong Co_1$.

\subsection{A geometry for the Monster with diagram}
$$\hbox{
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The idea of the construction is quite similar to the previous example.
Recall that $2\cdot BM=C_M(z)$ for an involution
$z\in M$ of type $2A$ in the notation of \cite{A} and that the objects of
the P-geometry for $BM$ can be identified with certain conjugacy classes of
elementary abelian 2-groups of orders $2,4,8,16$ resp. $32$ of $BM$ (see
e.g. \cite{IPS}). Therefore, we
define $\Gamma $ in the following way:
\begin{enumerate}
\item[--] $\Gamma^{(1)}=2A$,
\item[--] $\Gamma ^{(i)}=\{ zU| z\in 2A$, $U\le C_M(z)$, $U$ is elementary
abelian of order  $2^{i-1}$ and corresponds to an object of the P-geometry
for $C_M(z)/\langle z\rangle \cong BM\} $, for $2\le i\le 6$.
\end{enumerate}
Incidence is again defined by inclusion.\par
If $zU\in \Gamma ^{(i)}$ for some $i>1$ and $u\in U$, then $u$ is of type
$2B$ in $M$, i.e. $C_M(u)\cong 2^{1+24}Co_1$, and $\langle z,U\rangle \le
O_2(C_M(u))$. Since $O_2(C_M(u))$ is extraspecial and $N_M(U)\cap C_M(z)$ is
transitive on $U$, $N_M(\langle z, U\rangle )$ is transitive on $zU$. Now it
is straightforward to see that $\Gamma $ is a geometry of $c$-$P$-type. \\
\\

\section{Subgeometries}\label{sec1}

In this section $\Gamma $ always denotes a $c$-$P$-geometry of rank 
$n\ge 4$. The reader might take it as an exercise to generalize
the results to  any flag-transitive geometry
with diagram  \Diagn{1}{2}{n-2}{n-1}{n}{X}, where 
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\end{picture}
may be an arbitrary rank-2-geometry in which every object of the rightmost type  is
incident to exactly three objects of the other type. Further one has to assume that the
stabilizer of an element $x\in \Gamma $ of type $i\ge 2$ 
induces the full affine linear group $2^{i-1}L_{i-1}(2)$ on $res(x)^-$. In
particular, we need the fact  that  $G_x$  acts 3-fold
transitively on the set of points and 2-fold transitively 
on the set of partitions of $res(x)^-$ into classes of parallel lines.  
(This is automatically true if $X=P^*$.) 

\subsection{Shrinking}

In this subsection we  define a new incidence
structure $\cal{S}$ related to $\Gamma $. In general, the structure $\cal S$ might not be connected (in fact,
for most of the $c$-$P$-geometries it is disconnected), but it is
not dificult to see that its connected components are  
geometries with the same diagram as $\Gamma $ but of rank $n-1$. (Therefore,
we call this construction ``shrinking''.) \par
The structure ${\cal S}$ has the following sets of objects: 
\begin{enumerate} 
\item[--] ${\cal S}^{(1)}=\Gamma^{(2)}$,
\item[--] ${\cal S} ^{(i)}=\{ (x,P_x)| x\in \Gamma^{(i+1)}$, $P_x$ is a
partition of $res(x)^-$ into parallel lines$\}$,
\end{enumerate}
for  $2\le i\le n-1$.
The elements $(x,P_x)\in {\cal S} ^{(i)}$, $(y,P_y)\in {\cal S} ^{(j)}$ with
$i,j>1$, are  incident if $x$ and $y$ are incident in $\Gamma
$ and $P_x$ is compatible with $P_y$; $x\in {\cal S} ^{(1)}$ and $(y,P_y)\in
{\cal S} ^{(i)}$, $i>1$, are incident if $x$ is incident to $y$ in $\Gamma
$ and is part of the partition $P_y$ (recall that $x$ is a line).\par
\bigskip
In the following we derive some properties of $\cal S$. The first lemma is due to A. Pasini.

\begin{lem}\label{lem3} Let $\cal S$ be
disconnected. Then, for every $x\in \Gamma ^{(i)}$, $i\ge 
3$, and two different partitions $P_1$,$P_2$ of $res(x)^-$ into parallel lines, the
objects $(x,P_1)$ and $(x,P_2)$ belong to different connected components of
$\cal S$.
\end{lem}

{\it Proof.}~~The relation ``giving rise to objects in the same connected
component'' is clearly an equivalence relation on the set of partitions of $res(x)^-$
into parallel lines and it is preserved by the action of $G_x$ on that
set. But as mentioned above that action is 2-transitive and therefore
primitive. So either we 
have the statement or 
all partitions give rise to objects in the same connected component of $\cal
S$. However, in the latter case the connectedness of $\Gamma $ forces $\cal
S$ to be connected. \hfill $\Box $\\ 
 
We can also describe the shrinking in terms of the
corresponding parabolic subgroups: 
If $P_1,\ldots ,P_n$ are the minimal parabolics corresponding  to
a fixed maximal flag of $\Gamma $ and $R_1,\ldots ,R_{n-1}$ the minimal
parabolics belonging to a certain maximal flag in a 
suitable connected component $\Sigma $ of $\cal{S}$, then $R_i=\langle
P_1,P_{i+1}\rangle $ for $i\ge 2$ and $R_1=N_{\langle P_1,P_2\rangle }(P_1)
$. If $G_{\Sigma }$ denotes the stabilizer of $\Sigma $ in $G$,
then clearly $G_{\Sigma }=\langle R_1,\ldots ,R_{n-1}\rangle $ and $\cal S$
is connected iff $G=G_{\Sigma }$. For the Borel-subgroup
$B_{\Sigma }$ we have $B_{\Sigma }=R_1\cap\ldots\cap R_{n-1}=P_1$, hence
$|B_{\Sigma }:B|=2$.

\begin{lem}\label{lem2}
Let $G_{\Sigma }$ be the stabilizer of a connected component $\Sigma $ of
$\cal S$, $N\unlhd G_{\Sigma }$ the kernel of the action of $G_{\Sigma }$ on
$\Sigma $ and $H=G_{\Sigma }/N$. Suppose
\begin{enumerate}
\item[(a)] the kernel of the action of $H_p$ on the residue of $p$ in
  $\Sigma $ is trivial for $p\in
\Sigma $ of type 1,
\item[(b)] $G_2/K_2\cong {\bf Z}_2\times G_{12}/K_2$.
\end{enumerate}
Let $Q_2\unlhd G_2$ such that $|Q_2:K_2|=2$ and $G_2/Q_2\cong
G_{12}/K_2$. Then $N=Q_2$.
\end{lem}

