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\title{The non-canonical gluings of two affine spaces.}
\author{Barbara Baumeister, Gernot Stroth}
\begin{document}
\maketitle
\section{Introduction.}
Given two geometries each equipped with a suitable parallelism giving
rise to the same geometry at infinity, a new geometry can be constructed
by gluing the two geometries together along their geometry at infinity.
This method of constructing new geometries was introduced by Buekenhout,
Huybrechts and Pasini [BHP]. Suppose that the obtained geometry is
of rank $3$. Then each point is incident to each plane, i.e. $\Gamma$
is {\em flat}. Conversely, many flat geometries of rank three are 
gluings -- see for instance the classification of flat geometries of
type $Af.c^*$, $Af.Af^*,~c.c^*$ [BDMP,~DP, BP].
In [Pas2] Pasini examines the special case of gluing two affine spaces.

Let  $({\cal A}_0, ||_0)$ and  $({\cal A}_2, ||_2)$ be two affine spaces
isomorphic to $AG(n,q)$ and let $\Gamma^\infty = \Gamma_i^\infty$  be the
set of parallel classes of $||_i$ ($ i =0,2$). Then 
the gluing $AS(n,q) \circ_\alpha AS(n,q)$ of the points and lines of
${\cal A}_0$ and of ${\cal A}_2$ via the permutation $\alpha$ of 
$\Gamma^\infty$  is a geometry of rank three consisting of points, lines and planes
 defined as follows (see [Pas2]):\\
The points are the points of ${\cal A}_0$, the planes the points of
${\cal A}_2$ and the lines the pairs $(l_0,l_2)$ with $l_i$ a line 
in ${\cal A}_i$ such that $l_0^{\infty \alpha} = l_2^\infty$.
Any point is incident with any plane and a point $p$ (resp. plane $u$) is incident
with a line $(l_0,l_2)$ iff $p$ (resp. $u$) is incident to $l_0$ (resp. $l_2$)
in ${\cal A}_0$ (resp. ${\cal A}_2$).

The gluing is said to be {\em canonical}, if $\alpha \in P\Gamma L_n(q)$,
see [Pas2, 1.3.4]. In this paper we determine all the flag-transitive 
non-canonical gluings of two isomorphic desarguesian affine spaces $AS(n,q) \circ_\alpha AS(n,q)$.

Let $\Gamma = AS(n,q) \circ_\alpha AS(n,q)$, let 
 $A_i = Aut({\cal A}_i, ||_i))$ and let $K_i^\infty$ be the kernel
of the action of $A_i$ on $||_i$, for $i = 0,2$. Then $A_i^\infty
= A_i/K_i^\infty$ is the group induced by $A_i$ on  $\Gamma_i^\infty$ ($ i =0,2$).
Set $X = A_0^\infty \cap A_2^{\infty \alpha}$. Then $Aut(\Gamma) = (K_0^\infty \times K_2^\infty)X$,
see Lemma~\ref{auto}.

There are sixteen sporadic examples (nine  with $n = 2$, three with $n = 4$, two with $n = 6$) and two infinite series. More precisely we prove the following.

\begin{thm}\label{mthm}
Let $\Gamma$ be a non-canonical gluing $AS(n,q) \circ_\alpha AS(n,q)$ where $q = p^r$. Then  one of the following holds.
\begin{itemize}
\item[(1)] \begin{itemize}
            \item[(i)] $n = 2$,~ $q \in \{ 5,7,11,23\}$ and $X \cong S_4$;  
             \item[(ii)] $n = 2$,~ $q = 11$ and $X \cong A_4$;
           \end{itemize}
\item[(2)] $n = 2$, ~$q \in \{11,19,29,59\}$ and $X \cong A_5$;
\item[(3)] $n = 4$, ~$q = 3$, $O_2(X) \cong E_{2^4}$ and $\ZZ_5 \leq X/O_2(X)  \leq Frob(5:2)$;
\item[(4)] $n = 6$, ~$q = 3$ and $X \cong L_2(13)$;
\item[(5)] $n>2$ or $n = 2$ and $q \not\in \{2,3,4,5,8,9\}$; $X = N_{\Gamma L_n(q)}(T)$,
with $T$ a Singer cycle of $\Gamma L_n(q)$;
\item[(6)] $gcd(r \cdot n,(q^n-1)/(q-1)) \neq 1$, $X$ is naturally embedded
into $\Gamma L_1(q^n)$, but not of type (5).
\end{itemize}
In cases (1), (2) and (3) we may choose $\alpha$ such that $[X, \alpha ] = 1$.
In case (4) there exists just two examples: one with $[X,\alpha] = 1$ and the 
other with $X\la \alpha \ra \cong PGL_2(13)$.
\end{thm}

Moreover, we determine the universal covers of the $15$ exceptional gluings
of type (1) -- (4) as well as of the canonical gluings
$AS(n,q) \circ_\alpha AS(n,q)$.
We prove that the gluings in (1), (2) and (3) and in (4) with
$X\la \alpha \ra \cong PGL_2(13)$ are simply connected (see Propositions (5.4) -- (5.6)). In case (4), if $[X, \alpha] = 1$,
then the universal cover of $\Gamma$ is a triple cover (see Proposition (5.6)).
Let $\Gamma$ be the canonical gluing $\Gamma = AS(n,q)\circ_\alpha AS(n,q)$,
then
the universal cover is a $q^{{n \choose 2}}$ (resp. $q^{2^n-n-1}$) cover, if $q$
is odd (resp. $q$ is even) (see Theorem 6.4).

Theorem\ref{mthm} and the determination of certain universal covers solve some of the 
problems posed by Pasini in [Pas2, (5.4)].

Let $Aut(\Gamma) = (K_0^\infty \times K_2^\infty)X$. Then the group $X$ acts
faithfully on the set, say $\Gamma_0^\infty$, of parallel classes of lines of ${\cal A}_0$. Note, that $\Gamma_0^\infty$ is just the set of points
of the projective space $PG(n-1,q)$.
Moreover, $Aut(\Gamma)$ acts flag-transitively on $\Gamma$ if and
only if $X$ acts transitively on $\Gamma_0^\infty$.
In [Pas2, (4.3)] Pasini posed the question, if a group $X \leq Sym(\Gamma_0^\infty) = Sym((q^n-1)/(q-1))$
is transitive on the set of points of $PG(n-1,q)$ precisely, if $X$ contains 
a Singer-cycle. Since (1) - (4) and (6) of the main theorem do not contain a
Singer-cycle, our answer is negative. Pasini also asked, whether, if $X$ contains
 a Singer-cycle $T$ and if $q$ is large enough, $X$ is contained in the 
normalizer of $T$ in $P\Gamma L_n(q)$. Here our answer is affirmative.

Our notation is fairly standard, see for instance [Pas1].
For $G = Aut(\Gamma)$ and $\{a_0, a_1,a_2\}$ a maximal flag of $\Gamma$ we
denote by $G_i$ the stabilizer of $a_i$ in $G$. 

The proof of the Main Theorem runs as follows. Let $(\Gamma, Aut(\Gamma))$ be
a  flag-transitive non-canonical gluing $AS(n,q) \circ_\alpha AS(n,q)$.
Then $G_0/Z(G_0)$ is doubly transitive  of degree $q^n$.
Moreover, if $X$ is not isomorphic to a subgroup of $\Gamma L_1(q^n)$, 
then we prove that we may suppose that $\alpha$ normalizes $X$.
By definition $\Gamma$ is a non-canonical gluing if and only if $\alpha$ is in
$Sym((q^n-1)/(q-1))\sm{A_0^\infty}$. Hence we have just to determine those
doubly transitive permutation groups of degree $q^n$ for which the normalizer
of $X$ in $Sym((q^n-1)/(q-1))$ is not contained in $A_0^\infty$.

\section{Preliminaries.}

\begin{lemma}\label{centrinsym}
Let $G$ be a group which acts faithfully and transitively on a set $\Omega$.
Then $C_{Sym(\Omega)}(G)$ acts regularly on the set of fixed points, say
$\Lambda$, of $G_\omega$ on $\Omega$ for any $\omega \in \Omega$ and 
$C_{Sym(\Omega)}(G) \cong N_G(G_\omega)/G_\omega$.
\end{lemma}
\begin{bew}
The representations of $G$ on $\widetilde{\Omega} = \{ G_{\omega}g~|~g \in G\}$
and on $\Omega$ are equivalent: for $\psi:\widetilde{\Omega} \rightarrow
\Omega,~ G_\omega g \mapsto \omega^g$ holds $x^{\psi g} = x^{g\psi}$ for all
$x \in \widetilde{\Omega}$ and all $g \in G$. Therefore 
$C_{Sym(\Omega)}(G)$ acts regularly on $\Lambda$.

Suppose that $G$ acts regularly on $\Omega$. Then the action of $G$ 
on $\Omega$ and the action of $G$ on $G$ via right multiplication are
equivalent. Moreover,
$$\psi : G \rightarrow C_{Sym(\Omega)}(G),~g \mapsto c_g ~\mbox{with}~
\omega^{c_g} = \omega^{g^{-1}}$$ for all $\omega \in \Omega$ and all $g \in G$
is an isomorphism. As $C_{Sym(\Omega)}(G)$ and$ N_G(G_\omega)/G_\omega$ act regularly on $\Lambda$, we may embed $C_{Sym(\Omega)}(G)$ into $C_{Sym(\Lambda)}(N_G(G_\omega)/G_\omega)$. Application of the above reasonning
to  $N_G(G_\omega)/G_\omega$ and $\Lambda$  shows 
$C_{Sym(\Lambda)}(N_G(G_\omega)/G_\omega) \cong N_G(G_\omega)/G_\omega$.
Since $C_{Sym(\Omega)}(G)$ and $N_G(G_\omega)/G_\omega$ are of the same
order, also the last assertion of the lemma holds.
\end{bew}


Let $G$ be an affine doubly transitive permutation group of degree
$p^n,~p $ prime. Then $G = O_p(G):Y,~V = O_p(G)$ is elementary 
abelian and $Y$ acts transitively and faithfully on $V^\#$.

In the following we will use the notation introduced by Hering [He].
Let $L \subseteq Hom(V,V)$ be maximal in respect
with the conditions that 
\begin{itemize}
\item[(i)] $L$ is normalized by $Y$;
\item[(ii)] $L$ contains the identity and $L$ is a field with the addition and multiplication of $Hom(V,V)$.
\end{itemize}

Then $L$ is a field extension of $GF(p)$. Let $|L| = p^r =q$ and $n$ such that $|V| = q^n$. We say that
{\em $Y$ is naturally embedded into $\Gamma L_n(q)$}. 
The affine doubly transitive permutation groups are classifyed
by Hering and Liebeck [He,Li]. There are four infinite series:
$n = 1$ or $H \leq Y \leq \Gamma L_n(q), ~q = p^r$ with $H \in \{SL_n(q), Sp_{n}(q), n \geq 4, G_2(q)\}$ (see [He] or [Li]). We will call the remaining affine doubly transitive permutation 
 groups the {\em exceptional doubly transitive permutation groups}.

\begin{lemma}\label{conj}
Let $H = O_p(H):Y$ be an affine doubly transitive permutation group
and suppose that $Y$ is naturally embedded into $G = \Gamma L_n(q),~n \geq 2$.
Let $Z$ be a subgroup of $G$ which is isomorphic to $Y$ and 
which acts transitively on $O_p(H)^\#$. Then $Y$ and $Z$ are conjugate in
$\Gamma L_n(q)$.
\end{lemma}
\begin{bew}
Set $V = O_p(H)$.
As $Z$ acts transitively on $V^\#$, also $V:Z$ is a doubly transitive affine permutation groups of
degree $q^n$. Hence $Y$ and $Z$ are one of the groups listed by Hering 
(see [He] or [Li]).

