Weyl substructures of spherical buildings

Proof. We proceed by induction on \(n\), the result for \(n=3\) following directly from the definition of ovoid of a polar space of type \(\mathsf{D_3}\) (remember the Weyl diagram of \(\Sigma\) is \(\mathsf{A_{3;1}^2}=\mathsf{D_{3;1}^1}\) in this case).

Let \(n>3\). First suppose \(\Sigma\) is ovoidal of type 1. Then the Weyl diagram of \(\Sigma\) is \(\mathsf{A}_{2m+1;n}^2\), \(n=2m+1\). Let \(L\in \Sigma\) be a line of \(\mathsf{PG}(n,\mathbb{K})\). The induction hypothesis implies that each plane \(\pi\) containing \(L\) is contained in a unique subspace \(S\) of dimension \(3\) belonging to \(\Sigma\). Considering the residue of \(S\) we see that \(\pi\) does not contain any member of \(\Sigma\) except for \(L\), and that every point of \(\pi\) is contained in some line belonging to \(\Sigma\). This shows that the line set of \(\Sigma\) is a line spread of \(\mathsf{PG}(n,\mathbb{K})\).

Consider an arbitrary number of lines of \(\Sigma\). We may assume \(L\) is amongst these. Then, by induction, the subspaces of dimension \(3\) of \(\Sigma\) through \(L\) induce a spread in the residue at \(L\) of the subspace \(U\) generated by all these lines. Since the lines of \(\Sigma\) induce spreads in each of these subspaces of dimension \(3\), we see that the lines of \(\Sigma\) induce a spread in \(U\). Hence the lines of \(\Sigma\) define a composition line spread (with the mentioned additional property) and induction implies that the subspace generated by an arbitrary number of lines of \(\Sigma\) belongs to \(\Sigma\). Clearly, by taking residues, every member of \(\Sigma\) is generated by lines of \(\Sigma\), which completes the proof of the “only if” direction of the proposition.

Next assume that \(\Sigma\) is a composition line spread satisfying the mentioned additional property. First assume \(n=5\). We have to show that the solids (which is short for \(3\)-dimensional subspaces) of \(\Sigma\) through a given line \(L\in\Sigma\) form a bi-spread in the residue of \(L\). Let \(\pi\) be any plane through \(L\) and pick \(x\in\pi\setminus L\). Since \(\Sigma\) is a line spread, there is a line \(M\in\Sigma\) containing \(x\). Since \(\Sigma\) is a composition spread, the solid generated by \(L\) and \(M\) belongs to \(\Sigma\) and clearly contains \(\pi\). Now let \(U\) be a \(4\)-dimensional subspace of \(\mathsf{PG}(n,\mathbb{K})\) containing \(L\). Select two distinct solids \(S_1,S_2\in\Sigma\) containing \(L\). Next, select lines \(L_1\subseteq S_1\) and \(L_2\subseteq S_2\) distinct from \(L\). The property of \(\Sigma\) being a composition line spread implies that the solid \(S\) generated by \(L_1,L_2\) belongs to \(\Sigma\) and is disjoint from \(L\). Hence \(U\cap S\) is a plane, and since the set of lines of \(\Sigma\) in \(S\) defines a bi-spread of lines, the plane \(U\cap S\) contains a member \(N\) of it. Hence \(U\) contains the solid of \(\Sigma\) generated by \(L\) and \(N\). So \(\Sigma\) is an ovoidal subcomplex of class I.

Now assume \(n\geq 7\). We have to show that, given a \((2i-1)\)-dimensional subspace \(U\) belonging to \(\Sigma\), and a \((2i+3)\)-dimensional subspace \(W\in\Sigma\) containing \(U\), \(2i+3\leq n\), the members of dimension \(2i+1\) of \(\Sigma\) containing \(U\) and contained in \(W\) form a bi-spread of the residue of \(U\) in \(W\). If \(2i+3\leq n-2\), then this follows from the induction hypothesis, so assume \(2i+3=n\), that is, \(W\) is the whole space. Select a line \(L\in\Sigma\) in \(U\). Then in the residue of \(L\), the members of \(\Sigma\) define a composition line spread satisfying the additional property thanks to the case \(n=5\) above. The induction hypothesis now implies the assertion. \(\square\)

Proof. All spreads \(\mathcal{S}_S\) are classified in [37, Theorem 1]. There are four classes of examples, two of which are not rigid (the regular spread and the spread emerging from Galois descent). Let us handle only one of the two classes of rigid spreads; the other can be dealt with in a similar fashion.