{\it Proof.}~~We  identify $p\in \Sigma $ with $\alpha _2\in\Gamma
$ and $H_p$ with $G_2N/N$. Now $Q_2$ is trivial on the elements of types
$3,\ldots ,n$ of $res(\alpha _2)$ in $\Gamma $, hence it is trivial on
the residue of 
$p$ in $\Sigma $. By assumption (a) $Q_2\le N$. \par
Let $F_{\Sigma }$ be a maximal flag in $\Sigma $. Then
$F_{\Sigma }$ is stabilized by $N$. By
construction, $F_{\Sigma }$ comes from a unique flag $F$ of cotype 1 in
$\Gamma $. So $N$ stabilizes $F$. Since $F$ is contained in exactly two
maximal flags, we have $|N:N\cap B|\le 2$. On the other hand, 
our construction implies that $N\cap B\le K_2$. So the assertion follows
from $Q_2\le N$ and  $|Q_2:K_2|=2$.\hfill $\Box $\\

\begin{cor}\label{cor1}
Suppose the assumptions of Lemma \ref{lem2}. Then ${\cal S}$ is
disconnected.
\end{cor}

{\it Proof.}~~ By Lemma \ref{lem2} $Q_2=N\unlhd G_{\Sigma }$. Since $Q_2\le
G_2$, therefore $Q_2$ must stabilize each
element of $\{ \alpha _2^{G_{\Sigma }}\}$. Hence $G_{\Sigma }\neq G$.\hfill
$\Box $\\

If $n\ge 5$, we can repeat the  shrinking construction applying it to $\cal
S$. Then we get a new  incidence structure,  say $\cal T$, whose
connected components 
are contained in the connected components of $\cal S$ and constitute
$c$-$P$-geometries of rank $n-2$.
For  the set ${\cal T}^{(1)}$ of objects of $\cal T$ of type 1 we have by construction
$${\cal T}^{(1)}={\cal S}^{(2)}=\{ (x,P_x)| \hbox{$x\in \Gamma^{(3)}$, $P_x$ is a partition of $res(x)^-$ into parallel lines$\} $.} $$
Now, for each $x\in\Gamma ^{(3)}$, there are exactly three different
partitions $P_1,P_2,P_3$ of $res(x)^-$ into two parallel lines, and by Lemma
\ref{lem3} either $\cal S$ is connected or  each pair $(x,P_i)$ belongs to a
different connected component of $\cal S$, hence also to a different
connected component of $\cal T$. Define a relation $\equiv $ on the set of
connected components of $\cal T$ by
$\Theta _1:\equiv \Theta _2$ if there exist $x\in \Gamma^{(3)}$ and
partitions $P_1,P_2$ of $res(x)^-$ into parallel lines such that $(x,P_i)$ belongs
to $\Sigma _i$ for $i=1,2$. Then the following holds.
 
\begin{lem}
Suppose $\cal S$ is disconnected and define the relation $\equiv$ as just
described. 
If $\Theta _1\equiv \Theta _2$ then the elements of $\Theta _1^{(1)}$ and  
$\Theta _2^{(1)}$ arise from the same set of elements from
$\Gamma^{(3)}$. Moreover, 
$\equiv $ is an equivalence relation on the set of connected components of
$\cal T$,  each equivalence class is of size three and, if $\{\Theta
_1,\Theta _2,\Theta _3\}$ is such an equivalence class, then the $\Theta _i$
are contained in different connected components of $\cal S$.   
\end{lem}

{\it Proof.}~~ Obviously $\equiv$ is reflexive and symmetric. Moreover, if
the  first statement of  the lemma is shown, then it is clear that $\equiv $
is also transitive, hence an equivalence relation, and Lemma \ref{lem3}
together with the fact that, for any $x\in \Gamma ^{(3)}$, there are exactly
three partitions of $res(x)^-$ into parallel lines implies the rest of the
assertions. \par  
Let $\Theta _1\equiv \Theta _2$ and  $x\in
\Gamma^{(3)}$ such that $(x,P_i)\in \Theta _i$ for certain partitions
$P_1,P_2$ of $res(x)^-$. By $res(x,P_i)^{(2)}$ denote the set of  elements of
type 2 in the residue of $(x,P_i)$ in $\Theta _i$. It follows from the
construction of $\cal T$ that
\begin{enumerate}
\item[$res$]$(x,P_i)^{(2)}=\{ (l,P_{S,l})|l\in {\cal S}^{(3)}$, $l$ is
  incident to $(x,P_i)$ in ${\cal S}$, and $(x,P_i)$ is part of the
  partition $P_{S,l}$ of the residue of $(l,P_{S,l})$ into parallel lines of
  ${\cal S}\} $. 
\end{enumerate}
As $l\in {\cal S}^{(3)}$ and $l$ is incident to $(x,P_i)$, by
construction of $\cal S$ there is $l_0\in\Gamma ^{(4)}\cap res(x)$ and
a partition $P_0$ of $res(l_0)^-$ into parallel lines of $\Gamma $ such that
$l=(l_0,P_0)$ and $P_0$ is compatible with the partition $P_i$. In terms of $\Gamma $ then $x$ is a
plane of $l_0$  and $P_{S,l}$ induces a partition of $res(l_0)^-$ into two parallel
planes. If $(x',P_i')\in \Theta _i^{(1)}$ is the unique element incident to
$(l,P_{S,l})$ in $\Theta _i$ and different from $(x,P_i)$, then $x'$ is 
also a plane of $res(l_0)^-$ and $(x',P_i')$ must be part of
the partition $P_{S,l}$. So  $x'$ must be the
 plane in $res(l_0)^-$ parallel to $x$.  This means that 
 there is a bijection between the
 set of points collinear to $(x,P_i)$ in $\Theta _i$ and the planes parallel
 to $x$ in $\Gamma $. Since  $\Theta _i$ is connected, this implies the
 assertion. \hfill  
$\Box $\\  