Let  $Z$ be naturally embedded into $\Gamma L_{n_z}(p^{r_z})$.
Set $L_y = GF(q)$ and $L_z = GF(p^{r_z})$.

{\em Case 1.} Let $Y$ be non solvable. Then $E(Y) \in \{SL_n(q);
Sp_{n}(q), ~n \geq 4; G_2(q), n = 6; SL_2(5), n = 2,
q  = 9,11,19,29,59;n = 4,  A_6, q =  2; $\\$ A_7, n = 4,
q=2; SL_2(13), n = 6, q=  3; G_2(2)' \cong  U_3(3), n = 6, q  = 2\}$ or $Y \cong
2^{1+4}A_5$ or $2^{1+4}Sym(5)$ and $n = 4,q= 3$.

Suppose $E(Y) \cong SL_n(q)$. Then $q = p^{r_z}$ and $n = n_z$. Therefore
$L_y^\#$ and $L_z^\#$ are cyclic of order $q -1$. Moreover,
$L_y^\#$ (and  also $L_z^\#$) decompose $V$ into $GF(p)$-subspaces of
dimension $r$ such that 
$L_y^\#$ (resp. $L_z^\#$) acts as field multiplication on these
subspaces. Since $Y$ (resp. $Z$) acts irreducibly on $V$, these
subspaces are isomorphic $GF(q)L_y^\#$ (resp. $GF(q)L_z^\#$)-modules.
Therefore $L_y^\#$ and $L_z^\#$ are conjugate in $G$. This implies
that $E(C_G(L_y^\#)) = E(Y)$ and $E(C_G(L_z^\#)) = E(Z)$ are
conjugate in $G$. Let $E = E(Y)^g = E(Z)$. As $Y \cong Z$, there
is a $\phi \in Aut(E)$ such that $Y^{g\phi} = Z$.
Since we may assume $\phi \in \Gamma L_n(q)$, the groups $Y$ and $Z$ are conjugate in $G$.


Now let $E(Y) \cong Sp_{n}(q),~(n,q) \neq (4,2)$. Again $q = p^{r_z}$ and
$n = n_z$. $E(Y)$ and $E(Z)$ are contained in subgroups $S,T$ of $G$
isomorphic to $SL_{2n}(q)$. Since $E(Y)$ and $E(Z)$ act transitively
on $V$, also $S$ and $T$ act transitively on $V$ and therefore $S$ and $T$ are
conjugate in $G$ and we may assume $E(Y), E(Z) \leq S$. As there exist
up to equivalence only one symplectic scalar product on the $L$-vectorspace
$V$ (see for instance [As]), it follows that $E(Y)$ and $ E(Z)$ are
conjugate in $G$. As
above, we obtain that $Y$ and $Z$ are conjugate in $G$.


Let $E(Y) \cong G_2(q)$ or $G_2(2)'$. Then again $q = p^{r_z}$ and we may assume
$E(Y), E(Z) \leq H \leq G,~H \cong Sp_6(q)$. According to 
[Kl] $E(Y)$ and $E(Z)$ are conjugate in $N_G(H)$. Since the
outer automorphism group of $E(Y)$ is cyclic, the assertion also 
holds in this case.

So let $E(Y) \cong SL_2(5)$ and $q \in \{9,11,19,29,59\}$.
Again $q = p^{r_z}$. According to [Kl] there are two conjugacy
classes of subgroups isomorphic to $SL_2(5)$ in $SL_2(q)$. Since
$PGL_2(q)$ does not contain $S_5$ or $A_5 \times \ZZ_2$, $E(Y)$ and
$E(Z)$ are conjugate in $G$. The fact that $Y \leq E(Y)Z(G)$ and 
$Z \leq E(Z)Z(G)$ yields again the assertion.


In $L_4(2)\cong A_8$ there is just one conjugacy class of 
subgroups isomorphic to $A_6$. Therefore the assertion holds  for
$Y \cong A_6$ and $Y \cong A_7$.


If $Y \cong SL_2(13),~q = 3$ and $n = 6$, then $Y \leq H \leq G,~H \cong
 Sp_6(3)$. We may assume $Z \leq H$, as well. Now according to
[CCNPW] all subgroups of $Aut(Sp_6(3))$ isomorphic to $SL_2(13)$
are conjugate. 

Finally, if $(n,q) = (4,3)$ and $Y \cong 2^{1+4}A_5$ or $2^{1+4}S_5$,
then again according to [CCNPW] $Y$ and $Z$ are conjugate in $G$.
Thus the assertion holds in case 1.
\\

{\em Case 2.} $Y$ is solvable. Let $n = 2$. Then $Y$ contains a
subgroup isomorphic to $SL_2(3)$ [Hu] and $q \in \{3,5,7,11,23\}$. 
We have $\Gamma L_1(q^2) \cong \ZZ_8:\ZZ_2,~ \ZZ_{24}:\ZZ_2
,~ \ZZ_{48}:\ZZ_2,~ \ZZ_{120}:\ZZ_2,~ \ZZ_{528}:\ZZ_2$, respectively.
The facts that $|X| \not| |P\Gamma L_1(q^2)|$ or that 
$O_3(\Gamma L_1(q^2)) \neq 1$
imply $n_z = 2$.
Therefore $Z$ is of the same exceptional type as $Y$ and by [Hu]
$Y$ and $Z$ are conjugate in $G$.


If $(n,q) = (4,3)$ and $Y$ is containing a subgroup isomorphic to
$2^{1+4}\ZZ_5$, then according to [Hu] the assertion holds for
$Y$ and $Z$.
This proves the lemma.
\end{bew}


\begin{lemma}\label{normser}
Let $G = O_p(G):Y$ be an affine  doubly transitive permutation group
naturally embedded into $\Gamma L_n(q), ~q = p^r,~ n \geq 2$ 
such that $H \leq Y $ with $H \in \{SL_n(q), Sp_{n}(q), n \geq 4, G_2(q)\}$.
Let $m$ and $t$ be natural numbers with $n~|~m$ and $q^n = t^m$ and let
$\Gamma^\infty$ be the set of $1$-dimensional $GF(t)$-subspaces of
$O_p(G)$. Moreover, let $X$ be the group induced by $Y$ on $\Gamma^\infty$.
Then $C_{Sym(\Gamma^\infty)}(X) = C_{P\Gamma L_m(t)}(X)$ and $N_{Sym(\Gamma^\infty)}(X) =
N_{P\Gamma L_m(t)}(X)$.
\end{lemma}
\begin{bew}
Let $\omega \in \Gamma^\infty$ and let $M = stab_X(\omega)$. Then $N_X(M)/M \cong $
$C_{\Gamma L_m(t)}
(E(X))/Z(GL_n(q))$. Therefore, Lemma~\ref{centrinsym}
yields the first part of the assertion.

Since $\Gamma L_n(q)$ embeds into $\Gamma L_m(t)$ also the second part of the
assertion follows.
\end{bew}

\begin{lemma}\label{table}
Let $G = O_p(G):Y$ be an affine exceptional doubly transitive permutation group
with $Y \leq \Gamma L_n(q),~|O_p(G)| = q^n$. Let $\Gamma^\infty$ be the set
of $1$-spaces of $V = O_p(G)$ considering $V$ as a natural $GF(q)$-module
for $Y$. Then $Y$ induces a group $X$ on $\Gamma^\infty$ as listed in Table 1.
Moreover, $X_\omega,~ \omega \in \Gamma^\infty$, and $N_X(X_\omega)/X_\omega$
are as listed in Table 1.
\end{lemma}
\begin{bew}
In order to determine $N_{Sym(\Gamma^\infty)}(X)$ we are using the fact that by Lemma~\ref{centrinsym}
$C_{Sym(\Gamma^\infty)}(X) \cong
 N_X(X_\omega)/X_\omega$.

If $n = 2$, then it is easy to calculate $ N_X(X_\omega)/X_\omega$ and $N_{Sym(\Gamma^\infty)}(X)$.

If $n = 4$ and $q = 2$, then $|\Gamma^\infty | = 15$, $X \cong A_6$ or $A_7$ and in both
cases $C_{Sym(\Gamma^\infty)}(X) =1$. Since $Sym(7)$ has no permutation representation of 
degree $15$, we obtain $N_{Sym(\Gamma^\infty)}(X) \cong Sym(6)$ or $A_7$, as claimed.


Now let $n = 4$ and $q = 3$. If $Y \cong SL_2(5)*\ZZ_8$ (resp. $SL_2(5)2*\ZZ_8$), then
$X \cong A_5 \times \ZZ_4$ (resp. $Sym(5) \times \ZZ_4$), $X_\omega \cong Sym(3)$ (resp.
$Sym(3) \times \ZZ_2$) and therefore $C_{Sym(\Gamma^\infty)}(X) = Z(X) \cong \ZZ_4$ and
$N_{Sym(\Gamma^\infty)}(X) = Sym(5) \times \ZZ_4$.

