Line Spreads That Produce Projective Planes

1. Introduction

Let $V/Ł$ be a 3-dimensional right vector space over the skew field Ł. The 1- and 2-spaces of V form the points and lines of a Desarguesian projective plane $\mathsf{PG}(2,Ł)$. Suppose has a subfield $\mathbb{F}$ over which is a natural vector space of dimension 2 (with “natural” we mean using the scalar multiplication given by the multiplication of Ł). Then we may regard V as a 6-dimensional vector space $V/\mathbb{F}$ over $\mathbb{F}$, defining a 5-dimensional projective space $\mathsf{PG}(5,\mathbb{F})$. The 1-spaces of $V/Ł$ correspond to a selection $\mathfrak{L}$ of 2-spaces of $V/\mathbb{F}$ with the following properties:

(i)

Every 1-space of $V/\mathbb{F}$ is contained in a unique member of $\mathfrak{L}$;

(ii)

Two distinct members of $\mathfrak{L}$ generate a 4-space U of $V/\mathbb{F}$ with the property that every member of $\mathfrak{L}$ sharing at least a 1-space of $V/\mathbb{F}$ with U is entirely contained in U.

In $\mathsf{PG}(5,\mathbb{F})$, the set $\mathfrak{L}$ corresponds to a line spread (i.e., a set of lines, also denoted by $\mathfrak{L}$, partitioning the point set), which induces a line spread in every subspace spanned by two distinct but arbitrary members of $\mathfrak{L}$. We call such a line spread a composition line spread. The members of $\mathfrak{L}$ and all subspaces spanned by two of its members form the point set and line set, respectively, of the projective plane $\mathsf{PG}(2,Ł)$. We say that $\mathfrak{L}$ arises from the extension $Ł/\mathbb{F}$.

We can now reverse the procedure. We start with the projective space $\mathsf{PG}(5,\mathbb{F})$ over the field $\mathbb{F}$ and try to find a composition line spread. One way of achieving this is to find a fixed point free collineation $\theta$ of $\mathsf{PG}(5,\mathbb{F})$ with the property that, for each point p, the line spanned by p and $p^{\theta }$ is stabilised. Note that every fixed point free involution has that property. Then automatically, the fixed lines form a composition spread. In the present paper, we determine all composition line spreads of $\mathsf{PG}(5,\mathbb{F})$, with $\mathbb{F}$ a field, and determine their fix group. It will turn out that there exist such spreads whose fix group is trivial, that is, which can not be constructed as fix (line) structure of a fixed point free collineation of $\mathsf{PG}(5,\mathbb{F})$. More precisely, we show:

Theorem 1. 

Let $\mathfrak{L}$ be a composition line spread of $\mathsf{PG}(5,\mathbb{F})$. Then there exists a skew field containing $\mathbb{F}$ such that $\mathfrak{L}$ arises from the extension $Ł/\mathbb{F}$. Moreover, we have exactly one of the following situations, where we denote by T the fix group of $\mathfrak{L}$, that is, the group of all collineations of $\mathsf{PG}(5,\mathbb{F})$ stabilising each member of $\mathfrak{L}$.

(i)

Ł is a (separable or inseparable) quadratic extension field of $\mathbb{F}$ and T is a group abstractly isomorphic to ${Ł}^{\times }/{\mathbb{F}}^{\times }$, and as a permutation group acts sharply transitively on the set of points of each line of $\mathfrak{L}$;

(ii)

Ł is a quaternion algebra over a subfield ${\mathbb{F}}^{\prime }$ of $\mathbb{F}$. The latter is quadratic over ${\mathbb{F}}^{\prime }$. If $\mathbb{F}/{\mathbb{F}}^{\prime }$ is separable, then T has order 2 and its nontrivial member is a semi-linear involution corresponding to Galois descent. If $\mathbb{F}/{\mathbb{F}}^{\prime }$ is inseparable, then T is trivial.

In case $\left(i\right)$ above, the line spread $\mathcal{S}$ induced in a subspace of dimension 3 is regular, that is, for each triple of lines $L_1,L_2,L_3$ of $\mathcal{S}$, every line intersecting each transversal of $L_1,L_2,L_3$ belongs to $\mathcal{S}$ (a transversal of a set of lines is a line intersecting each line of the set in a point). If this condition is only satisfied for given lines $L_1,L_2$ and $L_3$ of $\mathcal{S}$, then we say that the triple $\{L_1,L_2,L_3\}$ is regular. If the triple $\{L_1,L_2,L_3\}\subseteq \mathcal{S}$ is regular for given $L_1,L_2\in \mathcal{S}$ and all $L_3\in \mathcal{S}\setminus \{L_1,L_2\}$, then we say that the pair $\{L_1,L_2\}$ is regular. We will show:

Theorem 2. 

A line spread $\mathcal{S}$ of $\mathsf{PG}(3,\mathbb{F})$ is regular if, and only if, there exists a regular pair $\{L_1,L_2\}\subseteq \mathcal{S}$ and a regular triple ${Ł}_1,L_3,L_4\}$ such that no point of $L_2$ is on any transversal of $\{L_1,L_3,L_4\}$.

This is a substantial weakening of the condition in the definition of regular spread. It is for instance satisfied as soon as there exist two different regular pairs!

Let $L_1,L_2,L_3$ be three members of a line spread $\mathcal{S}$ of $\mathsf{PG}(3,\mathbb{F})$. We define the perspectivity of $L_1$ to $L_2$ from $L_3$ as the map from the point set of $L_1$ to the point set of $L_2$ assigning to $x_1\in L_1$ the unique point $x_2\in L_2$ contained in the plane generated by $x_1$ and $L_3$ (or, in other words, such that the line $x_1{x}_2$ intersects $L_3$ in a point). The composition of a finite number of perspectivities is called a projectivity and if a projectivity has domain $L_1$ and target $L_1$, then we call it a self-projectivity of $L_1$. The set of all self-projectivities of $L_1$ forms a (permutation) group, called the projectivity group of $L_1$, denoted ${\Pi }_{\mathcal{S}}\left(L_1\right)$. The projectivity groups of all members of $\mathcal{S}$ are isomorphic, and so we can speak about the projectivity group $\Pi \left(\mathcal{S}\right)$ of $\mathcal{S}$. We will show:

Theorem 3. 