\subsection{From $c$-$P$- to $T$-geometries}\label{sec12}

In this subsection we show how, under certain conditions, starting from a
$c$-$P$-geometry $\Gamma $ we can
construct a  tilde geometry with the same automorphism group.\par
\bigskip
Let us first assume the rank of $\Gamma $ is 4 and
$G=Aut(\Gamma )\cong M_{24}$. We will identify  $\Gamma $
with the geometry from \cite[Example 2]{AC}. So $\Gamma $ has the
following sets of objects (where $\Omega $, $\cal T$, and $\cal O$ are as in
Example \ref{exm24} the points, trios and octads of the Steiner system): 
\begin{itemize}
\item[--] $\Gamma^{(1)}=\{ \{ p,q\} |p,q\in\Omega , p\neq q\} $
\item[--] $\Gamma^{(2)}=\{ \{ a,b\} |a,b\in \Gamma _1, a\cap b=\emptyset \}$
\item[--] $\Gamma^{(3)}=\{ (O_1,P,\{ O_2,O_3\} )| O_i\in{\cal O}, \{
O_1,O_2,O_3\}\in{\cal T}$, $P$ is a partition of $O_1$ into four parallel
lines in the affine geometry  on $\Omega\setminus O_2\} $
\item[--] $\Gamma^{(4)}=\{ (O,P) |O\in{\cal O}$, $P$ is a partition of
$\Omega\setminus O$ into parallel lines $\}$.
\end{itemize}
(Recall that for $O\in {\cal O}$   the complement
$\Omega\setminus O$ bears the structure of an affine space over $GF(2)$ and $G_O\cong 2^4A_8$ acts as affine group $AGL(4,2)$ on $\Omega\setminus O$.)\par
Let $\Sigma $ be a
connected component of the structure 
$\cal S$ obtained from $\Gamma $ by the shrinking construction described in
the previous section. Then $\Sigma $ has the diagram  
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}
\put(7,2.5){c}
\put(17,2.5){$\scriptsize P^*$}
\put(1,2.5){\scriptsize {1}}
\put(11,2.5){\scriptsize{2}}
\put(21,2.5){\scriptsize{3}}
\end{picture} .
We have $G_2\le G_{\Sigma }$ and $G_2\cong 2^{4+1}({\bf Z}_2\times\Sigma _5)$, and Lemma 
\ref{lem2} resp. Corollary \ref{cor1} yield that $\Sigma $ is disconnected
and that $O_2(G_2)$ is the kernel of the action of $G_{\Sigma }$ on  $\Sigma
$. In particular, $O_2(G_2)\unlhd G_{\Sigma}$, and the list of maximal
subgroups of $M_{24}$ in \cite{A} shows that $G_{\Sigma }\le G_S$, where
$G_S\cong 2^63\Sigma _6$ and $G_S$ is the stabilizer of a sextet $S$ in the
Steiner system. From the structure of $G_S$ we see
that  either $G_{\Sigma }=G_S$ or
$G_{\Sigma }\cong 2^6({\bf Z}_3\times \Sigma _5)$. But in the latter
$|G_{\Sigma }:G_2|=3$, so  $\Sigma $ would just contain 3 points, which
is not true (consider the residue of an element of type 3). Hence $G_{\Sigma }=G_S\cong 2^63\Sigma _6$. \par
Let $T^{(1)}$ be the set of connected components of $\cal S$, $T^{(2)}$ the
set of triples of elements from $\Gamma ^{(3)}$ of the shape
$$\{ (O_1,P_1,\{ O_2,O_3\} ), (O_2,P_2,\{ O_1,O_3\} ),
(O_3,P_3,\{ O_1,O_2\} )\} $$ 
such that $P_i\cup
P_j$ is a partition into parallel lines of the affine space $\Omega
\setminus O_k$ for all 
choices of $\{ i,j,k\} =\{ 1,2,3\} $, and let $T^{(3)}=\Gamma ^{(4)}$. Set
$T=T^{(1)}\cup T^{(2)}\cup T^{(3)}$ and define an incidence relation on $T$
in the way that $\Sigma \in T^{(1)}$ and $x\in T^{(2)}\cup T^{(3)}$ are
incident if $(x,P_x)$ (resp. $(y,P_y)$ for all $y\in x$) belong to $\Sigma $,
where   $P_x$, $P_y$ are suitable partitions of $res(x)^-$ resp. $res(y)^-$
into parallel lines of $\Gamma $. The elements in $T^{(3)}$ incident
to a typical element 
$$\{ (O_1,P_1,\{ O_2,O_3\} ), (O_2,P_2,\{ O_1,O_3\} ),
(O_3,P_3,\{ O_1,O_2\} )\} \in T^{(2)}$$ 
 are the three elements $(O_i,P_j\cup
P_k)$, $\{ i,j,k\} =\{ 1,2,3\} $.\par 
If $x=(O,P)\in T^{(3)}$ then $G_x\le G_O\cong 2^4A_8$. Since $G_x$ stabilizes
the partition $P$ we see $G_x\cong 2^{1+6}L_3(2)$. If $x\in T^{(2)}$, $x$ as above, then
$G_x\le G_t$ where $G_t$ is the stabilizer of the trio $t=\{ O_1,O_2,O_3\}$ and $G_t\cong [2^6](\Sigma _3\times L_3(2))$. The $\Sigma _3$-factor permutes the three octads in $t$, so it will stabilize $x$ as a set. The $L_3(2)$-factor can be viewed as the stabilizer of a partition of $\Omega\setminus O_i$ into two parallel planes for each of the $O_i$. Since $G_x$ also   
stabilizes the partitions $P_j\cup P_k$ we deduce $G_x\cong
[2^8](\Sigma _3\times \Sigma _3)$.  So we have 
shown that $G_x$ is of the right shape for all $x\in T$. We leave it to the
reader to convince himself that the incidence relation is defined in the
right way to get a model of the tilde geometry for $M_{24}$.\par
\bigskip
Now let $\Gamma $ be of arbitrary rank $n\ge 5$. Apply the shrinking
 construction $n-3$ times and call the obtained incidence structures ${\cal
 S}_1$, 
 ${\cal S}_2$, \ldots ,${\cal S}_{n-3}$. Suppose that ${\cal S}_1$ is
 disconnected and that the connected components of ${\cal S}_{i+1}$ are not
 equal to those of ${\cal S}_i$,  $i=1,\ldots
 ,n-4$. Then, similarly as in  section 2.1, for each $i\ge 2$, we can define an equivalence relation on the
 set of connected components of ${\cal S}_i$, whose equivalence classes
 correspond to the set of partitions into two parallel hyperplanes of an
 $i$-dimensional affine space over $GF(2)$. Hence they are of size
 $2^i-1$. Assume  the connected components of ${\cal
 S}_{n-4}$ are the $c$-$P$-geometries for $M_{24}$. Let
\begin{enumerate}
\item[--] $T^{(1)}=$ connected components of ${\cal S}_1$,
\item[--] $T^{(i)}=$ equivalence classes of connected components of ${\cal
S}_i$, for $2\le i\le n-3$,
\item[--] $T^{(n-2)}=$ triples of elements of $\Gamma ^{(n-1)}$ which, for
suitable partitions,  are contained in a common connected component of
${\cal S}_{n-2}$  and there form  a triple which corresponds to an element
of type 2 of the $T$-geometry for $M_{24}$,
\item[--] $T^{(n-1)}=\Gamma ^{(n)}$
\end{enumerate}
and $T$ be the  geometry on $T^{(1)}\cup\ldots \cup T^{(n-1)}$ with the following incidene relation:\par
Two equivalence classes $x\in T^{(i)}$, $y\in T^{(j)}$, $1<i<j<n-2$,
resp. $\Sigma \in T^{(1)}$ and $y\in T^{(i)}$ are incident if a
representative of $y$ is contained in a representative of $x$ resp. in
$\Sigma $. Elements $x\in T^{(i)}$, $i<n-2$, and $y\in T^{(n-2)}\cup T^{(n-1)}$ are incident if $y$ resp. all members of $y$ belong to (a representative of) $x$. Incidence between elements in $T^{(n-2)}$ and $T^{(n-1)}$  is defined by inclusion.\par 
By construction the diagram of $T$ is a string, the residue of an element of type $n-1$ is an $n-1$-dimensional  projective space over $GF(2)$, and  the geometry induced on the elements of types $n-2$ and $n-1$ in the residue of an element of type $n-3$ is the rank 2 tilde geometry for $3\Sigma _6$. So $T$ is  a tilde geometry of rank $n-1$. 
Since the stabilizer of a suitable connected component $\Sigma\in T^{(1)}$
contains $G_2$, since $T^{(n-1)}=\Gamma ^{(n)}$ and since $G=\langle
G_2,G_n\rangle $, it is straightforward to see that $G$ is a flag-transitive
automorphism group of it. We formulate this result in the
following

