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\def\runningheadeven{EMBEDDING PARTIAL GEOMETRIES}
\def\runningheadodd{BROUWER et al.}
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%7. Title and authors
\title{Embedding Partial Geometries in Steiner Designs}
\author{{\em Andries E. Brouwer}, {\em Willem H. Haemers}
\and {\em Vladimir D. Tonchev}}

\date{November, 3, 1996} 

\begin{document} 
\maketitle
\pagestyle{headings}


\begin{abstract}
We consider the following problem: given a partial geometry $\cal{P}$ 
with $v$ points and $k$ points on a line, can one add to the line set 
a set of $k$-subsets of points such that the extended family of 
$k$-subsets is a 2-$(v,k,1)$ design (or a Steiner system $S(2,k,v)$).
We give some necessary conditions for such embeddings and several
examples.
One of these is an embedding of the partial geometry
$PQ^+(7,2)$ into a 2-$(120,8,1)$ design.
\end{abstract}

\section{Introduction}

We consider the question whether, given a partial geometry
${\cal P} = (X,{\cal L})$, there is a Steiner 2-design
${\cal D} = (X,{\cal B})$ such that ${\cal L} \subseteq {\cal B}$.
Clearly, the existence of such an embedding of $\cal P$ does not
depend on the structure of $\cal P$, but only on its collinearity
graph and line size.

Troughout ${\cal P} = (X,{\cal L})$ will denote a partial geometry
with parameters $s$, $t$ and $\alpha$.
The number $v$ of points and the number $l$ of lines of $\cal P$ are
given by
\[ v=(s+1)(st+\alpha)/\alpha \ ,\ l=(t+1)(st+\alpha)/\alpha \ .\]
The collinearity graph (or point graph) is strongly regular having 
eigenvalues
$s(t+1)$, $s-\alpha$ and $-t-1$ with multiplicities
\[ 1,\ f=\frac{st(s+1)(t+1)}{\alpha(s+t+1-\alpha)},
\ g=\frac{s(s+1-\alpha)(st+\alpha)}{\alpha(s+t+1-\alpha)}, \]
respectively.
For these and other results on partial geometries we refer to the 
survey paper by De~Clerck and Van~Maldeghem~\cite{DV}.

Suppose we have a collection $\cal C$ of $(s+1)$-cocliques in the
point graph of $\cal P$ that cover all non-collinear pairs of points
exactly once.
Then $\cal P$ is embedded in the Steiner 2-design
${\cal D} = (X,{\cal C} \cup {\cal L})$ with parameters:
\[v,\ k=s+1,\ r=(st+t+\alpha)/\alpha,\ 
b=(st+\alpha)(st+t+\alpha)/{\alpha^2}.\]
Clearly a necessary condition for $\cal P$ to be embeddable is that 
$r$ and $b$ are integers.
A quick inspection of the parameters of the known partial geometries 
shows that these divisibility conditions are satisfied very often.
The following result excludes many more parameter sets.

\begin{thm}
\label{een}
Suppose $\cal P$ is embeddable in a Steiner $2$-design $\cal D$ and 
suppose that $\cal P$ is not a Steiner $2$-design or a net
(i.e. $\alpha \neq s+1$ and $\alpha \neq t$).
Then \[ \alpha \leq \frac{t(s+1)}{t+s+1} . \]
If equality holds then $\alpha^f (t-\alpha)^g$ is the square of an 
integer.
\end{thm}

\noindent
This result is an immediate corollary of the following `Fisher inequality'.

\begin{thm}
\label{twee}
Let $\Gamma$ be a strongly regular graph with parameters
$(v,k,\lambda,\mu)$. If $\Gamma$ is the collinearity graph
of a partial linear space with $l$ lines of size $s+1$, then
either the given partial linear space is a partial geometry
with parameters $(s,t,\alpha)$, or $l \ge v$.
If $l = v$, then ${\rm det} (A + (k/s)I)$ is a square,
where $A$ is the adjacency matrix of $\Gamma$.
\end{thm}

