article
Rieuwert J. Blok and Corneliu Hoffman
Department of Mathematics and Statistics
Bowling Green
OH 43403-1874
[5pt]
%Fill in We obtain the symplectic group as an amalgam of low rank subgroups akin to Levi components. We do this by having the group act flag-transitively on a new type of geometry and applying Tits' lemma. This provides a new way of recognizing the symplectic groups from a small collection of small subgroups.
math0604300.tex
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\begin{document}
\pagestyle{empty}
\title{A Quasi Curtis-Tits-Phan theorem for the symplectic group}
\author{Rieuwert J. Blok and Corneliu Hoffman 
\\ Department of Mathematics and Statistics 
\\ Bowling Green State
University \\ Bowling Green, OH 43403-1874\\[5pt]
%{\tt blokr@member.ams.org}\\[5pt]
}

\date{\today \\[1in]
    \begin{flushleft}
        Key Words: symplectic group, amalgam, opposite\\[1em]
    AMS subject classification (2000):
    Primary 51A50;%polar geometry
    Secondary 
    %20G15, %linear algebraic groups over arbitrary fields
    %20G40,% linear algebraic groups over finite fields
    %20E42,%Groups with a $BN$-pair; buildings (as method for classification)
    %51E24, % buildings and the geometry of diagrams
    57M07. %Topological methods in group theory
    \end{flushleft}
       }
\maketitle





%\begin{flushleft} Proposed running head: \end{flushleft}
%    \begin{center}
%A quasi Phan-Curtis-Tits theorem for the symplectic group
%    \end{center}

%

%Send proofs to:
%\begin{center}
%    Rieuwert J. Blok\\ Department of Mathematics and Statistics\\ 
%    Bowling Green State
%University \\ Bowling Green, OH 43403\\[5pt]
%Tel.: 419-372-7455\\
%FAX: 419-372-6092\\
%Email: blokr@member.ams.org
%\end{center}
%\tableofcontents
%\newpage
\begin{abstract}
%Fill in
We obtain the symplectic group as an amalgam of low rank subgroups akin to Levi components.
We do this by having the group act flag-transitively on a new type of geometry and applying Tits' lemma.
This provides a new way of recognizing the symplectic groups from a small collection of small subgroups.
\end{abstract}

\pagestyle{plain}

\section{Introduction}
In the revision of the classification  of finite simple groups one of the important steps requires one to prove that if  a simple group $G$ (the minimal couterexample)  contains a certain amalgam of subgroups that one normally finds in a known simple group $H$ then $G$ is isomorphic to $H$. A geometric approach to recognition theorems was initiated in \cite{BGHS03,BS04,GHS03}. The present paper provides a new recognition theorem for symplectic groups. It is a natural generalization of the Curtis-Tits-Phan theory from above.

Let us describe the general geometric approach.
We consider a group $G$ which is either semi-simple of Lie type or a Kac-Moody group.
Let $\cT=(B_+, B_-)$ be the associated twin building.
%For $\epsilon=\pm$, let $\dist_\epsilon\colon B_\epsilon\times B_\epsilon\to W$ be the distance function  from the chamber system into the Coxeter group $W$. Also, let $\dist_*$ be the co-distance function on $\cT$.
%If $G$ is of Lie type, we let $B_+=B_-=B$ be the spherical building associated with $G$ through its finite $BN$-pair twinned via the opposition relation. For any chamber $C$, let $\Opp(C)$ denote the collection of chambers opposite to $C$. 
We first define  a {\em flip} to be an involutory automorphism $\sigma$ of $\cT$ that interchanges the two halves, preserves distances and codistances and takes at least one chamber to an opposite.

Given a flip $\sigma$, construct $\cC_\sigma $ as the chamber system whose chambers are the pairs  of opposite chambers $(c, c^\sigma)$ of $\cT$.
%Note that by our assumption $\cC_\sigma$ is nonempty.  
Let $G_\sigma$ be the fixed subgroup under the $\sigma$-induced automorphism of $G$. 
%It acts flag-transitively on the geometry $\Gamma_\sigma$ induced on $\cC_\sigma$.
%Adopting the terminology of \cite{BGHS03}, we call $\Gamma_\sigma$  the {\em flip-flop} geometry of $\sigma$. 
We refer to \cite{BGHS03} for details on the construction.
Whenever the geometry $\Gamma_\sigma$ is simply connected one obtains $G_\sigma$ as the universal completion of the amalgam of maximal parabolics for the action of $G_\sigma$ on $\Gamma_\sigma$.
%Comparing the Curtis-Tits and Phan approach we see that these parabolics are $\sigma$-invariant subgroups of the Levi components of the standard parabolic subgroups. 
%Thus it is possible for instance to obtain a symplectic group as an amalgam of unitary subgroups~\cite{Horn}. 
 
Consider, the building of type $A_n$ associated to $G=\PSL_{n+1}(\FF)$ for some field $\FF$.
The objects of type $i\in\{1,2,\ldots,n\}$ are the $i$-spaces of a fixed $(n+1)$-dimensional vector space $V$ over $\FF$.
Two objects $X$ and $Y$ are opposite if $\langle X,Y\rangle=V$ and $X\cap Y$ is minimal.

Let $\sigma$ be a flip. Then it is immediate that $\sigma$ is induced by a polarity.
The requirement that $\sigma$-invariant pairs of opposite chambers exist enforces that at least one $1$-space $p$ does not intersect its polar hyperplane: that is, $p$ is not absolute. 
These polarities are well-known and classified to correspond to orthogonal or unitary forms.
The objects of the geometry $\Gamma_\sigma$ are the non-degenerate subspaces with respect to the polarity.
Since a symplectic polarity has no non-absolute points, the $A_n$ building does not have a symplectic flip. The aim of this paper is to deal with this exceptional situation.
We shall construct a geometry $\Gamma$ similar to $\Gamma_\sigma$ and obtain a presentation of the symplectic group. This geometry will have higher rank than the usual building geometry of the symplectic group and, in particular, will provide an amalgam presentation for $\Sp_4(q)$, a group that falls out of the scope of Curtis-Tits-Phan program.


Let $V$ be a vector space of dimension $2n\ge 4$ over a field $\FF$ endowed with a symplectic form $\sform$ of maximal rank. Let $I=\{1,2,\ldots,n\}$.  Moreover consider $\cH=\{e_i,f_i\}_{i\in I}$ a hyperbolic basis of $V$.
The group of linear automorpisms of $V$ preserving the form $\sform$ will be called the 
{\dfn symplectic group of $V$} and is denoted $G=\SP(V)$.
We shall define a geometry $\Gamma$ on the subspaces of $V$ whose radical has dimension at most $1$.
This geometry is transversal, residually connected, and simply connected. Moreover, $G$ acts flag-transitively on $\Gamma$. This then gives a presentation of $G$ as the universal completion of the amalgam $\cA$ of its maximal parabolic subgroups with respect to its action on $\Gamma$. Induction allows us to replace the amalgam $\cA$ of maximal parabolic subgroups by the amalgam $\cA_{\le 2}$ of parabolic subgroups of rank at most $2$. A refinement of the amalgam $\cA_{\le 2}$ then leads to the following setup.

Consider the following amalgam 
$\cA^\pi=\{\pM_i,\pS_j,\pM_{ik},\pS_{jl},\pQ_{ij}\}_{j,l\in I; i,k\in I-\{n\}}$, whose groups are characterized as follows:
\begin{itemize}
\item $\pS_i$ is the stabilizer in $G$ of all elements of $\cH-\{e_i,f_i\}$ and the subspace $\langle e_i,f_i\rangle$,
\item $\pM_i$ is the stabilizer in $G$ of all elements of $\cH-\{f_i,f_{i+1}\}$ and the subspace $\langle e_i,f_i,e_{i+1},f_{i+1}\rangle$,
\item $\pM_{ij}=\langle \pM_i,\pM_j\rangle$, $\pS_{ij}=\langle \pS_i,\pS_j\rangle$, $\pQ_{ij}=\langle \pM_i,\pS_j\rangle$.
\end{itemize}
% where 
%$$\pS_i=\cap_{j\ne i}(\mbox{Stab}_G(e_j)\cap \mbox{Stab}_G(f_j))$$
%$$\pM_i=\mbox{Stab}_G(e_i)\cap\mbox{Stab}_G(e_{i+1})\cap(\cap_{j\ne i,i+1 }(\mbox{Stab}_G(e_j)\cap\mbox{Stab}_G(f_j)))$$
%$$\pM_{ij}=\langle M_i,M_j\rangle, \pS_{ij}=\langle \pS_i,\pS_j\rangle, \pQ_{ij}=\langle \pM_i,\pS_j\rangle$$
We describe these groups in more detail in Section \ref{section:slim}. In particular, in Lemma~\ref{lem:structure sPij sQij sSij} we show that the groups in this amalgam are very small. For instance, 
$$\begin{array}{ll}
\pS_i & \cong \Sp_2(\FF),\\
\pM_i&\cong \FF^3.\\
\end{array}$$
Our main result is the following.
\bth\label{thm:main theorem1}
If $|\FF|>2$ then the symplectic group $\Sp(V)$ is the universal completion of the amalgam $\cA^\pi$. Moreover for any field $\FF$ the symplectic group is the universal completion of the amalgam $\cA$ from Section \ref{subsection:parabolics}
\eth

\bco\label{mainco:sp4}
The group $\Sp_4(\FF)$  is the universal completion of the following amalgam.
$\{\pS_1,\pS_2,\pM_1,\pS_{12},\pQ_{11},\pQ_{12}\}$.
Here, $\pS_{12}\cong\Sp_2(\FF)\times\Sp_2(\FF)$ and 
 $\pQ_{11}\cong\pQ_{12}\cong \FF^3\rtimes \Sp_2(\FF)$.
\eco

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\bigskip
The paper is organized as follows.
In Section ~\ref{section:preliminaries} we review some basic notions on geometries and some relevant facts about symplectic spaces.
In Section~\ref{section:geometry} we introduce a geometry $\Gamma$ on the almost non-degenerate subspaces of $V$ with respect to some symplectic form of maximal rank and describe its residues. We prove that this geometry is transversal and residually connected. 
In Section~\ref{section:simple connectedness} we show that the geometry and all residues of rank at least $3$ are simply connected with one exception. We show that in the exceptional case the residue has a simply connected $2$-cover.
In Section~\ref{section:group action} we describe the flag-transitive action of $\Sp(V)$ on $\Gamma$ and its rank $3$ residues. We describe the parabolic subgroups in some detail and prove that $\Sp(V)$ is the universal completion of the amalgam $\cA_{\le 2}$ of parabolics of rank at most $2$ for its action on $\Gamma$.
In Section~\ref{section:slim} we define a slim version of the amalgam $\cA_{\le 2}$ by removing most of the Borel subgroup from each of its groups.
Finally, in Section~\ref{section:concrete amalgam} we define the amalgam $\cA^\pi$ and prove Theorem~\ref{thm:main theorem1}.
%\newpage
\section{Preliminaries}\label{section:preliminaries}
We first need some notation to describe how $\sform$ restricts to the various subspaces of $V$.
Let $\perp$ denote the orthogonality relation between subspaces of $V$ induced by $\sform$.
Thus, for $U,W\le V$ we have
$$U\perp W \iff  \sform(u,w)=0\mbox{  for all }u\in U, w\in W.$$
We write 
$$U^\perp=\{ v\in V\mid \sform(u,v)=0\  \forall u\in U\}\le V.$$

The {\dfn radical} of a subspace $U$ is the subspace $\Rad(U)=U\cap U^\perp$.
The {\dfn rank} of $U$ is $\rank(U)=\dim U-\dim\Rad(U)$. 
Note that since $\sform$ is symplectic, we have $\sform(v,v)=0$ for all $v\in V$ and so $V$  has no anisotropic 
 part with respect to $\sform$.
%It is useful to have the following well-known fact at hand.

Let $2r=\rank(U)$ and $d=\dim(U)-2r$. A {\dfn hyperbolic basis} for $U$ is a basis $\{e_i,f_j\mid 1\le i\le r+d, 1\le j\le r\}$ such that
\begin{itemize}
\AI{i}
for all $1\le i,j\le r$,
$$\begin{array}{rl}
 \sform(e_i,e_j)=\sform(f_i,f_j)&=0,\\
 \sform(e_i,f_j)&=\delta_{ij},\mbox{ and }\\
 \end{array}$$
\AI{ii}
 $\{e_{r+i}\mid 1\le i\le d\}$ is a basis for $\Rad(U)$.
\end{itemize}


\ble\label{lem:hyperbolic basis}
Suppose that $W\le U-\Rad(U)$.
Then, any hyperbolic basis for $W$ extends to a hyperbolic basis for $U$.
\ele
\pf
This is in some sense Witt's theorem.
\qed

\ble\label{lem:radical complement}
Suppose that $W\le U-\Rad(U)$ and $r=\rank(U)$.
If $\dim(W)=2r$, then $U=W\oplus \Rad(U)$ and $W$ is non-degenerate.
\ele

\pf
That $U=W\oplus \Rad(U)$ is simple linear algebra. 
As a consequence, and since $\Rad(U)\perp U$, we have $\Rad(W)\le \Rad(U)$.
Thus $\Rad(W)\le \Rad(U)\cap W=\{0\}$.
\qed

\medskip

We now introduce some basic notions on geometries as we view them.

\Definition A {\dfn pre-geometry} is a triple $\Gamma=(\cO,\typ,I,\inc)$, where $\cO$ is a collection of objects, $I$ is a set of {\dfn types}, $\inc$ is a symmetric and reflexive relation, called the {\dfn incidence relation} and 
 $\typ\colon \cO\to I$  is a surjective {\dfn type function} such that whenever $X\inc Y$, then either $X=Y$ or $\typ(X)\ne\typ(Y)$. The {\dfn rank} of the geometry $\Gamma$ is the size of $I$.

\Definition
The {\dfn incidence graph} of $\Gamma$ is the graph whose vertices are the objects of $\Gamma$ and in which two objects are adjacent if they are incident in $\Gamma$.
This is a multipartite graph whose parts are indexed by $I$.

\Definition A {\dfn flag} $F$ is a collection of pairwise incident objects. The {\dfn rank} of $F$ is 
 $\rank(F)=|F|$. The {\dfn type} of $F$ is $\typ(F)=\{\typ(X)\mid X\in F\}$.
A {\dfn chamber} is a flag $C$ of type $I$.

A {\dfn residue} $R$ of a flag $F$ of type $I-J$ is the pre-geometry induced on the collection of all objects incident to $F$, but not belonging to $F$. We say that $R$ has type $J$.

\Definition Given subsets $J,K\sbe I$ and an $(I-J)$-flag $F$.
Then the {\dfn $K$-shadow} of $F$ is the collection of all $K$-flags incident to $F$.


\Definition A pre-geometry $\Gamma$ is {\dfn transversal} if any flag is contained in a chamber.

\Definition  A pre-geometry is {\dfn connected} if the incidence graph is connected.
A pre-geometry is {\dfn residually connected} if each of its residues of rank at least $2$ is connected.

\Definition A {\dfn geometry} is a pre-geometry that is transversal, connected and residually connected.

\Definition An {\dfn automorphism group} $G$ of a pre-geometry $\Gamma$ is a group of permutations of the collection of objects that preserves type and incidence.
We call $G$ {\dfn flag-transitive} if for any $J\sbe I$, $G$ is transitive on the collection of $J$-flags.

\Definition Let $G$ be a group of automorphisms of a geometry $\Gamma$ over an index set $I$.
Fix a maximal flag $C$. The {\dfn standard parabolic subgroup of type $J\sbe I$} is the stabilizer in $G$ of the residue of type $J$ on $C$.

\medskip
%\subsection{Simple connectedness}\label{subsection:simple connectedness}
Let us recall the definition of the fundamental group of a connected geometry $\Delta$.
A {\dfn path of length $k$} is a sequence of elements $x_0,\ldots,x_k$ such that $x_i$ and $x_{i+1}$ are incident for 
 $0\le i<k$. 
We do not allow repetitions, that is, $x_i\ne x_{i+1}$ for all $0\le i<k$.
A {\dfn cycle based at an element $x$} is a path $x_0,\ldots,x_k$ in which $x_0=x=x_k$.
Two paths $\gamma$ and $\delta$ are {\dfn homotopy equivalent} if one can be obtained from the other by inserting or eliminating cycles of length $2$ or $3$. We denote this by $\gamma\simeq \delta$.
The equivalence classes of cycles based at an element $x$ form a group under concatenation.
This group is called the {\em fundamental group of $\Delta$ based at $x$} and is denoted $\pi_1(\Delta,x)$. 
If $\Delta$ is (path) connected, then the isomorphism type of this group does not depend on $x$ and we call this group simply the {\em fundamental group} of $\Delta$ and denote it $\pi_1(\Delta)$.
We call $\Delta$ {\em simply connected} if $\pi_1(\Delta)$ is trivial.


The central tool for this paper is the following result by J. Tits.
\ble\label{lem:tits lemma}
Given a group $G$ acting flag-transitively on a geometry $\Gamma$. Fix a maximal flag $C$.
Then $G$ is the universal completion of the amalgam consisting of the standard maximal parabolics with respect to $C$ if and only if $\Gamma$ is simply connected.
\ele



In order to prove that $\Delta$ is simply connected, it suffices to show that any cycle based at a given element
 $x$ is homotopy equivalent to a cycle of length $0$. We call such a cycle {\em trivial} or {\em null homotopic}.

\ble\label{lem:common object}
If $\Delta$ is a pre-geometry, then any cycle all of whose elements are incident to a given element is null-homotopic.
\ele
\pf
Let $\gamma=(x_0,\ldots,x_k)$ be a cycle all of whose elements are incident to a given element $y$.
Then 
$$\begin{array}{rl}
\gamma&\simeq (x_0,y)\circ(y,x_0,x_1,y)\circ (y,x_1,x_2,y) \circ\cdots\circ (y,x_{k-1},x_k,y)\circ(y,x_0)\\
      &\simeq (x_0,y)\circ (y,x_1,x_2,y)\circ\cdots\circ (y,x_{k-1},x_k,y)\circ(y,x_0)\\
      &\simeq (x_0,y)\circ(y,x_0)\\
      &\simeq 0.\\
      \end{array}$$
\qed


%A {\em point path} of length $k$ is a path $x_0,\ldots,x_k$ of points such that $x_i$ and $x_{i+1}$ are collinear.
%That is, it is a path in the collinearity graph. Similarly, we define a point cycle.
All our pre-geometries will have a string diagram. We will give the following ad-hoc definition here.


\Definition
We say that a pre-geometry {\dfn has a string diagram} if there is a total ordering on its type set $I$ such that for any three types $i,j,k\in I$ with $i<j<k$ we have the following.
If $X,Y,Z$ are objects of type $i$,$j$, and $k$ respectively such that $X$ and $Z$ are incident with $Y$, then $X$ is incident with $Z$.

Note that if a pre-geometry has a string diagram, then so does every residue.

For a geometry with string diagram, by {\dfn points} we mean the objects whose type $p$ is minimal in $I$ and by {\dfn lines} we mean the objects whose type is minimal in $I-\{p\}$.

We now prove the following result, which is also a consequence of Theorem 12.64 in Pasini~\cite{Pas94}.

\ble\label{lem:point-line cycles only}
Let $\Delta$ be a geometry with a string diagram.
Then, every cycle based at $x$ is homotopy equivalent to a cycle consisting of points and lines only.
\ele
\pf
Consider an arbitrary cycle $\gamma=x_0,\ldots,x_k$ based at $x$.
By transversality, $\gamma$ is homotopic to a cycle based at the point incident to $x$.
We may therefore assume that $x$ itself is a point.

We prove that $\gamma$ is homotopic to a cycle of points and lines only by induction on $m$, the number of 
 elements other than points and lines on $\gamma$.
Clearly if $m=0$ we are done.

Let $i$ be minimal such that $x_i$ is not a point or a line.
If $\typ(x_{i+1})>\typ(x_i)$, then $\gamma$ is homotopy equivalent to $\gamma'=x_0,\ldots,x_{i-1},x_{i+1},\ldots,x_k$, since
 $x_{i-1},x_i,x_{i+1},x_{i-1}$ is a $3$-cycle.
Clearly $\gamma'$ contains $m-1$ elements that are not points or lines, so by induction we are done.