\begin{prop}\label{prop1}
Suppose $G$ acts flag-transitively on a $c$-$P$-geometry  $\Gamma $ of rank
$n\ge 5$. Apply the shrinking construction $n-3$ times on $\Gamma $ and call the obtained incidence structures ${\cal S}_1$, 
 ${\cal S}_2$, \ldots ,${\cal S}_{n-3}$. Suppose that ${\cal S}_1$ is
 disconnected, that the connected components of ${\cal S}_{i+1}$ are not
 equal to those of ${\cal S}_i$, $i=1,\ldots
 ,n-4$, and that the connected components of ${\cal S}_{n-4}$ are the
 $c$-$P$-geometries for $M_{24}$. Then $G$ acts flag-transitively on a
 tilde geometry of rank $n-1$. 
\end{prop}

{\bf Remark.}~~One can also start from the rank 3 case and first construct 
the $T$-geometry for $3\Sigma _6$ from the $c$-$P$-geometry. This can be
done, e.g., considering a model of the latter which exists inside $AGL(3,4)$
or getting a model via the shrinking process from the $c$-$P$-geometry for
$M_{24}$. We did 
not choose this approach here
because there was no obvious need  and it would have made the whole
construction even more technical.\par
Notice that it is essential for the construction of a tilde geometry that
the  connected 
components of ${\cal S}_{n-3}$ are the $c$-$P$-geometries for $3\Sigma
_6$. Otherwise their stabilizers might not act on a tilde geometry at all
and one will not be able to construct even a rank-2-tilde geometry. 
 
\section{Some preliminary lemmas}

The first two lemmas are maybe well known but as we had no reference
we provide the proofs. 

\begin{lem}\label{split}
Let $G=L_4(2)$ act on an 8-dimensional $GF(2)$-vectorspace  $V$ which
contains a $G$-invariant subspace $V_1$ such that $V_1$ and $V/V_1$ are
nontrivial 4-dimensional modules for $G$.  Suppose there exists $x\in
V\setminus V_1$ such that $|x^G|=15$. Then $V\neq\langle x^G\rangle $. 
\end{lem}

{\it Proof.}~~Since $|x^G|=15$, $C_G(x)\cong 2^3L_3(2)$ is a maximal
parabolic subgroup of $G$ and each nontrivial coset of $V/V_1$ contains
exactly one conjugate of $x$. Since $L_4(2)$ acts doubly transitively on the
4-dimensional module and hence also on the set of conjugates of $x$, it
suffices to show that for some conjugate $y$ of $x$ the element $xy$ is also
conjugate to $x$. Then $\langle x^G\rangle =\{ x^g|g\in G\} \cup\{ 1\} $,
the assertion.\par 
There exists an element $g\in G$ of order 3  acting fixed point freely on
both 4-dimensional modules, hence $g$ acts fixed point freely on $V$. Since
$xx^gx^{g^2}$ is centralized by $g$, we get $xx^gx^{g^2}=1$ and
$xx^g=x^{g^2}$ has the desired property.\hfill$\Box $\\ 

\begin{lem}\label{golay}
Let $G=M_{24}$ and $V$ be a 12-dimensional $GF(2)G$-module. If $V$ contains 
a $G$-submodule $V_1$ with $|V_1|=2^{11}$ and $V_1$ isomorphic to the (irreducible) Golay code module then $V$ splits over $V_1$.
\end{lem}

{\bf Proof.}~~Suppose not and consider the orbits of $G$ on $V\setminus V_1$. Since 2048 is not divisible by 23, each element of order 23 in $G$ must have a fixed point on $V\setminus V_1$. Now the list of maximal subgroups of $M_{24}$ in \cite{A} shows that $G$ has an orbit of length 24. Since any two subgroups of $G$ isomorphic to $M_{23}$ intersect in an $M_{22}$ then the sum of any two vectors in this orbit is fixed by a subgroup of $G$ isomorphic to $M_{22}$. But $M_{22}$ does not stabilize a nontrivial vector in $V_1$, a contradiction.
\hfill $\Box $\\ 

In the next three lemmas $G$ is a  flag-transitive automorphism group
of some 
$c$-$P$- or $c$-$T$-geometry $\Gamma $ satisfying ($*$)  and all other
notations are as  in the introduction.

\begin{lem}\label{reprgrp}
Suppose there exists an
involution $x\in G_2\setminus G_1$ such that $[G_{12},x]=1$.  
Let $X_1=\langle x^{G_1}\rangle $ and $X=\langle x^{G_{1n}}\rangle
$. If $X$ is elementary abelian and   $2^{n-1}\le |X|\le 2^n$ then
$|X|=2^{n-1}$ and $X_1$ is isomorphic to a factor group of the universal
representation group of $res(\alpha _1)$. 
\end{lem}

{\bf Proof.}~~If $p\in res(\alpha _1)\cap \Gamma ^{(2)}$ then
$p=\alpha _2^g$ for a suitable $g\in G_1$ and we set $u_p=x^g$.
Since $[G_{12},x]=1$ this is well-defined and it suffices to show that
$u_pu_qu_r=1$ whenever $p,q,r$ are incident to a common element $l\in 
res(\alpha _1)\cap \Gamma ^{(3)}$. (Then $X_1$ satisfies the axioms of
Definition 1.)\par
By flag-transitivity we may assume $l=\alpha _3$, $p=\alpha _2$. Then $l\in
res(\alpha _n)$, $q$ and $r$
are conjugate to $p$ in $G_{13n}$ and 
$\langle u_p,u_q,u_r\rangle \le X$. As $x\not\in K_n$ and $G_n/K_n\cong
2^{n-1}L_{n-1}(2)$ we have $G_n=XG_{1n}$. So either  $|X|=2^{n-1}$ or $|X\cap
K_n|=2$, and since $X\cap K_n\unlhd G_n$, in the latter we get $X\cap K_n\le
Z(G_n)$.  Furthermore, as
$C_{G_{1n}}(x)=G_{1n}\cap G_2$, the action of $G_{1n}$ on
$X/X\cap K_n$ corresponds to the action on the projective $GF(2)$-space
$res(\{\alpha _1,\alpha _n\} )$. Since $\alpha _3$ corresponds to a line in
this space the assertion holds in $X/X\cap K_n$, particularly it holds if
$|X|=2^{n-1}$. Suppose   $1\neq z=u_pu_qu_r\in X\cap K_n\le Z(G_n)$. Let
$X_3=\langle x^{G_3}\rangle $. Since $G_3=G_{123}G_{3\ldots n}$ and $[x,G_{123}]=1$
we have $X_3=\langle x^{G_{3\ldots n}}\rangle =\langle u_p,u_q,u_r\rangle $.
Now $X_3$ is invariant under $P_n$, $z\in X_3$, and as $z\not\in Z(G)$
we have $[z,P_n]\neq 1$. However,
we see in $G_1$ that
$z^{P_n}\subseteq B$. Since $X\cap B=X\cap K_n$ this yields  a
contradiction.\hfill $\Box $\\   