\noindent
{\bf Proof} (of Theorem \ref{twee}).
Let $N$ be the $v \times l$ point-line incidence matrix of the
partial linear space, and $A$ the adjacency matrix of $\Gamma$.
Then $N N^\top = A + (t+1)I$, where $t+1 := k/s$ is the number
of lines on each point.
If $l = v$, then $N$ is square, and
${\rm det} (A + (t+1)I) = ({\rm det\,} N)^2$.
If $l < v$, then $N N^\top$ has rank less than $v$, so that
$A$ has eigenvalue $-t-1$, i.e.,
$(t+1)^2 - (\mu - \lambda)(t+1) + \mu - k = 0$.
In this case, since $t+1$ divides $k$, it also divides $\mu$, say
$\mu = (t+1)\alpha$ for some nonnegative integer $\alpha$, and we
find $\lambda = s-1 + \alpha t$, so that $\Gamma$ has the parameters
of the point graph of a pg$(s,t,\alpha)$, and since the lines
are regular cliques, our partial linear space was in fact a partial geometry.
\qed

\noindent 
{\bf Proof} (of Theorem \ref{een}).
Apply Theorem \ref{twee} to the noncollinearity graph $\Gamma$
of $\cal P$. We find either $b-l \ge v$, which reduces to
$\alpha (s+t+1) \le t(s+1)$, or $\Gamma$ is the collinearity
graph of a partial geometry ${\cal P}'$ with parameters
$(s',t',\alpha')$, where $s' = s$, $t' = s-\alpha$,
$\alpha' = s-t$ and $(t-\alpha)(s+1-\alpha) = 0$.
The adjacency matrix $A$ of $\Gamma$ has eigenvalues
$st(s+1-\alpha)/\alpha$, $\alpha-1-s$, $t$ with respective
multiplicities 1, $f$, $g$, so if $b-l = v$ then
$\alpha^f (s+t+1)^g$ is a square, and since $t^2 = (t-\alpha)(s+t+1)$
also $\alpha^f (t-\alpha)^g$ is a square.
\qed

A strongly regular graph is called {\em imprimitive} when it or its
complement is a vertex disjoint union of cliques (i.e., when
$\mu = 0$ or $\mu = k$ or there are no (non)edges at all).
A union of $m$-cliques is the collinearity graph of a partial
linear space with lines of size $c$ if and only if a
2-$(m,c,1)$ design exists.
The complement of a union of $n$ $m$-cliques is the collinearity graph
of a partial linear space with lines of size $c$ if and only if a
group divisible design GD$(c,1,m;nm)$ exists.
Nothing nontrivial can be said here.

A partial geometry is called {\em improper} if $\alpha=1$, $s$, $s+1$,
$t$ or $t+1$.
For $\alpha=s$ or $\alpha=s+1$ the point graph is imprimitive.
Otherwise, if $\alpha=t+1$, $\cal P$ is a dual Steiner system,
which is not embeddable by Theorem~\ref{een}.
If $\alpha=t$, $\cal P$ is a net that we want to embed in a
2-$(k^2,k,1)$ design, i.e. an affine plane.
Therefore $\cal P$ is embeddable if and only if it is the
union of some parallel classes of an affine plane.

Finally we consider the case $\alpha=1$.
Then $\cal P$ is a generalized quadrangle $GQ(s,t)$.
In this case the divisibility conditions and the conditions of 
Theorem~\ref{een} are always fulfilled.
Several of the known generalized quadrangles are constructed as a
set of lines in a projective or affine space and hence are embeddable
by construction, see Payne and Thas~\cite{PT}.
The smallest (with respect to $v$) open case is a possible
embedding of $GQ(5,3)$ in a 2-($96,6,1$) design.

One might ask whether an embedding will be unique.
But when there is an embedding into some Steiner system with lots
of subsystems, like a projective or affine space, then by twisting
one or more subsystems one will in general get lots of embeddings.
For example, in the smallest non-trivial case, that of the generalized
quadrangle $GQ(2,2)$, naturally embedded into $PG(3,2)$ as the
$Sp(4,2)$ quadrangle, each plane contains three lines of the
quadrangle and there are two ways of extending that set of three
to a 2-(7,3,1) on that plane. Thus, a plane can be `flipped'.
Flipping one plane destroys all other planes except those that meet it
in a line from the quadrangle. An arbitrary 2-(15,3,1) containing
the lines of the quadrangle is obtained from $PG(3,2)$ by flipping
0, 1, 2 or 3 planes on a given line of the quadrangle, and we find
precisely four nonisomorphic 2-$(15,3,1)$ designs that contain $GQ(2,2)$.

\section{Proper partial geometries}

Let us examine the known families of proper partial geometries for
possible imbeddings.