A line spread $\mathcal{S}$ of $\mathsf{PG}(3,\mathbb{F})$ is regular if, and only if, ${\Pi }_{\mathcal{S}}\left(L\right)$ acts freely on L, for at least one and hence each $L\in \mathcal{S}$ if, and only if, ${\Pi }_{\mathcal{S}}\left(L\right)$ acts sharply transitively on L for at least one and hence each $L\in \mathcal{S}$ if, and only if, the restriction of the fix group T of $\mathcal{S}$ to the line L coincides with $\Pi \left(L\right)$, for at least one and hence each line $L\in \mathcal{S}$.

Finally, let us explain where the interest in the above problems comes from, without going into details of the various notions (referring to [6]). Recently, the author, together with Yannick Neyt and James Parkinson, classified all automorphisms of spherical Tits-buildings with the property that the Weyl distance between each chamber and its image lies in a given unique (possibly twisted) conjugacy class of the Weyl group (such automorphisms are called uniclass). For projective spaces, the uniclass collineations are exactly the members of the fix groups of line spreads. Hence the interest to determine these explicitly. Also, it is interesting the know that there exist composition line spreads with trivial fix group. That means that the geometric notion of composition line spread is not entirely equivalent with the notion of nontrivial uniclass collineation, in contrast to what happens for some other types of buildings. Note that our results carry over to arbitrary dimension at least 5 in an obvious way (every composition line spread restricts to a composition line spread in each subspace generated by three of its members not contained in the same 3-space—and hence generating a 5-dimensional subspace).

2. Preliminaries

In the present paper, our main objects are the Pappian projective spaces $\mathsf{PG}(n,\mathbb{F})$, that is, projective spaces originating from vector spaces $V_{n+1}$ of dimension $n+1$ defined over fields $\mathbb{F}$. Recall that the points of $\mathsf{PG}(n,\mathbb{F})$ are the 1-spaces of $V_{n+1}$. The set of 1-spaces in a given subspace of $V_{n+1}$ is also called a subspace of $\mathsf{PG}(n,\mathbb{F})$. The (projective) dimension of a subspace is one less than its corresponding vector space dimension. The 1-dimensional subspaces of $\mathsf{PG}(n,\mathbb{F})$ are also called lines, the 2-dimensional ones planes and the 3-dimensional ones solids. The $(n-1)$-dimensional subspaces are the hyperplanes and correspond to the points of the projective space defined by the dual vector space. If P is a set of points of $\mathsf{PG}(n,\mathbb{F})$, then the intersection of all subspaces containing P is called the span of P, denoted by $\langle P\rangle$ and we also say that P generates $\langle P\rangle$. If P has exactly two elements, then $\langle P\rangle$ is a line.

A coordinatisation of the projective space $\mathsf{PG}(n,\mathbb{F})$ consists of choosing a basis of $V_{n+1}$ and attach coordinates to each 1-space, determined up to a nonzero scalar multiple. Such a coordinatisation is equivalent to choosing $n+1$ points of $\mathsf{PG}(n,\mathbb{F})$ corresponding to $n+1$ distinct 1-spaces of $V_{n+1}$ generated by a basis $(e_1,\dots ,e_{n+1})$, and a unit point, that is, a 1-space of $V_{n+1}$ generated by a vector e that is linearly independent of every set of n basis vectors. Requiring that e has coordinates $(1,1,\dots ,1)$ determines the $e_i$ up to a common scalar multiplicative constant. We say that $(e_1,\dots ,e_{n+1};e)$ is a basic skeleton.

A (projective) line spread $\mathfrak{L}$ of $\mathsf{PG}(n,\mathbb{F})$ is a partition of the point set into lines. The seminal paper by Bruck & Bose [3] contains many fundamental results and conjectures, some of which have been proved or refuted in the meantime. However, over the past decades, spreads have mainly been investigated over either the finite fields, or the connected compact fields. Our results hold over arbitrary fields.

A composition line spread is a line spread with the property that the members of the spread contained in the subspace generated by any given pair of lines of the spread, form again a line spread. Composition line spreads are sometimes also called geometric line spreads (but this would interfere with our notion of geometric descent, see remark). For instance, composition line spreads in finite projective spaces of dimension at least 5 are classified, see [1]. A recovers this classification. Also, as shown in [3], the geometry with point set the lines of a composition line spread $\mathfrak{L}$ of $\mathsf{PG}(5,\mathbb{F})$ and line set the solids in which $\mathfrak{L}$ induces a line spread, is a projective plane which we denote by $\mathsf{PG}\left(\mathfrak{L}\right)$.

Let $S_1,S_2$ be two subspaces of $\mathsf{PG}(n,\mathbb{F})$ of the same dimension, and let T be a subspace complementary to both $S_1,S_2$, that is, T and $S_i$ generate the whole space, but are disjoint, $i=1,2$. Then we denote the map

$$S_1\to S_2:p\mapsto \langle p,T\rangle \cap S_2$$

by $S_1{⊼}_T{S}_2$ and call it the perspectivity of $S_1$ to $S_2$ from T. A (finite) product of perspectivities $S_1{⊼}_{T_1}{S}_2{⊼}_{T_2}{S}_3{⊼}_{T_3}\cdots$ is called a projectivity, and if the last subspace of that sequence is $S_1$ again, then we have a self-projectivity. The set of self-projectivities of $S_1$ is a group denoted by $\Pi \left(S_1\right)$ and called the projectivity group of $S_1$. If we restrict the subspaces $S_1,\cdots$ and $T_1,\dots$ to the members of a given line spread $\mathcal{S}$ of $\mathsf{PG}(3,\mathbb{F})$, then we obtain the projectivity group ${\Pi }_{\mathcal{S}}\left(S_1\right)$, which is clearly a subgroup of $\Pi \left(S_1\right)$.