\begin{lem}\label{kernels}
We have $K_i=K_{1i}$ for $i\ge 2$.
\end{lem}

{\it Proof.}~~Let $g\in K_{1i}$. Then $g$ fixes  all elements in $res(\alpha
_1)\cap res(\alpha _i)$. Let  $p\in \Gamma ^{(1)}$ be incident to $\alpha
_i$. Then there exists $l\in \Gamma ^{(2)}$ incident to $\alpha _i$ such
that $res(l)\cap \Gamma ^{(1)}=\{ \alpha _1,p\} $.  Now $g$ fixes $\alpha
_1$ and $l$, so $g$ must fix $p$. This implies that  $g$ acts trivially on
$res(\alpha _i)$ and hence $g\in K_i$. The other inclusion is trivial.\hfill
$\Box $\\

\begin{lem}\label{lem1}
If $n\le 5$  then $K_1=1$. If $n=6$ then $|K_1|\le 2$.
\end{lem}

{\it Proof.}~~By \cite{IS1} we know that, if $\Gamma $ is of $c$-$P$-type,
then $G_1/K_1\cong M_{23}$  
or $Co_2$, if $n=5$, and  $G_1/K_1\cong BM$, if $n=6$. If $\Gamma $ is of
$c$-$T$-type, then $G_1/K_1\cong M_{24}$, $Co_1$, $M$ for $n=4,5,6$,
respectively. We will make use of the descriptions of the corresponding
residual geometries as given in \cite{IPS}. \par
Let us call the elements of $\Gamma $ of type 1 ``points'' and the elements of
type 2 ``lines''. Then each line consists of 2 points and the relation
``having the same points'' is an equivalence relation on the set of lines on
the point $\alpha _1$, which is preserved by $G_1$. Since in the considered
cases $G_{12}/K_1$ is a maximal subgroup of $G_1/K_1$, the action of
$G_1/K_1$ on that set is primitive.  Hence there are no two lines with the same
points. In particular, $G_p\cap G_q\le G_l$ for two points $p,q$ on a line
$l$.\par 
Now let $g\in K_1$, $p$ a point collinear to $\alpha _1$ and $l$ the line on
$\alpha _1$ and $p$. Then $g$ fixes $l$ and $\alpha _1$, hence $g\in
G_p$. Let $k$ be a line on $p$ such that $l$ and $k$ are incident
to a common element $x$ of type 3 and $q$ the other point on $k$. Then $x$
is incident to $\alpha _1$ and there is a line through $\alpha _1$ and $q$ in
$res(x)$. So $g$ fixes $q$ and $g\in G_p\cap G_q\le G_k$. This means that
$g$ fixes all lines incident to $p$ which are coplanar with $l$. In terms of
the residual geometry  $res(p)$, where $l$ corresponds to a point, this
means that $g$ fixes all points collinear to $l$. Since the points  of the
$P$-geometry for 
$M_{23}$ are just the 23 points of the Steiner system ${\cal
  S}(23,7,4)$, the lines are all the pairs of different points, and
incidence is defined by inclusion, any two points are collinear in this
case. So $K_1$ fixes all points of the Steiner system  and  $K_1\le 
K_p$. Hence $K_1=K_p=1$. In the $M_{24}$-geometry $l$
corresponds to a sextet of the Steiner system and $g$ must fix all sextets
which refine a common trio with $l$. So we get $K_1\le K_p$ again. For the
geometries of the Conway groups one can deduce the same from the
descriptions in terms of the Leech lattice given in Example 2
resp. \cite{IPS}.\par 
Finally, also by \cite{IPS} the points of the geometries
for $BM$ and $M$ can be identified with the conjugacy class of 2-central
involutions and the involutions corresponding to the set of points collinear
to $l$ generate the subgroup $O_2(G_{pl}/K_p)$. Since this is an extraspecial
group of order $2^{1+22}$ resp. $2^{1+24}$ we get $|K_1:K_1\cap K_p|\le 2$ in
those cases. So it remains to show that $K_1\cap K_p=1$.  
From Lemma \ref{kernels} and $G_6/K_6\cong 2^5L_5(2)$ we see
that $K_1K_6/K_6$ and $K_1K_6/K_1$ are 2-groups. Hence $K_1$ and $K_6$
are 2-groups by the Frattini argument.
Since $B/K_1$
is also a 2-group, we get that $B$ and all $K_i$ are 2-groups.
Suppose first that $C_{G_1}(K_1)\le K_1$. Let $S\in Syl_2(G_1\cap
  G_2)$ and $Z_1=\langle \Omega
  _1(Z (S))^{G_1}\rangle $.  Notice that $|G_2:G_1\cap G_2|=2$, so
  $G_1\cap G_2\unlhd G_2$ and $G_2=(G_1\cap G_2)N_{G_2}(S)$. If
  $Z_1=\Omega _1(Z(S))$ then $Z_1\unlhd \langle G_1,N_{G_2}(S)\rangle
  =G$, which is impossible. So $Z_1\neq\Omega _1(Z(S))$ and $Z_1$
  involves a non-trivial $G_1/K_1$-module. By \cite[(11.1)]{Asch0} $BM$ and $M$  do not have  $F$-modules. Hence $J(S)\le K_1$ (where $J(S)$ is the Thompson subgroup of $S$). But then $J(S)=J(K_1)\unlhd\langle 
G_1,N_{G_2}(S)\rangle =G$, again a contradiction. So we 
have $C_{G_1}(K_1)\not\le K_1$, hence $G_1=C_{G_1}(K_1)K_1$ and, by symmetry, 
$G_p=C_{G_p}(K_p)K_p$. Therefore $K_1\cap K_p\unlhd \langle O^2(G_1),
O^2(G_p), K_1,K_p\rangle $. But as $G_6=\langle G_1\cap G_6,G_p\cap
G_6\rangle $ we have $G=\langle G_1,G_6\rangle \le \langle G_1,G_p\rangle =\langle O^2(G_1), O^2(G_p), K_1,K_p\rangle $. So $K_1\cap K_p\unlhd G$ and $K_1\cap K_p=1$. This proves the
assertion. \hfill $\Box $\\   