We use the notation of [37]. Let \(\mathbb{F}\) be a field with \(\mathop{\mathrm{\mathsf{char}}}\mathbb{F}=2\), let \(\mathbb{F}’\) be a subfield of \(\mathbb{F}\) and \(t\in\mathbb{F}\setminus\mathbb{F}’\) such that \(t^2\in\mathbb{F}’\) and assume that \(\mathbb{F}\) is \(2\)-dimensional over \(\mathbb{F}’\), that is, every element \(x\) of \(\mathbb{F}\) can be written as \(x=x_1+x_2t\), with \(x_1,x_2\in\mathbb{F}’\). Let \(F,G\in{\mathbb{F}’}^\times\) be such that the quadratic polynomial \(x^2+Gx+F\) is irreducible over \(\mathbb{F}’\). For each \(a=a_1+a_2t\) and \(b=b_1+b_2t\), \(a,i,b_i\in\mathbb{F}’\), \(i=1,2\), we define the points \(p(a,b)\) and \(q(a,b)\) in standard coordinates as \[p(a,b)=(a,b,1,0)\mbox{ and }q(a,b)=(Ga_2t+Fb,a+Gb_1,0,1).\]

Then \(\mathcal{S}_S=\{\left\langle p(a,b),q(a,b)\right\rangle\mid a,b\in \mathbb{F}\}\cup\{\left\langle (1,0,0,0),(0,1,0,0)\right\rangle\}\), where for clarity, the line generated by two points \(x,y\) is denoted as \(\left\langle x,y\right\rangle\). Now let \(L(a,b)\in\mathcal{S}_S\) denote the line generated by \(p(a,b)\) and \(q(a,b)\), and set \(L(\infty)=\left\langle 1,0,0,0),(0,1,0,0)\right\rangle\). One easily calculates that a set of equations for \(L(a,b)\) is given by (ordering the coordinates as \((X_1,X_2,X_3,X_4)\)) \[\begin{cases} X_1+aX_3+(Ga_2t+Fb)X_4=0,\\ X_2+bX_3+(a+Gb_1)X_4=0. \end{cases}\]

Hence in the dual projective space, a generic spread line is given in (dual) coordinates by \[\left\langle (1,0,a,Fb+Ga_2t),(0,1,b,a+Gb_1)\right\rangle.\]

Rearranging the order of the coordinates (writing them in opposite order), and replacing \(a_1\) with \(a1+Gb_1\), we obtain the projectively equivalent form \[\left\langle (a,b,1,0), (Fb+Ga_2t,a+Gb_1,0,1)\right\rangle,\] which is exactly a generic member of \(\mathcal{S}_S\) above. Hence \(\mathcal{S}_S\) is self-dual and consequently, it is a bi-spread.

Now suppose for a contradiction that \(\mathcal{S}_S\) is a symplectic spread. A symplectic form is given by \[\sum_{1\leq i<j\leq 4}\lambda_{ij}(X_iY_j+X_jY_i), \ \ \ \lambda_{ij}\in\mathbb{F}.\]

Expressing that \((1,0,0,0)\) and \((0,1,0,0)\) are perpendicular, and similarly for \((0,0,1,0)\) and \((0,0,0,1)\), we deduce \(\lambda_{12}=\lambda_{34}=0\). Now \(p(a,b)\) is perpendicular to \(q(a,b)\) if, and only if, \[\lambda{14}a+\lambda_{24}b+\lambda_{13}(Ga_2t+Fb)+\lambda_{23}(a+Gb_1)=0.\]