First, we consider the class $\cal S(\cal K)$.
These geometries exist whenever there is a maximal arc $\cal K$ of
degree $d$ in a projective plane of order $q=dc$.
The parameters are $s=d(c-1)$, $t=c(d-1)$, $\alpha=(c-1)(d-1)$.
Substitution in Theorem~\ref{een} gives $d \leq 1+c/(c-1)^2$, which is 
satisfied only if $d=c=2$.
Then $\cal P$ is $GQ(2,2)$, which has four embeddings, as we saw 
before.
Note that the same conclusion holds for the dual geometries, since
they belong to the same parameter family.

Next we consider the class ${\cal T}_2^*(\cal K)$ with parameters
$s=2^h-1$, $t=(2^h+1)(2^m-1)$, $\alpha=2^m-1$.
They exist whenever $m$ divides $h$ and consist of all the points and 
a subset of the lines of $AG(3,2^h)$, being all translates of the lines
through the origin that correspond to a maximal arc $\cal K$ of degree
$2^m$ in $PG(2,2^h)$.
So they all are embeddable.
For the duals, all our necessary conditions are fulfilled, but we 
don't know whether any embeddings exists (except, of course,
for the trivial case $m=h=1$).
For $m=1$, $h=2$ we get the open case $GQ(5,3)$, that we mentioned 
earlier.

The partial geometries $PQ^+(4n-1,2)$ were constructed by De Clerck,
Dye and Thas~\cite{DDT} using a non-singular hyperbolic quadric
$Q$ in $PG(4n-1,2)$ with a spread (i.e a partition of the point set
of $Q$ into maximal totally singular subspaces).
The points of the partial geometry are the points of $PG(4n-1,2)$ that
are not on $Q$ and the lines are the hyperplanes in the subspaces of
the spread.
A point $x$ is on a line $L$ if $x$ lies in the polar space of $L$ 
with respect to $Q$.
The parameters of $PQ^+(4n-1,2)$ are $s= 2^{2n-1}-1$, $t=2^{2n-1}$,
$\alpha=2^{2n-2}$.

%For $n=2$ there were earlier constructions of the geometry by Cohen, 
%and by Haemers and Van Lint,
%see~\cite{C} \cite{HvL}, \cite{K}, \cite{To}.
We found, first by computer and later by hand, that for $n=2$
this geometry indeed has an embedding. For an extensive discussion,
see the next section.

\begin{thm}
\label{drie}
The partial geometry $PQ^+(4n-1,2)$ is embeddable in a
Steiner $2$-design if and only if $n \leq 2$.
\end{thm}

\Proof
First suppose the geometry is embeddable.
Then the blocks of the embedding are cocliques of size $k=2^{2n-1}$ in
the point graph of $PQ^+(4n-1,2)$, which is the orthogonality
graph on the nonsingular points. The Gram matrix of the vectors
spanning the points of any such coclique is $J-I$, which is non-singular,
and hence these vectors are linearly independent and their number cannot
exceed the dimension of the space.
That is, $2^{2n-1} \leq 4n$, so $n \leq 2$.
The case $n = 1$ is trivial: $K_{3,3}$ can be extended to $K_6$.
It remains to show embeddibility in case $n = 2$.
That is, we have to construct a system of 8-cliques,
one on each edge, for the nonorthogonality graph on the nonsingular
points for $O^+_8(2)$. As follows: Pick a good system $\cal O$
of ovoids, one on each pair of nonorthogonal singular points,
and pick a good totally singular 4-space $V$. For each ovoid
$O$ in $\cal O$ we find a unique point $O \cap V = \langle p \rangle$,
and a {\em base} (basis consisting of 8 mutually nonorthogonal vectors)
$B = \{ a+p | \langle a \rangle \in O,~ a \ne p \}$.
The set of 120 bases thus obtained is the required system of 8-cliques.
For details on the choice of $\cal O$ and $V$ (not any $V$ will do),
see the next section.
\qed

Recently, Mathon and Street~\cite{MS} and De~Clerck~\cite{D} have
derived new partial geometries from $PQ^+(4n-1,2)$, but with the same
parameters.
We don't know if any of these admits an embedding.
The non-existence argument for $n > 2$ doesn't work anymore, because
these new geometries have other point graphs.

The parameter sets under consideration all meet the bound of 
Theorem~\ref{een}, but the condition there is always fulfilled.
For the related parameter sets $s=2^{2m}-1$, $t=2^{2m}$ and
$\alpha=2^{2m-1}$, no geometry is known, but for $m > 1$ they may 
exist.
The embedding however, can not exist by Theorem~\ref{een}
(indeed, $f+g=v-1$ is odd, so $2^{(2m-1)(f+g)}$ is not a square).
For $n \neq 1$, the dual parameters never satisfy the conditions of
Theorem~\ref{een}.