\section{Classification of the geometries}\label{sec2}
  
\begin{lem}\label{lem5}
If $\Gamma $ is a $c$-$P$-geometry of rank 5 and $G_1\cong M_{23}$, then
$G\cong M_{24}$. 
\end{lem}

{\it Proof.}~~We show that $|G:G_1|=|\Gamma ^{(1)}|=24$. Since $G_{12}\cong
M_{22}$, we have $|res(\alpha _1)^{(2)}|=|G_1:G_{12}|=23$. So there are 23
points which are collinear to $\alpha _1$ and it suffices to show that, if
$p$ and $q$ are collinear to $\alpha _1$, then $p$ is collinear to $q$. Let
$l_p,l_q\in \Gamma ^{(2)}$ be the lines through $\alpha _1$ and $p$
resp. $q$. Since  $l_p,l_q$ correspond to points in the geometry for
$M_{23}$ and any two points in this geometry are collinear by the
2-transitive action of $M_{23}$, we see in $res(\alpha _1)$ that 
there exists  $x\in \Gamma ^{(3)}$ such that   $x$ is 
incident to $l_p$ and $l_q$. But now $p$ and $q$ are incident to $x$, and
since the elements of  $res(x)^-$ form an affine space, the
claim follows. \hfill $\Box $\\  

For the rest of the section we fix the following situation: From Lemma
\ref{kernels} we know that $K_2=K_{12}$. Further we have $|G_2:G_{12}|=2$. So
$G_{12}\unlhd G_2$ and there exists $x\in G_2\setminus G_{12}$ such that
$x^2\in G_{12}$. Since $K_2=O_2(G_{12})$ we see that $x$ acts on
$G_{12}/K_2$. But except in one case, where we will have
$G_{12}/K_2\cong 3\Sigma _6$, the group $G_{12}/K_2$ will
not possess outer automorphisms of order 2, so  $G_2/K_2\cong {\bf
Z}_2\times G_{12}/K_2$ and we may assume $[x,G_{12}]\le K_2$.  If
$G_{12}/K_2\cong 3\Sigma _6$ then $|Out(G_{12}/K_2)|=2$ and the outer
automorphism of $3\Sigma _6$ interchanges the two conjugacy classes of
subgroups of type ${\bf Z}_2\times \Sigma _4$. But these correspond to
stabilizers of elements of different types in $res(\alpha _2)^+$ in $\Gamma
$, and as $x\in Aut(\Gamma )$, $x$  preserves the types. So
$x$ cannot act as outer automorphism and we get $[x,G_{12}]\le K_2$ in this
case as well. \par    
Unless stated otherwise, from now on $x$ will always denote such an
element. Then $[x,B]\le [x,G_{12}]\le K_2\le B$ and $x\in N_{G_2}(B)$. If $\Gamma $ is of $c$-$T$-type this yields $x\in P_1$, if $\Gamma $ is of $c$-$P$-type $x\in P_1P_n$. Furthermore, by the Three subgroup lemma  $G_{12}$ acts on
$[K_2,x]K_2'$. In particular, if $K_2$ is elementary abelian and irreducible as
$G_{12}/K_2$-module, then $[K_2,x]=1$. Further, since then $[G_{12},x^2]=1$
we get that $x$ is of order 2.  

\begin{lem}\label{m24}
If $\Gamma $ is a $c$-$T$-geometry of rank 4 and $G_1\cong M_{24}$, then
$G\cong 2^{11}\! :\! M_{24}$ and $O_2(G)$ is the universal representation group of the $T$-geometry for $M_{24}$ (so it is the Golay code module).
\end{lem}

{\it Proof.}~~We have $G_{12}\cong 2^63\Sigma _6$ where $K_2$ is an
irreducible $G_{12}/K_2$-module. So as just remarked,  $Q_2=\langle
K_2,x\rangle $ must be an elementary abelian group of order $2^7$. Let
$\rho\in O_{23}(G_{12})$, $o(\rho )=3$. Then $\rho $ acts fixed point freely on
$K_2$. So $C_{Q_2}(\rho )$ must be of order 2. Since $C_{Q_2}(\rho 
)$ is $G_{12}$-invariant we may assume $[x,G_{12}]=1$.
\par
Now consider $G_4$. We have $G_{14}\cong 2^{1+6}L_3(2)$, $G_4/K_4\cong
2^3L_3(2)$, and $G_4=G_{14}X$ where $X=\langle x^{G_{14}}\rangle $. Since
$K_4\le G_{12}$ we get $[X,K_4]=[x,K_4]^{G_{14}}=1$. As $K_4$ is
extraspecial this yields  $|X\cap K_4|\le 2$ and $|X|\le 16$.
So 
$X_1=\langle x ^{G_1}\rangle $ is a representation group for the tilde
geometry for $M_{24}$ by Lemma \ref{reprgrp} and $|X_1|=2^{11}$  by
\cite[(3.2)]{IPS} resp. \cite{RSm}. Since
$X_1$ is normalized by $G_1$ and $G=\langle x,G_1\rangle$, we get
$G=X_1G_1$ and the lemma is proved.  \hfill $\Box $\\ 

In the rank 5 and rank 6 cases we will consider $c$-$P$- and
$c$-$T$-geometries at the same time. This has the following reason: Many of
our calculations mainly  
depend on the 2-structures of some of the maximal parabolic subgroups. Since
the $P$-geometries for $Co_2$ and 
$BM$ can be defined as certain subgeometries in the $T$-geometries for
$Co_1$ resp. $M$, these structures are quite
similar for both classes of geometries. In particular, it turned out that
often  we just can use the same arguments without distinguishing between
$T$- and $P$-geometries.  

\begin{lem}\label{lem6} Let $rank(\Gamma )=5$.
\begin{enumerate}
\item[(a)] If $\Gamma $ is of $c$-$P$-type and $G_1\cong Co_2$, then
either $G\cong 2^{22}\! :\! Co_2$ resp. $2^{23}\! :\! Co_2$ and
$O_2(G)$ is  a representation group of the $P$-geometry for
$Co_2$ (so it is isomorphic to a section of the Leech lattice mod 2); or $G\cong Co_1$.  
\item[(b)] If $\Gamma $ is of $c$-$T$-type and $G_1\cong Co_1$, then
$G\cong 2^{24}\! :\! Co_1$ and $O_2(G)$ is the universal
representation group of the tilde geometry for $Co_1$ (so it is isomorphic to the Leech lattice mod 2).
\end{enumerate}
\end{lem}