If $X = O_2(X)N_X(S),~S \in Syl_5(X)$, then according
to the Theorem of Schur-Zassenhaus $N_{Sym(\Gamma^\infty)}(X) = O_2(X)N_M(N_X(S))$ with
$M = N_{Sym(\Gamma^\infty)}(O_2(X))$. If $N_{GL_4(3)}(N_X(S))$ induces the full automorphism
group on $N_X(S)$, then $N_M(N_X(S)) = C_{Sym(\Gamma^\infty)}(X)$\\$N_{GL_4(3)}(N_X(S))$. In 
all cases it is straightforward to determine $N_X(X_\omega)/X_\omega$.
Therefore, the assertion holds, if $X = O_2(X)N_X(S)$. If $X/O_2(X) \cong A_5$, then
$X_\omega \cong \ZZ_2A_4$ and therefore $N_X(X_\omega) = X_\omega$. This proves
the assertion for $X/O_2(X) \cong A_5$ or $Sym(5)$. Finally, let $n = 6$, $q = 3$ and $X \cong
L_2(13)$. Then $X_\omega \cong \ZZ_3$ which implies the assertion. This proves the
lemma.
\end{bew}
\begin{figure}
\leavevmode
\caption{Table 1.  }
\[\mbox{\scriptsize 
$\begin{array}{l|l|l|llll}  
n &   q  &  X &X_\omega & N_X(X_\omega)/X_\omega &N_{Sym(\Gamma^\infty)}(X)\\\hline\hline
  &      &    &         &   &\\
2 & 3    & A_4& \ZZ_3   & 1& Sym(4)\\
  &      & Sym(4) & Sym(3)&  1 &Sym(4)\\
  &      &    &         & &  \\
  &  5   & A_4& \ZZ_2   & \ZZ_2& Sym(4) \times \ZZ_2\\
  &      & Sym(4)& \ZZ_4   & \ZZ_2& Sym(4) \times \ZZ_2\\
  &      &    &         &  & \\
  & 7    & Sym(4)& \ZZ_3   & \ZZ_2& Sym(4) \times \ZZ_2\\
  &      &    &         & &  \\
  & 11   & A_4& 1       & A_4& (A_4 \times A_4)2\\
  &      & Sym(4)& \ZZ_2   &  \ZZ_2& Sym(4) \times \ZZ_2 \\
  &      &    &         &   &\\
  & 23   & Sym(4)& 1       & Sym(4)& Sym(4) \times Sym(4)\\
  &      &    &         &   & \\
  & 9    & A_5& Sym(3)     &1 & Sym(5)\\
  &      &  Sym(5)&Sym(3) \times \ZZ_2   &  1 & Sym(5)\\
  &      &    &         &  & \\
  & 11   & A_5& \ZZ_5  & \ZZ_2& Sym(5) \times \ZZ_2\\
  &      &    &         &   & \\
  & 19   & A_5& \ZZ_3   & \ZZ_2& Sym(5) \times \ZZ_2\\
  &      &    &         &   & \\
  & 29   & A_5& \ZZ_2   & \ZZ_2& Sym(5) \times \ZZ_2\\
  &      &    &         &  & \\
  & 59   & A_5& 1       & A_5 & (A_5 \times A_5)2\\
  &      &    &         &  & &\\
4 & 2    & A_6& Sym(4)     & 1& Sym(6)\\
  &      &    &         &  & \\
  & 2    & A_7& L_3(2)    & 1&A_7\\
  &      &    &         &  & \\
  & 3    & A_5 \times \ZZ_4& Sym(3)  & \ZZ_4 & Sym(5) \times \ZZ_4\\
  &      &    &         &  & \\
  &      & Sym(5) \times \ZZ_4& Sym(3) \times \ZZ_2 & \ZZ_4 & Sym(5) \times \ZZ_4\\
  &      &    &         &  & \\
  &      & E_{2^4}:\ZZ_5& \ZZ_2& E_{2^3}&((E_{2^4}:\ZZ_5) \times E_{2^3})\ZZ_3:\ZZ_4 \\
  &      &E_{2^4}:\ZZ_5:\ZZ_2  & \ZZ_2\ZZ_2 &\ZZ_2    &((E_{2^4}:(\ZZ_{15}:\ZZ_2)) \times \ZZ_2)\ZZ_2\\
  &      &E_{2^4}:\ZZ_5:\ZZ_4  & \ZZ_2\ZZ_4 & 1   &E_{2^4}:(\ZZ_5:\ZZ_4)\\
  &      & E_{2^4} :A_5 & \ZZ_2A_4 & 1& E_{2^4}:Sym(5)   \\
  &      & E_{2^4} :Sym(5) &\ZZ_2 Sym(4)& 1 & E_{2^4}:Sym(5) \\
  &      &    &         &  & \\
6 & 3    & L_2(13) & \ZZ_3 &E_{4} & (L_2(13) \times E_4)2\\
 \end{array} $}\]
\end{figure}

\begin{coro}\label{coro3}
Let $G = O_p(G):Y$ be an affine exceptional doubly transitive permutation group
with $Y \leq \Gamma L_n(q),~|O_p(G)| = q^n$. Let $\Gamma^\infty$ be the set
of $1$-spaces of $V = O_p(G)$ considering $V$ as a natural $GF(q)$-module
for $Y$. Let $X$ be the group induced by $Y$ on $\Gamma^\infty$.
If $N_{Sym(\Gamma^\infty)}(X) > N_{\Gamma L_n(q)}(X)$, then 
one of the following holds.
\begin{itemize}
\item[(1)] $n = 2$,~ $q \in \{ 5,7,11,23\}$ and $A_4 \leq X $; 
% 
\item[(2)] $n = 2$, ~$q \in \{11,19,29,59\}$ and $X \cong A_5$;
%
\item[(3)] $n = 4$, ~$q = 3$, $O_2(X) \cong 2^4$ and $\ZZ_5 \leq X/O_2(X)  \leq Frob(5:2)$;
%
\item[(4)] $n = 6$, ~$q = 3$ and $X \cong L_2(13)$;
\end{itemize}
\end{coro}

\begin{lemma}\label{order}
Let $q = p^r$ be a power of a prime $p$. Let $n > 2$ or $n = 2$ and $q \not=2,3,4,5,8,9$. Then $\varphi(q^n-1/q-1) > rn$.
\end{lemma}

{\it Proof.}  Let
$r = 1$, $n=2$. Then we claim $\varphi(p+1) > 2$, which is obviously true besides $p = 2,3,$ or $5$. Assume next $q^n = 64$. Then $q = 2,4,$ or $8$.
Hence $q^n-1/q-1 = 63,21,$ or $7$ and $\varphi(q^n-1/q-1) = 36,12$, or $6$.
So the conclusion holds in all these cases. Hence we may assume that there
is some Zsigmondy prime $t$ dividing $q^n-1$. In particular $\varphi(q^n-1/q-1) \geq t-1$. 

Assume by way of contradiction $\varphi(q^n-1/q-1) \leq rn$. Then we have $t-1 \leq rn$. But by the choice of $t$ we have $rn \mid t-1$. This now shows $t-1 =rn$ and $\varphi(q^n-1/q-1) =t-1$. Then we get $q^n-1/q-1 = t$ or $2t$. 
As $t$ is odd, we have an integer $u = (t-1)/2$ and $t \mid p^u+1$. As
$p^{t-1}-1 = t(q-1)$ or $2t(q-1)$, we see that $p^u-1$ divides $q-1$ or $2(q-1)$.
So in any case we get $p^u-1 \mid q-1$, as $2 \mid p^u+1$ in case $p$ odd. Then $u \mid r$. But $2u = nr$. This shows $n = 2$ and $u = r$. We now have
$p^{(t-1)/2} + 1 = t, (p = 2)$ or $2t , p$ odd. Let $p > 3$. Then $p^{(t-1)/2} > 2^{t-1}$. Hence we get $t \geq 2^{t-2}$. This shows $t = 3$. But then $u = f = 1$, and we are done. So we are left with $p = 2$ and $p = 3$.

Let $p = 2$. Then $2^{(t-1)/2} + 1 = t$. So $t \leq 5$. But then $q = 2$ or $4$, which are exceptional cases in the assertion. 
Let $p = 3$, then $3^{(t-1)/2}+1 = 2t$ and we get $t \leq 5$. Then $q = 3$ or $q = 9$ and again we have the exceptional cases.
This proves the lemma.\hfill \qed


\section{On gluings.}

\subsection{The construction.}

Let  $({\cal A}_0, ||_0)$ and  $({\cal A}_2, ||_2)$ be two affine spaces
isomorphic to $A(n,q)$ and let $\Gamma^\infty = \Gamma_i^\infty$  be the
set of parallel classes of $||_i$ ($ i =0,2$). Then Pasini introduced in [Pas2]
the gluing $AS(n,q) \circ_\alpha AS(n,q)$ of the points and lines of
${\cal A}_0$ and of ${\cal A}_2$ via the permutation $\alpha$ of 
$\Gamma^\infty$  as follows:\\
The points are the points of ${\cal A}_0$, the planes the points of
${\cal A}_2$ and the lines the pair of lines $(l_0,l_2)$ with $l_i$ a line 
in ${\cal A}_i$ such that $l_0^{\infty \alpha} = l_2^\infty$.
Any point is incident with any plane and a point $p$ (resp. plane $u$) is incident
with a line $(l_0,l_2)$ iff $p$ (resp. $u$) is incident to $l_0$ (resp. $l_2$)
in ${\cal A}_0$ (resp. ${\cal A}_2$).

Let  $\Gamma = AS(n,q) \circ_\alpha AS(n,q)$ be the gluing of two desargues affine
spaces and let 
 $A_i = Aut({\cal A}_i, ||_i))$,~ $K_i^\infty$ the kernel
of the action of $A_i$ on $||_i$  and $A_i^\infty$
the group induced by $A_i$ on  $\Gamma_i^\infty$ ($ i =0,2$).
Then $A_0^\infty = A_2^\infty = A^\infty$.
Moreover, let $K_i = O_p(K_i^\infty)$ and $Z_i$ be the stabilizer in 
$K_i^\infty$ of the zero-point of ${\cal A}_i$, for $i = 0,2$. Then $K_i \cong E_{q^n}$
and $Z_i \cong \ZZ_{q-1} (i = 0,2)$.

\begin{lemma}\label{auto}
Let $\Gamma = AS(n,q) \circ_\alpha AS(n,q)$. Then $G = Aut(\Gamma) = (K_0^\infty \times K_2^\infty)X$,
where $X = A^\infty \cap (A^\infty)^\alpha$,
$G_0 = (Z_0 \times K_2^\infty ) X, ~G_2 = (Z_2 \times K_0^\infty)X$ and 
$G_1 = ((\la x_1 \ra  \times (\la \alpha (x_1) \ra )(Z_0 \times Z_2)N_X(\la x_1\ra)$ with 
$\la x_1 \ra \leq K_2,~|\la x_1 \ra | = q$.
In particular, $G$ acts flag-transitively on $\Gamma$ iff $G_0/Z_0$
and $G_2/Z_2$ are doubly transitive permutation groups of degree $q^n$.
\end{lemma}
\begin{bew}
See [BHP, Theorem (3.6)].
\end{bew}

In the following we will use the  notation just introduced.
According to [BHP, Theorem (3.9)] we have

\begin{lemma}\label{auto2}
Let $\Gamma = AS(n,q) \circ_\alpha AS(n,q)$ and
 $\Gamma '=AS(n,q) \circ_{\alpha'} AS(n,q)$. Then $\Gamma$ and
$\Gamma'$ are isomorphic iff $\alpha' \in A^\infty \alpha A^\infty$.
\end{lemma}

\noindent
{\bf Remark}
Lemma~\ref{auto2} implies that, if $\Gamma$ is the canonical gluing
$AS(n,q) \circ_\alpha AS(n,q)$ we may assume that $\alpha$ is the identity on $\Gamma^\infty$ and that $\alpha$ induces a $GF(q)$-linear mapping from 
$K_2$ onto $K_0$.