The partial geometries $PQ^+(4n-1,3)$ (the construction method is due 
to Thas and is only known to work for $n=1$) have parameters
$s=3^{2n-1}-1$, $t=3^{2n-1}$, $\alpha=2\cdot 3^{2n-1}$.
These geometries and their duals have no embedding by Theorem~\ref{een}.

Finally we consider the two known sporadic proper partial geometries.
The one with parameters $s=4$, $t=17$ and $\alpha=2$, constructed by 
the second author, does not satisfy the divisibility conditions, so has no
embedding, but the dual may have one.
The other one due to Van~Lint and Schrijver with $s=t=5$ and 
$\alpha=2$ looks more interesting.
An embedding would lead to a 2-$(81,6,1)$ design and such a design is
not known to exist.
By (incomplete) computer search we were able to extend quite far, but
not far enough.
Probably the embedding does not exist.

\medskip
{\bf Remark}
The $120$ points and the $120$ blocks of the embedding of
$PQ^+(7,2)$ given in Theorem~\ref{drie}, form a (flag-transitive)
partial linear space, with an incidence graph that is not a
bipartite distance-regular graph of diameter 4 or 5, and yet,
both the point and the block graph are primitive
strongly regular graphs (in fact they are isomorphic).
This seems to be a remarkable property.
Examples with imprimitive strongly regular graphs are
given by the elliptic semiplanes.

\section{Triality, ovoids, spreads and bases}

Let $X$ be an 8-dimensional vector space over a field $K$,
provided with a nondegenerate quadratic form $Q$ of (maximal)
Witt index 4.
Let $L$ be the collection of totally singular (t.s.) lines,
and let $Z_0$, $Z_1$, $Z_2$ be the sets of singular points and of
t.s. 4-spaces of the first and second kind, respectively.
Put $Z = Z_0 \cup Z_1 \cup Z_2$.
Natural incidence (symmetrized containment) defines a bipartite
graph $\Gamma$ on $Z \cup L$ with bipartition $\{Z,L\}$.
This graph has automorphism group $G \isom O_8^+(K).{\rm Sym}(3)$.
The group $G$ is transitive on $Z$ and $L$ and preserves
$\{Z_0,Z_1,Z_2\}$.
The subgroup $G_0 \isom O_8^+(K)$ preserves the sets $Z_0$, $Z_1$, $Z_2$.
The phenomenon that the three sets $Z_0$, $Z_1$, $Z_2$ can be permuted
arbitrarily is called {\em triality}.

Let $N_0$ be the set of nonsingular points.
We need to interpret these in terms of the graph $\Gamma$ so that we can
apply triality and also get sets $N_1$, $N_2$.
One way of doing that is by representing a nonsingular point
$\langle n \rangle$ by the reflection
$$r_{\langle n \rangle} ~:~ x \mapsto x - { (x,n) \over Q(n) } n ~.$$
Let $R_0$ be this set of reflections.
There is a 1-1 correspondence between $N_0$ and $R_0$.
We have $R_0 \subset G$ and $R_0$ is closed under conjugation
by $G_0.2 \isom PGO_8^+(K)$. It follows that we can find three
sets $R_0$, $R_1$, $R_2$ of reflections under conjugation by $G$,
where $R_i$ consists of the reflections that preserve $Z_i$
and interchange $Z_j$ and $Z_k$ for $\{i,j,k\} = \{0,1,2\}$.

\begin{lem} {\rm (cf. Tits \cite{T}).}
Let $r \in R_i$ and $s \in R_j$ with $i \ne j$. Then $(rs)^3 = 1$.
\end{lem}

\Proof
We may suppose $r \in R_0$, $s \in R_1$.
Since $srsrs \in R_0$ it suffices to show that $r$ and $srsrs$
fix the same singular points. Let $p \in Z_0$ be fixed by $r$,
and put $W = sp$ and $V = rW$ so that $V \in Z_1$ and $W \in Z_2$.
Now $\pi := V \cap W$ is a plane containing $p$ and $s$ fixes each
line on $p$ in $\pi$ so that $V$ and $sV$ have at least a plane in
common. But $sV \in Z_1$, so $V = sV$, i.e., $rsp = srsp$.
\qed

So far the field was arbitrary, but from now on we take $K = {\bf F}_2$.
The property that only holds in this case is: {\em If $m,n$ are
nonsingular vectors orthogonal to the t.s. plane $\pi$, then $(m,n)=0$.}
Indeed, $\pi^\perp$ is the union of three totally isotropic (t.i.)
4-spaces on $\pi$, of which two are t.s., so $m$ and $n$ are both
contained in the third.