{\it Proof.}~~From the geometries of $Co_2$ and $Co_1$ we know that
$G_{12}\cong 2^{10}Aut(M_{22})$ resp. $2^{11}M_{24}$ and that $K_2$
is the irreducible Golay code module for $G_{12}/K_2$. So $[x,K_2]=1$ and
$O_2(G_2)=\langle x,K_2\rangle $ is an  elementary abelian group of order
$2^{11}$ resp. $2^{12}$ by the above remark.  Now if $G_{12}/K_{12}\cong
Aut(M_{22})$, then we 
have two general possibilities for the structure of $G_2$: 
\begin{enumerate}
\item[(a.1)] $G_2\cong {\bf Z}_2\times 2^{10}Aut(M_{22})$,
\item[(a.2)] $G_2\cong 2^{10+1}Aut(M_{22})$, where the action of $G_2/O_2(G_2)$
on $O_2(G_2)$ is non-split,
\end{enumerate}
whereas in (b) by Lemma \ref{golay} we must have $G_2\cong {\bf
  Z}_2\times 2^{11}M_{24}$.\par 

Next look at $G_5$. We have $G_5/K_5\cong
2^4L_4(2)$. From the structure of $G_1$ we know that in (a)
$G_{15}\cong (2^{1+6}\times 2^4)L_4(2)$, $|Z(K_{5})|=2^5$ and
$|K_{5}\cap K_2|=2^7$, and in (b) $G_{15}\cong
2^{1+8+6}L_4(2)$, $|Z(K_{5})|=2$, and $|K_{5}\cap K_2|=2^8$. \\

Let us first consider  (a.1) and  (b). In these cases $[x,K_5]=1$ and $x\in N_{G_2}(K_5)=G_2\cap G_5$. 
Hence if we set $X=\langle x^{G_{15}}\rangle $,
then $G_5=XG_{15}$, $[X,K_5]=1$, and $X\le Z(O_2(G_5))$. In (a.1) we apply
Lemma \ref{split} to see that $X$ splits over the 4-dimensional module
in $Z(K_{5})$. 
So in both cases we may assume $|X|\le 32$. Now  Lemma \ref{reprgrp}
yields that 
$X_1=\langle x^g|g\in G_1\rangle $ is a representation group for the
residual geometry of $G_1/K_1$. So $X_1$ is elementary abelian of order
$2^{22}$ or $2^{23}$ in (a.1) and  of order $2^{24}$ in (b) by
\cite[(5.7),(5.8)]{IPS}. Again because $X_1$ is normalized by $G_1$
and $G=\langle x,G_1\rangle $ we get that $G=X_1G_1$ is as stated. \\  
\\
Now consider (a.2). Suppose $[y,K_5]\le Z(G_{15})$ for all $y\in
O_2(G_2)\setminus K_2$. Because $O_2(G_2)=\langle y|y\in
O_2(G_2)\setminus K_2\rangle $ then we would get $[O_2(G_2),K_5]\le
Z(G_{15})$. But $K_5K_2/K_2$ does not induce transvections  on
$K_2$. So we get a contradiction. Hence we may choose $x$ with
$[x,K_5]\not\le Z(G_{15})$.  
Then, particularly, $[X,K_5]=[\langle x^{G_5}\rangle
,K_5]=[x,K_5]^{G_5}$ is not contained in  $K_5\cap O_2(G_2)$.\par
Now let $\Sigma $ be a connected component of the structure $\cal S$
obtained from $\Gamma $ by shrinking as described
in section \ref{sec1}. We can choose $\Sigma $  such that the flag 
$\{ \alpha _2,\ldots ,\alpha _n\}$ 
maps onto a maximal flag in $\Sigma $. Let $H=G_{\Sigma }$ and let $N$ be
the 
kernel of the action of $H$ on $\Sigma $.  By Lemma
\ref{lem2} we have $N=O_2(G_2)$ and by construction of $\cal S$
$O_2(G_5)N/N=O_2(H_4/N)$. Since $[X,K_5]\not\le N$, we see 
that $O_2(H_4/N)$ is non-abelian. So \cite{AC} 
implies that $H/N\cong M_{24}$. Now $G$ acts flag-transitively on a 
rank-4-tilde geometry by Proposition \ref{prop1}, and since $H$ is the
stabilizer of an 
object of this geometry, by \cite{IS1} we  get $G\cong Co_1$.\hfill$\Box $\\     

\begin{lem}\label{lem7}
Let $rank(\Gamma )=6$.
\begin{enumerate}
\item[(a)] If $\Gamma $ is of $c$-$P$-type and $G_1/K_1\cong BM$, then
either $G$ is isomorphic to a section of $(2\!\cdot\! BM\! *\! 2\!\cdot\!
BM)\! :\!{\bf Z}_2$ or $G\cong M$.  
\item[(b)] If $\Gamma $ is of $c$-$T$-type and $G_1/K_1\cong M$, then $G\cong
M\times M$ or $M\wr {\bf Z}_2$.
\end{enumerate}