3. Proofs

Introduction of coordinates—Let $\mathfrak{L}$ be a composition line spread of $\mathsf{PG}(5,\mathbb{F})$, with $\mathbb{F}$ a field. Select a a 3-space S spanned by two spread lines $L_1,L_2$ and select also a line $L_3\in \mathfrak{L}$ outside S, and a third line $L_{12}$ of $\mathfrak{L}$ in S (meaning $L_1\ne L_{12}\ne L_2$). Choose two points $e_1,e_2$ on $L_1$ and let $e_{i+2}$ be the unique point of $L_2$ with the property that $\langle e_1,e_{i+2}\rangle$ intersects $L_{12}$ nontrivally, say in the point $e_{i,i+2}$, $i=1,2$. In the solid $\langle L_2,L_3\rangle$ we select a third spread line $L_{23}$ and we consider the points $e_5$ and $e_6$ on $L_3$ such that the line $\langle e_i,e_{i+2}\rangle$ intersects the line $L_{23}$ nontrivally, say in the point $e_{i,i+2}$, $i=3,4$. Then the line $L_{123}:=\langle L_3,L_{12}\rangle \cap \langle L_1,L_{23}\rangle$ belongs to $\mathfrak{L}$. We may choose the unit point e on $L_{123}$ and then, taking $(e_1,e_2,\dots ,e_6;e)$ as a basic skeleton we have (with self-explaining shorthand notation, and with $L_{13}:=\langle L_1,L_3\rangle \cap \langle L_2,L_{123}\rangle$),

$$L_1=\langle 100000,010000\rangle ,L_2=\langle 001000,000100\rangle ,L_3=\langle 000010,000001\rangle ,$$

$$L_{23}=\langle 001010,000101\rangle ,L_{13}=\langle 100010,010001\rangle ,L_{12}=\langle 101000,010100\rangle .$$

Note that this coordinatisation depends on the choices for $e_1$ and $e_2$, and also on the choice of e. For instance, the following coordinate change preserves the above equalities:

$$\begin{array}{c}\hfill (x_1^{\prime },x_2^{\prime },\dots ,x_6^{\prime })=(x_1+k{x}_2,x_2,x_3+k{x}_4,x_4,x_5+k{x}_6,x_6),k\in \mathbb{F}.\end{array}$$

(1)

The spread $\mathcal{S}$ in coordinates—Let us represent the spread $\mathcal{S}$ in coordinates. For clarity, we leave out the last two coordinates. Every spread line distinct from $L_1$ intersects the plane $\langle L_1,e_3\rangle$ in a unique point $(a,b,1,0)$, $a,b\in \mathbb{F}$, and every such point lies on a unique spread line $L(a,b)$. The line $L(a,b)$ intersects the plane $\langle L_1,e_4\rangle$ in a unique point with coordinates $\left(f\right(a,b),g(a,b),0,1)$, where $f:\mathbb{F}\times \mathbb{F}\mapsto \mathbb{F}$ and $g:\mathbb{F}\times \mathbb{F}\to \mathbb{F}$ are two maps with $f(0,0)=0=g(0,0)$ and also $f(1,0)=0$ and $g(1,0)=1$.

Expressing that each point of $\langle L_1,L_2\rangle \setminus L_1$ lies on a unique line $L(a,b)$, we obtain the following sufficient and necessary condition for a set of lines of the form $L(a,b)$, together with $L_1$, to be a spread of $\langle L_1,L_2\rangle$: The system of equations

$$\begin{array}{c}\hfill \left\{\begin{array}{c}ax+bf(x,y)=c\hfill \\ ay+bg(x,y)=d\hfill \end{array}\right.\end{array}$$

(2)

has a unique solution for each $a,b,c,d\in \mathbb{F}$, $(a,b)\ne (0,0)$.

Regularity— Suppose for a moment that each line of $\mathcal{S}$ that intersects $\langle (1,0,0,0),(0,0,1,0)\rangle$ also intersects $\langle (0,1,0,0),(0,0,0,1)\rangle$. Then clearly $f(a,0)=0$. If moreover, each such line intersects each transversal of $L_1,L_2$ and $L_{12}$, then one calculates that $g(a,0)=a$.

Set $L:=L(0,1)$, and set $\alpha :=f(0,1)$ and $\beta =g(0,1)$. We express that $\{L_1,L\}$ is a regular pair. An arbitrary line $M_c$ in $\langle L_1,e_3\rangle$ through $(0,1,1,0)$ has single-parameter description $(cr,1-r,1,0)$, for some $c\in \mathbb{F}$, and we assume $c\notin \{0,1\}$, and r is the parameter, taking all values in $\mathbb{F}$. For $r=0$, we get $(0,1,1,0)$. For $r=1$, we get the point $(c,0,1,0)$, which lies on the line $L(c,0)$, which intersects $\langle L_1,e_4\rangle$ in the point $(0,c,0,1)$. The transversal $M_c^{\prime }$ of $L_1,L$ and $L(c,0)$ through $(\alpha ,\beta ,0,1)$ then goes through $(0,c,0,1)$ and has single-parameter description $(\alpha -\alpha r,\beta -r(\beta -c),0,1)$, where $r=0$ corresponds to $(\alpha ,\beta ,0,1)$ and $r=1$ corresponds to $(0,c,0,1)$. It can be checked that, due to regularity, common values of r in the descriptions of $M_c$ and $M_c^{\prime }$ above provide points on the same member of $\mathcal{S}$. Hence we conclude that

$$\left\{\begin{array}{c}f(cr,1-r)=\alpha -\alpha r,\hfill \\ g(cr,1-r)=\beta -r\beta +cr,\hfill \end{array}\right.$$

which is, after setting $r=1-b$ and $cr=a$, (for $r\ne 0$) equivalent to

$$\left\{\begin{array}{c}f(a,b)=\alpha b,\hfill \\ g(a,b)=\beta b+a.\hfill \end{array}\right.$$

This holds for all $a,b\in \mathbb{F}$, except for $b=1$ and $a\ne 0$. But these values correspond to the points $(t,1,1,0)$ of $\langle L_1,e_3\rangle$, and one can check that there is a unique line through such point not intersecting any spread line thus-far obtained, and it is given by setting $b=1$ in the above expressions.

Now one checks that the system of equations (2) has always a unique solution if, and only if, the quadratic polynomial $x^2+\beta x-\alpha$ is never zero, hence is irreducible. We will see in the next few paragraphs, that such a spread admits a 3-transitive group, hence each triple of lines of the spread is regular, which yields a regular spread. This shows B.

We now return to the general situation.