Now assume $\typ(x_{i+1})<\typ(x_i)$.
Note that $x_i$ is not a point or a line.
Let $y$ be a point incident to $x_{i+1}$. It exists by transversality.
Note that $y$ is incident to $x_i$ as well.

By residual connectedness of the residue of $x_i$, there is a path $\delta$ of points and lines on $x_i$ connecting $x_{i-1}$ to $y$.
Then, $x_{i-1},x_i,x_{i+1}$ is homotopic to $\delta\circ (y,x_{i+1})$ by Lemma~\ref{lem:common object}.
Hence $\gamma$ is homotopic to $\gamma'=(x_0,\ldots,x_{i-1})\circ\delta\circ(y,x_{i+1},x_{i+2},\ldots,x_k)$.
Again, $\gamma'$ contains $m-1$ elements that are not points or lines, so by induction we are done.
\qed


\Definition Given a geometry $\Gamma$ with type set $I$ and $k\in \NN_\ge 1$, 
a {\dfn $k$-cover} is a geometry $\bar{\Gamma}$ with type set $I$ together with an incidence and type preserving map $\pi\colon\bar{\Gamma}\to \Gamma$ with the following properties:
\begin{itemize}
\item For any $J$-flag $F$ in $\Gamma$ the fiber $\pi^{-1}(F)$ consists of exactly $k$ distinct and disjoint $J$-flags.
\item Given a flag $F$ in $\Gamma$ and some flag $\bar{F}$ in $\bar{\Gamma}$ such that $\pi(\bar{F})=F$, then the restriction
 $\pi\colon \Res(\bar{F})\to\Res(F)$ is an isomorphism. 
\end{itemize}
We employ the following ad-hoc definition. We call $(\bar{\Gamma},\pi)$ the {\dfn universal cover of $\Gamma$} if it is a cover of $\Gamma$ that is simply connected.

The fundamental group $\Pi_{x}(\Gamma)$ based at the point $x$ acts on $\pi^{-1}(x)$ as follows. As we know $\Pi_{x}(\Gamma)$ is generated by cycles that are not null-homotopic. Let $\delta=x=x_0,x_1,\ldots,x_k=x_0$ be such a cycle.
Pick a point $\bar{x}\in\pi^{-1}(x)$. 
By the covering properties, there is a unique object of type $\typ(x_1)$ incident to $\bar{x}$. Call this object $\bar{x_1}$.
Continuing in this way, we find a path $\bar{x_0},\bar{x_1}|,\ldots,\bar{x_k}$.
Note that $\bar{x_0}$ and $\bar{x_k}$ are (not necessarily equal) elements in $\pi^{-1}(x)$. Set $\delta\cdot \bar{x_0}=\bar{x_k}$.

The following is a special case of Lemma 12.2 of \cite{Pas94}.
\ble
Let $\Gamma$ be a geometry with universal cover $(\bar{\Gamma},\pi)$ and let $x$ be an object.
Then the fundamental group $\Pi_x(\Gamma)$ acts regularly on $\pi^{-1}(x)$.
\ele
In particular, the universal cover of $\Gamma$ is a $2$-cover if and only if 
 the fundamental group of $\Gamma$ is $\ZZ/2\ZZ$.
 
\section{A geometry for the symplectic group}\label{section:geometry}
Let $V$ be a vector space of dimension $n$ over a field $\FF$ endowed with a symplectic form $\sform$ of maximal rank.
The {\dfn quasi-Phan geometry} $\Gamma=\Gamma(V)$ is defined as follows. 
For $i\in I=\{1,2,\ldots,n-1\}$, the $i$-objects, or objects of type $i$, are the $i$-spaces $U\le V-\Rad(V)$ such that 
 $\dim(\Rad(U))\le 1$. 
More explicitly, since $\sform$ is symplectic, this means that 
$$\dim(\Rad(U))=\left\{
\begin{array}{rl}
0 &\mbox{ if $i$ is even}\\
1 &\mbox{ if $i$ is odd.}\\
\end{array}\right.$$

We say that two objects $X$ and $Y$ are {\dfn incident} whenever $X\sbe Y-\Rad(Y)$ or vice versa.

It is not too difficult to see that $\Sp(V)$ is an automorphism group for $\Gamma$. In fact, 
 Lemma~\ref{lem:flag-transitive} shows that it acts flag-transitively on $\Gamma$.

\bco\label{cor:hyperbolic bases and chambers}
Any hyperbolic basis for $V$ gives rise to a unique chamber of $\Gamma$ and, conversely, any chamber gives rise to a 
 (not necessarily unique) hyperbolic basis for $V$.
\eco
\pf
Let $C$ be a chamber. 
Then, for any two consecutive objects $W,U\in C\cup\{V\}$ we have $W\le U-\Rad(U)$.
Hence, using Lemma~\ref{lem:hyperbolic basis} repeatedly we find a hyperbolic basis $\cH$ for $V$ such that, for any $X\in C$, $\cH\cap X$ is a hyperbolic basis for $X$.

Conversely, let $\cH$ be a hyperbolic basis for $V$. Set $d=\dim(\Rad(V))$.
We have $\cH=\{e_i,f_j\mid 1\le i\le r+d, 1\le j\le r\}$.
For $1\le i\le r+d$, let $h_{2i-1}=e_i$ and, for $1\le j\le r$, let  
$h_{2j}=f_j$.
Then, setting $C_l=\langle h_1,h_2,\ldots,h_l\rangle_V$, the collection $C=\{C_l\}_{l=1}^{n-1}$ is a chamber of $\Gamma$.
\qed

\medskip


In the remainder of this paper we shall use the following standard setup.
Let $2r=\rank(V)$  and let $d=\dim(\Rad(V))\le 1$, so that $n=2r+d$.
We fix a hyperbolic basis 
$$\cH=\{e_i,f_j\mid 1\le i\le r+d, 1\le j\le r\}.$$
Alternatively we write
 $$\begin{array}{ll}
 \cH&=\{h_k\}_{k=1}^n\mbox{, where }\\
h_k&=\left\{\begin{array}{ll}
e_i&\mbox{ if }k={2i-1},\\
f_j & \mbox{ if }k=2j.\\
\end{array}\right.
\end{array}
$$
We call this the {\dfn standard hyperbolic basis}.
The {\dfn standard chamber} is the chamber $C=\{C_k\}_{k=1}^{n-1}$ associated to $\cH$ as in Corollary~\ref{cor:hyperbolic bases and chambers}.
That is, $C_k=\langle h_1,h_2,\ldots,h_k\rangle_V$, for all $1\le k\le n-1$.

\ble\label{lem:geometry transversal}
The pre-geometry $\Gamma$ is transversal and has a string diagram.
\ele
\pf
Let $F$ be a flag. 
Then, for any two consecutive objects $W,U\in (F\cup\{V\})$ we have $W\le U-\Rad(U)$.
Hence, using Lemma~\ref{lem:hyperbolic basis} repeatedly we find a hyperbolic basis $\cH$ for $V$ such that, for any $X\in C$, $\cH\cap X$ is a hyperbolic basis for $X$.
According to Corollary~\ref{cor:hyperbolic bases and chambers}, $\cH$ defines a unique chamber $C$.
One verifies that $F\sbe C$.
The natural ordering on $I$ provides $\Gamma$ with a string diagram.
\qed


\ble\label{lem:geometry connected}
The pre-geometry $\Gamma$ is connected.
More precisely, the $\{1,2\}$-shadow geometry has diameter at most $2$ with equality if $n\ge 3$.
\ele
\pf
Let $X,Z$ be $1$-spaces in $V-\Rad(V)$.
If $X$ and $Z$ span a non-degenerate $2$-space, then we are done.
In particular, if $n=2$, then the diameter is $1$.

Other wise let $W$ be a point on $\langle X,Z\rangle$ different from $X$ and $Z$.
In case $\langle X,Z\rangle\spe \Rad(V)$, let $W=\Rad(V)$.
Since $V/\Rad(V)$ is non-degenerate, there is a point $Y$ in $W^\perp$ that is not
 in $\langle X,Z\rangle^\perp$.
Then clearly $X,Z\not\sbe Y^\perp$ and so $X,Y,Z$ is a path in $\Gamma$ from $X$ to $Z$.
Thus the $\{1,2\}$-shadow geometry has diameter at most $2$. Clearly equality holds.
\qed

\subsection{Residual geometries}
Let $C=\{C_i\}_{i\in I}$ be the standard chamber of $\Gamma$ associated to the hyperbolic basis $\cH$.
For every $J\sbe I$ the {\dfn standard residue of type $J$}, denoted $R_J$, is the residue of the $(I-J)$-flag 
$\{C_i\}_{i\in I-J}$.
Let $\biguplus_{m=1}^M J_m$ be the partition of $J$ into maximal contiguous subsets.
(We call $K\sbe I$ contiguous if, whenever $i,k\in K$ and $i<j<k$, then $j\in K$.)

In this case, $R_J=R_{J_1}\times R_{J_2}\times\cdots\times R_{J_M}$ since $\Gamma$ has a string diagram.
It now suffices to describe $R_J$, where $J$ is contiguous.
Let $a=\min J$ and let $b=\max J$.
There are two cases according as $a$ is even or odd.

If $a$ is odd, then the residue is the geometry $\Gamma(C_{b+1}/C_{a-1})\cong \Gamma((C_{a-1}^\perp\cap C_{b+1})/C_{a-1})$ of rank $b-a+2$. We set $C_0=\{0\}$ and $C_n=V$ for convenience.

If $a$ is even, we may assume that $V=(C_{a-2}^\perp \cap C_{b+1})/C_{a-2}$ and that $a=2$.
%Clearly, if $i=n-1$, then $\Res_\Gamma(C_i)$ is simply the geometry $\Gamma(C_i)$ of all objects contained in $C_i$.
Thus we need to describe the residue of $C_1$.
We will show that $\Res_\Gamma(C_1)$ is isomorphic to a geometry $\Pi(p,H)$ defined as follows.

\Definition\label{dfn:PipH}
Note that $\dim(\Rad(V))\le 1$.
Let $p$ be a $1$-dimensional subspace of $V-\Rad(V)$ and let $H$ be some complement of $p$ in $V$ containing $\Rad(V)$.
Note that $\Rad(H)$ if it is non-trivial, is not contained in $p^\perp$.
Namely, $\Rad(H)\cap p^\perp\sbe \Rad(V)$, which is $0$ if $\dim(V)$ is even.

Then we define $\Pi(p,H)$ to be the geometry on the following collection of subspaces of $H$:
$$\{U\le H\mid \Rad(V)\not \subseteq U, \dim(\Rad(U))\le 2 \mbox{ and } \Rad(U) =\{0\}  \mbox{ or } \Rad(U)\not \subseteq  p^\perp\}.$$
Let $U$ and $W$ be in $\Pi(p,H)$ with $\dim U < \dim W$. We say that $U$ is incident to $W$ if either $\dim(W)$ is odd and $U\sbe W$ or $\dim(W)$ is even and $U\sbe W-\Rad(W\cap p^\perp)$. 


More precisely, the objects are subspaces $U \le H$ not containing $\Rad(V)$ with the following properties
\begin{enumerate}
\item If $U$ is odd dimensional then $\Rad(U)$ has dimension $1$ and does not lie in $p^\perp$.
\item If $U$ is even dimensional then $U$ is either non-degenerate or $\Rad(U)$ has dimension $2$ and is not contained in $p^\perp$.
\end{enumerate}

\ble The map  given by $$\begin{array}{ll}
\varphi\colon\Res_\Gamma(p)&\to\Pi(p,H)\\
X&\mapsto X\cap H.\\
\end{array}$$ is an isomorphism.
\ele
\pf
We first show that $X\in\Res_\Gamma(p)$ if and only if $X\cap H\in \Pi(p,H)$.
Note that since $p$ is isotropic, $p^\perp\cap H$ is a codimension $1$ subspace of $H$.
Note that if $X\in\Res_\Gamma(p)$, then since $X\in \Gamma$, $\Rad(V) \not \subset X$. Moreover  $X\cap H$ is a complement of $p$ in $X$ so $\Rad(X\cap H)$ cannot have  dimension more than two. Also $\Rad(X\cap H) \cap p^\perp \le \Rad(X)$ and so if $X$ is even dimensional  $\Rad(X\cap H) \cap p^\perp=\{0\}$ and if $X$ is odd dimensional and $X\cap H$ is degenerate then 
 $\Rad(X\cap H) \not\sbe p^\perp$.
 
 Conversely if $U \in \Pi(p,H)$ it is easy to see that $X=\langle U,p\rangle$ is in $\Res_\Gamma(p)$. Indeed if $U$ is odd dimensional, since $p$ is not perpendicular to $\Rad(U)$, the space $X$ is non-degenerate. If $U$ is even dimensional and non-degenerate then the space $X$ is of maximal possible rank, $p$ is not in $\Rad(X)$ and $\Rad(V)\not\subseteq X$. Finally, if $U$ is even dimensional and $\Rad(U)$ has dimension $2$, then, since $p$ is not perpendicular to the whole of $\Rad(U)$, we have $\Rad(X)=\Rad(U)\cap p^\perp\ne \Rad(V), p$ and so $X \in \Res_\Gamma(p)$.
 
 Suppose that $X$ and $Y$ are incident elements of $\Res_\Gamma(p)$ and $\dim(X)<\dim(Y)$. If $\dim(Y)$ is even then incidence is containment in both geometries. If $\dim(Y)$ is odd then we need to prove that $\Rad(Y)\not\sbe X$ iff $\Rad(Y^\varphi\cap p^\perp)\not\sbe X^\varphi$. We note that for any $Z\in \Res_\Gamma(p)$, we have 
  $\Rad(Z)^\varphi=\Rad(Z^\varphi\cap p^\perp)$ and so the conclusion follows. \qed

 \ble\label{lem:odd point-line residual is complete graph}
 Two points $p_1,p_2$ of $\Pi(p,H)$ are collinear iff $\Rad(V) \not\sbe\langle p_1,p_2\rangle$. In particular if $\dim(V)$ is even then the collinearity graph of $\Pi(p,H)$ is a complete graph and if $\dim(V)$ is odd then the collinearity graph has diameter two.
 \ele
 \pf
If $p_1, p_2$ are two points in $\Pi(p,H)$, then $\langle p_1,p_2\rangle$ is either totally isotropic but not contained in $p^\perp$ or non-degenerate.
So this is a line of $\Pi(p,H)$ if and only if $\Rad(V) \not\sbe \langle p_1,p_2\rangle$.
Therefore the conclusion follows.
 \qed
 

\ble\label{lem:geometry residually connected}
The pre-geometry $\Gamma$ is residually connected.
\ele
Let $J\sbe I$ and let $R_J$ be the residue of the $(I-J)$-flag $\{C_i\}_{i\in I-J}$.
Let $\biguplus_{m=1}^M J_m$ be the partition of $J$ into contiguous subsets.
If $M\ge 2$, then the residue is connected since it is a direct product of geometries.
Otherwise, the residue is isomorphic to $\Gamma(V)$ for some vector space $V$ of dimension at least $3$, or to 
 $\Pi(C_1,H)$ inside some $\Gamma(V)$ for some vector space $V$ of dimension at least $4$.
Thus the connectedness follows from Lemmas~\ref{lem:geometry connected} and~\ref{lem:odd point-line residual is complete graph}.
\qed

\bco\label{cor:geometry}
The pre-geometry $\Gamma$ is a geometry with a string diagram.
\eco
\pf
By Lemma~\ref{lem:geometry transversal}, $\Gamma$ is transversal and has a string diagram and by Lemma~\ref{lem:geometry residually connected}, it is residually connected.
\qed

%\ble\label{lem:geometry residually connected}
%Given any type set $J\sbe\{1,2,\ldots,n-1\}$ with $|J|\ge 2$, the geometry $\Gamma_J$ of objects of type $j\in J$ %with induced incidence, is connected.
%As a consequence, $\Gamma$ itself is connected.
%\ele
%\pf
%Clearly, it suffices to prove this in case $|J|=2$.
%So let $J=\{i,j\}$ with $i<j$.
%By transversality, it suffices to show that there is an $\{i,j\}$-path connecting any two objects $X$ and $Z$ of %type $i$.
%We prove this by induction on $s=i+j\ge 3$.
%If $s=3$, then $i=1$ and $j=2$.
%\qed



\section{Simple connectedness}\label{section:simple connectedness}
In this section we prove that the geometry $\Gamma$ and all of its residues of rank at least $3$ are simply connected.
Note that $\Gamma$ and all its residues are geometries with a string diagram.
Therefore by Lemma~\ref{lem:point-line cycles only}  it suffices to show that all point-line cycles are null-homotopic.
 
\ble\label{lem:point-line cycles are null-homotopic}
If $|\FF|\ge 3$ or $\dim(V)$ is even, then any point-line cycle of $\Gamma$ is null-homotopic.
\ele
\pf
Let $\gamma$ be a point-line cycle based at a point $x$.
We identify $\gamma$ with the sequence $x_0,\ldots,x_k$ of points on $\gamma$ (so $\gamma$ is in fact a 
$2k$-cycle).
We show by induction on $k$ that $\gamma$ is null-homotopic.

If $k\le 3$, then $U=\langle x_0,x_1,x_2\rangle_V$ is non-isotropic as it contains the hyperbolic line 
$\langle x_0,x_1\rangle_V$.
If $\Rad(V)\not\sbe U$, then it is an object of the geometry and so by Lemma~\ref{lem:common object}, $\gamma$ is null-homotopic.
If $\Rad(V)\sbe U$, then since $|\FF|\ge 3$ there is a point $x$ such that 
 $x$ is collinear to $x_0$, $x_1$, and $x_2$. Namely, consider the points
  $\Rad(V)x_i$ in the non-degenerate space $V/\Rad(V)$. These are $3$ distinct points on the non-degenerate $2$-space $U/\Rad(V)$. Take a fourth point $y$ on this line. Since $V/\Rad(V)$ is non-degenerate, there is a point $x$ orthogonal to $y$ but not to any other point on $U/\Rad(V)$.

Now let $k\ge 4$.
If two non-consecutive points $x_i$ and $x_j$ in $\gamma$ are collinear, then let 
$\gamma_1=\delta_1 \circ \delta_2\circ\delta_1^{-1}$ and $\gamma_2=\delta_1\circ\delta_3$, where
 $$\begin{array}{rl}
 \delta_1&=x_0,\ldots,x_i\\
 \delta_2&=x_i,x_{i+1},\ldots,x_{j-1},x_j,x_i.\\
 \delta_3&=x_i,x_j,\ldots,x_k.\\
 \end{array}$$
Clearly $\gamma_1\circ\gamma_2$ is homotopic to $\gamma$.
Also, the cycles $\delta_2$ and $\gamma_2$ are both shorter than $\gamma$.
By induction, $\delta_2$ and $\gamma_2$ are null-homotopic and hence so are $\gamma_1$ and $\gamma$ itself.

Therefore we may assume that no two non-consecutive elements in $\gamma$ are collinear.
In particular $x_1\perp x_{k-1}$ and $x_{k-2}\perp x_0$.
The line $\langle x_{k-1},x_k\rangle_V$ has at least three points.
Take any $y\in\langle x_{k-1},x_k\rangle-\{x_{k-1},x_k\}$.
Then, $y\not\perp x_1,x_{k-2}$.
Thus, since $x_1\not\perp x_k=x_0$ and $x_1\perp x_{k-1}$, the point $y$ is collinear to $x_1$.
By the same reasoning $y$ is collinear to $x_{k-2}$.