Since this must hold for every \(a_1,a_2,b_1,b_2\in\mathbb{F}’\), we easily deduce \[\begin{cases}\lambda_{14}+\lambda_{23}=0,\\ \lambda_{14}+G\lambda_{13}+\lambda_{23}=0,\\ \lambda_{24}+F\lambda_{13}+G\lambda_{23}=0,\\ \lambda_{24}+F\lambda_{13}=0. \end{cases}\]

We now see \(\lambda_{13}=\lambda_{14}=\lambda_{23}=\lambda_{24}=0\), a contradiction, which proves the lemma. \(\square\)

Proof. Let first \(\Sigma\) be the complex of all totally singular subspaces of a non-degenerate alternating form in \(\mathsf{PG}(2m+1,\mathbb{K})\). Let \(U\subseteq W\) be two totally singular subspace with \(2+\mathop{\mathrm{\mathsf{dim}}}U=\mathop{\mathrm{\mathsf{dim}}}W\leq m\). Let \(U^\theta\) and \(W^\theta\) be their respective images under the corresponding symplectic polarity. Then we have to show that the set of totally isotropic subspaces \(V\) with \(U\subseteq V\subseteq W\) and \(\mathop{\mathrm{\mathsf{dim}}}V=1+\mathop{\mathrm{\mathsf{dim}}}U\), together with the images \(V^\theta\), forms a simple ovoidal subcomplex. This defines indeed an ovoid-spread pairing of the corresponding generalized digon as every subspace \(V\) with \(U\subseteq V\subseteq W\) and \(\mathop{\mathrm{\mathsf{dim}}}V=1+\mathop{\mathrm{\mathsf{dim}}}U\) is totally isotropic. Likewise, if \(U\) is totally isotropic with \(\mathop{\mathrm{\mathsf{dim}}}U=m-1\), then every subspace \(V\) with \(U\leq V\leq U^\theta\) and \(\mathop{\mathrm{\mathsf{dim}}}V=m\) is totally isotropic and hence defines a simple ovoidal subcomplex of the corresponding residue.

Conversely, let \(\Sigma\) be an ovoidal subcomplex with Weyl diagram \(\mathsf{^2A}_{2m+1;m+1}\). We use induction on \(n\) to prove the assertion. We first claim that every point \(x\) of \(\mathsf{PG}(2m+1,\mathbb{K})\) is contained in a (point-hyperplane) flag of \(\Sigma\). We know that there is at least one point-hyperplane pair \((p,H)\) in \(\Sigma\). Induction implies that every line \(L\) through \(p\) in \(H\) is contained in a line-subhyperplane pair \((L,S)\) of \(\Sigma\) (and we fix such an \(L\)). By the definition of ovoidal subcomplex, every hyperplane \(G\) through \(S\) (in particular one that contains \(x\)) is contained in a unique point-hyperplane pair \((q,G)\) belonging to \(\Sigma\). Replacing \((p,H)\) by\((q,G)\) if necessary, this means that we can assume that \(x\in H\), and we may also assume that \(px=L\). As before, induction implies that \(x\) is contained in a unique point-hyperplane pair \((x,X)\) with \(S\subseteq X\) and \(X\neq H\). This not only proves our claim, but it also shows that, given \(H\), the point \(p\) is unique with \((p,H)\in\Sigma\) (as we may assume for a contradiction that also \((x,H)\in\Sigma\), and this leads, by the above argument, indeed to the contradiction \(X=H\)).

Now noting that in the previous argument \(p\in X\), we see that the relation \(p\sim q\) if \(q\) is contained in the unique hyperplane \(H\) such that \((p,H)\in\Sigma\) is symmetric and reflexive, and has the property that the set of points in relation to a given point is precisely a hyperplane. Hence we obtain a symplectic polarity, and using induction it is now routine to check that \(\Sigma\) consists precisely of the fixed flags of size \(2\). \(\square\)

Proof. Suppose first that \(\theta\) is non-trivial and pointwise fixes an ovoidal subcomplex \(\Sigma\). If \(\Sigma\) is of class 1, then this follows from Proposition 4.4. If \(\Sigma\) is of class 2, then it must be a duality since every point belongs to a member of \(\Sigma\) and hence would be fixed, leading to the identity. But then \(\theta\) must coincide with the polarity \(\rho\) associated to \(\Sigma\) by Proposition 4.3, because \(\theta\rho\) is type preserving and fixes \(\Sigma\) pointwise, and we just argued that only the identity can do this.