\subsection{Examples.}


Let $H = O_p(H):Y \leq A\Gamma L_n(q)$ such that $O_p(H)$ is
the translation group of $A = A\Gamma L_n(q)$ and such that $Y$ acts
transitively on $O_p(H)^\#$. Let $\Gamma^\infty$ be the set of
$1$-spaces of $O_p(H)$, $A^\infty$ the subgroup  of $Sym(\Gamma^\infty)$ induced
 by $A$ on $\Gamma^\infty$ and let $N$ be the normalizer of $Y^\infty$
in $Sym(\Gamma^\infty)$. Suppose $N > N_{A^\infty}(Y^\infty)$. Then
$\Gamma = AS(n,q) \circ_\alpha AS(n,q)$ is a non-canonical gluing of the two affine
spaces isomorphic to $AS(n,q)$ via the bijection $\alpha \in N\sm{N_{A^\infty}(Y^\infty)}$.
Moreover,
$$Aut(\Gamma) = (K_0^\infty \times K^\infty_2)X,~\mbox{where}~K_i^\infty \cong E_{q^n}\ZZ_{q-1}
~\mbox{and}~~Y^\infty \leq X < A^\infty,$$
$$\mbox{with}~X = A^\infty \cap (A^{\infty})^\alpha.$$
If $C_{A^\infty}(Y^\infty) = 1$, but $C_{Sym(\Gamma^\infty)}(Y^\infty) \neq 1$, then we may chose in particular $\alpha \in $\\$C_{Sym(\Gamma^\infty)}(Y^\infty)^\#$.
The following examples are constructed in this way.
\\

(a) Let $n = 2,~q \in \{5,7,11,23\}$ and let $Y$ be a group listed in [Hu].\\
By the Lemma of Schur $C_{A^\infty}(Y^\infty) = 1$, but by
Lemma~\ref{centrinsym} and Table 1  $C_{Sym(\Gamma^\infty)}(Y^\infty) \neq 1$.
Let $\alpha \in C_{Sym(\Gamma^\infty)}(Y^\infty)^\#$.
According to Table 1 either $X \cong Sym(4)$ or $q = 11$ and $X \cong A_4$.
\\

(b) Let $n = 2,~q \in\{ 11,19,29,59\}$ and let $Y$ be a group listed in [He, Paragraph
5, E1]. Then $Y^\infty \cong A_5$ and $C_{A^\infty}(Y^\infty) = 1$.
By Lemma~\ref{centrinsym} and Table 1 
$C_{Sym(\Gamma^\infty)}(Y^\infty) \neq 1$ and for $\alpha \in C_{Sym(\Gamma^\infty)}(Y^\infty)^\#$ we have $X \cong A_5$.
\\

(c) Let $n = 4,~q = 3$ and let $Y \cong 2^{1+4}\ZZ_5$.\\
Again $C_{P \Gamma L_4(3)}(Y^\infty) = 1$. By Table 1 
$ C_{Sym(\Gamma^\infty)}(Y^\infty)\cong 2^3$
and $X/O_2(X) \cong \ZZ_5$ or $Frob(5:2)$.
\\

(d) Let $n = 6,~q = 3$ and let $Y \cong SL_2(13)$.
Then $C_{A^\infty}(Y^\infty) = 1$ and 
according to Table 1 $N_{Sym(\Gamma^\infty)}(Y^\infty) \cong (L_2(13) \times 2^2)2$. So we may chose $\alpha$ either such that $\alpha$ commutes with
$Y^\infty$ or such that $Y^\infty \la \alpha \ra \cong PGL_2(13)$. 
In both cases $X \cong L_2(13)$.
\\

(e) Let $Y$ be naturally embedded into $\Gamma L_1(q^n)$ and assume that $Y^\infty$
contains a Singer-cycle $T$ of $A^\infty$. Moreover, let 
$n>2$ or $n = 2$ and $q \not\in \{2,3,4,5,8,9\}$.
Recall that $T$ is cyclic of order $(q^n-1)/(q-1)$. Let $q = p^r$.
By Lemma~\ref{order} $|Aut(T)| > r\cdot n$. Therefore
$N_{A^\infty}(T) \cong T:\ZZ_{r\cdot n}$ is a proper subgroup
of $N_{Sym(\Gamma^\infty)}(T) \cong T:Aut(T)$ (see Lemma~\ref{centrinsym}).
Let $\alpha \in N_{Sym(\Gamma^\infty)}(T)\sm{A^\infty}$.
Since $N_{Sym(\Gamma^\infty)}(T)/T$ is abelian,  $X$ is the normalizer
of $T$ in $A^\infty$.
\\

(f) Let $T$ be a Singer-cycle of $A^\infty$. Let $q = p^r $
and set $m = r \cdot n$. Suppose  $gcd(m,(q^n-1)/(q-1)) \neq 1$.\\
Let $s$ be a prime which divides $gcd(m,(q^n-1)/(q-1))$ and let
$U$ be a subgroup of $T$ of order $|T|/s$. Let $t$ be an element of $T$
of order $|T|_s = s^b$ and let $g$ be an element of $N_{A^\infty}(T)$ of order $s$ which fixes a point in $\Gamma^\infty$. 
Set $Y^\infty = U \la tg\ra$. Then  $|Y^\infty| = |T|$.

We claim that $Y^\infty$ acts regularly on $\Gamma^\infty$
(and
thereby the preimage $Y$ of $Y^\infty$ in $\Gamma L_n(q)$ acts
transitively on $O_p(A\Gamma L_n(q))^\#$ ).
If $\la t,g \ra$ is an abelian group, then $tg$ is of order $o(t) = s^b$.
If $\la t,g \ra$ is not an abelian group and if $s$ is odd, then 
$\la t,g \ra$ is isomorphic to $Mod_{s^{b+1}}$ (for the definition
of the modular group $Mod_{s^{b+1}}$ see [As, p. 107])
and therefore $tg$ is
again of order $s^b$. Hence any Sylow $s$-subgroup of $Y^\infty$ is cyclic.
Therefore, any element of order $s$ of $Y^\infty$ is contained in $U$
which implies that $Y^\infty$ acts regularly on $\Gamma^\infty$.
If $\la t,g \ra$ is not an abelian group and $s = 2$, then $g$ and $tg$
are not conjugate. Thus also in this case  $Y^\infty$ acts regularly, as
claimed.

By Lemma~\ref{centrinsym}
$C = C_{Sym(\Gamma^\infty)}(Y^\infty) \cong Y^\infty$ and $N_{Sym(\Gamma^\infty)}(Y^\infty) \leq C Aut(T)$. 
We claim that $C_{A^\infty}(Y^\infty)$ is a proper subgroup of $C$.
If there exists a Zsigmondy 
prime $t$ for $p^m - 1 = q^n - 1$, then according to [HuI, II. (7.3)] $C_{A^\infty}(U) = T$
which implies $C_{A^\infty}(Y^\infty) = C_T(g) < C$.
If there exists no Zsigmondy prime for $p^m - 1$, then $(m,p) = (6,2)$
or $m = 2$ and $p$ is a Mersenne prime [Zsig]. If $(m,p) = (6,2)$, 
then $s = 3$, $U \cong \ZZ_{21}$ and $C_{A^\infty}(Y^\infty) = Z(Y^\infty)
< C$. Now let $m = 2$ and $p$ be a Mersenne prime. Then $m = n = s =2$,
$U \cong \ZZ_{p+1/2}$ and $C_{A^\infty}(Y^\infty) = Z(Y^\infty)
< C$. Thus our claim turns out to be true in all cases and 
therefore,  there exists a non-canonical gluing
via $\alpha \in C\sm{A^\infty}$ such that $Y^\infty \leq X$.

Clearly, the above construction can be generalized in such a way that
 $[Y^\infty :T \cap Y^\infty]$ will not be a prime. 


\section{The Main Theorem.}

\begin{thm}
Let $\Gamma$ be a non-canonical gluing $AS(n,q) \circ_\alpha AS(n,q)$, where $q = p^r$ . Then $\Gamma$ is one of the examples 
listed above, i.e. one of the following holds.
\begin{itemize}
\item[(1)] \begin{itemize}
            \item[(i)] $n = 2$,~ $q \in \{ 5,7,11,23\}$ and $X \cong S_4$;  
             \item[(ii)] $n = 2$,~ $q = 11$ and $X \cong A_4$;
           \end{itemize}
\item[(2)] $n = 2$, ~$q \in \{11,19,29,59\}$ and $X \cong A_5$;
\item[(3)] $n = 4$, ~$q = 3$, $O_2(X) \cong 2^4$ and $\ZZ_5 \leq X/O_2(X)  \leq Frob(5:2)$;
\item[(4)] $n = 6$, ~$q = 3$ and $X \cong L_2(13)$;
\item[(5)] $n>2$ or $n = 2$ and $q \not\in \{2,3,4,5,8,9\}$; $X = N_{\Gamma L_n(q)}(T)$,
with $T$ a Singer cycle of $\Gamma L_n(q)$;
\item[(6)] $gcd(r \cdot n,(q^n-1)/(q-1)) \neq 1$, $X$ is naturally embedded
into $\Gamma L_1(q^n)$, but not of type (5).
\end{itemize}
In cases (1), (2) and (3) we may choose $\alpha$ such that $[X, \alpha ] = 1$.
In case (4) there exists just two examples: one with $[X,\alpha] = 1$ and the 
other with $X\la \alpha \ra \cong PGL_2(13)$.
\end{thm}
\begin{bew}
By Lemma~\ref{auto} $G_0/Z_0  \cong K_2^\infty X$ is a doubly transitive permutation group of
degree $q^n$. Therefore $X$ and $X^\alpha$ are subgroups of $A^\infty \cong P\Gamma
L_n(q)$ which act both transitively on $\Gamma^\infty$ and the preimages 
$\hat{X}$ and $\hat{X}^\alpha$ of $X$ and $X^\alpha$ in $\Gamma L_n(q)$ act
transitively on $O_p(G_2)^\#$.

Notice, since $\Gamma$ is a non-canonical gluing  $n$ is at least $2$.
If moreover, $\alpha$ normalizes $X$, then $N_{Sym(\Gamma^\infty)}(X)$ is not contained in 
$N_{A^\infty }(X)$. Suppose that $\hat{X}$ is naturally embedded into
$\Gamma L_m(t)$. Then $m$ divides $n$ and $q^n = t^m$.

Suppose first $m \geq 2$. According to Lemma~\ref{conj} $\hat{X}$
and $\hat{X}^\alpha$ are conjugate in $\Gamma L_n(q)$ which implies 
$X = X^{\alpha g}$ for some $g \in A^\infty$. Hence by Lemma~\ref{auto2}
we may assume $\alpha \in N_{Sym(\Gamma^\infty)}(X)$. Since $\Gamma$
is a non-canonical gluing Lemma~\ref{auto2} yields $N_{A^\infty}(X)$ is a 
proper subgroup of $N_{Sym(\Gamma^\infty)}(X)$. Now Lemma~\ref{normser} and
Corollary~\ref{coro3} imply (1) -- (4) of the Theorem.

Now suppose that $m = 1$. We distinguish the cases that $X$
contains a Singer-cycle $T$ or not.
First suppose $T \leq X$. Then $X \leq N_{A^\infty}(T)$. As $X^\alpha$ has also
a Singer-cycle $\widetilde{T}$ and is also contained in  $N_{A^\infty}(\widetilde{T})$,
$X$ and $X^\alpha$ are conjugate in $A^\infty$. Thus we may
assume $\alpha \in N_{Sym(\Gamma^\infty)}(X)$.
Since $N_{Sym(\Gamma^\infty)}(T)$ is not
contained in $N_{A^\infty}(T)$ Lemma~\ref{order} implies $n > 2$ or $n = 2$ 
and $q \not\in \{2,3,4,5,8,9\}$. 
Hence (5) of the theorem holds.
Finally suppose that $X$ does not contain a Singer-cycle.
Since $X \leq \Gamma L_1(q^n)$ acts transitively on $\Gamma^\infty$, we obtain
that $gcd(an,(q^n-1)/(q-1)) \neq 1$ and that  $\Gamma$ is as in (6). 