Now let $\gamma_1=\delta_1\circ\delta_2$ and $\gamma_2=\delta_2^{-1}\circ\delta_3\circ\delta_2$ and $\gamma_3=\delta_2^{-1}\circ\delta_4\circ\delta_2$, where
 $$\begin{array}{rl}
 \delta_1&=x_0,x_1,y\\
 \delta_2&=y,x_0\\
 \delta_3&=y,x_1,x_2,\ldots,x_{k-2},y\\
 \delta_4&=y,x_{k-2},x_{k-1},y\\
 \end{array}$$
Then, 
$$\gamma\simeq \gamma_1\circ\gamma_2\circ\gamma_3\simeq \gamma_2\simeq 0,$$
where the second equivalence holds since $\gamma_1$ and $\gamma_3$ are triangles, and the third equivalence holds since $\gamma_2\simeq 0$ by induction.
\qed

\bpr\label{prop:gamma simply connected}
The geometry $\Gamma$ is simply connected.
\epr
\pf
First of all, $\Gamma$ is connected by Lemma~\ref{lem:geometry connected}.
Thus it suffices to show that any cycle is null-homotopic.
By Lemma~\ref{lem:point-line cycles only} such a cycle is homotopic to a point-line cycle.
Finally, by Lemma~\ref{lem:point-line cycles are null-homotopic} point-line cycles are null-homotopic and the result follows.
\qed

\subsection{Residual geometries}
 \medskip
We shall now prove that, apart from one exception, all residues of rank at least $3$ are simply connected.
As we saw above any residue is either isomorphic to $\Gamma(V)$ for some $V$, or to $\Pi(p,H)$ for some point $p$ inside $\Gamma(V)$ for some $V$ or it is a direct product of such geometries and possibly rank $1$ residues.

We already proved that $\Gamma(V)$ is simply connected. We shall now prove that, apart from one exception, $\Pi(p,H)$ is a simply connected geometry. By Lemma~\ref{lem:odd point-line residual is complete graph}, if $n$ is even, then the collinearity graph of $\Pi(p,H)$ is a complete graph and so we only have to show that all triangles are null-homotopic.
 If $n$ is odd, then the diameter of the collinearity graph is $2$ and so all $k$-cycles can be decomposed into cycles of length $3$, $4$, and $5$.
 \ble Suppose that $\dim(V)$ is odd. Then any cycle of length $4$ or $5$ can be decomposed into triangles.
 \ele
 \pf
We first note that two points $p_1$ and $p_2$ are at distance $2$ only if $\langle p_1,p_2\rangle\spe \Rad(V)$. Therefore, if $p_1,p_2,p_3,p_4$ is a $4$-cycle, then $\Rad(V)\sbe \langle p_1, p_3\rangle, \langle p_2,p_4\rangle$. Since $|\FF|\ge 2$ there exists a line $L$ on $\Rad(V)$ different from these two lines and any point of $L-p^\perp$ is collinear to all of $p_1,p_2,p_3,p_4$. Thus we decompose the $4$-cycle into triangles. 
Now suppose $p_1,p_2,p_3,p_4,p_5$ is a  $5$-cycle. Then $\Rad(R)$ lies on at most $1$ of $p_1p_3$ and $p_1p_4$ and so one of these lines is in fact a line of the geometry. Thus, we can decompose the $5$-cycle into shorter cycles. 
 \qed
 
 \ble\label{lem:hyperplanes of Pi} Consider a hyperplane $W$ of $H$.
 If $\dim(V)$ is even, then $W\in\Pi(p,H)$.
 If $\dim(V)$ is odd, then $W\in\Pi(p,H)$ if and only if $\Rad(V)\not\sbe W$.
 \ele
 \pf
 Let $\dim(V)$ be even and let $S=\Rad(H)$. Then $\dim(S)=1$ and $S\not\sbe p^\perp$.
 If $S\not\sbe W$, then $W$ is non-degenerate and so it belongs to $\Pi(p,H)$.
 If $S\sbe W$, then $S\sbe \Rad(W)$, which has dimension $2$ and so $\Rad(W)\not\sbe p^\perp$.
 So again $W\in \Pi(p,H)$.
 
 Now let $\dim(V)$ be odd and let $R=\Rad(V)$. Then $R\sbe \Rad(H)$, which has dimension $2$ and is not included in $p^\perp$. Now since $W$ is a hyperplane of $H$ it either contains $\Rad(H)$ or it intersects it in a $1$-dimensional space. In the former case, $R\sbe \Rad(W)$ and $W\not\in\Pi(p,H)$.
 In the latter case, either $R\sbe \Rad(W)$ and $W\not\in\Pi(p,H)$, or $\Rad(W)\not\sbe p^\perp$ and 
  $W\in\Pi(p,H)$.
 \qed
 
 
 \ble\label{lem:Pi simply connected n odd}
  Suppose $\dim(V)$ is odd.
 Then any triangle of $\Pi(p,H)$  is null-homotopic.
 \ele
 \pf
 Take a triangle on the points $p_1,p_2,p_3$. Note that this means that $R=\Rad(V)$ does not lie on any of the lines $p_ip_j$. Let $U=\langle p_1,p_2,p_3\rangle$.
 We have two cases:
 1) $R\not\sbe U$. Then pick a hyperplane $W$ of $H$ containing $U$ but not $R$.
 Then by Lemma~\ref{lem:hyperplanes of Pi}, $W\in \Pi(p,H)$. Moreover, since $\dim(W)$ is odd it is incident to each of the lines $p_ip_j$.
 
 2) $R\sbe U$. Let $S$ be a point of $\Rad(H)-R$. Then $S\not\sbe p^\perp$.
 Also note that for any $i$ and $j$, $p_i,p_j,S$ is a triangle of type (1), so the triangle  $p_1,p_2,p_3$ is null-homotopic.  
 \qed
 
 \ble\label{lem:>6 or >2 is null}
 Suppose $\dim(V)\ge 7$ or $|\FF|\ge 3$.
 Then any triangle of $\Pi(p,H)$  is null-homotopic. 
 \ele
% \pf
% If $\dim(V)$ is odd, then we are done by Lemma~\ref{lem:Pi simply connected n odd}
% So now let $\dim(V)=n$ be even so that $\dim(V)\ge 8$.
% Take a triangle on the points $p_1,p_2,p_3$ and let $U=\langle p_1,p_2,p_3\rangle$.
%If $T=\Rad(U)\cap p^\perp=0$, then $U\in \Pi(p,H)$ and we're done.
%Now suppose $T\sbe p^\perp\cap H$.
% Let $L=U\cap p^\perp$ and let $q_{ij}=p_ip_j\cap L$.
%If  $|\FF|\ge 3$ we can find a point $r\in L-\{q_{12},q_{23},q_{13}\}$.
%If $\dim(V)\ge 8$, $\dim(U^\perp\cap p^\perp\cap H)\ge 3$ and so again we can find a point 
% $r\in H\cap p^\perp-\{q_{12},q_{13},q_{23}\}$.

% Consider the subgeometry $\Gamma_r=r^\perp$ of $\Gamma$. Note that it contains $p,p_1,p_2,p_3$.
% Then consider $\Pi(p,H\cap r^\perp)$ inside $\Gamma_r$.
% Note that the lines $p_ip_j$ are still lines of $\Pi(p,H\cap r^\perp)$ since they do not intersect the radical $r$ of $r^\perp$.
% Now since $\dim(r^\perp)$ is odd we are done by Lemma~\ref{lem:Pi simply connected n odd}. 
% \qed
% 
% 
% 
%\pf
%If $\dim(V)$ is odd, then we are done by Lemma~\ref{lem:Pi simply connected n odd}
%So now let $\dim(V)=n$ be even.
%Let $Q=\Rad(H)$.
%Take a triangle on the points $p_1,p_2,p_3$ and let $U=\langle p_1,p_2,p_3\rangle$.
%We consider the case where $U$ is non-isotropic.
%Let $T=\Rad(U)$.
%If $T$ belongs to $H-p^\perp$, then $U\in \Pi(p,H)$ and the triangle is null-homotopic.

%Let $L=U\cap p^\perp$. For $1\le i<j\le 3$, let 
%$L_{ij}=p_ip_j$ and $q_{ij}=L_{ij}\cap L$.

%
%Now first suppose that all lines $L_{ij}$ are non-degenerate.
%Then the planes $\langle p_i,p_j,Q\rangle$ belong to $\Pi(p,H)$ since their radical is $Q\sbe H-p^\perp$. Hence we can decompose the triangle $p_1,p_2,p_3$ into triangles
% $p_i,p_j,Q$ that are all null-homotopic and we're done.
%Next assume that two lines, say $L_{12}$ and $L_{13}$ are isotropic.
%Then $T=p_1\sbe H-p^\perp$ and again the triangle is null-homotopic.

%Finally we deal with the case where exactly one line, say $L_{12}$ is isotropic.
%Then $T\in L_{12}$. If $T\sbe H-p^\perp$ then we're done, so assume that 
%$T\sbe p^\perp\cap H$.

%
%There are two cases.
%(1) $n=6$ and $|\FF|\ge 3$.
%Note that $q_{12}=T$ and so since $|\FF|\ge 3$ we can find a point $r\in L-\{T,q_{12}.q_{13}.q_{23}\}$.
%Let $W=\langle r^\perp\cap p^\perp\cap H,U\rangle$.
%First note that $\dim(r^\perp\cap p^\perp\cap H)=n-3$.
%This is because $p^\perp\cap H$ is a non-degenerate symplectic space of dimension $n-2$. 
%Note that this space contains $L$ since it is totally isotropic (it contains $T$) and contains $r$. Also, since $W\not\sbe p^\perp\cap H$ and $U\cap p^\perp = L$ we have
% $\dim(W)=\dim(r^\perp\cap p^\perp\cap H)+\dim(U)-\dim(L)=n-2$.
%Thus $W$ is an object of $\Pi(p,H)$ by Lemma~\ref{lem:hyperplanes of Pi}.
%Moreover, since $p^\perp\cap H$ is non-degenerate, $W\cap p^\perp=r^\perp\cap p^\perp\cap H$ of odd dimension $n-3$ has a radical of dimension $1$, so this must be $r$.
%By choice of $r$, none of the lines $L_{ij}$ meets $r$ and so these lines are incident with $W$.
%By Lemma~\ref{lem:common object} therefore, the triangle is null-homotopic.

%(2) $n\ge 7$. Note that in this case in fact $n\ge 8$ since we assume that $n$ is even.
%In this case we find a point $r$ in $L^\perp\cap p^\perp\cap H-L$.
%This is possible since $p^\perp\cap H$ is non-degenerate and $L$ is a totally isotropic $2$-space so that $\dim(L^\perp\cap p^\perp\cap H)=(n-2)-2\ge 4>\dim(L)=2$.
%Again consider $W=\langle r^\perp\cap p^\perp\cap H,U\rangle$.
%For the same reasons as above, $W$ has dimension $n-2$ and therefore belongs to $\Pi(p,H)$.
%Also $W$ intersects $p^\perp$ in $r^\perp\cap p^\perp\cap H$, which is of odd dimension $n-3$ and has radical $r$.
%Since $r\not\in L$, none of the lines $L_{ij}$ meets $r$ and so these lines are incident with $W$.
%By Lemma~\ref{lem:common object} therefore, the triangle is null-homotopic.
%\qed

%%A good proof.
%\pf
%If $\dim(V)$ is odd, then we are done by Lemma~\ref{lem:Pi simply connected n odd}
%So now let $\dim(V)=n$ be even.
%Let $Q=\Rad(H)$.
%Take a triangle on the points $p_1,p_2,p_3$ and let $U=\langle p_1,p_2,p_3\rangle$.
%In case $U$ is non-isotropic, let $T=\Rad(U)$.
%Let $L=U\cap p^\perp$. For $1\le i<j\le 3$, let 
%$L_{ij}=p_ip_j$ and $q_{ij}=L_{ij}\cap L$.

%
%First consider the case that $U$ is non-isotropic.
%We note that if $T$ belongs to $H-p^\perp$, then $U\in \Pi(p,H)$ and the triangle is null-homotopic.

%Now first suppose that all lines $L_{ij}$ are non-degenerate.
%Then the planes $\langle p_i,p_j,Q\rangle$ belong to $\Pi(p,H)$ since their radical is $Q\sbe H-p^\perp$. Hence we can decompose the triangle $p_1,p_2,p_3$ into triangles
% $p_i,p_j,Q$ that are all null-homotopic and we're done.
%Next assume that exactly two lines, say $L_{12}$ and $L_{13}$ are isotropic.
%Then $T=p_1\sbe H-p^\perp$ and again the triangle is null-homotopic.

%Suppose that only $L_{12}$ is isotropic.
%Then $T\in L_{12}$. If $T\sbe H-p^\perp$ then we're done, so assume that 
%$T\sbe p^\perp\cap H$. 
%There are two cases left. Namely, either $U$ is isotropic, or exactly one line of the triangle is isotropic and it intersects $p^\perp\cap H$ in $T$.
%Note that in both cases $L$ is totally isotropic.

%Suppose we can find a point $r$ in $p^\perp\cap H$ such that $r^\perp\spe L$.
%For such a point let $W=\langle r^\perp\cap p^\perp\cap H, U\rangle $.
%First note that $\dim(r^\perp\cap p^\perp\cap H)=n-3$.
%This is because $p^\perp\cap H$ is a non-degenerate symplectic space of even dimension $n-2$. 
%Note that $L\sbe  r^\perp\cap p^\perp\cap H$.
%Also, since $U\not\sbe p^\perp\cap H$ and $U\cap p^\perp = L$ we have
% $U\cap (W\cap p^\perp)=L$ and $W\cap p^\perp=r^\perp\cap p^\perp\cap H$.
%Hence $W=\langle (W\cap p^\perp),U\rangle$ and so $\dim(W)=\dim(W\cap p^\perp)+\dim(U)-\dim(L)=n-2$.
%Thus $W$ is an object of $\Pi(p,H)$ by Lemma~\ref{lem:hyperplanes of Pi}.
%Moreover, since $p^\perp\cap H$ is non-degenerate, $W\cap p^\perp=r^\perp\cap p^\perp\cap H$ of odd dimension $n-3$ has a radical of dimension $1$, so this must be $r$.

%We now show that we can find such a point $r$ that is different from $q_{12}$ , $q_{13}$, and $q_{23}$. Then, all lines of the triangle $p_1,p_2,p_3$ are incident to $W$ and so 
% the triangle is null-homotopic by Lemma~\ref{lem:common object} .

%There are two cases.
%(1) $n=6$ and $|\FF|\ge 3$.

%Since $|\FF|\ge 3$ we can find a point $r\in L-\{q_{12}.q_{13}.q_{23}\}$.
%In both cases $L$ is totally isotropic and so $L\sbe r^\perp$, as required.

%(2) $n\ge 7$. Note that in this case in fact $n\ge 8$ since we assume that $n$ is even.
%We find a point $r$ in $L^\perp\cap p^\perp\cap H-L$.
%This is possible since $p^\perp\cap H$ is non-degenerate and $L$ is a $2$-space so that $\dim(L^\perp\cap p^\perp\cap H)=(n-2)-2\ge 4>\dim(L)=2$.
%\qed

\pf
If $\dim(V)$ is odd, then we are done by Lemma~\ref{lem:Pi simply connected n odd}.
So now let $\dim(V)=n$ be even.
Let $Q=\Rad(H)$.
Take a triangle on the points $p_1,p_2,p_3$ and let $U=\langle p_1,p_2,p_3\rangle$.
If $U$ is a plane, then we're done by Lemma~\ref{lem:common object}.
From now on assume this is not the case. So $U$ is either totally isotropic, or it has rank $2$ and  $T=\Rad(U)$ is contained in $p^\perp\cap H$.

Let $L=U\cap p^\perp$. For $1\le i<j\le 3$, let $L_{ij}=p_ip_j$ and $q_{ij}=L_{ij}\cap L$.
Our aim is to find an object $W$ of $\Pi(p,H)$ that is incident to all lines $L_{ij}$.
Then, by Lemma~\ref{lem:common object} we're done.

Suppose we can find a point $r$ in $p^\perp\cap H$ such that $r^\perp\spe L$.
For such a point let $W=\langle r^\perp\cap p^\perp\cap H, U\rangle $.
First note that $\dim(r^\perp\cap p^\perp\cap H)=n-3$.
This is because $p^\perp\cap H$ is a non-degenerate symplectic space of even dimension $n-2$. Clearly $r^\perp\cap p^\perp\cap H$ has radical $r$.

Note that $L\sbe  r^\perp\cap p^\perp\cap H$.
Also, since $U\not\sbe p^\perp\cap H$ and $U\cap p^\perp = L$ we have
 $U\cap (W\cap p^\perp)=L$ and $W\cap p^\perp=r^\perp\cap p^\perp\cap H$.
Hence $W=\langle (W\cap p^\perp),U\rangle$ and so $\dim(W)=\dim(W\cap p^\perp)+\dim(U)-\dim(L)=n-2$.
Thus $W$ is an object of $\Pi(p,H)$ by Lemma~\ref{lem:hyperplanes of Pi} as required.

We now show that we can find such a point $r$ that is different from $q_{12}$ , $q_{13}$, and $q_{23}$. Then, since $r=\Rad(W\cap p^\perp)$, all lines $L_{ij}$ are incident to $W$ as required.

There are two cases.
(1) $n=6$ and $|\FF|\ge 3$.
Since $|\FF|\ge 3$ we can find a point $r\in L-\{q_{12},q_{13},q_{23}\}$.
Note that $L$ is totally isotropic and so $L\sbe r^\perp$, as required.

(2) $n\ge 7$. Note that in this case in fact $n\ge 8$ since we assume that $n$ is even.
We find a point $r$ in $L^\perp\cap p^\perp\cap H-L$.
This is possible since $p^\perp\cap H$ is non-degenerate and $L$ is a $2$-space so that $\dim(L^\perp\cap p^\perp\cap H)=(n-2)-2\ge 4>\dim(L)=2$. Again we have found $r$ as required.
\qed


\bpr\label{prop:residues simply connected}
If $|\FF|\ge 3$ then the residues of rank at least $3$ are simply connected. \epr
\pf
Let $J\sbe I$ and let $R_{I-J}$ be the residue of the $J$-flag $\{C_j\}_{j\in J}$.
Set $r=|I-J|$.
Let $\biguplus_{m=1}^M I_m$ be the partition of $I-J$ into contiguous subsets.
If $M\ge 2$, then the residue is a direct product of two geometries, at least one of which is a residue of rank at least $2$. Such rank $2$ residues are connected by Lemma~\ref{lem:geometry residually connected}.
Therefore $R$ is simply connected in this case. 
Otherwise, the residue is isomorphic to $\Gamma(V)$ for some vector space $V$ of dimension at least $3$, or to 
 $\Pi(p,H)$ for some point $p$ inside some $\Gamma(V)$ for some vector space $V$ of dimension at least $4$.
Therefore the simple connectedness follows from ~\ref{lem:>6 or >2 is null}, \ref{lem:Pi simply connected n odd} and ~\ref{prop:gamma simply connected}.
\qed
 
\medskip
\subsection{The exceptional residue}
We are now left with the intriguing case $n=6, q=2$. Let us first describe the geometry $\Pi(p,H)$.
The points are the $16$ points of $H-p^\perp$.
The lines are those lines of $H$ not contained in $p^\perp$.
Thus there are two types of lines, totally isotropic and hyperbolic ones. Each line has exactly two points and any two points are on exactly one line.

The planes of $\Pi(p,H)$ are those non-isotropic planes of $H$ whose radical is not contained in $p^\perp$.
Such a plane has rank $2$ and its radical is a point of $\Pi(p,H)$, i.e.\ it is in $H-p^\perp$.
Each plane has exactly $4$ points. A plane may or may not contain $\Rad(H)$. Any point or line contained in a plane is incident to that plane.

The $4$-spaces of $\Pi(p,H)$ are those $4$-spaces of $H$ that are either non-degenerate or have a radical of dimension $2$ that is not contained in $p^\perp$.
A $4$-space $W$ is incident to all $8$ points it contains and is incident to any line or plane it contains that doesn't pass through the point $\Rad(W\cap p^\perp)$. 
Thus in fact $W$ is incident to all planes $Y$ of $\Pi(p,H)$ contained in $W$. Namely, if $\Rad(W\cap p^\perp)\le Y$, then $p^\perp\cap Y$ is a totally isotropic line, implying that $\Rad(Y)\le p^\perp$, a contradiction.

The geometry $\Pi(p,H)$ is not simply connected. 
To see this, we construct a simply connected rank $4$ geometry $\spi$ which is a degree two cover of $\Pi(p,H)$.
Let us now describe the geometry $\spi$.
A point $q\in \Gamma$ is a point of $\spi$ if the line $pq$ is non-degenerate. In particular the points of $\Pi(p,H)$ are among the points of $\spi$. We construct  a map, which on the point set of $\spi$ is given by
$$\begin{array}{rl}
\psi\colon \spi &\to \Pi(p,H)\\
q &\mapsto\langle p,q\rangle\cap H.
\\
\end{array}$$
This map is two-to-one and for any $q\in \Pi(p,H)$ we set $q^-=q$ and denote by $q^+$ the point of $\spi$ such that  $\psi^{-1}(q)=\{q^-,q^+\}$. We extend this notation to arbitrary point sets $S$ of $\Pi(p,H)$ by setting $S^\epsilon=\{q^\epsilon\mid q\in S\}$ for $\epsilon=\pm$.