Now that $K = {\bf F}_2$, let us use $+$ between projective points
instead of the spanning vectors, and write
$\langle a \rangle + \langle b \rangle := \langle a+b \rangle$.

Let a {\em base} be a set of 8 mutually nonorthogonal nonsingular points.
Let an {\em ovoid} be a set of 9 mutually nonorthogonal singular points.
Let a {\em spread} be a set of 9 pairwise disjoint t.s. 4-spaces.
All elements of a spread are of the same kind, and we talk about
a {\em $j$-spread} when the spread is a subset of $Z_j$ ($j=1,2$).
If we call ovoids {\em $0$-spreads}, then $i$-spreads ($i=0,1,2$)
correspond under triality.

If $B$ is a base, then $b_0 := \sum_{b \in B} b$ is singular,
and $O_B := \{ b_0 \} \cup \{ b_0 + b | b \in B \}$ is an ovoid.
Conversely, if $O$ is an ovoid, and $p \in O$, then
$B_{O,p} := \{ a+p | a \in O, p \ne a \}$ is a base.
Thus, we find a 9-1 correspondence between bases and ovoids.

\begin{prop}
Let $S$ be a 1-spread. Then ${\cal O} := \{ rS | r \in R_2 \}$
is a system of ovoids, one on each pair of nonorthogonal singular points.
\end{prop}

\noindent
These are the sets ${\cal O}$ called `good' in the previous section.
\medskip

\Proof
The numbers fit, so we have to show that no pair of noncollinear points
is covered twice. Interchanging types 0 and 2, we have to show that no
two disjoint 4-spaces $W, W'$ are contained in both $rS$ and $r'S$
for $r,r' \in R_0$.
Let $r = r_m$ and $r' = r_n$. Then $m^\perp$ and $n^\perp$ meet
$W$ and $W'$ in the same plane $\pi$ and $\pi'$, respectively.
Both $m$ and $n$ lie in the t.i. but not t.s. 4-spaces $Y$ and $Y'$
on $\pi$ and $\pi'$. But then $Y$ and $Y'$ have the line $l$ spanned
by $m$ and $n$ in common, and $l$ hits the disjoint planes $\pi$,
$\pi'$ in distinct points, so has at least 4 points, contradiction.
\qed

\begin{prop}
Take $S$ and $\cal O$ as above, and fix an element $V \in S$.
The set $\cal B$ of 120 bases $B_{O,p}$ with $O \in {\cal O}$ and
$p = O \cap V$ is a system of 8-cliques, one on each edge, for the
nonorthogonality graph on the 120 nonsingular points.
\end{prop}

Of course the existence of $\cal B$ is the whole point of this section
(in fact, of this paper).

The above description does not show the symmetry between the 120
nonsingular points and the 120 bases. A more symmetric description
of the same configuration: Take a base $B$ and put $O := O_B$.
Let ${\cal S}_1 = \{ rO | r \in R_2 \}$ and
${\cal S}_2 = \{ rO | r \in R_1 \}$, and join $S \in {\cal S}_1$
to the 8 spreads $r_b S$ $(b \in B)$ in ${\cal S}_2$.

Or, in terms of reflections: Take a base $B$ and join $r \in R_2$
to the 8 reflections $r_b r r_b$ in $R_1$.

\Proof
We have to show that both descriptions are equivalent, and that
they actually work. As to the former, interchanging types 0 and 1
we see that the spread $S$ with fixed element $V$, the collection
${\cal O} = \{ rS | r \in R_2 \}$, an ovoid $O = rS$ in ${\cal O}$,
the point $p = O \cap V$ and the set of 8 reflections fixing 7 points
of $O$ and interchanging $p$ with the ninth point
correspond to, respectively, the ovoid $O$ with fixed element $b_0$,
the collection ${\cal S}_1 = \{ rO | r \in R_2 \}$, a spread
$S = rO$ in ${\cal S}_1$, the 4-space $rb_0 \in S$ containing $b_0$
and the set of 8 reflections $\{ r r_b r | b \in B \}$.
Since $r r_b r = r_b r r_b$, and $r_b O = O$, this shows that
both descriptions are equivalent.