Next suppose that \(\theta\) is uniclass. If \(\theta\) is a duality, then Proposition 4.4 implies it is a symplectic polarity, and Proposition 4.3 shows that it pointwise fixes an ovoidal subcomplex of type 2. If \(\theta\) is non-trivial and type preserving, then by Proposition 4.4, it elementwise fixes a composition line spread \(\mathcal{S}\). In view of Proposition 4.1, it suffices to show that the spread induced by two members \(L_1,L_2\in\mathcal{S}\) in \(\left\langle L_1,L_2\right\rangle\), is a bi-spread. Let \(\pi\) be any plane i \(\left\langle L_1,L_2\right\rangle\). Pick \(x\in \pi\cap\pi^\theta\) and note that, by the comment following Proposition 4.4, \(x\neq x^\theta\). Since \(\left\langle x,x^\theta\right\rangle\) is fixed, \(x^\theta\) also belongs to \(\pi\cap\pi^\theta\), and so the \(\left\langle x,x^\theta\right\rangle\subseteq\pi\) with obviously \(\left\langle x,x^\theta\right\rangle\in\mathcal{S}\).

Remember that we obtain the identity by considering the Weyl structure consisting of the whole building. The corresponding Weyl diagram is full. \(\square\)

Proof. From Proposition 7.2.5 and Definitions 7.2.6 of [35], we deduce that the fix structure of a polarity is always an ovoid-spread pairing.

Now let \(\theta\) be a duality of \(\Gamma\) elementwise fixing an ovoid-spread pairing \((\mathcal{O},\mathcal{S})\). Let \(p\) be any point. Then by the definitions of ovoid and spread, \(p\) is at distance at most \(d\) from at least one member of \(\mathcal{O}\cup\mathcal{S}\), and we can make the corresponding shortest path longer so that, without loss of generality, we may assume that there is a unique flag \(\{x,L\}\), with \(x\in\mathcal{O}\) and \(L\in\mathcal{S}\) such that \(p\) has distance \(d-1\) to one of \(x\) or \(L\), and distance \(d\) to the other. Let us, to fix the ideas, assume that \(\delta_\Gamma(p,L)=d=\delta_\Gamma(p,x)+1\) (then \(d\) is odd; the case \(d\) even is handled similarly). Consider an arbitrary line \(M\) through \(p\) distinct from the one at distance \(d-1\) from \(L\). Then there is a unique flag \(\{x’,L’\}\), with \(x’\in\mathcal{O}\) and \(L’\in\mathcal{S}\) with \(\delta_\Gamma(M,L’)=d-1\) and \(\delta_\Gamma(M,x’)=d\). It follows that \(\theta\) interchanges the unique shortest path connecting \(x\) and \(L’\) with the unique shortest path connecting \(L\) and \(x’\). In particular \(\theta\) acts as an involution on the union of these paths, hence \(p^{\theta^2}=p\). The proposition follows. \(\square\)

Proof. The statement is trivial for \(d=1\), so we assume \(d\geq 2\).