We now note that every object of $\Pi(p,H)$ can be identified with its point-shadow. It follows from the above description of $\Pi(p,H)$ however that inclusion of point-shadows does not always imply incidence.

In order to describe $\spi$, we shall identify objects in $\Pi(p,H)$ and $\spi$ with their point-shadow. We denote point-shadows by roman capitals. If we need to make the distinction between objects and their point-shadows explicit, we'll use calligraphic capitals for the objects and the related roman capitals for their point-shadows.
Thus $\cX$ may denote an object of $\Pi(p,H)$ (or $\spi$) whose point-shadow is $X$.

For any object $\cX$ of $\Pi(p,H)$ we will define exactly two objects $\cX_-$ and $\cX_+$ in $\spi$ such that $\psi(X_-)=\psi(X_+)=X$ as point-sets. 
We then define objects $\cX$ and $\cY$ of $\spi$ to be incident whenever 
 $X\sbe Y$ or $Y\sbe X$ and $\psi(\cX)$ and $\psi(\cY)$ are incident in $\Pi(p,H)$.

We shall obtain $\cX_-$ and $\cX_+$ by defining a partition $X_0\uplus X_1$ of the point-set of $\cX$ and setting  
 $X_+=X_0^+\uplus X_1^-$ and $X_-=X_0^-\uplus X_1^+$.

First let $\cX$ be a line. If $\cX$ is non-degenerate, then $X_0=X$ and $X_1=\emptyset$ so that $X_+=X^+$ and $X_-=X^-$. If $X$ is isotropic, then $X=\{x_0\}$ and $X_1=\{x_1\}$ so that $X_-=\{x_0^-,x_1^+\}$ and $X_+=\{x_0^+,x_1^-\}$.

Next, let $\cX$ be a plane. Then $X_0=\{\Rad(X)\}$ and $X_1=X-X_0$.
Note that for every line $\cY$ incident to $\cX$ the partition $Y_0\uplus Y_1$ agrees with the partition $X_0\uplus X_1$.
Hence if $\cY$ is a line and $\cX$ is a plane in $\spi$ such that $\psi(\cY)$ is incident to $\psi(\cX)$, then either $Y\sbe X$ or $Y\cap X=\emptyset$.
%The planes of $\spi$ are defined as follows.Given a plane in $\Pi(p,H)$, it contains four points $r,q_1,q_2,q_3$ %where $r$ is its radical. We consider two planes in $\spi$ consisting of $r,q_1^+,q_2^+,q_3^+$  and respectively %$r^+,q_1,q_2,q_3$.

Finally, let $\cX$ be a $4$-space. Let $r=\Rad(X\cap p^\perp)$. Note that the projective lines $L_i$ ($i=1,2,3,4$) of $\cX$ on $r$ meeting $H-p^\perp$ are not incident to $\cX$ in $\Pi(p,H)$.

First let $\cX$ be non-degenerate. Then $L_i$ is non-degenerate for all $i$.
For $i=1,2,3,4$, let $L_i=\{r,p_i,q_i\}$ such that $q_i=p_1^\perp\cap L_i$ for $i=2,3,4$. Note that this implies the fact that if $i\ne j$, then  $p_i\perp q_j$ but $q_i\not\perp q_j$ and $p_i\not\perp p_j$.
Then $X_0=\{p_1,\ldots,p_4\}$ and $X_1=\{q_1,\ldots,q_4\}$.
We now claim that for every line $\cY$ incident to $\cX$ the partition $Y_0\uplus Y_1$ agrees with $X_0\uplus X_1$.
Thus we must show that if $\cY$ is non-degenerate, then $Y\sbe X_0$ or $Y\sbe X_1$ and if $\cY$ is isotropic, then
 $Y$ intersects $X_0$ and $X_1$ non-trivially.
First, since $\cX$ is non-degenerate, the lines $q_iq_j$ are all non-degenerate for $2\le i<j\le 4$.
Moreover, $q_1q_i$ is non-degenerate as well, for $i=2,3,4$ since otherwise $L_i$ must be totally isotropic, a contradiction.  As a consequence, $q_1p_i$ is totally isotropic. Hence by interchanging the $p_i$'s for $q_i$'s we see that $p_ip_j$ is non-degenerate for all $1\le i<j\le 4$.
Considering the non-degenerate lines $L_i$ and $L_j$ we see that $p_iq_j$ must be isotropic for all $1\le i\ne j\le 4$.
We have exhausted all lines incident to $\cX$ and it is clear that the claim holds.


If $\cX$ is degenerate, then it has a radical $R=\Rad(X)$ of dimension $2$ passing through the radical $r=\Rad(X\cap p^\perp)$. In this case the lines $L_i$ are all totally isotropic. Let $R=L_1$. Then $X_0=\{p_1,q_1\}$ and $X_1=\{p_i,q_i\mid i=2,3,4\}$.
We now claim that for every line $\cY$ incident to $\cX$ the partition $Y_0\uplus Y_1$ agrees with $X_0\uplus X_1$.
Clearly every line meeting $X_0$ and $X_1$ is totally isotropic. Next consider a line $\cY$ meeting $L_i$ and $L_j$
 in $X_1$. If $\cY$ were totally isotropic, then $X=\langle L_1,L_i,L_j\rangle$ is totally isotropic, a contradiction.
We have exhausted all lines incident to $X$ and it is clear that the claim holds.

Next we claim that for every plane $\cY$ incident to $\cX$ the partition $Y_0\uplus Y_1$ agrees with $X_0\uplus X_1$.
First note that $\cY$ doesn't contain $\Rad(X\cap p^\perp)$, because that would make $Y\cap p^\perp$ totally isotropic and $\Rad(Y)\le p^\perp$ contrary to the description of planes of $\Pi(p,H)$.
As a consequence, if $\cL$ is a line incident to $\cY$, then it doesn't meet $\Rad(X\cap p^\perp)$ and so is incident to $\cX$ as well.
Hence since for every line $\cL$ incident to $\cY$, the partition $L_0\uplus L_1$ agrees with the partition $X_0\uplus X_1$ as well as with the partition $Y_0\uplus Y_1$, also the latter two partitions agree.
We can conclude that if $\cY$ is a point, line or plane and $\cX$ is a $4$-space of $\spi$ such that 
 $\psi(\cY)$ and $\psi(\cX)$ are incident, then either $Y\sbe X$ or $Y\cap X=\emptyset$.
 
% Finally given a four space of $\Pi(p,H)$, the set of its points can be separated into two subsets $X_1$ and $X_2$ such that if two points are in different sets then the line generated by them is isotropic and if two pints are in the same set the line is non-degenerate. The sets  $X_1\cup\{q^+\mid q \in X_2\}$ and respectively $\{q^+\mid q \in X_1\}\cup X_2$ are $4$-spaces in $\spi$.

%There is a certain ambiguity regarding the difference between the objects of $\spi$ and the sets of points incident to these objects. To minimise this for the rest of the section we will use the following notation. The set of points incident to the object  $\cal{X}$ is $X$.
Let $\Psi$ be the extension of $\psi$ to the entire collection of objects in $\spi$.
\ble\label{lem:2-cover}
\begin{itemize}
\AI{a} The map $\Psi\colon\spi\to\Pi(p,H)$ is $2$-to-$1$ on the objects.
For any object $\cX$ of $\Pi(p,H)$, if $\Psi^{-1}(\cX)=\{\cX_-,\cX_+\}$, then $\psi^{-1}(X)=X_-\uplus X_+$.
\AI{b} Two objects $\cal{X}$ and $\cal{Y}$ of $\spi$ are incident if and only if $\Psi(\cal{X})$ and $\Psi(\cal{Y})$ are incident and $X\cap Y\ne\emptyset$.
\AI{c} 
For any flag $F_\bullet$ in $\spi$, the map $\Psi\colon\Res(F_\bullet)\to\Res(\Psi(F_\bullet))$ is an isomorphism of geometries.
\AI{d} The map $\Psi\colon\spi\to\Pi(p,H)$ is a $2$-cover.
\AI{e} The pre-geometry $\spi$ is transversal.
\end{itemize}
\ele
\pf
(a) This clear from the construction of $\spi$.

(b) By definition of incidence in $\spi$, $\cX$ and $\cY$ can only be incident if $\Psi(\cX)$ and $\Psi(\cY)$ are incident. The preceding discussion has shown that in this case either $X\cap Y=\emptyset$ or $X\sbe Y$ or $Y\sbe X$. Now $\cX$ and $\cY$ are defined to be incident precisely in the latter case.
 
(c)
Let $\cX_\bullet$ and $\cY_\bullet$ denote objects of $\spi$.
Also, let $\cX=\Psi(\cX_\bullet)$, $\cY=\Psi(\cY_\bullet)$ and $F=\Psi(F_\bullet)$.
By definition of the objects in $\spi$, $\psi\colon X_\bullet\to X$ is a bijection of points.
Therefore if $\cY$ is incident with $\cX$, then $\cX_\bullet$ is incident with exactly one object in $\Psi^{-1}(\cY)$.
Let $\cY$ be incident with $F$. Then by the same token $F_\bullet$ is incident with at most one object in $\Psi^{-1}(\cY)$.
We now show that there is at least one such object.
Suppose that $\cY_\bullet$ is incident to at least one object $\cZ_\bullet$ of $F_\bullet$.
Without loss of generality assume $Y_\bullet\sbe Z_\bullet$.
If $X_\bullet$ is an element of $F_\bullet$ and $Z_\bullet\sbe X_\bullet$ then $\cY_\bullet$ is incident to $\cX_\bullet$ as well.
Now assume $X_\bullet \sbe Z_\bullet$.
Since $\cX$ and $\cY$ are incident, $\cY_\bullet$ must be either incident to $\cX_\bullet$ or to the object in $\Psi^{-1}(\cX)$ different from $\cX_\bullet$.
In the latter case it follows that $\cZ_\bullet$ is incident to both objects in $\Psi^{-1}(\cX)$, a contradiction.
A similar argument holds when $Z_\bullet\sbe Y_\bullet$.
We conclude that $\Psi\colon \Res(F_\bullet)\to \Res(F)$ is a bijection.
Clearly incidence is preserved by $\Psi$, but we must show the same holds for $\Psi^{-1}\colon \Res(F)\to\Res(F_\bullet)$. Let $\cX,\cY\in\Res(F)$ be incident and let $\cX_\bullet, \cY_\bullet\in \Res(F_\bullet)$. 
Then there is a point $q$ incident to $\cX$, $\cY$ and $\cF$.
Suppose $q^\epsilon\in F_\bullet$. Then $q^\epsilon\in X_\bullet\cap Y_\bullet$ and by (b) we find that $\cX_\bullet$ and $\cY_\bullet$ are incident.

(d) This follows from (a) and (c).

(e) This is immediate from (c).  
\qed 
 
%The adjacency is the one induced from $\Pi(p,H)$, it is inclusion for all but the $4$ spaces. Two objects in $\spi$ are adjacent iff their omages in $\Pi(p,H)$ are.
%Note that transversality is rather obvious and that this geometry has a string diagram.
\ble\label{lem:spi connected} 
The pre-geometry $\spi$ is  connected. Any two points are at distance at most $2$, except the points 
 $\Rad(H)^\pm$, which are at distance $3$ from one another.
\ele
\pf
Let $\epsilon\in\{+,-\}$.
Let $Q=\Rad(H)$.
Then for any point $q\ne Q$, since the line $qQ$ is totally isotropic, $Q^\epsilon$ is collinear to $q^{-\epsilon}$ but not to $q^\epsilon$.
In particular any two points with the same sign are at distance at most $2$. It is also clear that $Q^+$ and $Q^-$ have no common neighbors and are at distance at least $3$.

Now consider two points $q_1,q_2\ne Q$. If the line $q_1q_2$ is totally isotropic, then $q_1^\epsilon$ is collinear to $q_2^{-\epsilon}$ in $\spi$.
If the line $q_1q_2$ is non-degenerate, we claim that there exists $q_3\in\Pi(p,H)$ with $q_1\perp q_3\not\perp q_2$.
Namely, we must show that $q_1^\perp-(q_2^\perp\cup p^\perp)\ne\emptyset$. However, this is clear since both 
 $p^\perp$ and $q_2^\perp$ define proper hyperplanes of the $4$-space $q_1^\perp$. Since no linear subspace of $V$ is the union of two of its hyperplanes our claim follows.
In $\spi$ we find both $q_1^\epsilon$ and $q_2^{-\epsilon}$ collinear to $q_3^{-\epsilon}$ so again $q_1$ and $q_2$ are at distance at most $2$.

Finally consider $Q^+$ and $Q^-$. Let $q_1$ be a point of $\Pi(p,H)$ and let $q_2$ be a point of $\Pi(p,H)$
 in $q_1^\perp-\{Q\}$. Then $Q^-,q_1^+,q_2^-,Q^+$ is a path of length $3$.
%
%Note also that if $q_1,q_2 \in \Pi(p,H)$ then either $q_1q_2$ is nondegenerate or there exists a point $q\in %\Pi(p,H)$ such that both $qq_1$ and $qq_2$ are nondegenerate. This means that any two points on the same level of %$\spi$ are at distance at most two in the collinearity graph. Consider now $q_1$ and $q_2^+$. If $q_1q_2$ is %isotropic then the two are colinear in $\spi$. If $q_1q_2$ is nondegenerate pick a point $q$ such that $qq_1$ is %nondegenerate and $qq_2$ is isotropic. Then $q$ and $q_1$  respectively $q_2^+$ and $q$ are collinear in $\spi$. %This means that the collinearity graph of $\spi$ has diameter two and in particular $\spi$ is connected. 
\qed

\ble\label{lem:spi residually connected}
The  pre-geometry $\spi$  is  residually  connected.
\ele
\pf 
By Lemma~\ref{lem:spi connected}, $\spi$ is connected so it suffices to show that every residue of rank at least $2$ is connected. This follows immediately from Lemmas~\ref{lem:2-cover}~and~\ref{lem:geometry residually connected}.
%Firsts  since  the  pre-geometry  has  a string  diagram the  so  do  its residues .  Moreover  the  residues  %whose  strings  are  disconnected  will  be  product  geometries  so  they  will  be  connected.
%Note  that  the  set of  points  of  a plane  or  a $4$-space  form  a  a clique  in  the  collinearyity  graph. %Therefore   any  residue  of  a flag  that  does  not  contain  points  is  connected. We  therefore  need  only  %consider  residues  of  a  that  do  not  contain  points  and  whose  string  diagram  is  connected. There  are  %two  such, the  residue  of  a point  and  the  residue  of  a point-$4.$-space  flag. Consider  $R$  the  residue  %of  a  point.  If  two  lines  in  this  residue   are  such that their respective images in $\Pi(p,H)$ are both  %isotropic  or  nondegenerate  then  they  will  be  in  a common  plane. If one projection is isotropic and one is %nondegenerate then since they contain the common point there is also a common plane. The same argument applies for %the smaller residue.
\qed

\ble\label{lem:spi geometry}
The pre-geometry $\spi$ is a geometry with a string diagram.
\ele
\pf
By Lemma~\ref{lem:2-cover} it is transversal and by Lemma~\ref{lem:spi residually connected} it is residually connected. Therefore it is a geometry. That it has a string diagram is clear since it is a cover of $\Pi(p,H)$, which does have a string diagram.
\qed

\ble\label{lem:spi simply connected}
The geometry $\spi$ is simply connected.
\ele
\pf 
By Lemma~\ref{lem:spi geometry} and ~\ref{lem:point-line cycles only} it suffices to  show that any point-line cycle is null-homotopic.

Let $Q=\Rad(H)$. For any point $q\in \Pi(p,H)$, let $q^*$ denote one of $q^+$, $q^-$.
We claim that any $k$-cycle with $k\ge 5$ can be decomposed into triangles, quadrangles, and pentagons.
Namely, let $\gamma=q_1^*,q_2^*,\ldots,q_k^*,q_1^*$ be a $k$-cycle in $\spi$.
If $q_1$ and $q_4$ are not both $Q$, then they are at distance at most $2$ by Lemma~\ref{lem:spi connected} and so 
 we can decompose $\gamma$ as $(q_1^*,q_2^*,q_3^*,q_4^*)\after \delta \after\delta^{-1}\after
  (q_4^*,\ldots,q_k^*,q_1^*)$, where $\delta$ is a path from $q_1^*$ to $q_4^*$ of length at most $2$.
Thus  we can decompose the $k$-cycle into a $(k-1)$-cycle and a quadrangle or pentagon.
If $q_1$ and $q_2$ are both $Q$, then replacing $q_1$ and $q_4$ by $q_2$ and $q_5$, we can again decompose the $k$-cycle into a $(k-1)$-cycle and a quadrangle or pentagon.

We shall now analyze the triangles, quadrangles, and pentagons case by case.

\paragraph{Triangles}
The points of a triangle $q_1^*,q_2^*,q_3^*$ either all have the same sign or one has a sign different from the others. Note that a point that is collinear to another point with the same sign can not  cover $Q$. In both cases $X=\langle q_1,q_2,q_3\rangle$ has dimension $3$.
Also, $X$ is non-isotropic because at least one of the lines is. 
We'll show that $r=\Rad(X)$ does not lie in $p^\perp$.
In the former case $q_1,q_2,q_3$ form a triangle in $\Pi(p,H)$ whose lines are non-degenerate. In particular $r$ does not lie on any of these lines. The three remaining points on these lines are $X\cap p^\perp$.
In the latter case this is because $r$ is covered by the point of the triangle with the deviating sign.
Thus $X$ is a plane of $\Pi(p,H)$ and $q_1^*,q_2^*,q_3^*$ belong to a plane of $\spi$.
%, 
%There are two types of triangles, those who sit on one level only and those that have one point on one level and two on the other. Both such trinagles are homotopically trivial because they all lie in a plane. 

%Let us denote by $q$  the radical of $H$.

\paragraph{Quadrangles}
Next we consider a quadrangle $q_1^*,q_2^*,q_3^*,q_4^*$. 
There are four cases.
Let $\epsilon\in\{+,-\}$.


%They are of type $q_1^\epsilon q_2q_3q_4$, $q_1q_2^+q_3q_4^+$, $q_1q_2q_3q_4^+$ and $q_1q_2^+q_3^+q_4$. 
(1) All points have the same sign, say $\epsilon$. As we saw with the triangles, none of these is $Q^\epsilon$. Hence $Q^{-\epsilon}$ is connected to all $4$ points and so the quadrangle decomposes into triangles.

(2) All points but one, have the same sign, say $+$. Note that again the points on the same level do not cover $Q$, but are all collinear to $Q^-$. This decomposes the cycle into triangles and a quadrangle with two points of each sign.

(3) There are two points of each sign and these points are consecutive.
Without loss of generality let $q_1^-,q_2^-,q_3^+,q_4^+$ be the quadrangle.
We first note that the points $q_1,q_2,q_3,q_4$ are all distinct.
This is because no point of $\spi$ is collinear to both covers of the same point in 
 $\Pi(p,H)$.
Also note that $q_1q_4$ and $q_2q_3$ are totally isotropic, but $q_1q_2$ and $q_3q_4$ are not. We may also assume that $q_1q_3$ and $q_2q_4$ are non-degenerate, for otherwise we can decompose the quadrangle into triangles.
Consider the space $Y=q_1^\perp\cap q_2^\perp\cap H$. It is a $3$-dimensional space  of rank $2$ whose radical is $Q$.
 Now both $q_3^\perp\cap Y$ and $q_4^\perp\cap Y$ are lines of $Y$ through $Q$.
 Note that $p^\perp\cap Y$ is a line of $Y$ not through $Q$. 
Hence there is a point $q\in Y-q_3^\perp-q_4^\perp-p^\perp$.
We find that $q^+$ is collinear to all points on the quadrangle $q_1^-,q_2^-,q_3^+,q_4^+$, which therefore decomposes into triangles.