An additive automorphism group of $\mathcal{S}$—For every $a,b\in \mathbb{F}$, the line $L^{\prime }(a,b)=:\langle L_3,L(a,b)\rangle \cap \langle L_1,L_{23}\rangle$ belongs to $\mathfrak{L}$. An elementary calculation shows that

$$L^{\prime }(a,b)=\langle (a,b,1,0,1,0),(f(a,b),g(a,b),0,1,0,1).$$

Likewise, the line $L^{\prime \prime }(a,b)=:\langle L_2,L^{\prime }(a,b)\rangle \cap \langle L_1,L_3\rangle$ belongs to $\mathfrak{L}$. In coordinates:

$$L^{\prime \prime }(a,b)=\langle (a,b,0,0,1,0),(f(a,b),g(a,b),0,0,0,1)\rangle .$$

Now we define the following projectivity $\rho (a,b)$ of S: we project S onto $\langle L_1,L_{23}\rangle$ from the line $L_3$ and then project $\langle L_1,L_{23}\rangle$ back onto S from the line $L^{\prime \prime }(a,b)$. In coordinates we have (leaving out the last two coordinates again):

$$\rho (a,b):\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)\mapsto \left(\begin{array}{cccc}1& 0& -a& -f(a,b)\\ 0& 1& -b& -g(a,b)\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right).$$

Now denote the matrix $\left(\begin{array}{cc}a& f(a,b)\\ b& g(a,b)\end{array}\right)$ by $M(a,b)$ and let $\mathcal{M}$ be the set of all such matrices. Then $M(0,0)$ is the 0-matrix, also denoted by $\mathsf{0}$, and $M(1,0)$ the identity matrix, also denoted by $\mathsf{I}$. Furthermore, since $\rho (a,b)$ preserves $\mathcal{S}$, the set of $(4\times 4)$-matrices

$$\left(\begin{array}{cc}\mathsf{I}& -M(a,b)\\ \mathsf{0}& \mathsf{I}\end{array}\right),a,b\in \mathbb{F},$$

forms a group A acting on the left sharply transitivily on the set $\left\{L\right(a,b\left)\right|a,b\in \mathbb{K}\}$. Applying $\rho (a,b)$ to $L(0,0)$, we deduce $-M(a,b)=M(-a,-b)$. Consequently A consists of the linear collineations with matrix $\left(\begin{array}{cc}\mathsf{I}& M(a,b)\\ \mathsf{0}& \mathsf{I}\end{array}\right)$. Since $\rho (a,b)\left(L\right(x,y\left)\right)=L(a+x,b+y)$, we see that $\mathcal{M}$ is an additive group isomorphic to A.

Additivity of f and g—Let $a,b,c,d\in \mathbb{F}$ be arbitrary. Expressing that $M(a,b)+M(c,d)\in \mathcal{M}$ we deduce $f(a+c,b+d)=f(a,b)+f(c,d)$ and likewise for g. In particular $f(a,d)=f(a,0)+f(0,d)$ and likewise $g(a,d)=g(a,0)+g(0,d)$, for all $a,d\in \mathbb{F}$. We may set $f(a,0):=f_1\left(a\right)$ and $f(0,b)=f_2\left(b\right)$; likewise we set $g(a,0)=g_1\left(a\right)$ and $g(0,b)=g_2\left(b\right)$. Then $f(a,b)=f_1\left(a\right)+f_2\left(b\right)$ and $g(a,b)=g_1\left(a\right)+g_2\left(b\right)$. Note that $f_1(a+c)=f(a+c,0)=f(a,0)+f(c,0)=f_1\left(a\right)+f_1\left(c\right)$, hence $f_1$ is additive. Similarly, $f_2,g_1,g_2$ are additive maps.

A multiplicative automorphism group of $\mathcal{S}$—For every $a,b\in \mathcal{L}$, the line $L^{\prime \prime \prime }(a,b)=:\langle L_{13},L(-a,-b)\rangle \cap \langle L_2,L_3\rangle$ belongs to $\mathfrak{L}$. With coordinates,

$$L^{\prime \prime \prime }(a,b)=\langle (0,0,1,0,a,b),(0,0,0,1,f(a,b),g(a,b)\rangle .$$

Now we define the following projectivity $\xi (a,b)$ of S: we project S onto $\langle L_2,L_3\rangle$ from the line $L_{13}$ and then project $\langle L_2,L_3\rangle$ back onto S from the line $L^{\prime \prime }(a,b)$. In coordinates we have (leaving out the last two coordinates again):

$$\xi (a,b):\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)\mapsto \left(\begin{array}{cccc}a& f(a,b)& 0& 0\\ b& g(a,b)& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right).$$

Hence we obtain a group of (linear) collineations with matrices $\left(\begin{array}{cc}M(a,b)& \mathsf{0}\\ \mathsf{0}& \mathsf{I}\end{array}\right)$. We deduce immediately that all nontrivial members of $\mathcal{M}$ are nonsingular and that $\mathcal{M}$ is closed, not only under addition, but also under multiplication. Hence it defines a skew field. This also implies that the automorphism group of $\mathcal{S}$ is triply transitive, as mentioned earlier.

$g_1$ is a field endomorphism|The fact that for all $a,x\in \mathbb{K}$ the matrix $M(a,0)M(x,0)$ belongs to $\mathcal{M}$ is equivalent to the identities

$$\begin{array}{ccc}\hfill f_1\left(ax\right)& =& a{f}_1\left(x\right)+g_1\left(x\right)f_1\left(a\right),\hfill \end{array}$$

(3)

$$\begin{array}{ccc}\hfill g_1\left(ax\right)& =& g_1\left(a\right)g_1\left(x\right).\hfill \end{array}$$

(4)

Hence $g_1$ is a field endomorphism. Since every member of $\mathcal{M}$ is invertible, $g_1$ is injective. For clarity, we denote $g_1\left(x\right)=:\overline{x}$. The identity automorphism of $\mathbb{F}$ shall be denoted by $\mathrm{id}$.