Since the bases by definition are 8-cliques in the nonorthogonality
graph on the nonsingular points, and the numbers fit, we only have
to check that no two bases $B_{O,p}$ have a pair in common, or,
equivalently, that no two sets $\{ r r_b r | b \in B \}$ and
$\{ s r_b s | b \in B \}$ have a pair in common (for $r,s \in R_2$).
But if $r r_a r = s r_b s$ and $r r_c r = s r_d s$
($a,b,c,d \in B$, $a \ne c$, $b \ne d$) then
$r_a r r_a = r_b s r_b$, $r_c r r_c = r_d s r_d$, i.e.,
$r = r_a r_b s r_b r_a = r_a r_b r_d r_c r r_c r_d r_b r_a$.
With $S = rO$ this means that $r_a r_b r_d r_c S = S$.
But if $a,b,c,d$ are all distinct, then $r_a r_b r_d r_c$ has
order 5 (as is seen by its action on $O$) hence must fix some
element $V \in S$. On the other hand, both
$\langle a,b,c,d \rangle$ and ${\langle a,b,c,d \rangle}^\perp$
are elliptic quadrics, and
$r_a r_b r_d r_c x = x + (x,c)c + (x,c+d)d + (x,b+d)b + (x,a+b)a$
while $a,b,c,d$ and sums of two of them are not in $Z_0$,
so when two of the inner products $(x,c)$, $(x,c+d)$, $(x,b+d)$,
$(x,a+b)$ vanish for $x \in V$, all do. But
$V \cap {\langle a,b,c,d \rangle}^\perp$ does not contain a line,
contradiction.
If $a,b,c,d$ are not all distinct, say $a = d$, then
$r_a r_b r_d r_c = r_a r_b r_a r_c = r_e r_c$ for $e = r_a b = a+b$.
But if $r_e r_c S = S$, then, since $r_e r_c$ fixes at least a line
on each element of $S$, $r_e r_c$ must fix all elements of $S$.
But $c \ne e$ and $r_e r_c x = x + (x,c)c + (x,e)e$, so we find
that $(x,c) = (x,e)$ on each $V \in S$, so on all of $X$, so $c = e$,
contradiction.
\qed

What about the automorphism group of these constructions?
The 8 reflections $r_b$ ($r \in B$) generate a ${\rm Sym}(9)$
and clearly this is the full stabilizer of $O$ in $G$.
The stabilizer of $O$ in $G_0$ is ${\rm Alt}(9)$, and that is
also the full group of the system $\cal O$.
It follows that the full group of the system $\cal B$ is the
stabilizer of $b_0$ in the previous, i.e., is ${\rm Alt}(8)$.

As mentioned earlier, De Clerck-Dye-Thas construct a partial geometry
pg(7,8,4) on the nonsingular points by fixing a spread $S$ and
taking the sets of nonsingular points on the t.i. but not t.s.
4-spaces meeting some element of $S$ in a plane.
(This is the dual of the pg(8,7,4) obtained from $\cal O$.
Indeed, the DDT system has point set $R_0$, and its lines are
the planes in some element of $S$, where $r$ is incident with
$\pi$ if $r \pi = \pi$. If we let $S$ be a 1-spread, then these
planes can be identified with the elements of $Z_2$ containing them,
and interchanging types 0 and 2 we find the description of $\cal O$.)

If we join our system $\cal B$ to the set of lines of this partial
geometry, we get a Steiner system $S(2,8,120)$ with automorphism
group (at least) ${\rm Alt}(8)$, when both systems were constructed
starting from the same spread $S$.
(No doubt several nonisomorphic $S(2,8,120)$'s arise in this way,
but we have not investigated the details. Several nonisomorphic
$S(2,8,120)$'s were known already - obtained as the exterior lines
and the points off a hyperoval in some projective plane
of order 16, cf. \cite{PRS}.)

\begin{thebibliography}{99}

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\bigskip
\noindent{\bf Andries E. Brouwer}
Department of Mathematics,
Eindhoven University of Technology,
P.O. Box 513, 5600 MB \ Eindhoven, The Netherlands.\\
 e-mail: aeb@cwi.nl

\bigskip
\noindent{\bf Willem H. Haemers}
Department of Econometrics,
Tilburg University,
P.O. Box 90153, 5000 LE \ Tilburg, The Netherlands.\\
 e-mail: haemers@kub.nl

\bigskip
\noindent{\bf Vladimir D. Tonchev}
Department of Mathematical Sciences,
Michigan Technological University,
Houghton, Michigan 49931, USA.\\
 e-mail: tonchev@mtu.edu

\end{document}