(4) There are two points of each sign and these points are not consecutive.
Without loss of generality let $q_1^-,q_2^+,q_3^-,q_4^+$ be the quadrangle.
Let $X=\langle q_1,q_2,q_3,q_4\rangle$.
We first note that we can assume that the points $q_1,q_2,q_3,q_4$ are all distinct. No two consecutive ones can be the same so if for example $q_1=q_3$ then the quadrangle is just a return.
 Hence $\dim(X)=3,4$.
Note that if either $q_1q_3$ or $q_2q_4$ is non-degenerate, then we can decompose the quadrangle into triangles. Therefore all lines $q_iq_j$ are totally isotropic and it follows that $X$ is a totally isotropic $3$-space.
This means that $Q=q_i$ for some $i$, which we may assume to be $4$.
Consider a totally isotropic $3$-space $Y$ on $q_2q_4$ different from $X$.
Then $q_2q_4$ and $p^\perp \cap Y$ are intersecting lines of $Y$ and so there is a point 
 $q\in Y-q_2q_4-p^\perp$.
Note that $q\not \perp q_1,q_3$ for otherwise $\langle q, X\rangle$ is a totally isotropic $4$-space.
We find that $q^-$ is collinear to all points of the quadrangle  $q_1^-,q_2^+,q_3^-,q_4^+$, which therefore decomposes into triangles.

%Finally in the case $q_1q_2^+q_3^+q_4$, we can assume that $q_1q_3$ and $q_2q_4$ are both nondegenerate lines. Consider the space $W=q_2^\perp\cap q_3^\perp\cap H$. It is a 3 dimensional space of rank two. Pick $q' \in W-p^\perp$ and note that if $q'$ is orthogonal to $q_1$ then the space $\langle q,q',q_1,q_2\rangle$ is totally isotropic of dimension $4$ which is a contradiction. Similarly we obtain that $q'\not\perp q_4$ and so $q'$ is collinear to $q_1,q_2^+,q_3^+,q_4$ and the $4$-gon is homotopically trivial.

%
%Then introducing $Q^-$ 
%Without loss of generality assume the quadrangle is $qSuppose $
%In the case $q_1q_2^+q_3q_4^+$, we note that either the lines $q_1q_3$ and  $q_2q_4$ are isotropic or the $4$-gon decomposes into triangles. An imediate consequence is that  $\langle q_1,q_2,q_3,q_4\rangle$ is totally isotropic so it is at most 3 dimensional. If it is 2 dimensional then the $4$-gon is degenerate and contained in a plane. 

%Let $W$  be a 3 dimensional totally isotropic subspace of $H$, $q_1\in W-p^\perp$ other than the radical of $H$ and $q'\in (q_1^\perp\cap H)-W$. Then $q'$ is not orthogonal to any element of $W$ that does not belong to the space generated by $q_1$ and $q$. In particular if $q_1$ and $q_3$ are not the radical of $H$ then there exists a point $q'$ such that it is incident to $q_1,q_3,q_2^+,q_4^+$ and so the $4$-gon is homtopically trivial.

%In the case  $q_1q_2q_3q_4^+$  we note that if $q_1\ne q_3$, then $\langle q_1,q_3,q_4\rangle$ is tottaly isotropic and so it contains $q$ the radical of $H$. Surely $q_1\ne q and q_3\ne q$. If $q=q_4$ then  $q_4^+$ is collinear to $q_1$ so the $4$-gon decomposes into triangles. If $q\ne q_1$ we are back in the case of the previous paragraph.

\paragraph{Pentagons}
We first note that we may assume that a pentagon has no more than $2$ consecutive points of the same sign.
If it contains $4$ or more of sign $\epsilon$, then these points are all collinear to $Q^{-\epsilon}$ which then yields a decomposition of the pentagon into triangles and quadrangles. If it contains  exactly $3$ consecutive points at the same level   the same argument decomposes it into $2$ triangles and a pentagon that contains no more than $2$ consecutive points at the same level.

Therefore we may assume without loss of generality that the pentagon is 
 $q_1^-$, $q_2^-$, $q_3^+$, $q_4^-$, $q_5^+$. If the point $q_4=Q$ then we can pick $q_4'$ to be the fourth point of $\langle q_3, q_4,q_5\rangle - p^\perp$ and decompose the pentagon into the quadrangle $q_4'^-, q_5^+,q_4^-,q_3^+$ and the pentagon $q_1^-$, $q_2^-$, $q_3^+$, $q_4'^-$, $q_5^+$. We therefore can assume that $q_4\ne Q$.
Moreover modifying this pentagon, if necessary, by the quadrangle 
 $q_2^-,q_3^+,q_4^-,Q^+$, we may assume that $q_3=Q$. 
But then $q_3^+$ is collinear to $q_1^-$ as well and we can decompose the pentagon into the triangle $q_1^-,q_2^-,q_3^+$ and the quadrangle
 $q_1^-,q_3^+,q_4^-,q_5^+$.
%
%Note that in any pentagon at least 3 of the vertices are on the same level. there are two cases, depending on whether the vertices of the same level are be consecutive or not. 
%
%Consider the pentagon vertices to be $x_1,x_2,x_3,x_4,x_5$.  Assume that $x_1,x_2,x_3$ are on the same level (say in $\Pi(p,H)$ and $x_4,x_5$ are in the other level. Replacing $x_2$ by $q^+$ the inverse image of the radical of $H$ we can decompose the pentagon into a fourgon and a pentagon in which   $x_2,x_4,x_5$ are on the same level. 
%
%Therefore  we can assume that $x_2,x_4,x_5$ are in the same level and without loss of generality we assume that they are all in $H$. If none of them equal $q$ the radical of $H$ then they are all connected to $q^+$ and so the pentagon decomposes into a triangle and two $4$-gons. If $x_2=q$ then we can pick an element $q_1\in \Pi(p,H)$ such that it is orthogonal to both $x_4$ and $x_5$. It means that $q_1^+$ is collinear to $x_2,x_4,x_5$ and so the pentagon again decomposes into a triangle and two $4$-gons.
\qed

\bco If $q=2$ and $n=6$ then the fundamental group of $\Pi(p,H)$ is $\ZZ/2\ZZ$. 
\eco
\pf
The geometry $\Pi(p,H)$ has a $2$-cover $\spi$ by Lemma~\ref{lem:2-cover}.
This $2$-cover is simply connected by Lemma~\ref{lem:spi simply connected}.
Therefore $\spi$ is the universal cover of $\Pi(p,H)$ and the fundamental group
 is $\ZZ/2\ZZ$.
\qed
% We will consider the case $n =5$ separately. We concentrate to the case $n \ge 6$.
%  Note that a three dimensional space $W\le H$ is in $\Pi(C_0,H)$ if and only if $\Rad(W)\not\subseteq p^\perp$. Also $\dim \Rad(H) =2$ and $\Rad(H) \cap p^\perp = \Rad(V)$.

%
% 
% \begin{prop}
% If $n=6$ then   $\Pi(C_0,H)$ is simply connected.
% \end{prop}
%\pf
% 
%We need to show that any triangle is null-homotopic. Let $(p_1,p_2,p_3)$ be a triangle.

%
% We assume first that $R=\Rad(V) \not\subseteq<p_1,p_2,p_3>$. There are two possible cases.
% 
%  If the space $W=<p_1,p_2,p_3>$ has rank 2 then let $S=\Rad(W)$. If $S \subseteq \Rad(H)$ since $S\ne \Rad(V)$, $S  \not\subseteq p^\perp$ then $W \in \Pi(C_0,H)$ and the triangle is null-homotopic. If $S\not\subseteq \Rad(H)$ then we can pick a point $q\in H - p^\perp$ such that $q\not\perp S$ and the space $<p_1,p_2,p_3,q>$ contains $W$ and it is an object of  $\Pi(C_0,H)$.
%  
%  If the space $W$ is totally isotropic take a point  $S\in W\cap p^\perp$ and pick a point $q\in H \ p^\perp$ such that $q \not\perp S$. Then the space $<p_1,p_2,p_3,q>$ is 4 dimensional and its radical cannot be contained in $W$. Moreover since $q\not\perp S$, $S$ cannot belong to the radical of $<p_1,p_2,p_3,q>$ and so this radical is not contained in $p^\perp$. it follows that $<p_1,p_2,p_3,q>$ is an object of $\Pi(C_0,H)$.
%  
% \qed
% 
%\medskip
% 
%We now deal with the case $n=5$. One immediately  notices that in this case $H$ is of dimension four and rank two. Let us denote by $R=\Rad(V)$ and let $\Rad(H)=<R,T>$. As before $T \not\subseteq p^\perp$.
% 
% \begin{lem} Let $n=5$ and let $ l \subseteq H$ be a non-isotropic line. Then, the only $3$-dimensional subspace $W \le H$ such that $l \le W$ and $W \not\in \Pi(C_0,H)$ is $W=\langle l,R\rangle$.
% \end{lem}
% \pf 
%If  $W$ is such a plane then $S =\Rad(W)\le p^\perp$. If $S \ne R$ then $\langle R,S\rangle = \Rad(H)$ which leads to a contradiction.
%\qed

%\begin{cor}
%Any triangle in $\Pi(C_0,H)$ whose sides are non-degenerate lines is null-homotopic.
%\end{cor}
%\pf Either the triangle generates an element of $\Pi(C_0,H)$ then we are done. Otherwise $R$ is in this plane so if we take a point of $H \setminus p^\perp$ that lies outside of the space generated by the vertices of the triangle then it is collinear to each vertex and using the Lemma each one of the three newly formed triangles will generate planes of $\Pi(C_1,H)$.
%\qed

%\begin{cor} Any triangle of $\Pi(C_0,H)$ whose sides are all totally isotropic is null-homotopic.
%\end{cor}

%\pf 
%Let the triangle be $(p_1,p_2,p_3)$. Since it generates a totally isotropic 3-dimensional space $\Rad(H)\subseteq <p_1,p_2,p_3>$ but $R \not \subseteq <p_i,p_j>$. Therefore on each line $<p_i,p_j>$ there is at most one point in $\Rad(H)$ and so at least two of the $p_i$ are not in the radical. Assume that $p_1, p_2$ are not in $\rad(H)$. Then the two differ by an element in the radical and so  any point $q \in H$ will have the same product with $p_1$ and $p_2$. Let $q$ be a point that is not orthogonal to $p_1$ and $p_2$ ( since $H$ is not isotropic such a point exists). Then $Q \not\in <p_1,p_2,p_3>$. therefore the triangles $(q,p_i,p_j)$ contain a non-isotropic line and do not contain $R$. Therefore they generate a 3-space that belongs to $\Pi(C_0,H)$ and the initial triangle is null-homotopic.
%\qed

%\begin{prop}\label{prop:odd residues simply connected}
%All triangles of $\Pi(C_0,H)$ are null-homotopic.
%\end{prop}

%\pf
%If all the sides are isotropic or all the sides are non-degenerate we are done. So we need to consider the case where one or two of the sides are isotropic.

%Assume that only $<p_1,p_2>$ and $<p_1,p_3>$ are totally isotropic. It then follows that $<p_1>$ is the radical of the 3-dimensional space generated by the $p_i$'s. since $p_1 \not \in p^\perp$ it follows that $<p_1,p_2,p_3>\in \Pi(C_0,H)$.

%Finally assume that only $<p_1,p_2>$ is totaly isotropic. then the radical of the plane lies on $<p_1,p_2>$ and so it cannot be equal to $R$ therefore $<p_1,p_2,p_3>\in \Pi(C_0,H)$.
%\qed



%\newpage
\section{The group action}\label{section:group action}
\subsection{The classical amalgam}\label{sec:classical amalgam}
In this section we shall prove that $G=\Sp(V)$ is the universal completion of subgroups of small rank.
Our first aim is to prove Theorem~\ref{thm:maxparam} using Tits' Lemma~\ref{lem:tits lemma}. To this end, we proved in Section~\ref{section:simple connectedness} that the geometry $\Gamma$ is simply connected. The other result we shall need is the following.
  
\ble\label{lem:flag-transitive}
The symplectic group $\Sp(V)$ acts flag-transitively on $\Gamma$.
\ele
\pf
Let $F_1$ and $F_2$ be two flags of the same type.
We prove the lemma by induction on $|F_1|=|F_2|$.
Let $M_i$ be the object of maximal type in $F_i$ for $i=1,2$.
These objects have the same isometry type since their dimensions are equal.
If $V$ is non-degenerate, then by Witt's theorem there is an isometry $g\in \Sp(V)$ with 
 $gM_1=M_2$.
Now let $V$ be degenerate and let $R=\Rad(V)$.
Choose a complement $V'$ to $R$ containing $M_1$.
If $M_2\le V'$ as well, then again by Witt's theorem there is a $g\in\Sp(V')\le \Sp(V)$ with 
$gM_1=M_2$.
Finally, if $M_2\not\le V'$, then $M_2=M_2'\oplus \langle v+\lambda r\rangle_V$, where 
 $M_2'=M_2\cap V'$ , $v\in V'-M_2'$, $\langle r\rangle=R$, and $\lambda\ne 0$.
Then the transvection $t$ that is the identity on $M_2'\oplus R$ and maps $v\mapsto v+\lambda r$ belongs to $\Sp(V)$ and satisfies $t M_2\le V'$.
Thus, by the preceding case, there is an element $g\in\Sp(V)$ sending $M_1$ to $M_2$.

By induction, there is an element in $\Sp(M_2)$ sending $gF_1$ to $F_2$.
Considering a complement $V''$ of $R$ containing $M_2$ we see that 
 $\Sp(M_2)\le \Sp(V'')$.
Clearly any element in $\Sp(V'')$ can be extended to an element of $\Sp(V)$ by fixing every vector in $R$.
Thus, there is an element of $\Sp(V)$ sending $F_1$ to $F_2$, as desired.
\qed

\medskip
\Definition



Our next aim is to describe the stabilizers of flags of $\Gamma$ in $G=\Sp(V)$.
Let $I=\{1,2,\ldots,n-1\}$.
Let $C=\{C_i\}_{i\in I}$ be a chamber of $\Gamma$.
For any $J\sbe I$, let $R_J$ be the $J$-residue on $C$ and let $F_J$ be the flag of cotype $J$ on $C$.
We set 
$$\begin{array}{ll}
P_J&=\Stab_G(R_J)\\
 B&=\Stab_G(C).\\
 \end{array}$$
Note that $P_J=\Stab_G(F_J)$ and $B=P_\emptyset = \bigcap_{J\sbe I}P_J$.
The group $B$ is called the {\dfn Borel group} for the action of $G$ on $\Gamma$.

\bth\label{thm:maxparam}
The group $\Sp(V)$ is the universal completion of the amalgam of the maximal parabolic subgroups for the action on $\Gamma$.
\eth
\pf
The group $G=\Sp(V)$ acts flag-transitively on $\Gamma$ by Lemma~\ref{lem:flag-transitive}. 
By Proposition~\ref{prop:gamma simply connected}, $\Gamma$ is simply connected and now the result follows from
 Tits' Lemma~\ref{lem:tits lemma}.
\qed

\bco\label{cor:parabolics flag-transitive}
The parabolic subgroup $P_J$ acts flag-transitively on the residue $R_J$.
\eco
\pf
This follows immediately from Lemma~\ref{lem:flag-transitive} and the definition of $P_J$ and $R_J$.
\qed

\bth\label{thm:maxparminparam}
Assume $|\FF|\ge 3$. Let $J\sbe I$ with $|J|\ge 3$.
Then, the parabolic $P_J$ of $\Sp(V)$ is the universal completion of the amalgam  $\{P_{J-\{j\}}\mid j\in J\}$
 of rank $(r-1)$ parabolic subgroups contained in $P_J$. 
\eth
\pf
The group $P_J$ acts flag-transitively on the residue $R_J$ by Corollary~\ref{cor:parabolics flag-transitive}.
By Proposition~\ref{prop:residues simply connected} the residue $R_J$ is simply connected if $|J|\ge 3$.
Again the result follows from Tits' Lemma~\ref{lem:tits lemma}.
\qed

\bco\label{cor:minparam}
Let $|\FF|\ge 3$ and $\dim(V)\ge 4$. Then, $\Sp(V)$ is the universal completion of the amalgam $\cA_{\le 2}=\{P_J\mid |J|\le2\}$ of rank $\le 2$ parabolic subgroups for the action on $\Gamma$. 
\eco
\pf
This follows from Theorems~\ref{thm:maxparam}~and~\ref{thm:maxparminparam} by induction on the rank.
\qed

\medskip

The remainder of this paper is devoted to replacing even the rank $\le 2$ parabolics in the amalgamation result above
 by smaller groups. 

\subsection{Parabolic subgroups}\label{subsection:parabolics}
In this section we analyze the parabolic subgroups of rank $\le 2$ of $\Sp(V)$, where $V$ is non-degenerate over a field $\FF$ with $|\FF|\ge 3$. These are the groups in the amalgam $\cA_{\le 2}$ of Corollary~\ref{cor:minparam}.
In  order to study these groups in some detail we will use the following setup.
Let  $n=2r$.
%and $d=\dim(\Rad(V))\le 1$ so that $n=2$.
and $\cH=\{e_i,f_j\mid 1\le i\le r, 1\le j\le r\}$ be a hyperbolic basis corresponding to $C$ 
 as in Corollary~\ref{cor:hyperbolic bases and chambers}.
That is, if we relabel $\cH$ such that 
$$\begin{array}{lll}
h_{2i-1}&=e_i & \mbox{ for }1\le i\le r,\\
h_{2j}  &=f_j & \mbox{ for }1\le j\le r,\\
\end{array}$$
then, $C=\{C_l\}_{l\in I}$, where $C_l=\langle h_1,h_2,\ldots,h_l\rangle_V$.
For $i=1,2,\ldots,r$, let $H_i=\langle e_i,f_i\rangle_V$
We will use this setup throughout the remainder of the paper.

Let 
$$E=\left(\begin{array}{@{}rr@{}} 0 & 1 \\ -1 & 0 \\\end{array}\right).$$
The matrix defining the symplectic form with respect to the basis $\cH$ is 
$$S=%\left\{\begin{array}{ll} 
                   \left(\begin{array}{@{}cccc@{}} E       & 0      & \cdots & 0      \\
                                                 0       & \ddots & \ddots & \vdots \\ 
                                                 \vdots  & \ddots & \ddots & 0      \\
                                                 0       & \cdots & 0      & E      \\
                                                 \end{array}\right) 
                                                 %& 
%                                                 \mbox{ if }d=0 
%                                                 %\\
%     \mbox{  and }
%                   S=\left(\begin{array}{@{}cccc|c@{}} E    & 0      & \cdots & 0      & 0 \\
%                                                 0       & \ddots & \ddots & \vdots & \vdots \\ 
%                                                 \vdots  & \ddots & \ddots & 0      & \vdots \\
%                                                 0       & \cdots & 0      & E      & 0 \\
%                                               \hline
%                                                 0       & \cdots & \cdots      & 0      & 0 \end{array}\right) 
%                                                 %& 
%                                                 \mbox{ if }d=1
%                                                 %\\
%                   %\end{array}\right.
                   $$

Since the Borel group is contained in all parabolic subgroups, even the ones of rank $\le 2$, we must know exactly what it looks like.

%\ble\label{lem:borel}
%We have 
%$$B\cong \left\{\begin{array}{ll}
%(\FF\rtimes\GL_1(\FF))^{(n/2)} & \mbox{ if  $n$ is even}\\
%(\FF\rtimes \GL_1(\FF))^{\lfloor n/2\rfloor}\times \GL_1(\FF) & \mbox{ if  $n$ is odd},\\
%\end{array}\right.$$
%where, for $j=1,2,\ldots,r$, $\FF\rtimes \GL_1(\FF)$ is realized on $\langle e_j,f_j\rangle$ as  
%$$B_j=\{\left(\begin{array}{@{}ll@{}} a_j & b_j \\ 0 & a_j^{-1}\end{array}\right)\mid a_j\in\FF^*,b_j\in \FF\}.$$
%Also, the kernel of the action of $G$ on $\Gamma$ is 
% $$H=\left\{\begin{array}{ll} \{\pm 1\}&\mbox{ if }d=0\\
%                              \{\pm 1\}\times \GL_1(\FF) & \mbox{ if }d=1.\\
%                              \end{array}\right.$$
%\ele
\ble\label{lem:borel}
We have 
$$B\cong (\FF\rtimes\GL_1(\FF))^{(n/2)}$$
where, for $j=1,2,\ldots,r$, $\FF\rtimes \GL_1(\FF)$ is realized on $\langle e_j,f_j\rangle$ as  
$$B_j=\{\left(\begin{array}{@{}ll@{}} a_j & b_j \\ 0 & a_j^{-1}\end{array}\right)\mid a_j\in\FF^*,b_j\in \FF\}.$$
Also, the kernel of the action of $G$ on $\Gamma$ is 
 $H=\{\pm 1\}$
\ele

\pf
First note that if $n=2$, then $G=\SL_2(\FF)$ and the stabilizer of $C$ is the usual Borel group
$$B_1=\{\left(\begin{array}{@{}ll@{}} a_1 & b_1 \\ 0 & a_1^{-1}\end{array}\right)\mid a_1\in\FF^*,b_1\in \FF\}.$$
Clearly, $B_1\cong \FF\rtimes \GL_1(\FF)$.