Let $\Gamma$ be an example of type (1), (2) or (3). 
In all these examples we have $N_{Sym(\Gamma^\infty)}(X) =
C_{Sym(\Gamma^\infty)}(X) N_{A^\infty}(X)$. Therefore the last assertion
follows with Lemma~\ref{auto2}.
This proves the theorem.
\end{bew}

\begin{coro}
Let $\Gamma = AS(n,q) \circ_\alpha AS(n,q)$ be a non-canonical gluing
not of type (4),(5) or (6) of Theorem~\ref{mthm}. Then we may choose $\alpha$ such that $[X, \alpha ] = 1$.
\hfill \qed
\end{coro}

\section{Exceptional coverings}
In this chapter we are going to describe the universal covers of the non-canonical  
gluings obtained in the previous chapter. Here we just determine those which are not related to the affine operating group. The question for these groups is still open. 
\medskip

The basic idea reads as follows. Let $G$ be the group of the universal covering. Then $G$ is generated by the three parabolic subgroups $G_0$, $G_1$, $G_2$, where $G_0 = (Z_0 \times K_2^\infty) X$ and $G_2 = (Z_2 \times 
K_0^\infty) X^\alpha$. As has been seen in (4.2) we may choose 
$\alpha$ as a permutation inside of $\Sigma_n$ which centralizes $X$ besides of the case $X \cong SL_2(13)$. From the structure of $G_1 = \langle \langle x_1 \rangle, \alpha (\langle x_1 \rangle) \rangle X_1$, where $\langle x_1 \rangle$ is a 1-space in $K_2$, we see that $[\langle x_1 \rangle,  \alpha (\langle x_1 \rangle)] = 1$. Now $G = \langle K_0^\infty, K_2^\infty, X \rangle$, where the action of $X$ on $K_0^\infty$ and $K_2^\infty$ are known. The only additional relation we have is that there is some bijective mapping $\alpha$ from  the 1-spaces of $K_2$ onto the 1-spaces of $K_0$ which commutes with the action of $X$ and satisfies $[U,\alpha (U)] = 1$ for any 1-space $U$ in $K_2$. Hence we are going to determine precisely those groups.

\begin{lemma} \label{commutator}

Let $p$ be a prime, $K_0\cong K_2 \cong E_{q^2}, q = p^r,$ be subgroups of a group
$G$, $G = \langle K_0,K_2 \rangle$.
 Suppose that $K_0$ and $K_2$ admit $GF(q)$-vector space structures such that there is an  $GF(q)$-linear isomorphism $f:K_2 \rightarrow K_0$. Moreover,  suppose that $[U, f(U)] = 1$ for any $1$-dimensional subspace $U$ of $K_2$.   
Then $G^\prime \leq Z(G)$ and $|G^\prime| \leq q$.

\end{lemma}

\begin{bew} Let $U$ and $V$ be two different subspaces of $K_2$. Then
$K_2 = U\times V$. Observe that $[xy, f(x) f(y)] = 1 ~(x, y \in K_2)$
implies $[x,f(y)] = [y, f(x)]^{-1}$. Let $u_1, u_2 \in U$ and $v_1, v_2 \in
V$. As $[v_1, f(v_2)] = 1$, we see $[u_1,f(v_1), v_2] = 1$. As $[u_1, f(u_2) ] =1$,
we see $[v_1, f(u_1),u_2] = 1$. So $[u_1, f(v_1), K_2] = 1$. The same
argument also shows $[u_1,f(v_1),K_0] = 1$. So $[u_1, f(v_1)] \in Z(G)$
and $G^\prime \leq Z(G)$.  Furthermore $ [u_1,f(v_2)]^p = [u_1^p,f(v_2)] = 1$.

Let $\lambda \in GF(q)$ such that $\langle \lambda \rangle = GF(q)^\#$
and let $x_1 \in U^\#$ and $y_1 \in V^\#$. Set $x_{i+1} = x_1^{\lambda^i}$
and $y_{i+1} = y_1^{\lambda^i}$ for $i = 1, \ldots , r-1$. Then 
$\{x_1, \ldots , x_r\}$ and $\{y_1, \ldots , y_r\}$ are a basis of
the $GF(p)$-vector spaces $U$ and $V$, respectively.

Set $D = \langle [x_1, f(y_i)]~|~i = 1, \ldots, r\rangle $.
We claim that $G^\prime = D$. Since $G^\prime \leq Z(G)$, we have
$G^\prime = [U,f(V)]$. The fact that $x_iy_i = (x_1y_1)^{\lambda^{i-1}}$
implies $[x_1y_1, f(x_i)f(y_i)] = 1 ~(i = 1, \ldots, r$). Therefore
$[y_1, f(x_i)] = [x_1, f(y_i)]^{-1}$ and, as $[x_i,f(y_1)] = 
[y_1, f(x_i)]^{-1}$, it follows $[x_i, f(y_1)] \in D$ ~ ($i = 1, \ldots, r$).
Now suppose that $[x_i, f(y_j)] \in D$ for $i = 1, \ldots, r$ and 
$j \leq k$ and $[x_i, f(y_{k+1})] \in D$ for $i = 1, \ldots, l$. Since 
$(x_l  y_k)^{\lambda} =  x_l^{\lambda}  y_{k+1}$ we obtain
$[x_ly_k, f(x_l^\lambda )f(y_{k+1})] = 1$ and $[x_l, f(y_{k+1})] =
[y_k, f( x_l^\lambda)]^{-1}$.

If $l < r$, then $x_l^\lambda  = x_{l+1}$ and $[x_l, f(y_{k+1})]= 
 [y_k, f(x_{l+1})]^{-1} = [x_{l+1}, f(y_k)] \in D$. 

If $l = r$, then $ x_r^\lambda = \prod_{i = 1}^r x_i^{a_i}, ~a_i \in GF(p)$, and
$[x_r, f(y_{k+1})] = \prod_{i = 1}^r ([y_k, f(x_i)]^{a_i})^{-1} = \prod_{i= 1}^r 
[x_i, f(y_k)]^{a_i} \in D$. This proves $G^\prime = D$. So $|G^\prime| \leq q$. 
\end{bew}

Using the same method as in the first paragraph of the previous proof we
obtain the following more general result in the  case $q = p$.

\begin{lemma} \label{commutator2}

Let $p$ be a prime and $K_0 \cong K_2 \cong E_{p^2}$ be subgroups of a group $G$, $G = \langle K_0, K_2 \rangle$ and $f$ be an isomorphism from $K_2$ onto $K_0$. Let $K_2 = \langle x_1, x_2 \rangle$ and suppose $[x_1,f(x_1)] = [x_2,f(x_2)] = [x_1x_2,f(x_1)f(x_2)] = 1$, then $|G^\prime| \mid p$.

\end{lemma}

\begin{lemma} \label{fextra}

Let $p$ be a prime and $K_0 \cong K_2 \cong E_{p^2}$ be subgroups of a group $G$, $G = \langle K_0, K_2 \rangle$, where $G$ is extraspecial of order $p^5$. Then there is some isomorphism $f : K_2 \rightarrow K_0$ with $[x,f(x)] = 1$ for all $x \in K_2$.

\end{lemma}

\begin{bew} Let $K_2 = \langle x_1,x_2 \rangle$ and $K_2 = \langle y_1,y_2 \rangle$ with $[x_1,y_1] = [x_2,y_2] = z$ and $[x_1,y_2] = [x_2,y_1] = 1$, $\langle z \rangle = Z(G)$. Now define $f$ as follows $f(x_1^ix_2^j) = y_2^iy_1^{-j}$, $i,j \in \{0, \ldots , p-1\}$. Obviously $f$ is an isomorphism. Using the fact $G^\prime \leq Z(G)$ we now have

$\begin{array}{ll}
& [x_1^ix_2^j,f(x_1^ix_2^j)] = [x_1^i,f(x_1^ix_2^j)][[x_2^j,f(x_1^ix_2^j)] \\
= & [x_1,f(x_1^ix_2^j)]^i[x_2,f(x_1^ix_2^j)]^j \\
= & [x_1,f(x_1)]^{i^2}[x_1,f(x_2)]^{ji}[x_2,f(x_1)]^{ij}[x_2,f(x_2)]^{j^2} \\
= & [x_1,y_2]^{i^2}[x_1,y_1]^{-ij}[x_2,y_2]^{ij}[x_2,y_1]^{-j^2} = 1.
\end{array}$

\end{bew}
 

\begin{lemma} \label{aextra}

Let $p$ be a prime and $K_0 \cong K_2 \cong E_{p^2}$ be subgroups of a group $G$, $G = \langle K_0, K_2 \rangle$, where $G$ is extraspecial of order $p^5$. Let $\alpha$ be a bijective mapping from the 1-spaces of $K_2$ onto the 1-spaces of $K_0$ such that $[U,\alpha (U)] = 1$ for each 1-space $U$ of $K_2$. Then there is an isomorphism $f :  K_2 \rightarrow K_0$ with $[x,f(x)] = 1$ for $x \in K_2$, such that $f(U) = \alpha(U)$ for all 1-spaces $U$ of $K_2$.

\end{lemma}

\begin{bew} Let $K_2 = \langle x_1,x_2 \rangle$ and $K_0 = \langle y_1,y_2 \rangle$ with $[x_1,y_1] = [x_2,y_2] = z$ and $[x_1,y_2] = [x_2,y_1] = 1$, $\langle z \rangle = Z(G)$. Then for each $x \in K_2$ we see that $\alpha (\langle x \rangle)$ is uniquely determined. So $\alpha (\langle x \rangle) = f(\langle x \rangle)$, where $f$ is the mapping from \ref{fextra}.
\end{bew}

\begin{proposition} \label{p2cover} 
Let $\Gamma$ be a non-canonical gluing $AS(2,p) \circ_\alpha AS(2,p)$, $p$ an odd prime. Then $\Gamma$ is simply connected.
\end{proposition}

\begin{bew} By (3.1) we have $G_0 = Z_0K_2^\infty X$, $G_2 = Z_2K_0^\infty X$ and $G_1 = Y(Z_0 \times Z_2)X_1$, where $X_1 \leq X$ and stabilizes in the elementary abelian group $K_2$ a point $\langle x_1 \rangle$ and in $K_0$ a point $\alpha (\langle x_1 \rangle)$. Hence $Y = \langle \langle x_1 \rangle , \alpha (\langle x_1 \rangle) \rangle$ is elementary abelian of order $p^2$.  So we get $[\langle x \rangle , \alpha (\langle x \rangle)] = 1$ for all $x \in K_2$. 