Suppose that \(\theta\) fixes some point or line, say \(x\), not belonging to the ovoid. Let \(i\) be the minimal distance from \(x\) to some element of \(\mathcal{O}\), say \(\delta_\Gamma(x,p_x)=i\), \(p_x\in\mathcal{O}\). By the definition of ovoid, \(i\leq d\). We claim that \(\theta\) fixes each element \(y\) incident with \(x\). If \(i=d\), then, by parity, \(y\) is at distance \(d-1\) from some member \(p_y\in\mathcal{O}\). By minimality of \(i\), we have \(\delta_\Gamma(p_y,x)=i=d\), implying that the path of length \(d\) joining \(p_y\) with \(x\), which contains \(y\), is unique; hence \(y\) is fixed. Now assume \(1\leq i<d\). If \(\delta_\Gamma(p_x,y)=i-1\), then \(y\) is fixed by uniqueness of such a point with the given distances. Suppose now \(\delta_\Gamma(p_x,y)=i+1\). We select an element \(z\) at distance \(d-i\) from \(y\) and distance \(d+1\) from \(p_x\); then \(\delta_\Gamma(z,x)=d+1-i\). Also, again by parity, \(z\) is at distance at most \(d-1\) from some point \(p_z\in\mathcal{O}\). Then \(\delta_\Gamma(p_z,x)\leq (d-1)+(d+1-i)=2d-i<2d\). This implies that every line or point on a minimal path from \(p_z\) to \(x\) is fixed. In particular \(y\) is fixed and the claim is proved.

We can now interchange the roles of \(x\) and an arbitrary element incident with \(x\). Then all elements at distance \(2\) from \(x\) are fixed. Going on like this, all points and lines are fixed by connectivity, and \(\theta\) is the identity. \(\square\)

Proof. We show this by induction on \(n-i\geq 1\). For \(n-i=1\), this is by definition the case, since \(\Sigma\) is then just a simple ovoidal subcomplex, hence an ovoid. Now assume \(n-i\geq 2\). Let \(M\) be a maximal singular subspace of \(\Gamma\) and pick a point \(p\in\Sigma\). If \(p\in M\), then looking in the residue at \(p\), the elements of \(\Sigma\) containing \(p\) form an ovoidal subcomplex of the residue, and hence the induction hypothesis yields a subspace of dimension \(n-i-1\) in \(M\). If \(p\notin M\), then there is a unique maximal singular subspace \(M’\) containing \(p\) and intersecting \(M\) in a subspace \(H\) of dimension \(n-2\). The previous argument yields a subspace of dimension \(n-i-1\geq 1\) through \(p\) in \(M’\), which necessarily intersects \(H\), and hence \(M\), in at least a point \(p’\). With \(p’\) in the role of \(p\), the previous argument now concludes the proof. \(\square\)

Proof. We first claim that \(U(\Sigma)\) is a subspace. Let \(x_1,x_2\in U(\Sigma)\) be two collinear points. Select a maximal singular subspace \(M\) containing \(x_1\) and \(x_2\). Lemma 4.10 yields singular subspaces \(U_i\ni x_i\) of dimension \(n-2\) contained in \(M\) and belonging to \(\Sigma\). A dimension argument implies that these have a subspace \(U\) of dimension \(n-3\geq 0\) in common. The definition of ovoidal subcomplex requires that \(U_1=U_2\); hence \(x_1\) and \(x_2\) are contained in a common higher dimensional member of \(\Sigma\), showing the claim.

Now by definition of ovoidal subcomplex, \(U(\Sigma)\) is ideal, and it is of corank \(1\) by Lemma 4.10. \(\square\)

Proof. If \(U(\Sigma)\) is a subspace, then the definition of ovoidal subcomplex implies that \(U(\Sigma)\) is ideal, and it is of corank \(i\) by Lemma 4.10.

Let the Weyl diagram of \(\Sigma\) be \(\mathsf{B}_{n;n-i}^1\) or \(\mathsf{D}_{n;n-i}^1\). If \(U(\Sigma)\) is not a subspace, then arguing as in the first paragraph of the proof of Proposition 4.12, we obtain two distinct singular subspaces \(U_1\) and \(U_2\) of dimension \(n-i-1\) in a maximal singular subspace \(M\). The fact that the span of \(U_1\) and \(U_2\) is contained in the convex closure of \(\Sigma\) contradicts now condition (WCC) and the weaker condition mentioned in the proposition. \(\square\)

Proof. We may restrict the vector space \(V\) to \(\left\langle A,B\right\rangle\), and afterwords extend the linear map over \(W\) with the identity on a complementary subspace. Hence we replace \(W\) with \(W\cap\left\langle A,B\right\rangle\).