Now let $n\ge 3$.
We observe that, if $g\in G$ stabilizes $C$, then it also stabilizes $C_l^\perp\cap C_m$ for any $1\le l\le m\le n-1$. 
Hence, $g$ stabilizes the subspaces spanned by $e_i$ and it stabilizes the subspaces spanned by 
 $\{e_i,f_i\}$ for all $1\le i\le r$.
Thus 

$$g=               \left(\begin{array}{@{}cccc@{}} g_1       & 0      & \cdots & 0      \\
                                                 0       & \ddots & \ddots & \vdots \\ 
                                                 \vdots  & \ddots & \ddots & 0      \\
                                                 0       & \cdots & 0      & g_r      \\
                                                 \end{array}\right) 
%                                                 \mbox{ if }d=0 
%                                                 %\\
%     \mbox{  and }
%                   g=\left(\begin{array}{@{}cccc|c@{}} g_1    & 0      & \cdots & 0      & 0 \\
%                                                 0       & \ddots & \ddots & \vdots & \vdots \\ 
%                                                 \vdots  & \ddots & \ddots & 0      & \vdots \\
%                                                 0       & \cdots & 0      & g_r      & 0 \\
%                                               \hline
%                                                 0       & \cdots & \cdots      & 0      & g_{r+1} \end{array}\right) 
%                                                 %& 
%                                                 \mbox{ if }d=1,
%                                                 %\\
%                   %\end{array}\right.
$$    
where if  $j=1,2,\ldots,r$ we have 
$$g_j=\left(\begin{array}{@{}ll@{}} a_j & b_j \\ 0 & a_j^{-1}\end{array}\right)$$
for some $a_j\in\FF^*$ and $b_j\in \FF$.

The kernel of the action is given by $\bigcap_{x\in G} xBx^{-1}$.
This means that if $g$ is described as above, then, for all $1\le j\le r$ we have 
$b_j=0$  and $a_j=a_j^{-1}=a$ for some fixed $a\in \FF^*$. This means $a=\pm 1$.  The result follows. 
\qed

\medskip

%Since any parabolic subgroup is the intersection of maximal parabolic subgroups we now determine the shape of these.

%\ble\label{lem:maxparabolics}
%Let $U_1$ be the $i$-object of the chamber $C$ in $\Gamma$ and let $U_2=U_1^\perp$.
%For $i=1,2$, let $M_i$ be a complement of the radical $T_i=\Rad(U_i)$.
%Let $R=\Rad(V)$.

%Let $g\in \Stab_G(U_1)$.
%There are four cases
%\begin{itemize}
%\AI{i} $n=2r$ and $i=2s$,
%$$g=\left(\begin{array}{@{}ll@{}} A_1 & 0 \\
%                                                0   & A_2 \\
%                                                \end{array}\right)
%%\mid A_1\in\Sp_{2s}(\FF), A_2\in\Sp_{2(r-s)}(\FF)\}
%$$
%\AI{ii} $n=2r$ and $i=2s+1$, 
%$$g=\left(\begin{array}{@{}llll@{}} A_1 & 0 & w_1    & 0   \\
%                                                v_1^T & a & b      & v_3^T \\
%                                                0   & 0 & a^{-1} & 0   \\  
%                                                0   & 0 & w_3    & A_3 \\ 
%                                                \end{array}\right)
%%\mid A_1\in\Sp_{2s}(\FF), A_2\in\Sp_{2(r-s-1)}(\FF), v_1,w_1\in\FF^{2s}, v_2,w_2\in\FF^{2(r-s-1)}\}
%$$
%\AI{iii} $n=2r+1$ and $i=2s$,
%$$g=\left(\begin{array}{@{}lll@{}} A_1  & 0    & 0 \\
%                                                0   & A_2  & 0 \\
%                                                0   & z_2^T  & A_4\\
%                                                \end{array}\right)
%%\mid A_1\in\Sp_{2s}(\FF), A_2\in\Sp_{2(r-s)}(\FF),v_2,w_2\in \FF^{2(r-s)}\}
%$$
%\AI{iv} $n=2r+1$, $i=2s+1$, 
%$$g=\left(\begin{array}{@{}lllll@{}}A_1 & 0 & w_1    & 0   & 0\\
%                                                v_1^T & a & b      & v_3^T & 0\\
%                                                0   & 0 & a^{-1} & 0   & 0\\  
%                                                0   & 0 & w_3    & A_3 & 0\\ 
%                                                0   & 0 & z_1    & z_2^T & A_4\\
%                                                \end{array}\right)
%%\mid A_1\in\Sp_{2s}(\FF), A_2\in\Sp_{2(r-s-1)}(\FF), v_1,w_1\in\FF^{2s}, v_2,w_2\in\FF^{2(r-s-1)}, x_1\in\FF,x_2\in\FF^{2(r-s-1)},z\in\GL_1(\FF)\}
%$$
%\end{itemize}
%Here $$\begin{array}{ll}
%A_1       & \in\Sp_{2s}(\FF),\\
%v_1,w_1   & \in\FF^{2s},\\ 
%A_2       & \in\Sp_{2(r-s)}(\FF),\\
%v_2,w_2   & \in\FF^{2(r-s)},\\ 
%A_3       &    \in\Sp_{2(r-2-1)}(\FF),\\
%v_3,w_3   & \in\FF^{2(r-s-1)},\\
%z_1       & \in\FF, \\
%z_2       & \in\FF^{2(r-s-1)},\\
%A_4       &\in\GL_1(\FF)\\
%\end{array}$$
%\ele
%\pf
%Clearly any element $g\in P_{I-\{i\}}$ stabilizes both $U_1$, $U_2$, and $T_1=\Rad(U_1)=U_1\cap U_2$.
%It also stabilizes $R=\Rad(V)$.

%First, let $n=2r$. Then $R=\{0\}$.
%If $U_1$ is non-degenerate, then $T_1=\{0\}$ and $U_1\cap U_2=\{0\}$ and $V=U_1\oplus U_2$.
%Hence $g$ is of the form indicated in (i).

%If $U_1$ is degenerate, then $T_1=U_1\cap U_2=\langle e_{s+1}\rangle_V$.
%The fact that $g$ stabilizes $U_1$ explains the first two block-columns of the matrix in (ii).
%The fact that $g$ stabilizes $U_2$ explains the second and last block columns of the matrix in (ii).
%Now let us consider the third $n\times 1$ block column.
%The $a^{-1}$ is to ensure that $\sform(g(e_{s+1}),g(f_{s+1}))=1$.

%The requirement that $g\in\Sp(V)$ is equivalent to requiring that, for $i=1,2$, we have 
%$$aw_i=A_iS_iv_i,$$
%where $S_i$ is the matrix for the symplectic form induced on $M_i$.
%The choice of $b$ is free.

%Next, let $n=2r+1$. Then $R=\langle e_{r+1}\rangle_V$.
%If $U_1$ is non-degenerate, then $T_1=\{0\}$ and $U_1\cap U_2=\{0\}$ and $V=U_1\oplus U_2$.
%Note that $T_2=R$.
%Hence $g$ is of the form indicated in (iii).

%If $U_1$ is degenerate, then $T_1=U_1\cap U_2=\langle e_{s+1}\rangle_V$.
%Note that $T_2=T_1\oplus R$.
%The fact that $g$ stabilizes $U_1$ explains the first two block-columns of the matrix in (iv).
%The fact that $g$ stabilizes $U_2$ explains the second and last block columns of the matrix in (iv).

%Now let us consider the third $n\times 1$ block column.
%The $a^{-1}$ is to ensure that $\sform(g(e_{s+1}),g(f_{s+1}))=1$.
%The choice of $z_1$ is free because it changes the vector $f_{s+1}$ by a vector in the radical.
%The choice of $b$ is free because it changes the vector $f_{s+1}$ by a vector in $T_1$, which is fixed by $g$.
%The requirement that $g\in\Sp(V)$ is equivalent to requiring that, for $i=1,3$, we have 
%$$aw_i=A_iS_iv_i,$$
%where $S_i$ is the matrix for the symplectic form induced on $M_i$.
%\qed

\medskip

We now focus on the parabolic subgroups of rank $\le 2$.
%Recall that $P_{\{i\}}$ (also written $P_i$) is the parabolic subgroup fixing the residue of type $\{i\}$ and hence the flag of type $I-\{i\}$ on $C=\{C_i\}_{i\in I}$.
Since we want to make a distinction between the various rank $\le 2$ parabolic subgroups, we shall give them an individual name.

%We first focus on the rank $1$ parabolics.
%Let $\cH=\{e_i,f_j\mid 1\le i\le r+d, 1\le j\le r\}$ be a hyperbolic basis corresponding to $C$ 
% as in Corollary~\ref{cor:hyperbolic bases and chambers}.

\Definition\label{dfn:sPij sSij sQij}
We assign the following names to the various rank $\le 2$ parabolic subgroups:
$$\begin{array}{lll}
S_j &= P_{2j-1} & \mbox { for } 1\le j\le r\\
M_i &=P_{2i}    & \mbox{ for } 1\le i \le r-1 \\
S_{ij}&=P_{2i-1,2j-1} & \mbox{ for } 1\le i<j\le r\\
M_{ij}&=P_{2i,2j}       & \mbox{ for }1\le i<j\le r-1\\
Q_{ij}&= P_{2i,2j-1}   & \mbox{ for }1\le i\le r-1\mbox{ and } 1\le j\le r\\
\end{array}$$
Thus the collection of groups in the amalgam $\cA_{\le 2}$ is 
 $\{M_i,S_j, S_{jl},M_{ik},Q_{ij}\mid 1\le i,k\le r-1\mbox{ and } 1\le j,l\le r\}$.
 
\Definition\label{dfn: abstract S&M}
In order to describe the groups in $\cA_{\le 2}$ abstractly and as matrix groups we 
 define the following matrix groups:

$$M=\{\left(\begin{array}{@{}llll@{}} a_1 & b_1         & 0 & w \\
                                                               0      & a_1^{-1} & 0 &  0      \\
                                                               0      & a_1^{-1} w a_2 & a_2 & b_2\\
                                                               0      & 0                               & 0     & a_2^{-1}\\
                                                               \end{array}\right)\mid a_1,a_2\in\FF^*,b_1,b_2,w\in \FF\},
$$
 $$S=\SL_2(\FF)\cong \Sp_2(\FF)=\{\left(\begin{array}{@{}ll@{}} a & b\\
                                                               c      & d\\
                                                               \end{array}\right)\mid a,b,c,d\in\FF,\mbox{ where }ad-bc=1\},
$$
 $$
M_*=
\left\{
\left(\begin{array}{@{}ll|ll|ll@{}} a_1 & b_1 & 0 & w_1 & 0 & 0 \\
                                  0 & a_1^{-1} & 0 & 0 & 0 & 0 \\
                                  \hline
                                  0 & a_1^{-1}w_1a_2 & a_2 & b_2 & 0 & w_3 \\
                                   0  & 0   & 0 & a_2^{-1} & 0 & 0  \\
                                  \hline
                                  0 & 0 & 0  & a_2^{-1}w_3a_3    & a_3       & b_3  \\
                                   0 & 0 & 0   & 0 & 0           &  a_3^{-1} \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l}
   a_1,a_2,a_3\in \FF^* \\
   b_1,b_2, b_3,w_1,w_3\in\FF\\
\end{array}
\right\},
$$
 $$
Q_-=
\left\{
\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & w_1 \\
                                  c_1 & d_1 & 0 & w_2\\
                                  v_2 & v_1 & a_2 & b_2\\
                                   0  & 0   & 0 & a_2^{-1}  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1,w_1,w_2,v_1,& v_2, a_2,b_2\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
         a_1w_2-c_1w_1+v_2a_2^{-1}&=0\\
         b_1w_2-d_1w_1+v_1a_2^{-1}&=0\\
\end{array}
\right\},
$$
and
$$
Q_+=
\left\{
\left(\begin{array}{@{}llll@{}} 
                                  a_2     & b_2 & v_2    & v_1 \\
                                  0     & a_2^{-1}     & 0        & 0      \\
                                  0     & w_1 & a_1     & b_1\\
                                   0    & w_2 & c_1     & d_1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1,w_1,w_2,v_1,& v_2, a_2, b_2\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
         a_1w_2-c_1w_1+v_2a_2^{-1}&=0\\
         b_1w_2-d_1w_1+v_1a_2^{-1}&=0\\
\end{array}
\right\}.
$$

%\{\left(\begin{array}{@{}ll@{}} a & b\\
%                                                c & d  \\
%                                            \end{array}\right)\mid a,b,c,d\in\FF, ad-bc=1\}.
%$$

\ble\label{lem:structure parabolics}
\begin{itemize}
\AI{a}
For all indices that apply, we have the following isomorphisms:
$$\begin{array}{ll}
S_j &\cong S\times \Pi_{i\ne j} B_i, \\
M_i &\cong M\times\Pi_{j\ne i,i+1} B_j, \\
S_{ij}&=\langle S_i, S_j\rangle_G, \\
M_{ij}&=\langle M_i, M_j\rangle_G, \\
Q_{ij}&=\langle M_i, S_j\rangle_G.\\
\end{array}$$
Moreover, we have the following isomorphisms,
\AI{b}
$$M_{ij}\cong\left\{
\begin{array}{ll}
M_*\times \Pi_{k\ne i,i+1,j,j+1} B_k&\mbox{ if }|i-j|=1\\
M\times M \times\Pi_{k\ne i,i+1,j,j+1} B_k &\mbox{ if } |i-j|\ge 2\\
\end{array}\right.,$$
\AI{c}
$$S_{ij}\cong (S\times S)\times  \Pi_{k\ne i,j} B_k,$$
\AI{d}
If $j\not\in\{i,i+1\}$, then 
$$Q_{ij}\cong (M\times S)\times \Pi_{k\ne i,i+1,j} B_k.$$
Furthermore,
$$
Q_{ii}\cong Q_-\times\Pi_{k\ne i,i+1} B_k,
$$
and
$$
Q_{i i+1}\cong
Q_+\times\Pi_{k\ne i,i+1} B_k.
$$
\end{itemize}
\ele 
\pf
(a) By definition $S_i$ is the stabilizer of the flag of type $I-\{2i-1\}$ on the standard chamber $C$.
Therefore it acts on $H_j$ as $B_j$ for all $j\ne i$ and it stabilizes and acts as $S\cong\Sp_2(\FF)$ on $H_i$.
Similarly, by definition, $M_i$ acts on $H_j$ as $B_j$ for all $j\ne i,i+1$ and it stabilizes $\langle e_i\rangle$ and 
 $\langle e_i,f_i,e_{i+1}\rangle$ and $H_i\oplus H_{i+1}$. Therefore it acts as $M$ on $H_i\oplus H_{i+1}$.
The last three equalities follow from the fact that rank $2$ residues are connected.

The remaining isomorphisms follow from similar considerations.
The parabolic subgroup under consideration acts as $B_j$ on $H_j$ for all but $2$,$3$, or $4$ values of $j$.
Then the action on the subspace generated by the remaining $H_j$ determines the non-Borel part ($S$, $M$, $M_*$, $Q_-$, $Q_+$) of the group.
\qed
%
%Part (a) and (c) are straightforward.
%As for part (d), if $j\not\in\{i,i=1\}$, then clearly $M_i$ and $S_j$ commute and intersect trivially.

%We now turn to the cases $Q_{ii}$ and $Q_{i i+1}$.
%First note that conjugation by the permutation matrix that switches $(e_i,f_i)$ and $(e_{i+1},f_{i+1})$, interchanges
% $S_i$ and $S_{i+1}$ and stabilizes $M_i$.
%Thus it suffices to prove the claim for $Q_{ii}$. 

%
%We claim that $Q_-$ is the stabilizer of the $1$-space $E$ spanned by the vector $e_{i+1}$ in 
%$\Sp(H_i\oplus H_{i+1})$.
%It is clear from the shape of the third column that $Q_-$ stabilizes $E$.
%On the other hand, any matrix $A$ in $\Sp(H_i\oplus H_{i+1})$ stabilizing $E$ must have such a
% third column.
%It must also have zeroes in the last row as in $Q$ since in fixing $E$ it must also stabilize $E^\perp$.
%Any such matrix $A$ must satisfy the conditions on the entries as indicated in the description of $Q_+$ since 
%$\{Ae_i,Af_i, e_{i+1},Af_{i+1}\}$ must be isometric to $\{e_i,f_i,e_{i+1},f_{i+1}\}$ in order for $A$ to be symplectic.
%Therefore $Q_-$ equals this stabilizer and hence is a group.

%Clearly $Q_{ii}$ contains $S_i$ and $M_i$. We now show that $\langle S_i,M_i\rangle=Q_{ii}$.
%First we note that by taking $s'sms^{-1}$ with $s,s'\in S_i$ and $m\in M_i$ we obtain all matrices of 
% the form

%$$\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & w_1 \\
%                                  c_1 & d_1 & 0 & w_2 \\
%                                  v_2 & v_1 & 1 & 0   \\
%                                   0  & 0   & 0 & 1   \\
%      \end{array}
%\right),$$
%where the entries are as in the lemma.
%Namely, taking $s'=1$, we obtain all such matrices with $a_1=d_1=1$ and $b_1=c_1=0$.
%This is because $S_i\cong\Sp_2(\FF)\cong\SL_2(\FF)$ is transitive on vectors of $\FF^2$.
%Then allowing $s'$ to be any other element in $S_i$ produces the elements as described.

%It now suffices to show that the matrices in $Q_{ii}$ such that all entries are zero, except $a_1=d_1=1$ and $b_2\in\FF$ is arbitrary also belong to $Q_{ii}$.
%As we saw above, for any vector $\vec{w}=(w_1,w_2)^T\in\FF^2$ there is an element 
% $$A(\vec{w})=\left(\begin{array}{@{}llll@{}} 
%                                   1   & 0   & 0  & w_1 \\
%                                   0   & 1   & 0  & w_2 \\
%                                  -w_2 & w_1 & 1  & 0   \\
%                                   0   & 0   & 0  & 1   \\
%      \end{array}
%\right) \in Q_{ii}$$

%One easily calculates that, for $\vec{v},\vec{w}\in\FF^2$ the matrix 
% $A(\vec{w})A(\vec{v})A(-\vec{w}-\vec{v})$ is of the form
% $$K(b_2)=\left(\begin{array}{@{}llll@{}} 
%                                   1   & 0   & 0  & 0 \\
%                                   0   & 1   & 0  & 0 \\
%                                   0   & 0   & 1  & b_2   \\
%                                   0   & 0   & 0  & 1   \\
%      \end{array}
%\right)
%\mbox{, where }b_2=\sform(\vec{w},\vec{v}).
%$$
%Since we can choose $\vec{w}$ and $\vec{v}$ arbitrarily, $b_2$ can take any value in $\FF$.
%Since $K=\{K(b_2)\mid b_2\in \FF\}\le Z(Q_{ii})$ it follows that $Q_{ii}$ is as described in the lemma.
%\qed



%\Definition\label{dfn:sPi sSj}
%For $1\le i\le r-1$, the group $\sM_i$ fixes every vector in $H_k$, for every $k\ne i,i+1$ and 
%a generic element acts on $H_i\oplus H_{i+1}$ as
%$$P(b_1,w,b_2)=\left(\begin{array}{@{}llll@{}} 1 & b_1 & 0 & w \\
%                                      0 & 1 & 0 & 0 \\
%                                      0 & w & 1 & b_2 \\
%                                      0 & 0 & 0 & 1 \\
%            \end{array}\right), \mbox{ where } w,b_1,b_2\in \FF.
%$$

%

%\Note We note that $\sS_j$ is generated by two opposite long-root groups.