Let $K_2 = \langle x_1,x_2 \rangle$ and $y_1,y_2 \in K_2$ with $\alpha (\langle x_i \rangle) = \langle y_i \rangle$, $i = 1,2$. Let $\alpha( \langle x_1x_2 \rangle) = \langle y_1^ay_2^b \rangle$. Define the linear mapping $f : K_2 \rightarrow K_0$ by $f(x_1) = y_1^a, f(x_2) = y_2^b$. Then by \ref{commutator2} we
have $|\langle K_0, K_2 \rangle^\prime| \mid p$. If $\langle K_0, K_2 \rangle$ is nonabelian, then it is extraspecial of order $p^5$. Now by \ref{aextra} there is some linear mapping $g : K_2 \rightarrow K_0$ with $g(\langle x \rangle) = \alpha (\langle x \rangle)$ for all $x\in K_2$. But this means that $g = \alpha$ is in $\Gamma L_2(p) \leq \Sigma_{p+1}$, which contradicts (4.2) and example (3.2)(a),(b). So we have $[K_0,K_2] = 1$ and $\Gamma$ is simply connected. 
\end{bew}

\begin{proposition} \label{81cover}

 Let $\Gamma$ be a non-canonical gluing $AS(4,3) \circ_\alpha AS(4,3)$, then $\Gamma$ is simply connected.
\end{proposition}

\begin{bew} By (3.1) $G_0$ contains a subgroup $H_0 = K_2Y$, where $K_2$ is elementary abelian of order 81 and $Y$ is an extention of an extraspecial group of order 32 by $Z_5$. Furthermore $G_2$ contains $H_2 = K_0Y$, where $K_0$ is elementary abelian of order 81 too. The action of $Y$ on $K_2$ is described in \cite{Hu}. Let $\{x_1,x_2,x_3,x_4\}$ be the basis used there and $\{y_1,y_2,y_3,y_4\}$ the corresponding basis of $K_0$. We have $[\langle x_1 \rangle, \alpha(\langle x_1 \rangle)] = 1$. As $\alpha(\langle x_1 \rangle)$ has to be fixed by the one point stabilizer of $Y$ on $K_0$ and $\alpha$ centralizes $Y$ by (4.2) we see that $\alpha (\langle x_1 \rangle) \subseteq \langle y_1, y_3 \rangle$. As $\alpha$ is not the identity we see that $\alpha (\langle x_1 \rangle) \in \{\langle  y_3 \rangle, \langle y_1y_3 \rangle, \langle y_1^{-1}y_3 \rangle \}$. Set $Y_1 = G_1 \cap Y$ and $H_1 = \langle \langle x_1 \rangle, \alpha (\langle x_1 \rangle) \rangle Y_1$. Let $H = \langle H_1,H_2,H_3 \rangle$. Using the
 relations from $H_1$ and $H_3$ and the additional relation $[x_1, y] = 1$, for some $y \in \{y_3, y_1y_3, y_1y_3^{-1}\}$, we get in any case by the Todd-Coxeter algorithm that $H = (K_0 \times K_2)Y$. This shows that $G = K_0^\infty K_2^\infty X$ and so $\Gamma$ is simply connected. 
\end{bew}

\begin{proposition} \label{729cover}

   Let $\Gamma$ be a non-canonical gluing $AS(6,3) \circ_\alpha AS(6,3)$. Then one of the following holds

(i) If $\alpha$ induces an outer automorphism on $X$, then $\Gamma$ is simply connected.

(ii) If $[X,\alpha] = 1$, then the universal cover of $\Gamma$ is a 3-fold covering.
\end{proposition}

\begin{bew} Let $H_0 = K_2X$, where $K_2$ is elementary abelian of order 729 and $X \cong SL_2(13)$. We first determine the structure of $H_0$. According to \cite{Ples} the following is a presentation of $X$ :\\
\\
$\langle a,x,y,r~|~x^{13} = r^3 = 1, x^r = x^3, y^4 = 1, x^y = x^{-1}, [y,r] = 1, a^2 = y^2, y^a = y^{-1},\\ r^a = r^{-1},
 (ax)^3 = 1, x^6ax^{-1}axax^7a = (ry)^{-1}\rangle$
\\
\\
We are going to calculate the action of $X$ on $K_2$. First of all, $\langle x,r \rangle$ decomposes $K_2$ into two 3-dimensional modules. As $GL(3,3)$ does not contain a subgroup of order $13\cdot 12$, we see that $y$ has to interchange these two modules. So we get the following matrices according to a basis $\{x_1,x_2,x_3,x_4,x_5,x_6\}$ of $K_2$\\
\\
$$\begin{array}[t]{c}
x\rightarrow\\
\end{array}\left( \begin{array}{rrrrrr}
0&1&0&0&0&0\\
0&0&1&0&0&0\\
1&-1&-1&0&0&0\\
0&0&0&1&1&1\\
0&0&0&1&0&0\\
0&0&0&0&1&0
\end{array}\right), \begin{array}[t]{c}
r\rightarrow\\
\end{array}\left(\begin{array}{rrrrrr}
1&0&0&0&0&0\\
1&-1&-1&0&0&0\\
-1&1&0&0&0&0\\
0&0&0&1&0&0\\
0&0&0&1&-1&-1\\
0&0&0&-1&1&0
\end{array}\right)$$

$$\begin{array}[t]{c}
y\rightarrow\\
\end{array}\left(\begin{array}{rrrrrr}
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1\\
-1&0&0&0&0&0\\
0&-1&0&0&0&0\\
0&0&-1&0&0&0
\end{array}\right)$$

Now we have that $a$ has to act on the fixed points of $r$ i.e. $\langle x_1,x_4 \rangle$. So $\langle y,a \rangle$ acts as a quaternion group on $\langle x_1,x_4 \rangle$ and so it is uniquely determined. Using the relations above we see $x_1^a = x_1^{-1}x_4$ and $x_4^a = x_1x_4$. This then yields
$$\begin{array}[t]{c}
a\rightarrow\\
\end{array}
\left(\begin{array}{rrrrrr}
-1&0&0&1&0&0\\
1&-1&0&1&1&0\\
1&1&1&1&-1&-1\\
1&0&0&1&0&0\\
1&1&0&-1&1&0\\
1&-1&-1&-1&-1&-1
\end{array}\right)$$

So let now $K_0 = \langle y_1,y_2,y_3,y_4,y_5,y_6 \rangle$. If $[X,\alpha] = 1$ then $X$ acts by the same matrices on $K_2$ as on $K_0$.  Let $X_1$ be the stabilizer of $\langle x_1 \rangle$ in $X$. Then $\alpha (\langle x_1 \rangle)$  is contained in the set of fixed points of $X_1$ on $K_0$ and so is some 1-space in $\langle y_1,y_4 \rangle$. Hence we get $\alpha (\langle x_1 \rangle) \in \{\langle y_4 \rangle, \langle y_1^{-1}y_4 \rangle , \langle y_1y_4 \rangle \}$. Let $H = \langle K_0X,K_2X \rangle$ with the defining relations of $K_0X$ and $K_2X$ and the additional relation $[x_1,t] = 1$ for some $t \in \{y_4, y_1^{-1}y_4, y_1y_4 \}$. Then application of the Todd-Coxeter algorithm in all three cases shows that $H = \langle K_0,K_2 \rangle X$ where $\langle K_0, K_2 \rangle$ is extraspecial of order $3^{13}$, the assertion.

So assume now that $\alpha$ induces an outer automorphism on $X$. Then $x$ acts as $x^8$ and $a$ as $ay$ on $K_0$, while the action of $r,y$ is unchanged. Hence the fixed points of $X_1$ on $K_0$ are the same as before. So we now choose $t \in \{y_1, y_4, y_1^{-1}y_4, y_1y_4 \}$ and add the relation $[x_1,t]$. Then we apply the Todd-Coxeter algorithm to the group $\langle K_0,K_2, X \rangle$ again. This then shows that $\langle K_0, K_2 \rangle$ is elementary abelian in all four cases, hence $\Gamma$ is simply connected. 
\end{bew}

\section{Canonical coverings}

In this section we are going to determine the universal coverings of the canonical gluings.

\begin{lemma} \label{comgr}
Let $G = \langle K_0, K_2 \rangle$ with elementary abelian $p$-groups $K_0, K_2$ of order at most $p^3$, $p$ an odd prime. Let $f : K_2 \rightarrow K_0$ be an isomorphism with $[x,f(x)] = 1$ for all $x \in K_2$. Then $G^\prime \leq Z(G)$ and $|G^\prime| \leq |K_2|$.
\end{lemma}

\begin{bew}
 The assertion is trivial for $|K_2| = p$. Let $|K_2| = p^2$. Then the assertion follows from \ref{commutator}. So we may assume that $|K_2| = p^3$. Set $K_2 = \langle x_1,x_2,x_3 \rangle$ and $K_0 = \langle y_1,y_2,y_3 \rangle$, where $f(x_i) = y_i$, $i = 1,2,3$. Let $X$ be the free group in the generators $x_1,x_2,x_3,y_1,y_2,y_3$. We extend $f$ to a mapping  $g$ of $\{x_1,x_2,x_3,y_1,y_2,y_3\}$ by setting $g(x_i) = y_i$ and $g(y_i) = x_i$, $i = 1,2,3$. As $X$ is free $g$ can be extended to an automorphism of $X$ which we will denote by $g$ too. Let $N = \langle x_i^p, y_i^p, [x_i,x_j], [y_i,y_j], i,j = 1,2,3, [x,f(x)], [f(x),x], x \in K_2 \rangle$.  Then we see that $N$ is invariant under $g$. So $g$ is an automorphism of $H = X/N$. As $G$ is a factorgroup of $H$ it is enough to prove the assertion for $H$. Hence we may assume that $g$ is an automorphism of $G$. By definition we have $g^2 = 1$. 