$f_2$ is a multiple of $g_1$|The fact that for all $a,x\in \mathbb{K}$ the matrix $M(0,a)M(x,0)$ belongs to $\mathcal{M}$ is equivalent to the identities

$$\begin{array}{ccc}\hfill f_2\left(ax\right)& =& f_2\left(a\right)\overline{x},\hfill \end{array}$$

(5)

$$\begin{array}{ccc}\hfill g_2\left(ax\right)& =& a{f}_1\left(x\right)+\overline{x}{g}_2\left(a\right).\hfill \end{array}$$

(6)

It immediately follows from Identity (5), setting $f_2\left(1\right)=F$, that $f_2\left(x\right)=F\overline{x}$. Comparing Identities (3) and () we obtain, taking into account $f\left(1\right)=0$ and setting $g_2\left(1\right):=G$:

$$\begin{array}{ccc}\hfill g_2\left(x\right)& =& f_1\left(x\right)+G\overline{x}.\hfill \end{array}$$

(7)

More identities—The fact that for all $a,x\in \mathbb{K}$ the matrix $M(a,0)M(0,x)$ belongs to $\mathcal{M}$ is equivalent to the identities (taking into account the above expressions for $f_2$ and $g_2$ in function of $f_1$ and $g_1$):

$$\begin{array}{ccc}\hfill f_1\left(x{f}_1\left(a\right)\right)+F\overline{\overline{a}}\overline{x}& =& Fa\overline{x}+f_1\left(a\right)f_1\left(x\right)+G\overline{x}{f}_1\left(a\right),\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill \overline{x}\overline{f_1\left(a\right)}+f_1\left(\overline{a}x\right)+G\overline{\overline{a}}\overline{x}& =& \overline{a}{f}_1\left(x\right)+G\overline{a}\overline{x}.\hfill \end{array}$$

(9)

Finally, the fact that for all $a,x\in \mathbb{K}$ the matrix $M(0,a)M(0,x)$ belongs to $\mathcal{M}$ is equivalent to the identities (taking into account the above expressions for $f_2$ in function of $g_1$):

$$\begin{array}{ccc}\hfill f_1\left(Fx\overline{a}\right)+F\overline{x}\overline{g_2\left(a\right)}& =& F\overline{a}{g}_2\left(x\right),\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill \overline{x}\overline{F}\overline{\overline{a}}+g_2\left(x{g}_2\left(a\right)\right)& =& \overline{x}Fa+g_2\left(a\right)g_2\left(x\right).\hfill \end{array}$$

(11)

The case of $g_1\equiv \mathrm{id}$—Suppose for a moment that $x=\overline{x}$, for all $x\in \mathbb{F}$. Then Identities (3) and () imply $2x{f}_1\left(a\right)=0$, for all $x,a\in \mathbb{K}$. Hence, if $\mathrm{char}\mathbb{F}\ne 2$, then $f_1\equiv 0$ and $\mathcal{S}$ is a regular spread with $f(a,b)=Fb$ and $g(a,b)=a+Gb$ (corresponding to the irreducible quadratic polynomial $x^2-Gx-F$; the projective plane $\mathsf{PG}\left(\mathfrak{L}\right)$ is isomorphic to $\mathsf{PG}(2,\mathbb{K})$, where $\mathbb{K}=\mathbb{F}\left(\alpha \right)$, with $\alpha$ a root of the said polynomial). This also holds if $\mathrm{char}\mathbb{F}=2$ and $f_1\equiv 0$.

We now claim that, in the above case, the spread $\mathcal{S}$, and hence also $\mathfrak{L}$, arises from the field extension $\mathbb{K}/\mathbb{F}$. First note that

$$\mathcal{S}=\left\{\langle e_1,e_2\rangle \right\}\cup \{\langle (a,b,1,0),(Fb,a+Gb,0,1)\rangle \mid a,b\in \mathbb{F}\}.$$

Write a generic member of $\mathbb{K}$ as $a+b\alpha$ and consider the 1-space $(a+b\alpha ,1)\mathbb{K}$. We select the two particular vectors

$$(a+b\alpha ,1)and(a\alpha +b{\alpha }^2,\alpha )=(Fb+(a+Gb)\alpha ,\alpha )$$

and write these as vectors of ${\mathbb{F}}^4$ with respect to the basis $\left(\right(1,0),(\alpha ,0),(0,1),(0,\alpha \left)\right)$. This yields the two vectors $(a,b,1,0)$ and $(Fb,a+Gb,0,1)$ and the claim follows.

Now suppose $\mathrm{char}\mathbb{F}=2$ and $f_1\neg \equiv 0$. Assume first that $G\ne 0$. Let ${\mathbb{F}}^{\prime }$ be the set of all elements $x\in \mathbb{F}$ such that $f_1\left(x\right)=0$. If $x,y\in {\mathbb{F}}^{\prime }$, then by Identity (3), we see that $xy\in {\mathbb{F}}^{\prime }$ and by linearity also $x+y\in {\mathbb{F}}^{\prime }$. Hence ${\mathbb{F}}^{\prime }$ is a subfield of $\mathbb{F}$. Since we assume $f_1\neg \equiv 0$, there exists $t\in \mathbb{F}$ with $t=f_1\left(t^{\prime }\right)\ne 0$ for some $t^{\prime }\in \mathbb{F}$, and we fix such t and $t^{\prime }$. Also, ${\mathbb{F}}^2\subseteq {\mathbb{F}}^{\prime }$ as $f\left(x^2\right)=xf\left(x\right)+xf\left(x\right)=0$, by Identity (3).

Identity (8) says $f_1\left(f_1\left(a\right)\right)=G{f}_1\left(a\right)$. Let, for all $a\in \mathbb{F}$, the map $f:\mathbb{F}\to \mathbb{F}$ be defined as $f\left(a\right)=f_1\left(a\right)G^{-1}$. Then one easily checks $f\left(f\right(a\left)\right)=f\left(a\right)$, for all $a\in \mathbb{F}$.

Now we can write an arbitrary element $x\in \mathbb{F}$ as $x=f\left(x\right)+(x+f(x\left)\right)$. The element $x+f\left(x\right)$ lies in ${\mathbb{F}}^{\prime }$, as $f(x+f(x\left)\right)=f\left(x\right)+f\left(f\right(x\left)\right)=f\left(x\right)+f\left(x\right)=0$. Moreover, the element $f\left(x\right)t^{-1}$ belongs to ${\mathbb{F}}^{\prime }$; indeed, $f\left(f\left(x\right)t^{-1}\right)=f\left(f\left(x\right)\right)t^{-1}+f\left(x\right)f\left(t^{-1}\right)=f\left(x\right)(t^{-1}+f\left(t^{-1}\right))=0$, since $f\left(t^{-1}\right)=t^{-2}f\left(t\right)=t^{-2}f\left(f\left(t^{\prime }\right)\right)=t^{-2}f\left(t^{\prime }\right)=t^{-2}t=t^{-1}$. Hence we can write every element $x\in \mathbb{F}$ as $x=x^{\prime }+x^{\prime \prime }t$, with $x^{\prime },x^{\prime \prime }\in {\mathbb{F}}^{\prime }$. This decomposition is unique since, if x would also be written as $x_0^{\prime }+x_0^{\prime \prime }t$, with $x_0^{\prime },x_0^{\prime \prime }\in {\mathbb{F}}^{\prime }$, then $(x^{\prime \prime }+x_0^{\prime \prime })t\in {\mathbb{F}}^{\prime }$, which means, again using Identity (3) (translated to f, i.e., $f\left(ab\right)=af\left(b\right)+bf\left(a\right)$), that $(x^{\prime \prime }+x_0^{\prime \prime })f\left(t\right)=0$, implying $0=f\left(t\right)=t$, a contradiction, or $x^{\prime \prime }=x_0^{\prime \prime }$, which we had to prove. Hence $\mathbb{F}$ is a quadratic extension of ${\mathbb{F}}^{\prime }$; more exactly, $\mathbb{F}={\mathbb{F}}^{\prime }\left(t\right)$.