%\ble\label{lem:sPi and sS}
%For $i=1,2,\ldots,r-1$ and $j=1,2,\ldots,r$, the sets $\sM_i$ and $\sS_j$ are subgroups of $\Sp(V)$ and  
%\begin{itemize}
%\AI{a} $\sM_i\cong \FF^3$,
%\AI{b} $\sS_j\cong\Sp_2(\FF)$.
%\end{itemize}
%\ele
%\pf
%Setting $P=P(b_1,w,b_2)$ one verifies easily that $(P e_i, P f_i)$ and $(P e_{i+1}, P f_{i+1})$
% are two orthogonal hyperbolic pairs. Hence $P$ is a symplectic matrix.
%Clearly $\sS_j=\Stab_{\Sp(V)}(H_j)$ is a subgroup of $\Sp(V)$.

%(a) It is straightforward to check that $\sM_i$ is an abelian group isomorphic to $\FF^3$.

%(b) This is true by definition.
%\qed

%

%We now determine these minimal parabolic subgroups.

%\ble\label{lem:minparabolics}
%If $n$ is even, then
%$$P_i\cong \left\{\begin{array}{ll}
%\Sp_2(\FF)\times (\FF\rtimes\GL_1(\FF))^{(n/2)-1} & \mbox{ for  $i$ odd}\\
%P\times (\FF\rtimes\GL_1(\FF))^{(n/2)-2} & \mbox{ for  $i$ even}\\
%\end{array}\right.$$
%where 

%$$P=\{\left(\begin{array}{@{}llll@{}} a_1 & b_1         & 0 & w \\
%                                                               0      & a_1^{-1} & 0 &  0      \\
%                                                               0      & a_1^{-1} w a_2 & a_2 & b_2\\
%                                                               0      & 0                               & 0     & a_2^{-1}\\
%                                                               \end{array}\right)\mid a_1,a_2\in\FF^*,b_1,b_2,w\in \FF\}.
%$$
%If $n$ is odd, then

%$$P_i\cong \left\{\begin{array}{ll}
%\Sp_2(\FF)\times (\FF\rtimes \GL_1(\FF))^{\lfloor n/2\rfloor-1}\times \GL_1(\FF) & \mbox{ if  $i$ is odd},\\
%P\times (\FF\rtimes\GL_1(\FF))^{(n/2)-2} \times \GL_1(\FF) & \mbox{ for  $i<n-1$ even}\\
%Q\times (\FF\rtimes\GL_1(\FF))^{(n/2)-1} & \mbox{ for  $i=n-1$}\\
%\end{array}\right.$$
%where 
%$$\FF\rtimes \GL_1(\FF)=\{\left(\begin{array}{@{}ll@{}} a & b \\ 0 & a^{-1}\end{array}\right)\mid a\in\FF^*,b\in \FF\}.$$
%$$Q=\{\left(\begin{array}{@{}lll@{}} a_1 & b_1         & 0 \\
%                                                               0      & a_1^{-1} & 0 \\
%                                                               0      &  v       & r \\
%                                                             \end{array}\right)\mid a_1,a_2,r\in\FF^*,v,b_1\in \FF\}.
%$$

%\ele
%\pf
%First let $i$ be odd.
%Then $P_i$ fixes all objects $C_j$ with $j\le i-1$ and $j\ge i+1$.
%Thus $P_i$ acts as the Borel group on $C_{i-1}$ and on $C_{i+1}^\perp$, whereas it acts as 
% $\Sp_2(\FF)$ on $C_{i-1}^\perp\cap C_{i+1}$.
%The result follows.
% 
%Next, let  $i$ be even.
%Then   $P_i$ fixes all objects $C_j$ with $j\le i-1$ and $j\ge i+1$.
%In particular $P_i$ acts as the Borel group on $C_{i-2}$ and on $C_{i+2}^\perp$, whereas it acts as 
% a minimal parabolic $P$ fixing the flag $(C_{i-1}^\perp\cap C_i, C_{i-1}^\perp\cap C_{i+1})$ of type $\{1,3\}$ inside $\Sp(C_{i-2}^\perp\cap C_{i+2})\cong\Sp_4(\FF)$.
%That $P$ is as indicated, follows from Lemma~\ref{lem:maxparabolics}.
%We can also see this directly. Without loss of generality assume $n=4$.
%Then $C_1=\langle e_1\rangle$ and $\Rad(C_3)=\langle e_2\rangle$ are stabilized. 
%This accounts for the first and third column.
%Also, $C_3=\langle e_1,f_1,e_2\rangle$ and $C_1^\perp=\langle e_1,e_2,f_2\rangle$ are stabilized.
%This accounts for the zeroes in rows $2$ and $4$.
%The fact that $P=(p_{ij})_{i,j=1}^4$ preserves the symplectic form induces the following conditions on the remaining entries:
%$$\begin{array}{rl}
%p_{11}p_{22}&=1\\
%p_{22}p_{14}&=p_{32}p_{44}\\
%p_{33}p_{44}&=1\\
%\end{array}$$
%The result follows.
%\qed


%\newpage
\section{The slim amalgam}\label{section:slim}
%Let $\cH=\{e_i,f_i\mid 1\le i\le r\}$ be a hyperbolic basis corresponding to $C$ 
% as in Corollary~\ref{cor:hyperbolic bases and chambers}.
%For $i=1,2,\ldots,r$, let $H_i=\langle e_i,f_i\rangle_V$
In this section we define a slim version $\pcA$ of the amalgam $\cA_{\le 2}$ by eliminating 
 a large part of the Borel group from each of its groups.
More precisely, the collection of groups in $\pcA$ is $\{\pX\mid X\in \cA_{\le 2}\}$, where 
 $\pX$ is given in Definitions~\ref{dfn:sPi sSj} and~\ref{dfn:pA}.
Note that the groups $\pX$ are subgroups of $G$ and the inclusion maps for the amalgam $\pcA$ are the inclusion maps induced by $G$.
In Section~\ref{section:concrete amalgam} we shall construct an abstract version of $\pcA$, whose groups are not considered to be subgroups of $G$.

\Definition\label{dfn:sPi sSj}
For $1\le i\le r-1$, the group $\pM_i$ fixes every vector in $H_k$, for every $k\ne i,i+1$ and 
a generic element acts on $H_i\oplus H_{i+1}$ as
$$m(b_1,w,b_2)=\left(\begin{array}{@{}llll@{}} 1 & b_1 & 0 & w \\
                                      0 & 1 & 0 & 0 \\
                                      0 & w & 1 & b_2 \\
                                      0 & 0 & 0 & 1 \\
            \end{array}\right), \mbox{ where } w,b_1,b_2\in \FF.
$$
For $1\le j\le r$, the group $\pS_j$ fixes every vector in $H_k$, for every $k\ne j$ and acts on $H_j$ as $\Sp(H_j)\cong\Sp_2(\FF)$. 
A generic element of $\pS_j$ is denoted  
$$s(a,b,c,d)= \left(\begin{array}{@{}ll@{}} a & b \\
                                      c & d\\
            \end{array}\right),$$ 
where the matrix defines the action on $H_j$ with respect to the basis $\{e_j,f_j\}$.

\medskip

\Note We note that $\pS_j$ is generated by two opposite long-root groups.

\Definition\label{dfn:pA}
For $1\le i<j\le r$, let $$\pS_{ij}=\langle\pS_i,\pS_j\rangle_G.$$
For $1\le i<j\le r-1$, let $$\pM_{ij}=\langle\pM_i,\pM_j\rangle_G.$$
For $1\le i\le r-1$ and $1\le j\le r$, let $$\pQ_{ij}=\langle\pM_i,\pS_j\rangle_G.$$

\Definition
For any $i=1,2,\ldots,r$, we set
$$\pU_i=\left\{
\begin{array}{ll}
\pM_i\cap \pS_i &\mbox{ if }1\le i<r\\
\pM_{i-1}\cap \pS_i & \mbox{ if }i=r\\
\end{array}\right.,$$
and 
$$\pB_i=N_{\pS_i}(\pU_i).$$
Moreover, 
$$\pB=\langle \pB_1,\ldots,\pB_r\rangle\cong\pB_1\times\cdots\times\pB_r.$$


\Definition\label{dfn: abstract slim S&M}
In order to describe the groups in $\pcA$ abstractly and as matrix groups we 
 define the following matrix groups:
$$\pM=\{\left(\begin{array}{@{}llll@{}} 1 & b_1         & 0 & w \\
                                                               0      & 1 & 0 &  0      \\
                                                               0      & w & 1 & b_2\\
                                                               0      & 0         & 0     & 1\\
                                                               \end{array}\right)\mid b_1,b_2,w\in \FF\},
$$
 
 $$\pS=S,$$



  $$
\pM_*=
\left\{
\left(\begin{array}{@{}ll|ll|ll@{}} 1 & b_1 & 0 & w_1 & 0 & 0 \\
                                  0 & 1 & 0 & 0 & 0 & 0 \\
                                  \hline
                                  0 & w_1 & 1 & b_2 & 0 & w_3 \\
                                   0  & 0   & 0 & 1 & 0 & 0  \\
                                  \hline
                                  0 & 0 & 0  & w_3   & 1       & b_3  \\
                                   0 & 0 & 0   & 0 & 0           &  1 \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l}
      b_1,b_2, b_3,w_1,w_3\in\FF\\
\end{array}
\right\},
$$

$$
\pQ_-=
\left\{
\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & w_1 \\
                                  c_1 & d_1 & 0 & w_2\\
                                  v_2 & v_1 & 1 & b_2\\
                                   0  & 0   & 0 & 1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1,w_1,w_2,v_1,& v_2, b_2\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
    v_2&=    -a_1w_2+c_1w_1\\
      v_1&=   - b_1w_2+d_1w_1\\
\end{array}
\right\},
$$
and
$$
\pQ_+=
\left\{
\left(\begin{array}{@{}llll@{}} 
                                  1     & b_2 & v_2    & v_1 \\
                                  0     & 1     & 0        & 0      \\
                                  0     & w_1 & a_1     & b_1\\
                                   0    & w_2 & c_1     & d_1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1,w_1,w_2,v_1,& v_2, b_2\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
        v_2&=    -a_1w_2+c_1w_1\\
      v_1&=   - b_1w_2+d_1w_1\\
\end{array}
\right\}.
$$



\ble\label{lem:sPi and sS}
For $i=1,2,\ldots,r-1$ and $j=1,2,\ldots,r$, the sets $\pM_i$ and $\pS_j$ are subgroups of $\Sp(V)$ and  
\begin{itemize}
\AI{a} $\pM_i\cong \pM\cong \FF^3$,
\AI{b} $\pS_j\cong \pS\cong \Sp_2(\FF)$.
\end{itemize}
\ele
\pf
Setting $m=m(b_1,w,b_2)$ one verifies easily that $(m e_i, m f_i)$ and $(m e_{i+1}, m f_{i+1})$
 are two orthogonal hyperbolic pairs. Hence $m$ is a symplectic matrix.
Clearly $\pS_j=\Stab_{\Sp(V)}(H_j)$ is a subgroup of $\Sp(V)$.

(a) It is straightforward to check that $\pM_i$ is an abelian group isomorphic to $\FF^3$.

(b) This is true by definition.
\qed

\medskip



\ble\label{lem:sPij sQij sSij}
For all indices $i\ne j$ that apply, we have 
\begin{itemize}
\AI{a}
$$\pM_{ij}\cong\left\{
\begin{array}{ll}
\pM_* &\mbox{ if }|i-j|=1\\
\pM_i\times\pM_j &\mbox{ if } |i-j|\ge 2\\
\end{array}\right.,$$
\AI{b}
$$\pS_{ij}\cong\pS_i\times \pS_j.$$
\AI{c}
If $j\not\in\{i,i+1\}$, then 
$$\pQ_{ij}\cong\pM_i\times \pS_j.$$
Furthermore,
$$\begin{array}{ll}
\pQ_{ii}&\cong\pQ_-,\\
\pQ_{i i+1}&\cong\pQ_+.\\
\end{array}
$$
\end{itemize}
\ele 
\pf
Part (a) and (b) are straightforward.
As for part (c), if $j\not\in\{i,i=1\}$, then clearly $\pM_i$ and $\pS_j$ commute and intersect trivially.

We now turn to the cases $\pQ_{ii}$ and $\pQ_{i i+1}$.
First note that conjugation by the permutation matrix that switches $(e_i,f_i)$ and $(e_{i+1},f_{i+1})$, interchanges
 $\pS_i$ and $\pS_{i+1}$, but fixes every element in $\pM_i$.
Thus it suffices to prove the claim for $\pQ_{ii}$. 

We consider $\pQ_-$ to be represented as a matrix group with respect to the basis $\{e_i,f_i,e_{i+1},f_{i+1}\}$.
We shall now prove that with this identification $\pQ_-=\pQ_{ii}$.
To this end we show that $\pQ_-$ is the stabilizer of the vector $e_{i+1}$ in $\Sp(H_i\oplus H_{i+1})$.
It is clear from the shape of the third column that $\pQ_-$ stabilizes $e_{i+1}$.
On the other hand, any matrix $A$ in $\Sp(H_i\oplus H_{i+1})$ stabilizing $e_{i+1}$ must have such a
 third column.
It must also have zeroes in the last row as in $\pQ_-$ since in fixing $e_{i+1}$ it must also stabilize $e_{i+1}^\perp$.
Any such matrix $A$ must satisfy the conditions on the entries as indicated in the description of $\pQ_-$ since 
$\{Ae_i,Af_i, e_{i+1},Af_{i+1}\}$ must be isometric to $\{e_i,f_i,e_{i+1},f_{i+1}\}$ in order for $A$ to be symplectic.
Therefore $\pQ_-$ equals this stabilizer and hence is a group.

Clearly $\pQ_-$ contains $\pS_i$ and $\pM_i$. We now show that $\langle \pS_i,\pM_i\rangle=\pQ_-$.
We note that if $m=m(0,w,b_2)$ and $s=s(a,b,c,d), s'=s(a_1,b_1,c_1,d_1)$ then $sms^{-1}s'$ is the following matrix:
$$\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & aw \\
                                  c_1 & d_1 & 0 & cw \\
                                  v_2 & v_1 & 1 & b_2  \\
                                   0  & 0   & 0 & 1   \\
      \end{array}
\right),$$
where $v_1,v_2$ are as in the definition of $\pQ_-$. Therefore all the elements of $\pQ_-$ can be obtained that way.

\qed

\ble\label{lem:structure sPij sQij sSij}
We have 
\begin{itemize}
\AI{a}
$$\pM_{ij}\cong\left\{
\begin{array}{ll}
\FF^5 &\mbox{ if }|i-j|=1\\
\FF^6 &\mbox{ if } |i-j|\ge 2\\
\end{array}\right.,$$
\AI{b}
$$\pS_{ij}\cong\Sp_2(\FF)\times \Sp_2(\FF),$$
\AI{c}
If $j\not\in\{i,i+1\}$, then 
$$\pQ_{ij}\cong\FF^3\times \Sp_2(\FF).$$
Furthermore,
$$\pQ_{ii}\cong (U\times V)\rtimes \pS_i\cong \FF^3\rtimes \Sp_2(\FF)
.$$
Here, taking the labelling as in Lemma~\ref{lem:sPij sQij sSij} we have
$$U=
\left\{
\left(\begin{array}{@{}llll@{}} 1 & 0 & 0 & 0 \\
                                0 & 1 & 0 & 0\\
                                0 & 0 & 1 & b_2\\
                                0 & 0 & 0 & 1  \\
      \end{array}
\right)
\Big|%\mid 
   b_2\in\FF
\right\}
\cong \FF,$$

$$
V=
\left\{
\left(\begin{array}{@{}llll@{}}    1 & 0 & 0 & w_1\\
                                   0 & 1 & 0 & w_2\\
                                  v_2 & v_1& 1 & 0 \\
                                   0  & 0  & 0 & 1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   w_1,w_2,v_1,& v_2\in\FF\mbox{ such that}\\
         v_2&=-w_2\\
         v_1&=w_1\\
\end{array}
\right\}
\cong \FF^2,$$

$$
\pS_i=
\left\{
\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & 0 \\
                                  c_1 & d_1 & 0 & 0\\
                                  0 & 0 & 1 & 0\\
                                   0  & 0   & 0 & 1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
\end{array}
\right\}
\cong\Sp_2(\FF).$$
Moreover, the action of $\pS_i$ on $V$ by conjugation is the natural action from the left of $\Sp_2(\FF)$ on $V$.
Also $U=Z(\pQ_{ii})$.

Finally, 
$$\pQ_{i i+1}\cong\pQ_{ii}$$
where the isomorphism is given by the labelling of the entries.
\end{itemize}
\ele
\pf
(a) and (b): This is straightforward.
As for (c), if $j\not\in\{i,i+1\}$, then the result follows from the corresponding part in Lemma~\ref{lem:sPij sQij sSij}.

Since $\pM_i\le V\le \pQ_{ii}$ and $U\le \pQ_{ii}$, we have 
$\pQ_{ii}=\langle \pM_i,\pS_i\rangle=\langle V,\pS_i\rangle=\langle U, V,\pS_i\rangle$.
Also, $U\cap V=1$ and $\langle U,V\rangle$ is abelian so that $\langle U,V\rangle\cong U\times V$.
Furthermore, $\langle U,V\rangle\cap \pS_i=1$ and $\pS_i\le N_{\pQ_{ii}}(\langle U,V\rangle)$.
In fact $\pS_i\le C_{\pQ_{ii}}(U)$ and one calculates that 
 conjugation of $V$ by $\pS_i$ is the natural action from the left of $\Sp_2(\FF)$ on $\FF^2=V$.

We now show that $U=Z(\pQ_{ii})$. This follows froma a general argument about semidirect products using the fact that $\pS_i$ acts faithfully and fixed point free on $V$.
%We clearly have $U\le Z(\pQ_{ii})$.
%Let $z\in Z(\pQ_{ii})$. Then $z=uvs$, where $u\in U$, $v\in V$, and $s\in \pS_i$.
%Clearly, we may assume $u=1$.
%Taking some $v'\in V-\{0\}$ we must have $v'vs=vsv'=vs(v')s= s(v')vs$, that is, $v'$ must be a fixed vector
% for the symplectic transformation $s\in\pS_i$ in its natural action on $V$.
%Clearly, this forces $s=1$ and $z=v$.
%Since $sv=s(v)s\ne vs$, by the argument that forced $s=1$, we now find $v=1$.
%Thus, $Z(\pQ_{ii})\le U$ as well.


The isomorphism $\pQ_{ii}\cong\pQ_{i i+1}$ is given by conjugation as in the proof of Lemma~\ref{lem:sPij sQij sSij}.
\qed

\medskip



\section{The concrete amalgam}\label{section:concrete amalgam}
\Definition
Let $J$ be some index set.
A {\em concrete amalgam} of finite rank $s\in\ZZ_{\ge 0}$ over $J$ is a collection $\cA=\{A_K\mid K\sbe J,\ |K|\le s\}$ of groups together with inclusion homomorphisms
 $\varphi_{M,K}\colon A_K\to A_M$ for every pair $(M,K)$ with $M\sbe K$ satisfying
 $\varphi_{L,M}\after\varphi_{M,K}=\varphi_{L,K}$ whenever $L\sbe M\sbe K$.

The {\em universal completion} of ${\cA}$ is then a group $\hat{{G}}$ whose elements are words in the elements of the groups in $\cA$ subject to the relations between the elements of $A_K$ for any $K\sbe J$ and in which for each $K\sbe M\sbe J$ each $a\in A_K$ is identified with $\varphi_{M,K}(a)\in A_M$.

Let $\hat{\cdot}\ \colon \cA\to \hat{{G}}$ be the embedding of $\cA$ into $\hat{{G}}$.