\end{lem}

{\it Proof.}~~
By Lemma \ref{lem1} we have $|K_1|\le 2$. Hence we get the following
possibilities for $G_1$: 
\begin{enumerate}
\item[(a.1)] $K_1=1$, $G_1\cong BM$,
\item[(a.2)] $|K_1|=2$, $G_1\cong {\bf Z}_2\times BM$,
\item[(a.3)] $|K_1|=2$, $G_1\cong 2\!\cdot\! BM$, a non-split extension,
\end{enumerate}
respectively
\begin{enumerate}
\item[(b.1)] $K_1=1$, $G_1\cong M$,
\item[(b.2)] $|K_1|=2$, $G_1\cong {\bf Z}_2\times M$,
(as the Schurmultiplier of $M$ is trivial by \cite{A}). 
\end{enumerate}
We set $\langle z\rangle =Z(G_1)=K_1$. Then $G_{12}/\langle z\rangle \cong
2^{1+22}Co_2$ resp. $2^{1+24}Co_1$.  \par
Let us first assume we are not in case (a.3). Then $Q_2=[K_2,G_{12}]$ is 
extraspecial and $[K_2,Q_2]=Q_2'$. Since
$[x,G_{12},Q_2]\le [K_2,Q_2]=Q_2'$ and $[Q_2,x,G_{12}]\le [K_2,G_{12}]=Q_2$,
the Three-Subgroup-Lemma yields $[Q_2,x]=[G_{12},Q_2,x]\le Q_2$. Hence
$Q_2\unlhd G_2$. Now also $C_{G_2}(Q_2)\unlhd G_2$ and
$[C_{G_2}(Q_2),O^2(G_{12})]=1$. Notice that $|C_{G_2}(Q_2)|=4$
resp. $8$ (depending whether we have $K_1=1$ or not) and that this group
is not necessarily abelian but it has exponent at most 4.\par
From now on we assume $x\in C_{G_2}(Q_2)$. Then $o(x)\le 4$ and
$x$ centralizes a Sylow-2-subgroup $S\le G_1'$. In particular $x$
centralizes $O_2(H_{16})$, where $H_{16}=G_6\cap G_1'$. So $x\in N_{G_2}(O_2(H_{16}))=G_2\cap G_6$. Notice that
$H_{16}/O_2(H_{16})\cong L_5(2)$ with $|O_2(H_{16})|=2^{30}$ resp. $2^{36}$,
and  that
$G_6=\langle z\rangle H_{16}X$ where $X=\langle x^{H_{16}}\rangle $. Now
$XK_6/K_6$ is an elementary abelian group of order $2^5$ and with $x$ also
$X$  centralizes $O_2(H_{16})$. Hence
$X\le Z(O_2(H_{16})X)$. Let $Z=Z(O_2(H_{16}))$. Then $|Z|=2^5$ and
$|Z(O_2(H_{16})X)|=2^{10}$. Now if $o(x)=4$ then $X=Z(O_2(H_{16})X)\cong {\bf
Z}_4^5$, but by  \cite{Bei}  $GL_5({\bf Z}_4)$ does not contain a subgroup isomorphic to
$L_5(2)$. So $o(x)=2$.\par
 We claim that $XZ$ splits over $Z$. We know that $Z$ and $XK_6/K_6\cong X/(X\cap K_6)=X/(X\cap Z)\cong XZ/Z$ are isomorphic as
$L_5(2)$-modules. 
Consider a subgroup $L$ of $L_5(2)$ of order $31\cdot 5$.  
Since $L$ acts completely reducibly on $XZ$, we see that $L$ has three orbits of length $31$ and $6$ orbits of length $155$ on $(XZ)^{\scriptstyle{\#}}$ where each of the three orbits of length $31$ generates a 5-dimensional subspace of $XZ$. As $x$ has exactly 31 conjugates under $H_6$ the claim follows, i.e., $X$ is elementary abelian 
of order $2^5$. Replacing $G_1$ by $G_1'$ in  Lemma \ref{reprgrp} now we get that $X_1=\langle X^{G_1'}\rangle $ is a 
representation group for the residual
geometry for $G_1'\cong G_1/K_1$. Hence $X_1\cong BM$ or $2\!\cdot BM$
resp. $M$ by \cite{IPS}. \par
Consider the semidirect product $X_1G_1'$. If $z=1$ then $X_1G_1'=X_1G_1=G$
and 
since $Out(G_1)=1$ we get $G\cong G_1\times G_1$. If $z\neq 1$ then
$[z,x]\neq 1$ (otherwise we would get $z\in Z(\langle x,G_1\rangle
)=Z(G)$. But as $z\in K_6$ we have $[z,X]\le K_6$ and $[z,X_1]\le
X_1G_1'$. So $z$ normalizes $X_1G_1'\cong X_1\! *\! X_1$ but not $X_1$ and we
get the first case of (a) resp. (b). (Notice that $G_1$ is the diagonal
subgroup of $G$).\par  
\bigskip
Now consider case (a.3).
Let $\Sigma $ be a connected component of the structure $\cal S$ as
described in Section \ref{sec1}. Let $H=G_{\Sigma }$ and $N$ the kernel of
the action of $H$ on $\Sigma $. By Lemma \ref{lem2} we have
$N=K_2\langle x\rangle $, and by Lemma \ref{lem6} we have $H/N\cong Co_1$ or
$2^{22}Co_2$ (resp. $2^{23}Co_2$).\par 
If $H/N\cong Co_1$, then Proposition \ref{prop1} yields that $G$ acts flag-transitively
on a  rank-5-tilde geometry. So $G\cong M$ by \cite{IS1} in this
case.\par 
Suppose $H/N\cong 2^{22}Co_2$ (resp. $2^{23}Co_2$). Since $G_{12}$ acts on
$[K_2,x]K_2'$, the structure of $K_2$ as $G_{12}/K_2$-module implies that
$[x,K_2]\le 
Z(G_{12})$. Hence $|K_2:C_{K_2}(x)|\le 4$. If $[K_2,x]\not\le K_2'$, then
$|K_2/K_2':C_{K_2/K_2'}(x)|=|[K_2/K_2',x]|=2$, but $K_2/K_2'$ does not contain
a $G_{12}$-submodule of index 2. So $[K_2,x]\le K_2'$, and since $N=\langle x, K_2\rangle $, $N'=K_2'$ is of
order 2. As $x^2$ centralizes $K_2$  we get $o(x)\le 4$ again. As above
$[x,z]\neq 1$, so   $Z(N)=N'$ and $N$ is 
extraspecial of order $2^{1+24}$.\par  
Let $N_1=O_2(H)$. Then $N_1$ acts on $N/N'$ and $C_{N/N'}(N_1)\neq 1$. Let
$C$ be the full preimage in $N$ of $C_{N/N'}(N_1)$. Then $C\unlhd H$
and the action of $Co_2$ on $N/N'$ implies
$z\in C$. Therefore, $[z,N_1]\le N'$ and $|N_1:C_{N_1}(z)|\le 2$. Since
$N_1/N$ does not contain a $G_{12}$-submodule of index 2, we get
$N_1=C_{N_1}(z)N$. 
Now consider the action of $N_1/N$ on $\Sigma $. Because
$N$ acts trivially on $\Sigma $ and $N_1/N$ acts regularly on 
the points of $\Sigma $, i.e., on $\Sigma\cap\Gamma ^{(2)}$, we see that
even $C_{N_1}(z)$ acts regularly. We may assume
$\alpha _2\in\Sigma $. Let $\beta _2\in\Gamma ^{(2)}$ be the line parallel
to $\alpha _2$ in $res(\alpha _3)$. Our construction implies that $\beta
_2\in \Sigma $. Choose $n\in C_{N_1}(z)$ such that $\beta _2=\alpha
_2^n$ and let $\beta _1=\alpha _1^n$. (Notice that $\beta _1\neq \alpha _1$
because $\beta _1$ is incident to $\beta _2$, and $\alpha _2$ and
$\beta _2$ are parallel.)  Then 
$$K_{\beta _1}=K_{\alpha _1^n}=K_1^n=\langle z\rangle ^n=\langle z\rangle
=K_1.$$ 
But $\alpha _1,\beta _1\in res(\alpha _3)$, so $\alpha _1$ and $\beta _1$
lie on a common line  $l\in\Gamma ^{(2)}\cap res(\alpha _3)$. By
flag-transitivity $\alpha _1$ and $\beta _1$ are conjugate in $G_l$, so
$G_l\le N_G(K_1)$ and $K_1\unlhd \langle G_1,G_l\rangle =G$, a
contradiction.
\hfill $\Box $\\ 



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\end{document}