Note that $f\left(x\right)t^{-1}\in {\mathbb{F}}^{\prime }$, for all $x\in \mathbb{F}$, implies in particular $G^{-1}=\left(t{G}^{-1}\right)t^{-1}=\left(f_1\left(t^{\prime }\right)g^{-1}\right)t^{-1}=f\left(t^{\prime }\right)t^{-1}\in {\mathbb{F}}^{\prime }$. Hence $G\in {\mathbb{F}}^{\prime }$. Also, putting $a=x=1$ in Identity (10), we deduce $f_1\left(F\right)=0$, hence $F\in {\mathbb{F}}^{\prime }$.

We conclude that, if we write every element $a\in \mathbb{F}$ as $a_1+a_{2}t$, with $a_1,a_2\in {\mathbb{F}}^{\prime }$, then

$$M(a,b)=\left(\begin{array}{cc}{a}_1+a_{2}t& G{a}_{2}t+F(b_1+b_{2}t)\\ b_1+b_{2}t& a_1+a_{2}t+G{b}_1\end{array}\right).$$

The determinant of $M(a,b)$ is $(a_1^2+G{a}_1{b}_1+F{b}_1^2)+t^2(a_2^2+G{a}_2{b}_2+F{b}_2^2)$. Since $F,G\in {\mathbb{F}}^{\prime }$, this is the norm of a quaternion algebra $\mathbb{H}$ over ${\mathbb{F}}^{\prime }$, with basis $\{1,\alpha ,t,\alpha t\}$, with $\alpha$ a root of $x^2+Gx+F=0$, and $\alpha t=t(\alpha +G)$. Writing out the multiplication explicitly, one indeed sees that $\mathcal{M}$ is a quaternion algebra over ${\mathbb{F}}^{\prime }$ with above norm form and given multiplication rule for $\alpha$ times t.

To see that $\mathcal{S}$, and hence $\mathfrak{L}$, is obtained from the extension of $\mathbb{F}$ to $\mathbb{H}$, we write every element of $\mathbb{H}$ in the form $(a_1+b_1\alpha )+(a_2+b_2\alpha )t$ and associate it to the vector $(a_1+a_{2}t,b_1+b_{2}t)\in \mathbb{F}\times \mathbb{F}$. The rest is similar to the arguments above for the case $f_1\equiv 0$, taking into account that we now must multiply with $\alpha$ from the right to obtain the second vector.

Now suppose $\mathrm{char}\mathbb{F}=2$ and $G=0$. Identity (8) says $f_1\left(f_1\left(a\right)\right)=0$, for all $a\in \mathbb{F}$ (keeping in mind that $f_1\left(1\right)=0$). This time one calculates, using Equation (3) that, for an arbitrary $t\in \mathbb{F}\setminus {\mathbb{F}}^{\prime }$ (where ${\mathbb{F}}^{\prime }$ is again the subfield consisting of those elements x of $\mathbb{F}$ for which $f_1\left(x\right)=0$), one has $f_1\left(t{f}_1{\left(t\right)}^{-1}\right)=1$. So we set $u=:t{f}_1{\left(t\right)}^{-1}$. Then we can write every element a of $\mathbb{F}$ uniquely as a sum $a_1+a_{2}u$, with $a_1,a_2\in {\mathbb{F}}^{\prime }$. Moreover, $a_2=f_1\left(a\right)$, since $f_1(a+f_1\left(a\right)u)=f_1\left(a\right)+f_1\left(a\right)f_1\left(u\right)=0$. Hence we have

$$M(a,b)=\left(\begin{array}{cc}{a}_1+a_{2}u& a_2+F(b_1+b_{2}u)\\ b_1+b_{2}u& a_1+a_{2}u+b_2\end{array}\right).$$

This again defines a quaternion algebra $\mathbb{H}$ with norm form

$$\left(a_1^2+a_1{b}_2+\left(F{u}^2\right)b_2^2\right)+u^{-2}\left({\left(a_2{u}^2\right)}^2+\left(a_2{u}^2\right)b_1+\left(F{u}^2\right)b_1^2\right).$$

Similarly as before one shows that $\mathcal{S}$ is obtained from the extension of $\mathbb{F}$ to $\mathbb{H}$.

This completes the analysis for the case $g_1\equiv \mathrm{id}$. From now we assume that $g_1$ is not the identity.

Reduction—We start by reducing the number of identities. From Identity (3) follows that $(x-\overline{x})f_1\left(a\right)=(a-\overline{a})f_1\left(x\right)$, for all $a,x\in \mathbb{F}$. Hence there is a constant C such that $f_1\left(x\right)=C(x-\overline{x})$, for all $x\in \mathbb{F}$ (note that possibly $C=0$). This determines all the maps $f_1,f_2,g_1,g_2$ in function of the constants $C,F,G$ and the (nontrivial) field endomorphism $x\mapsto \overline{x}$. Indeed,

$$\left\{\begin{array}{c}{f}_1\left(x\right)=C(x-\overline{x}),\hfill \\ f_2\left(x\right)=F\overline{x},\hfill \\ g_1\left(x\right)=\overline{x},\hfill \\ g_2\left(x\right)=C(x-\overline{x})+G\overline{x}.\hfill \end{array}\right.$$

This replaces Identities (3)–() above.