\medskip

\Note 
\begin{itemize}
\AI{i}
For each $K\sbe J$ with $|J|\le s$, $\hat{\cdot}\ \colon A_k\to \hat{A_K}\le \hat{G}$ is a homomorphism, which is surjective, but not necessarily injective.
\AI{ii}
We have $\hat{A_K}\le \hat{A_M}$ whenever $K\sbe M$.
\AI{iii}
Although it may be the case in some situations, we do not assume that 
 $\hat{A_{K\cap M}}=\hat{A_K}\cap \hat{A_M}$.
\end{itemize}
 

\Definition We will define a concrete amalgam $\scA$. Its set of subgroups is $\{\sM_{ij},\sQ_{ik},\sS_{kl},\sM_i,\sS_k\mid 1\le i,j\le r-1, 1\le k,l\le r\}$, where for each $X\in \cA_{\le 2}$,  $\sX$ is a copy of $\pX$ and
 the inclusion homomorphisms are as follows:
$$\begin{array}{ll}
\varphi^{P}_{i,\{i,j\}}\colon &\sM_i\to\sM_{i j}\\
\varphi^{P}_{j,\{i,j\}}\colon &\sM_j\to\sM_{i j}\\
\varphi^{S}_{k,\{k,l\}}\colon &\sS_k\to\sS_{k l}\\
\varphi^{S}_{l,\{k,l\}}\colon &\sS_l\to\sS_{k l}\\
\varphi^{PQ}_{i,\{i,k\}}\colon &\sM_i\to\sQ_{i k}\\
\varphi^{SQ}_{k,\{i,k\}}\colon &\sS_k\to\sQ_{i k}.\\
\end{array}$$
These inclusions are given by the presentations of $\pX$ as matrix groups as given in 
 Definitions~\ref{dfn:sPi sSj} and~\ref{dfn:pA}.
We denote the universal completion of $\scA$ by $\sG$.

%Recall the amalgam $\cA=\{M_{ij},Q_{ik},S_{kl},M_i,S_k\mid 1\le i,j\le r-1, 1\le k,l\le r\}$  of parabolics of rank at most two as in \ref{subsection:parabolics}.

\Definition
We now define the map $\pi\colon\scA\to G$.
For any $X\in\cA_{\le 2}$, it identifies $\sX$ with its isomorphic copy,  $\pX$, in $G$.
Thus the image of $\scA$ under $\pi$ is $\pcA$.

\ble\label{lem:pi = iso}
For any element $X\in\cA_{\le 2}$, the map
 $\pi\colon \sX\to\pX$ is an isomorphism. 
\ele
\pf
This is true by the definition of $\pi$.
\qed

\medskip

It then follows that the map $\pi$ extends to a surjective map from the universal cover $G^\circ$ to $G$.

\ble\label{lem:intersection of Pi}
If $i\ge2$ then $\sU_i=\sS_i\cap \sM_i=\sS_i\cap\sM_{i-1}=\sM_i\cap \sM_{i-1}$.
\ele
\pf
Note that the above are true for the images under $\pi$. By definition the group $\pM_i$ stablizes the spaces $H_k$ if $k\ne i,i+1$ and it also stabilizes $e_i$ and $e_{i+1}$. Moreover $\pS_i$ stabilizes all the $H_k$ for $k \ne i$. Therefore  $\pS_i\cap \pM_i$ stabilizes all vectors in $H_k$ for $k\ne i$ and it also stabilizes $e_i$. Similarly for $\pS_i\cap\pM_{i-1}$ and $\pM_i\cap \pM_{i-1}$. 
We notice that  $\pi$ is an isomorphism when restricted to $\cA^\circ$ and so the conclusion follows.
\qed



\Definition
For $i=1,2,\ldots,r$, we define the following subgroup of $\sS_i\in\scA$:
$\sB_i=N_{\sS_i}(\sU_i)$.
Furthermore, we set 
 $\sB=\langle\sB_i\mid i=1,2,\ldots,r\rangle$ as a subgroup of $\sB$.

\ble
For $i=1,2,\ldots, r$, we have 
\begin{itemize}
\AI{a} $\pU_i=\pi(\sU_i)$ and $\pB_i=\pi(\sB_i)$.
\AI{b} $$\sU_i=\left\{
\left(\begin{array}{@{}ll@{}} 1 & b_1 \\
                              0 & 1 \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   b_1\in\FF\\
\end{array}
\right\}
\cong\FF.$$

\AI{c} $$\sB_i=\left\{
\left(\begin{array}{@{}ll@{}} a_1 & b_1 \\
                                  0 & d_1 \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,d_1\in\FF\mbox{ such that }\\
   a_1d_1=1\\
\end{array}
\right\}
\cong\FF\rtimes \FF^*.$$

\end{itemize}

\ele



\ble \label{lem:B normalizes}
For $X\in\cA_{\le 2}$, $\sB$ normalizes $\sX$. The action is described by the following
\begin{enumerate}
\item $[\sB_i, \sS_j]=1$ if $i\ne j$ and $\sB_i$ acts on $\sS_i$ as inner automorphisms.
\item $[\sB_i,\sM_j]=1$ if $j \ne i,i+1$. $\sB_i$ acts on $\sM_i$ as the conjugation in $\sQ_{ii}$ and $\sB_{i+1}$ acts on $\sM_i$ as the conjugation in $\sQ_{ii+1}$.
\item The action on the rank 2 parabolics is determined by the action above on the rank one parabolics.
\end{enumerate}
\ele
\pf
Note that de actions are as above for the $\pB_i$ acting on the various $\pX$. Moreover the map $\pi$ is an isomorphism when restricted to the various $\sX\in\cA^\circ$. Since $\sB_i, \sS_j \le \sS_{ij}$, the action of $\sB_i$ on $\sS_j$ is the same as the action of $\pB_i$ on $\pS_j$. For part 2 we note that $\sB_i, \sM_j \le Q_{ij}$ and so the action of $\sB_i$ on $\sM_j$ corresponds to the action of $\pB_i$ on $\pM_j$.

Part 3 follows imediately since the groups in $\cA^\circ$ are all embedded in $\sG$.
\qed

\ble\label{lem:structure sB}
\begin{itemize}
\AI{a}
The group $\sB$ is the internal direct product
 $$\sB=\sB_1\times \sB_2\times\cdots\times \sB_r,$$
\AI{b}
 $$\pi\colon\sB\to\pB\mbox{ is an isomorphism.}$$
\end{itemize}
\ele
\pf
For any $1\le i<j\le r$, we have $\sB_i\le \sS_i$ and since $\sS_{ij}=\sS_i\times \sS_j$ we have
 $\langle \sB_1,\sB_j\rangle=\sB_i\times \sB_j$.
\qed

\ble\label{lem: pBpX=X}
For $X\in\cA_{\le 2}$ we have 
 $\langle B^\pi,X^\pi\rangle_G=X$.
\ele 
\pf
This is an easy calculation inside $G$.
\qed

\ble\label{lem:pi preserves intersection}
For any $X\in\cA_{\le 2}$, $\pi(\sX\cap\sB)=X^\pi\cap B^\pi$.
\ele
\pf
The inclusion $\sbe$ is trivial. 
We now prove $\spe$.
We do this case by case for any $X\in\cA$.

Let $X=S_i$.
Then $\pX\cap\pB=\pB_i$ and since $\sB_i\le \sS_i\cap\sB$, we find $\pi(\sX\cap\sB)\spe \pi(\sB_i)=\pB_i$.

Let $X=P_i$.
Then $\pX\cap\pB=\pU_i\times \pU_{i+1}$ and since $\sU_i\times \sU_{i+1}\le \sM_i\cap\sB$, we find 
$\pi(\sX\cap\sB)\spe \pi(\sU_i\times\sU_{i+1})=\pU_i\times \pU_{i+1}$.

Let $X=S_{ij}$.
Then $\pX\cap\pB=\pB_i\times \pB_j$ and since $\sB_i\times \sB_j\le \sS_{ij}\cap\sB$, we find 
$\pi(\sX\cap\sB)\spe \pi(\sB_i\times\sB_j)=\pB_i\times \pB_j$.

Let $X=Q_{ij}$.
Then $\pX\cap\pB=\langle\pU_i, \pU_{i+1}, \pB_j\rangle$ and since $\langle\sU_i, \sU_{i+1}, \sB_j\rangle\le \sQ_{ij}\cap\sB$, we find 
$\pi(\sX\cap\sB)\spe \pi(\langle\sU_i, \sU_{i+1}, \sB_j\rangle)=\langle\pU_i, \pU_{i+1}, \pB_j\rangle$.

Let $X=P_{ij}$.
Then $\pX\cap\pB=\langle\pU_i, \pU_{i+1}, \pU_j, \pU_{j+1}\rangle$ and since $\langle\sU_i, \sU_{i+1}, \sU_j, \sU_{j+1}\rangle\le \sM_{ij}\cap\sB$, we find 
$\pi(\sX\cap\sB)\spe \pi(\langle\sU_i, \sU_{i+1}, \sU_j, \sU_{j+1}\rangle)=\langle\pU_i, \pU_{i+1}, \pU_j, \pU_{j+1}\rangle$.




\qed

Our next aim is to define a map $\chi\colon \cA_{\le 2}\to \sG$ extending $\pi^{-1}\colon \pX\to \sX$ for every
 $X\in\cA_{\le 2}$.

\ble\label{lem:well defined on intersections}
$\chi$ is well-defined on $X^\pi\cap B^\pi$ for all $X\in\cA_{\le 2}$.
\ele

\pf
This follows from Lemma~\ref{lem:pi preserves intersection}
\qed

Define $\chi$ on $X=B^\pi X^\pi$ for any $X\in\cA_{\le 2}$ as follows:
$\chi(bx)=\chi(b)\chi(x)$.

\ble\label{lem:chi injective}
$\chi$ is well-defined and injective on $X$.
\ele
\pf
Note that if $b,b' \in \pB$ and $x, x'\in \pX$ then $bx=b'x' \Rightarrow b^{-1}b'=xx'^{-1} \in \pB\cap \pX$. Moreover by Lemma \ref{lem:well defined on intersections} $\chi(b^{-1}b')=\chi(xx'^{-1})$ and so $\chi(bx)=\chi(b'x')$ and $\chi$ is well defined.

Also if $bx\in X$ with $\chi(bx)=1$ it follows that $\chi(b)=\chi(x)^{-1} \in \sB\cap\sX$ (because $\chi$ is the inverse of $\pi$ when restricted to $\pB$ and $\pX$). Therefore $b=\pi(\chi(b))$ and $x=\pi(\chi(x))$ are both in $\pi(\sX\cap\sB)=X^\pi\cap B^\pi$. But $\chi$ is a bijection when restricted to $\pX\cap \pB$ and so $b=x^{-1}$ and $\chi$ is injective. 
\qed

\ble\label{lem:chi amalgam hom}
$\chi$ is an embedding of the amalgam $\cA_{\le 2}$ into $\sG$.
\ele
\pf
Let $X \in \cA$ then by lemma \ref{lem: pBpX=X} and Lemma \ref{lem:B normalizes} we know that $X=\pB\pX$ and so if $bx, b'x' \in X$ then $\chi(b'x'bx)=\chi(b'b)\chi(b^{-1}x'bx)=\chi(b'b)\chi(b^{-1}x'b)\chi(x)$. Also $\chi(b'x')\chi(bx)=\chi(b')\chi(x')\chi(b)\chi(x)=\chi(b')\chi(b)\chi(b^{-1})\chi(x')\chi(b)\chi(x)$. so we only need to prove that $\chi(b^{-1})\chi(x')\chi(b)=\chi(b^{-1}x'b)$ which is equivalent to the fact that the action of $\pB$ on $\pX$ is the same as the action of $\sB$ on $\sX$. This follows from Lemma \ref{lem:B normalizes}.
\qed

As a consequence, we find the following result.

\bpr\label{prop:chi surjective}
The map $\chi$ extends to a surjective homomorphism $G\to \sG$ which we also denote by $\chi$.
\epr
\pf
By Lemma~\ref{lem:chi amalgam hom} the map $\chi$ extends to a homomorphism $G\to\sG$ whose image contains
 the subgroup of $\sG$ generated by the subgroups in $\cA^\circ$. 
Since $\langle\scA\rangle=\sG$, the conclusion follows.
\qed

\medskip

\pf Of Theorem~\ref{thm:main theorem1}.
By Corollary~\ref{cor:minparam} $G$ is the universal completion of the amalgam $\cA_{\le 2}$. 

The map $\pi\colon \cA^\circ\to G$ given by 
 $\pi\colon X^circ\to X^\pi$ for all $X\in\cA_{\le 2}$ extends to surjection 
   $\pi\colon G^\circ\to H^\pi$ where $H^\pi$ is the subgroup of $G$ generated by the subgroups in $\cA^\pi$.
By Lemma~\ref{lem: pBpX=X} $H^\pi=G$, and so $\pi\colon G^\circ\to G$ is surjective.


By Proposition~\ref{prop:chi surjective} there is a surjective map $\chi\colon G\to \sG$, which by Lemma~\ref{lem:chi injective} is injective on the subgroups of 
 $\cA_{\le 2}$.

The composition $\chi\after \pi\colon \sG\to\sG$ is a surjective homomorphism which is the identity on every subgroup in 
 the amalgam $\cA^\circ$.
Since $\sG$ is the universal completion of $\cA^\circ$, the only such map is the identity.

Hence $\pi\colon\sG\to G$ is an isomorphism.
\qed

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\end{document}


\subsection{The first slim amalgam}
Let $\cH=\{e_i,f_j\mid 1\le i\le r+d, 1\le j\le r\}$ be a hyperbolic basis corresponding to $C$ 
 as in Corollary~\ref{cor:hyperbolic bases and chambers}.
For $i=1,2,\ldots,r$, let $H_i=\langle e_i,f_i\rangle_V$

We now introduce the subgroups $\sS_i$, $\sM_j$ of $\Sp(V)$, where $1\le i\le r$ and $1\le j\le r-1$.
The group $\sS_i$ fixes every vector in $H_k$, for every $k\ne i$ and acts on $H_i$
 as $\Sp(H_i)\cong\Sp_2(\FF)$.

\Note We note that $\sS_i$ is generated by two opposite long-root groups.

A generic element of $\sM_i$ fixes every vector in $H_k$, for every $k\ne i,i+1$ and 
 acts on $H_i\oplus H_{i+1}$ as
$$\{\left(\begin{array}{@{}llll@{}} 1 & 0 & 0 & w \\
                                      0 & 1 & 0 & 0 \\
                                      0 & w & 1 & 0 \\
                                      0 & 0 & 0 & 1 \\
            \end{array}\right)\mid w\in \FF\}.
$$

\Note We note that $\sM_i$ is a short-root group.

For $1\le i<j\le r$, let $\sS_{ij}=\langle\sS_i,\sS_j\rangle_G$.
For $1\le i<j\le r-1$, let $\sM_{ij}=\langle\sM_i,\sM_j\rangle_G$.
For $1\le i\le r-1$ and $1\le j\le r$, let $\sQ_{ij}=\langle\sM_i,\sS_j\rangle_G$.
\ble
For all indices $i,j$ that apply, we have 
\begin{itemize}
\AI{a}
$$\begin{array}{ll}
\sS_{ij}&\cong\sS_i\times \sS_j\\
\sM_{ij}&\cong\sM_i\times \sM_j.\\
\end{array}$$
\AI{b}
If $j<i$ or $i+1<j$, then 
$$\sQ_{ij}\cong\sM_i\times \sS_j.$$
Furthermore,
$$
\sQ_{ii}\cong
\left\{
\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & w_1 \\
                                  c_1 & d_1 & 0 & w_2\\
                                  v_2 & v_1 & 1 & b_2\\
                                   0  & 0   & 0 & 1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1,w_1,w_2,v_1,& v_2, b_2\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
         a_1w_2-c_1w_1+v_2&=0\\
         b_1w_2-d_1w_1+v_1&=0\\
\end{array}
\right\}
$$
and
$$
\sQ_{i i+1}\cong
\left\{
\left(\begin{array}{@{}llll@{}} 
                                  1     & b_2 & v_2    & v_1 \\
                                  0     & 1     & 0        & 0      \\
                                  0     & w_1 & a_1     & b_1\\
                                   0    & w_2 & c_1     & d_1  \\
      \end{array}
\right)
\Big|%\mid 
\begin{array}{l@{}l}
   a_1,b_1,c_1,d_1,w_1,w_2,v_1,& v_2, b_2\in\FF\mbox{ such that}\\
         a_1d_1-b_1c_1&=1\\
         a_1w_2-c_1w_1+v_2&=0\\
         b_1w_2-d_1w_1+v_1&=0\\
\end{array}
\right\}
$$
\end{itemize}
\ele 
\pf
Looking at the shape of the groups $\sM_i$ and $\sS_j$, it is easy to see that these components intersect trivially and commute.
We now turn to the cases $\sQ_{ii}$ and $\sQ_{i i+1}$.
First note that conjugation by the permutation matrix that switches $(e_i,f_i)$ and $(e_{i+1},f_{i+1})$, interchanges
 $\sS_i$ and $\sS_{i+1}$, but fixes every element in $\sM_i$.
Thus it suffices to prove the claim for $\sQ_{ii}$. 

Let $Q$ be the proposed collection of matrices.
We claim that $Q$ is the stabilizer of the vector $e_{i+1}$ in $\Sp(H_i\oplus H_{i+1})$.
It is clear from the shape of the third column that $Q$ stabilizes $e_{i+1}$.
On the other hand, any matrix $A$ in $\Sp(H_i\oplus H_{i+1})$ stabilizing $e_{i+1}$ must have such a
 third column.
It must also have zeroes in the last row as in $Q$ since in fixing $e_{i=1}$ it must also stabilize $e_{i+1}^\perp$.
Any such matrix $A$ must satisfy the conditions on the entries as indicated in the description of $Q$ since 
$\{Ae_i,Af_i, e_{i+1},Af_{i+1}\}$ must be isometric to $\{e_i,f_i,e_{i+1},f_{i+1}\}$ in order for $A$ to be symplectic.
Therefore $Q$ equals this stabilizer and hence is a group.

Clearly $Q$ contains $\sS_i$ and $\sM_i$. We now show that $\langle \sS_i,\sM_i\rangle=Q$.
First we note that by taking $S'SPS^{-1}$ with $S,S'\in \sS_i$ and $P\in\sM_i$ we obtain all matrices of 
 the form

$$\left(\begin{array}{@{}llll@{}} a_1 & b_1 & 0 & w_1 \\
                                  c_1 & d_1 & 0 & w_2 \\
                                  v_2 & v_1 & 1 & 0   \\
                                   0  & 0   & 0 & 1   \\
      \end{array}
\right),$$
where the entries are as in the lemma.
Namely, taking $S'=1$, we obtain all such matrices with $a_1=d_1=1$ and $b_1=c_1=0$.
This is because $\sS_i\cong\Sp_2(\FF)\cong\SL_2(\FF)$ is transitive on vectors of $\FF^2$.
Then allowing $S'$ to be any other element in $\sS_i$ produces the elements as described.

It now suffices to show that the matrices in $Q$ such that all entries are zero, except $a_1=d_1=1$ and $b_2\in\FF$ is arbitrary also belong to $\sQ_{ii}$.
As we saw above, for any vector $\vec{w}=(w_1,w_2)^T\in\FF^2$ there is an element 
 $$A(\vec{w})=\left(\begin{array}{@{}llll@{}} 
                                   1   & 0   & 0  & w_1 \\
                                   0   & 1   & 0  & w_2 \\
                                  -w_2 & w_1 & 1  & 0   \\
                                   0   & 0   & 0  & 1   \\
      \end{array}
\right) \in \sQ_{ii}$$

One easily calculates that, for $\vec{v},\vec{w}\in\FF^2$ the matrix 
 $A(\vec{w})A(\vec{v})A(-\vec{w}-\vec{v})$ is of the form
 $$K(b_2)=\left(\begin{array}{@{}llll@{}} 
                                   1   & 0   & 0  & 0 \\
                                   0   & 1   & 0  & 0 \\
                                   0   & 0   & 1  & b_2   \\
                                   0   & 0   & 0  & 1   \\
      \end{array}
\right)
\mbox{, where }b_2=\sform(\vec{w},\vec{v}).
$$
Since we can choose $\vec{w}$ and $\vec{v}$ arbitrarily, $b_2$ can take any value in $\FF$.
Since $K=\{K(b_2)\mid b_2\in \FF\}\le Z(\sQ_{ii})$ it follows that $\sQ_{ii}$ is as described in the lemma.
\qed 
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