\medskip

As $[ab,g(a)g(b)] = 1$, we see $[a,g(b)] = [b,g(a)]^{-1}$ for all $a,b \in K_2$. Now $g([b,g(a)]) = [g(b),a] = [b,g(a)]$ and so $[G^\prime,g] = 1$. Set $L = \langle x_1g(x_1)^{-1}, x_2g(x_2)^{-1}, x_3g(x_3)^{-1} \rangle$. Then $L \leq [G,g]$. Let $a \in G$ with $g(a) = a^{-1}$. Let $x \in G^\prime$. Then $[x,a] = g([x,a]) = [x,g(a)] = [x,a^{-1}]$. This implies $x^{-1}a^{-1}xa = x^{-1}axa^{-1}$ or $[a^2,x] = 1$. If $a$ is of odd order we get $[a,x] = 1$. Now choose for $a$ one of the elements $x_ig(x_i)^{-1}$. Then $o(a) = p$ is odd and so we have $[a,G^\prime] = 1$. In particular $[L,G^\prime] = 1$. This shows 

\begin{itemize}
\item[(1)]~~~~~~~~$L^\prime \leq Z(L)$.  
\end{itemize}
We have
\begin{itemize}
\item[(2)] $[x_1g(x_1)^{-1},x_2g(x_2)^{-1}] = [x_1,x_2g(x_2)^{-1}]^{g(x_1)^{-1}}[g(x_1)^{-1},x_2g(x_2)^{-1}]$\\ $ = [x_1,g(x_2)^{-1}][g(x_1)^{-1},x_2].$
\end{itemize}
By \ref{commutator} we have $[g(x_i),x_j]^{g(x_i)} = [g(x_i),x_j]$ for all $i,j$. So we get
\begin{itemize}
\item[(3)] $[g(x_i)^n,x_j] = [g(x_i),x_j]^n$ for all $n \in \Z$ and all $i,j$.
\end{itemize}
This now shows
$$[x_1,g(x_2)^{-1}][g(x_1)^{-1},x_2] = [x_1,g(x_2)]^{-1}[g(x_1),x_2]^{-1} = [x_1,g(x_2)]^{-2}.$$
So (1) - (3) imply
\begin{itemize}
\item[(4)] $L^\prime = \langle [x_i,g(x_j)]^{-2} ~|~ i,j = 1,2,3 \rangle$
\end{itemize}
By \ref{commutator} we have that $o([x_i,g(x_j)]) \mid p$. Hence we get
\begin{itemize}
\item[(5)] $L^\prime = \langle [g(x_1),x_2], [g(x_1), x_3], [g(x_2),x_3] \rangle$.
\end{itemize}
In particular, we have $|L^\prime| \leq p^3$.

It remains to show that $L^\prime = G^\prime \leq Z(G)$.  Obviously we have $\langle x_jg(x_j)^{-1}, x_k \rangle\leq$ $\langle x_j,x_k, g(x_j), g(x_k) \rangle = Y$. By \ref{commutator} we see that $Y^\prime = \langle [x_j,g(x_k)] \rangle$. So by (5) $L$ is normal in $G$. We have that $\langle x_1, x_3 \rangle$ acts on $\langle [g(x_1),x_2], [g(x_3),x_2] \rangle = C_{L^\prime}(x_2)$. Hence there are $i,j$ with $x_1^ix_3^j \not= 1$ and $[x_1^ix_3^j, C_{L^\prime}(x_2)] = 1$. This shows that $[g(x_1),x_2,x_3^j] = 1 = [g(x_3),x_2,x_1^i]$ by \ref{commutator}. We may assume $[g(x_1),x_2,x_3] = 1$. By the Witt identity we have $1 = [g(x_1), x_2,x_3]^{x_2^{-1}}[x_3,g(x_1),x_2]^{g(x_1)^{-1}}$ and so $[g(x_1),x_3,x_2] = 1$. Also we get $1 = [g(x_2),x_3,x_1]^{x_3^{-1}}[x_1,g(x_2),x_3]^{g(x_2)^{-1}}$. As $[x_1,g(x_2)] = [x_2,g(x_1)]^{-1}$ and $[[x_2,g(x_1)]^{-1},x_3] = 1$, we get $[g(x_2),x_3,x_1] = 1$. Hence we have $[K_2,L^\prime] = 1$. As $[g,L^\prime] = 1$, we now get $[G,L^\prime] = 1$. This implies $G^\prime = L^\prime \leq Z(G)$. 
\end{bew}

\begin{lemma} \label{pcover1}

Let $\Gamma$ be a canonical gluing $AS(n,q) \circ_\alpha AS(n,q)$, $q = p^r$,~ $p$ an odd
prime,  with group $G$. Let $\tilde{G}$ be the group belonging to the universal covering. Then $G = \tilde{G}/N$, where $N$ is an elementary abelian $p$-group of order at most $q^{{n \choose 2}}$.

\end{lemma}

\begin{bew}
 We have $\tilde{G_0} = (Z_0 \times K_2^\infty) X$, $\tilde{G_2} = (Z_2 \times K_0^\infty) X$ and $\tilde{G_1} = Y(Z_0 \times Z_2)X_1$, where $K_0 = O_p(K_0^\infty)$ and $K_2 = O_p(K_2^\infty)$ are elementary abelian of order $q^n$. Further $|Y \cap K_0| = |K_2 \cap Y| = q$, and $Y = (K_0 \cap Y)(K_2 \cap Y)$ is elementary abelian of order $q^2$. 
Finally there is an $GF(q)$-linear isomorphism $\alpha : K_2 \rightarrow K_0$ which commutes with the action of $X$ and maps $K_0 \cap Y$ onto $K_2 \cap Y$. Hence we have $[U,\alpha (U)] = 1$ for every $1$-dimensional $GF(q)$-subspace $U$ of $K_2$.

\medskip

Set $H = \langle K_0 ,K_2 \rangle$. Let $u,v,w \in K_2$. Then $[u,\alpha(v),w]
\in \langle u,v,w, \alpha(u), \alpha(v), \alpha(w) \rangle$. Therefore
Lemma~\ref{comgr} implies $[u,\alpha(v),w] = 1$, so $[u, \alpha(v),K_2] = 1$
for all $u,v \in K_2$. Using the same argument we also obtain
$[u, \alpha(v),K_0] = 1$ for all $u,v \in K_2$.
This shows $H^\prime \leq Z(H)$. In particular,
$H^\prime$ is elementary abelian and $H^\prime = \langle [u,\alpha(v)]~|~u,v \in K_2
\rangle$. 

Let $U_1, \ldots , U_n$ be $1$-dimensional $GF(q)$-subspaces of $K_2$ with
$K_2 = U_1 \times \ldots \times U_n$. Since $[U_i,\alpha(U_j)] \leq
\langle U_i, U_j, \alpha(U_i), \alpha(U_j) \rangle$ ($1 \leq i,j \leq n$),
Lemma~\ref{commutator} implies $[U_i,\alpha(U_j)] = [U_j,\alpha(U_i)]$
is a group of order at most $q$ ($1 \leq i,j \leq n$). Hence $H^\prime = \langle [U_i,\alpha(U_j)]
~|~i < j \leq n \rangle$ is of order at most $q^{{n \choose 2}}$. As $\tilde{G} = H(Z_0 \times Z_2)X$, the assertion follows. 
\end{bew}

\newpage

{\LARGE
\begin{proposition} \label{pcover}
Let $\Gamma$ be a canonical gluing $AS(n,q) \circ_\alpha AS(n,q)$, $q = p^r, ~p$ an odd prime,  with group $G$. Let $\tilde{G}$ be the group belonging to the universal covering. Then $G = \tilde{G}/N$, where $N$ is an elementary abelian $p$-group of order $q^{{n \choose 2}}$.

\end{proposition}

\begin{bew}
Let $W$ be an $(n+1)$-dimensional vectorspace over $GF(q)$ with basis
$\{e_1, \ldots , e_{n+1} \}$ and let $V = \langle e_1, \ldots , e_n \rangle$.
Let $S = SL(W) \cong SL_{n+1}(q)$. Then $N_S(V) = C_S(V)N_S(\langle
e_{n+1} \rangle )$ with $K_2 = C_S(V)$ an elementary abelian group
of order $q^n$ and $X = N_S(\langle e_{n+1} \rangle ) \cong GL_n(q)$.
Set $U = W \wedge W$ and $N = V \wedge V$. Then $N$ is elementary abelian 
of order $ q^{{n \choose 2}}$ and $U/N \cong \langle e_i \wedge e_{n+1}~|~i = 1, \ldots, n\rangle$ elementary abelian of order $q^n$.

 Set
$K_0 = \langle e_i \wedge e_{n+1}~|~i = 1, \ldots , n \rangle$.
Let $H = U:K_2X$. Then $H$ is generated by $H_0 = K_2X$ and $H_2 = K_0X$.

Let $\alpha$ be the isomorphism from $K_2$ onto $K_0$ with $\alpha(a)
= [a, e_{n+1}] \wedge e_{n+1}$ for all $a \in K_2$. Then $\alpha$ is
an $GF(q)X$-isomorphism. Observe that $[U, \alpha(U)] = 1$ for all
$1$-dimensional subspaces $U$ of $K_2$. Set $H_1 = (T \times \alpha(T))
N_X(T)$, where $T$ is a $GF(q)$-subspace of $K_2$. Then  $\{H_0,H_1,H_2\}$ generates a cover of $\Gamma$. By \ref{pcover1} this is the universal cover of $\Gamma$. 
\end{bew}
}

\newpage

Now we want to determine the universal cover of the canonical gluing
assuming that $q$ is even.

\begin{lemma}\label{2cover2}
Let $\Gamma$ be a canonical gluing $AS(n,q) \circ_\alpha AS(n,q),~q = 2^r$.
Then the universal cover $\overline{\Gamma }$ of $\Gamma$ has at most
$q^{2^n-1}$ points.
\end{lemma}
\begin{bew}
Since $\Gamma$ is a canonical gluing, the subgroup $G = (K_2 \times K_0)X
$ of $Aut(\Gamma)$ acts flag-transitively on $\Gamma$, where $X \cong GL_n(q),~
K_0 \cong K_2 \cong E_{q^n}$ are isomorphic $GF(q)X$-modules with isomorphism
$\alpha:K_2 \rightarrow K_0$. We may
choose notation such that $G_0 = K_2X,~G_2 = K_0X$ and
$G_1 = (\langle u \rangle _{GF(q)} \times \langle \alpha(u) \rangle _{GF(q)})
N_X(\langle u \rangle _{GF(q)})$, where $u \in K_2$ and where $\langle u \rangle _{GF(q)}$
is the $GF(q)$-span of $u$ in $K_2$.
Then $G_0 \cap G_1 = \langle u \rangle _{GF(q)}N_X(\langle u \rangle _{GF(q)})$,
$G_0 \cap G_2 = X$, $G_1 \cap G_2 = \langle \alpha( u)\rangle _{GF(q)}N_X(\langle u \rangle _{GF(q)})$ and $B = N_X(\langle u \rangle _{GF(q)})$. 

The universal cover of $\Gamma$  is the group geometry $\overline{\Gamma} =
\Gamma(\overline{G}, (G_0,G_1, G_2))$, with $\overline{G}$ the universal completion
of the amalgam $(G_i, G_i \cap G_j, B~|~ i,j = 0,1,2)$.

Since $K_0$ and $K_2$ are $GF(q)X$-modules, we may also read them as $GF(2)X$-modules.

Consider the following amalgam:
\[H_0 = G_0,~H_2 = G_2,~H_1 = (\langle u \rangle _{GF(2)} \times \langle \alpha(u) \rangle _{GF(2)})
N_X(\langle u \rangle _{GF(2)})\] with
\[H_0 \cap H_1 = \langle u \rangle _{GF(2)}N_X(\langle u \rangle _{GF(2)}),~
H_0 \cap H_2 = X,~ H_1 \cap H_2 = \langle \alpha( u)\rangle _{GF(2)}N_X(\langle u \rangle _{GF(2)})\]
 and \[B = N_X(\langle u \rangle _{GF(2)}).\]
For $H = \langle H_0,H_1,H_2\rangle = G$ the group geometry $\widetilde{\Gamma}=
\Gamma (H,(H_0,H_1,H_2))$ is a canonical gluing $AS(nr,2) \circ_\alpha AS(nr,2)$
with flag-transitive group $H= G$. 