We can now rewrite Identity () as

$$\begin{array}{ccc}\hfill (\overline{C}+C-G)(\overline{a}-\overline{\overline{a}})& =& 0,\hfill \end{array}$$

(12)

which readily implies $G=\overline{C}+C$, and hence, by Identity (7), $g_2\left(x\right)=Cx+\overline{C}\overline{x}$. Now Identity (8) can be rewritten as

$$\begin{array}{ccc}\hfill (F+C\overline{C})(a-\overline{\overline{a}})& =& 0.\hfill \end{array}$$

(13)

Identity (10) reduces to

$$\begin{array}{ccc}\hfill C\overline{F}& =& \overline{\overline{C}}F,\hfill \end{array}$$

(14)

whereas Identity () reduces to, taking into account Identity (13),

$$\begin{array}{ccc}\hfill F+C\overline{C}& =& \overline{F}+\overline{C}\overline{\overline{C}}.\hfill \end{array}$$

(15)

The case where $g_1$ is an involution—Suppose for a moment that $a=\overline{\overline{a}}$, for all $a\in \mathbb{F}$. Then $g_1$ is surjective. Identity (15) implies $F=\overline{F}$ and we have

$$M(a,b)=\left(\begin{array}{cc}a& C(a-\overline{a})+F\overline{b}\\ b& \overline{a}+Cb+\overline{C}\overline{b}\end{array}\right).$$

We perform the coordinate change mentioned in Formula (1) with $k=C$. This transforms $M(a,b)$ into (and we use the same notation $M(a,b)$ and set $K=F+C\overline{C}$)

$$M(a,b)=\left(\begin{array}{cc}a& K\overline{b}\\ b& \overline{a}\end{array}\right).$$

Let $\mathbb{K}$ be the fix field of $g_1$. Then $F=\overline{F}$ belongs to $\mathbb{K}$ and hence, so does K. The latter cannot be written as $z\overline{z}$, for any $z\in \mathbb{F}$, as otherwise $M(z,1)$ is singular, a contradiction. Hence this defines a quaternion algebra $\mathbb{H}$ over $\mathbb{K}$ with norm form $a\overline{a}-Kb\overline{b}$, with both $a,b\in \mathbb{F}$ considered as pairs of $\mathbb{K}$ in the natural way with respect to the field extension $\mathbb{F}/\mathbb{K}$.

Similarly (but even simpler) as before one shows that $\mathcal{S}$ is obtained from the extension of $\mathbb{F}$ to $\mathbb{H}$.

The case where $g_1$ has order at least 3—Hence from now on we may assume that $g_1$ is not an involution. This implies, by Identity (13), that $F=-C\overline{C}$. Identities (14) and (15) become redundant. So we have $M(a,b)=\left(\begin{array}{cc}a& C(a-\overline{a})-C\overline{C}\overline{b}\\ b& \overline{a}+Cb+\overline{C}\overline{b}\end{array}\right).$ Setting $a=-Cb$, we obtain $M(Cb,b)=\left(\begin{array}{cc}-Cb& -C^{2}b\\ b& Cb\end{array}\right),$ which has determinant 0 and hence does not define any legal member of $\mathcal{M}$.

Fix groups—We now determine the fix groups of the spreads found in the previous paragraphs.

Let $\theta (M,\sigma )$ be a semi-linear transformation in the vector space underlying S, with matrix M and field automorphism $\sigma$. Suppose $\theta (M,\sigma )$ stabilises each line of $\mathcal{S}$. Then $e_1$ and $e_2$ are mapped to points of $L_1=\langle e_1,e_2\rangle$, and $e_3$ and $e_4$ are mapped to points of $L_2=\langle e_3,e_4\rangle$. So M looks like

$$M=\left(\begin{array}{cccc}x& y& 0& 0\\ z& u& 0& 0\\ 0& 0& x^{\prime }& y^{\prime }\\ 0& 0& z^{\prime }& u^{\prime }\end{array}\right).$$

Expressing that $\theta (M,\sigma )$ stabilises each member $\langle (a,b,1,0),\left(f\right(a,b),g(a,b),0,1)\rangle$ of $\mathcal{S}$ results, by linear algebra, in the equalities

$$\left\{\begin{array}{c}{a}^{\sigma }x+b^{\sigma }y=a{x}^{\prime }+f(a,b)z^{\prime },\hfill \\ a^{\sigma }z+b^{\sigma }u=b{x}^{\prime }+g(a,b)z^{\prime },\hfill \\ f{(a,b)}^{\sigma }x+g{(a,b)}^{\sigma }y=a{y}^{\prime }+f(a,b)u^{\prime },\hfill \\ f{(a,b)}^{\sigma }z+g{(a,b)}^{\sigma }u=b{y}^{\prime }+g(a,b)u^{\prime },\hfill \end{array}\right.$$

which must hold for all $a,b\in \mathbb{F}$. Setting $a=1$ and $b=0$, taking into account $f(1,0)=0$ and $g(1,0)=1$, we deduce $(x,y,z,u)=(x^{\prime },y^{\prime },z^{\prime },u^{\prime })$. This implies

$$\left\{\begin{array}{c}(a^{\sigma }-a)x+b^{\sigma }y-f(a,b)z=0,\hfill \\ bx+(g(a,b)-a^{\sigma })z-b^{\sigma }u=0,\hfill \\ f{(a,b)}^{\sigma }x+(g{(a,b)}^{\sigma }-a)y-f(a,b)u=0,\hfill \\ by-f{(a,b)}^{\sigma }z+(g(a,b)-g{(a,b)}^{\sigma })u=0,\hfill \end{array}\right.$$

for all $a,b\in \mathbb{F}$.

Suppose now first that $\sigma =\mathrm{id}$. Then the first (and also the last) equation implies that, if $b\ne 0$, then $f(a,b)b^{-1}$ is independent of $a,b$. This is only the case if $f_1\equiv 0$, which in our examples only holds in Case $\left(i\right)$ of A (if in the case $g_1\ne \mathrm{id}$, $C=0$, with above notation, then the inverse coordinate change as given above transforms the matrices to a case where $f_1\neg \equiv 0$). Hence $f(a,b)=Fb$ and $g(a,b)=a+Gb$. If $z=0$, then $y=0$ by the first equation, and $x=u$ by the second, hence we have the identity. So we may assume $z=1$. Then $y=F$ by the first equation, and $u=x-G$. Hence we get a group consisting of the identity and linear maps with $4\times 4$ block matrices having two identical $2\times 2$ blocks $\left(\begin{array}{cc}x& F\\ 1& x-G\end{array}\right)$ on the diagonal, and 0 elsewhere. This group clearly acts sharply transitively on $L_1$ (and hence on every line of $\mathcal{S}$).