We may lift $H$ to a flag-transitive group of automorphism $\overline{H}$ of the 
universal cover of $\widetilde{\Gamma}$. According to [BMP, Corollary 9] $\overline{H}$
contains a normal subgroup $N$, $N \cong \sum_{i=1}^{rn}\Lambda^i(K_2)$, and $\overline{H} =
N:H_2 = N:G_2$.

Any relation which holds in the amalgam $(H_i, ~H_i \cap H_j,~H_0 \cap H_1 \cap H_2~|~
i,j = 0,1,2)$ also holds in the amalgam $(G_i, G_i \cap G_j, B~|~ i,j = 0,1,2)$.
Therefore $\overline{G}$ is a factor group of $\overline{H}$, so $\overline{G} =
(N:G_2)/M$ for a normal subgroup $M$ of $\overline{H}$.

Moreover, according to  [BMP] $[K_2,K_0] = N^\prime$ and $N/N^\prime \cong 
\Lambda (K_2) \cong K_2$. Hence, for $v,w \in K_2$ we have $[v, \alpha(w)] = v \wedge
w ~\mbox{mod}~ M^{\prime \prime}$, i.e. $[v, \alpha(w)] + v \wedge
w \in M^{\prime \prime}$. In $\overline{G}$ for any $v \in K_2$ we have $[v,\alpha(\lambda v)] = 1$. This
implies $v \wedge \lambda v = 0$ for all $v \in K_2$.
Therefore $(v + w)\wedge \lambda(v+ w) = 0$ for all $v , w \in K_2$. But this 
implies $0 = v \wedge \lambda w + w \wedge \lambda v$, so
$v \wedge \lambda w =  w \wedge \lambda v = \lambda v \wedge w$.
Thus $NM/M$ becomes an $GF(q)$-module for $G_2$. In particular, $\wedge_{i = 1}^k v_i = 0,
~v_i \in K_2$, whenever $v_j$ and $v_l$ are linear dependant over $GF(q)$ for 
some $1 \leq j < l \leq k$. Notice, if $v_k$ lies in the $GF(q)$-span of
$v_1, \ldots , v_{k-1}$, then $\wedge_{i = 1}^k  v_i = 0$, as well.

In order to determine the order of $NM/M$ we count for any $k$, $1 \leq k \leq rn$, the
elements of the subspace $\Lambda^k(K_2)_q$ of $NM/M$ which is generated by $\wedge_{i = 1}^k v_i,~
v_i \in K_2$.
Let $\{w_1, \ldots , w_n\}$ be a $GF(q)$-basis of $K_2$. Then $\{w_{i_1} \wedge \ldots \wedge
w_{i_k}~|~1 \leq i_1 <i_2< \ldots <i_k \leq n\}$ is a $GF(q)$-basis of $\Lambda^k(K_2)_q$, so 
$|\Lambda^k(K_2)_q| = q^{n \choose k}$. Therefore 
\[|NM/M| = \prod_{k = 1}^n q^{n \choose k} = q^{\sum_{k = 1}^n{n \choose k}} = q^{2^n-1}.\]
This implies that $\overline{\Gamma}$ has at most $q^{2^n-1}$ points, as asserted.
\end{bew}

Let $\Gamma$ be the canonical gluing $AS(n,q) \circ_\alpha AS(n,q)$, $q = 2^r$. 
Let $K_2:X \leq Aut(\Gamma)_0$ ( $K_0:X \leq Aut(\Gamma)_2$), 
where $K_2 \cong E_{q^n}$ ($K_0 \cong E_{q^n}$) and $X \cong GL_n(q)$.
Then $G = (K_0 \times K_2)X$ acts flag-transitively on $\Gamma$.
According to Lemma~\ref{auto2}
 we may assume that $\alpha$ is a $GF(q)X$-isomorphism
between $K_2$ and $K_0$.
Let $N = \sum_{i = 1}^n \Lambda^i(K_0)$. Then there is a natural action of $X$
on $N$. We may also define an action of $K_2$ on $N$.
Let $\{v_1, \ldots , v_n\}$ be a $GF(q)$-basis of $K_0$.
Let $u \in K_2$. Then $u = \alpha(v)$ for some $v \in K_0$.
Let $a_1, \ldots , a_n \in GF(q)$ with $v = \sum_{i = 1}^n a_i v_i$.
For $w \in \bigwedge^k(K_0)$ set
\[w^u = \sum_{J \subseteq \{1, \ldots ,n\}} (w \wedge_{i \in J} {a_i}v_i).\]
Since $q$ is even, this defines an action of $K_2$ on $N$.
 

\begin{proposition} \label{2cover}
Let $\overline{\Gamma}$ be the universal cover of the  canonical gluing $\Gamma = AS(n,q) \circ_\alpha AS(n,q)$, $q = 2^r$. Then $\overline{\Gamma}$ is a $q^{2^n-n-1}$-fold cover of $\Gamma$ and $\overline{G} = N:K_2:X$ acts flag-transitively on $\overline{\Gamma}$.
\end{proposition}
\begin{bew}
Let $V \leq \overline{G}$ be the diagonal $V = \{u \alpha(u)~|~u \in K_2\}$.
Then $V$ is invariant under the action of $X$. Set
$\overline{G}_0 = V:X$, $\overline{G}_2 = K_2:X$ and $\overline{G}_1 =
(\langle u \rangle_{GF(q)} \times \langle \alpha (u)_{GF(q)} \rangle )N_X(\langle u \rangle_{GF(q)})$, where $u \in K_2^\#$.
Clearly, $\overline{G} = \langle \overline{G}_0, \overline{G}_1, \overline{G}_2 \rangle$
and $\Gamma(\overline{G},(\overline{G}_0, \overline{G}_1, \overline{G}_2))$
is a cover of $\Gamma$.
 By \ref{2cover2} this is the universal cover of $\Gamma$. 
\end{bew}


\begin{theorem} \label{ucover}

Let $\Gamma$ be a canonical gluing $AS(n,q) \circ_\alpha AS(n,q),~q = p^r$ with group $G$ and $\tilde{\Gamma}$ the universal covering with group $\tilde{G}$. Then $\tilde{G}/N = G$, where $N$ is an elementary abelian $p$-groups and
\begin{itemize}
\item[(1)] If $q$ is odd $|N| = q^{{n \choose 2}}$
\item[(2)] If  $q$ is even, then $|N| = q^{2^n-n-1}$
\end{itemize}

\end{theorem}

\begin{bew}
 The assertion (1) follows from \ref{pcover} and (2) follows from \ref{2cover}.
\end{bew} 

\noindent
{\bf Remark} 
The covers of $AS(n,q) \circ_\alpha AS(n,q),~q$ odd or $q = 2$, which turned
out to be universal (Theorem~\ref{ucover}), were already constructed in [Pas2].
For the convenience of the reader we include the constructions of the 
universal covers. Moreover we give a geometric description of the universal cover
for $q= 2^r, ~r >1$.

\medskip

First let $q = 2$. Let $\Gamma_m$ be the Coxeter complex of type $D_m,~m = 2^n$
and let $Tr(\Gamma _m)$ be the $\{+,0,-\}$-truncation of $\Gamma_m$ formed by
the elements of type $+,0$ and $-$, where $+,0$ and $-$ are as follows:

\begin{picture}(300,70)(0,0)
\put(7,9){$\bullet$}
\put(7,51){$\bullet$}
\put(9,10){\line(2,1){44}}
\put(9,54){\line(2,-1){44}}
\put(52,30){$\bullet$}
\put(52,20){$0$}
\put(55,32){\line(1,0){47}}
\put(102,30){$\bullet$}
\put(105,32){\line(1,0){12}}
\put(120,32){.....}
\put(138,32){\line(1,0){12}}
\put(150,30){$\bullet$}
\put(153,32){\line(1,0){47}}
\put(200,30){$\bullet$}
\put(0,9){$-$}
\put(0,53){$+$}
\end{picture}

Then $Tr(\Gamma_m)$ is the universal cover of the canonical gluing
$AS(n,2) \circ_\alpha AS(n,2)$.

This universal cover is isomorphic
to the $GF(2)$-expansion $D(A)$, where $A$ is the affine space $AG(n,2)$,
see [BPM, Corollary 9].
Similiar, we may define the $GF(q)$-expansion $D(A)$, where $A$
is the affine space $AG(n,q)$. Then $D(A)$ is a cover of the canonical
gluing which is by Lemma~\ref{2cover2} the universal cover.

\medskip

Now let $q$ be odd. Let $\Delta$ be the building of type $D_{n+1},~n \geq 2$.
As above take the types as follows:

\begin{picture}(300,70)(0,0)
\put(7,9){$\bullet$}
\put(7,51){$\bullet$}
\put(9,10){\line(2,1){44}}
\put(9,54){\line(2,-1){44}}
\put(52,30){$\bullet$}
\put(52,20){$0$}
\put(55,32){\line(1,0){47}}
\put(102,30){$\bullet$}
\put(102,20){$1$}
\put(105,32){\line(1,0){12}}
\put(120,32){.....}
\put(138,32){\line(1,0){12}}
\put(150,30){$\bullet$}
\put(150,20){$n-3$}
\put(153,32){\line(1,0){47}}
\put(200,30){$\bullet$}
\put(200,20){$n-2$}
\put(0,9){$-$}
\put(0,53){$+$}
\end{picture}

Let us write $\varepsilon$ to denote any of the two types $+$ or $-$. 
For $\varepsilon \in \{+,-\}$ let $\Delta^\varepsilon$ be the half-spin geometry relative to the type $\varepsilon$, that is $\Delta^\varepsilon$ is the 
geometry of rank $2$ having the elements of $\Delta$ of type $\varepsilon$
as points and those of type $0$ as lines, with the incidence inherited
from $\Delta$. The diameter of $\Delta^\varepsilon$ is $n/2$ (resp. $(n+1)/2$) for $n$ even (resp. $n$ odd). Let $a^+$ and $a^-$ be elements of type
$+$ and $-$ respectively which are incident. Let $H^\varepsilon$ be the 
set of those points of $\Delta^\varepsilon$ which are not at 
maximal distance from $a^\varepsilon$ for $\varepsilon \in \{+,-\}$.
Then $H^\varepsilon$ is a hyperplane in $\Delta^\varepsilon$.

Let $\overline{\Delta}$ be the following geometry:
The elements of $\overline{\Delta}$ are the elements of type $\varepsilon$
which are not in $H^\varepsilon$, $\varepsilon \in \{+,-\}$, and
the elements of type $0$ which meet $H^\varepsilon$ in exactly one element
for $\varepsilon \in \{+,-\}$; the incidence is inherited from $\Delta$.

Then $\overline{\Delta}$ is the universal cover of the canonical gluing
$AS(n,q) \circ_\alpha AS(n,q),~q$ odd.


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\noindent
Fachbereich Mathematik und Informatik\\
Martin-Luther Universit\"at Halle Wittenberg\\
06099 Halle\\
Germany\\
\\
baumeis@coxeter.mathematik.uni-halle.de\\
stroth@coxeter.mathematik.uni-halle.de 


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