Now suppose $\sigma \ne \mathrm{id}$. The second equality implies, setting $b=0$, that either $z=0$, or $\overline{a}=a^{\sigma }$, for all $a\in \mathbb{F}$. If $z=0$, then the first equation implies first (setting $a=0$) that $y=0$, and then (for general a) $a=a^{\sigma }$ (as $x\ne 0$), for all $a\in \mathbb{F}$, a contradiction. Hence $\overline{a}=a^{\sigma }$, for all $a\in \mathbb{F}$ and we are in the Galois case. Then we may assume $f(a,b)=K\overline{b}$ and $g(a,b)=\overline{a}$. With this it is now easy to calculate $x=u=0$ and $y=Kz$. This yields a unique involution (the Galois involution).

This completes the proof of A.

Remark 1.(We again refer to [6] for undefined notions in the theory of buildings.) A illustrates three phenomena that can occur in order to construct subcomplexes of spherical buildings that are also buildings. The first phenomenon is Galois descent, where one considers the fixed complex of a Galois group (here this group is the one generated by $g_1$). This phenomenon is completely understood; a classification can be found in [5]. The second is an analogue of this, but then using a linear group: one considers the fixed complex of a linear automorphism group. Usually this group is larger than its Galois analogue (and, remarkably, als the subcomplex is—dimensionwise in the sense of algebraic groups—usually larger). Also in the situation of the present paper, we can observe this: in the linear case, the group acts transitively on each spread line. One could call this linear descent. This phenomenon is less well understood, and there is no classification, but only partial results available. We refer to [4] for a lot of background and a systematic treatment of these two phenomena. The third does not use a group, but is simply a subgeometry constructed in an algebraic (here using a subfield of a quaternion algebra) or geometric way; its fix group is trivial. We could call this geometric descent. As geometric descent seems to be a rare phenomenon, it would be interesting to determine other examples of the third phenomenon, and perhaps classify under mild conditions. At present, and also inspired by the results of the present paper, the author is tempted to think that geometric descent is a characteristic 2 or 3 phenomenon. Is this really true?

4. Groups of Projectivities

With the notation of sec2, we have seen that an arbitrary regular line spread $\mathcal{S}$ of $\mathsf{PG}(3,\mathbb{F})$ can be represented as a set of lines

$$\{\langle (1,0,0,0),(0,1,0,0)\rangle \}\cup \{\langle (a,b,1,0),(\alpha b,a+\beta b,0,1)\rangle \mid a,b\in \mathbb{F}\},$$

for some constants $\alpha ,\beta \in {\mathbb{F}}^{\times }$ such that the polynomial $x^2-\beta x-\alpha$ is irreducible over $\mathbb{F}$. We now determine the projectivity group ${\Pi }_{\mathcal{S}}\left(L_1\right)$, with $\langle 1,0,0,0),(0,1,0,0)\rangle =L_1\in \mathcal{S}$ as before.

We first consider a special case. Let $L\in \mathcal{S}\setminus \{L_1,L_2\}$ be arbitrary. Then we calculate $L_1{⊼}_{L_{12}}{L}_2{⊼}_L{L}_1$. The first perspectivity maps $(x,y,0,0)$ to $(0,0,x,y)$, and the second maps $(0,0,x,y)$ back to the point $(ax+\alpha by,bx+(a+\beta b)y,0,0)$. In binary coordinates, this yields the map

$$\left(\begin{array}{c}x\\ y\end{array}\right)\mapsto \left(\begin{array}{cc}a& \alpha b\\ b& a+\beta b\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=:M(a,b)\left(\begin{array}{c}x\\ y\end{array}\right),$$

using similar notation as in sec2. Note that this already defines a sharply transitive group G acting on the points of $L_1$.

Since the matrices $M(a,b)$, $(a,b)\ne (0,0)$, form a multiplicative group, each projectivity $L_1{⊼}_K{L}_2{⊼}_L{L}_1$, $K,L\in \mathcal{S}\setminus \{L_1,L_2\}$, has this form, as this can be written as $L_1{⊼}_K{L}_2{⊼}_{L_{12}}{L}_1{⊼}_{L_{12}}{L}_2{⊼}_L{L}_1$.

Moreover, since the matrices $M(a,b)$ form an additive group acting sharply transitively on $\mathcal{S}\setminus \left\{L_1\right\}$ (as deduced in sec2), the same remains true if we substitute $L_2$ by an arbitrary member of $\mathcal{L}\setminus \left\{L_1\right\}$. Now we can break up any sequence of projectivities

$$L_1{⊼}_{K_1}{M}_1{⊼}_{K_2}{M}_2{⊼}_{K_3}{M}_3{⊼}_{K_4}{M}_4{⊼}_{K_5}\cdots$$

into subsequences of self-projectivities as follows

$$\left(L_1{⊼}_{K_1}{M}_1{⊼}_{K_2}{L}_1\right)·\left(L_1{⊼}_{K_2}{M}_2{⊼}_{K_3}{L}_1\right)·\left(L_1{⊼}_{K_3}{K}_4{⊼}_{K_5}{L}_1\right)·\cdots ,$$

which shows that the full group of projectivities ${\Pi }_{\mathcal{S}}\left(L_1\right)$ of $L_1$ is exactly G. This shows that, if a line spread of $\mathsf{PG}(3,\mathbb{F})$ is regular, then all other conclusions in C hold.

Now assume a spread $\mathcal{S}$ of $\mathsf{PG}(3,\mathbb{F})$ is not regular. Then there exist four lines $K,L,M,N\in \mathcal{S}$ admitting a common transversal X such that the lines $K,L,M$ admit a transversal Y not intersecting N. Clearly the self-projectivity $K{⊼}_{L}M{⊼}_{N}K$ fixes $K\cap X$ but moves $K\cap Y$. Hence $\Pi \left(K\right)$ does not act freely on K and hence also not sharply transitively. Moreover, if the fix group T of $\mathcal{S}$ fixed $x:=K\cap X$, then we claim it would be the identity. Indeed, T then fixes the plane $\langle L,x\rangle$, and hence fixes it pointwise as every point of the plane off L is fixed (because each such point is the intersection of $\langle L,x\rangle$ with a spread line). This holds for arbitrary $L\in \mathcal{S}$ and the claim follows.

This completes the proof of C.