Set \(\ell=\mathop{\mathrm{\mathsf{dim}}}W\cap U\). Let \(\{e_j\mid 1\leq j\leq\ell\}\) be a basis for \(W\cap U\), and extend this to a basis \(\{e_j\mid 1\leq j\leq d\}\) of \(W\). For each \(j \in\{1,2,\ldots,\ell\}\), there exist unique \(a_j\in A\) and \(b_j\in B\) such that \(e_j=a_j+b_j\) (by complementarity of \(A\) and \(B\) in \(\left\langle A,B\right\rangle\supseteq U\)). Similarly, there exist \(u_j,u’_j\in U\) such that \(e_j=u_j+a_j=u’_j+b_j\), \(\ell+1\leq j\leq d\). We claim that \(\{a_j\mid 1\leq j\leq d\}\) is a basis for \(A\). Indeed, the projection of \(\left\langle e_{\ell+1},\ldots,e_d\right\rangle\) from \(U\) onto \(A\) is a subspace of \(A\) of dimension \(d-\ell\), which coincides precisely with \(\left\langle W,U\right\rangle\cap A\). Also, clearly, since \(A_\ell=:\left\langle a_j\mid 1\leq j\leq \ell\right\rangle\) is the projection of \(W\cap U\) onto \(A\) from \(B\), we deduce that \(\mathop{\mathrm{\mathsf{dim}}}A_\ell=\ell\). If some vector of \(A_\ell\), say \(k_1a_1+\cdots + k_\ell a_\ell\) would belong to \(\left\langle W,U\right\rangle\), then the vectors \(k_1e_1+\cdots k_\ell e_\ell\), \(k_1a_1+\cdots + k_\ell a_\ell\) and \(k_1b_1+\cdots + k_\ell b_\ell\) are linearly dependent vectors in \(\left\langle W,U\right\rangle\), contrary to our assumptions. The claim now follows. Likewise, \(\{b_j\mid 1\leq j\leq d\}\) is a basis for \(B\). Hence, by the complementarity of \(W\) and both \(A\) and \(B\), The sets \(\{a_j,e_j \mid 1\leq j\leq d\}\) and \(\{b_j,e_j \mid 1\leq j\leq d\}\) are bases for \(V\). Then the linear map defined on the basis by \(e_j\mapsto e_j\) and \(a_j\mapsto b_j\), for all \(j\in\{1,2,\ldots,d\}\), fixes \(W\) pointwise and maps \(A\) to \(B\). From our definitions also follows that \(u_j\) is mapped onto \(u’_j\), for all \(j\in\{\ell+1,\ldots,d\}\), hence \(U\) is stabilised. It also easily follows from the definitions of the base vectors that \(\left\langle U,a_j\right\rangle=\left\langle U,b_j\right\rangle\), for \(1\leq j\leq d\). Hence all subspaces containing \(U\) are stabilised and the lemma is proved. \(\square\)

Proof. By [23] and [24, Main Result 1], there do not exist ideal subspaces of corank 1 in non-embeddable polar spaces. Hence we may assume that \(\Gamma\) is embeddable, say in the projective space \(\Pi\). Let \(X\) be the point set of \(\Gamma\) and \(Y\) the ideal subspace of corank \(i\) at most \(n-2\). Then our assumption and the main result of [4] yield a subspace \(S\) of \(\Pi\) such that \(S\cap X=Y\). Let \(T=S^\rho\), with \(\rho\) the underlying polarity, which is non-degenerate by uniqueness of the embedding. We claim that \(T\cap X=\varnothing\). Indeed, suppose for a contradiction that \(x\in T\cap X\). Then \(x\in T\setminus S\) as \(Y\) is non-degenerate. Pick a singular subspace \(A\) of \(Y\) of dimension \(n-i-2\). Then \(\mathop{\mathrm{\mathsf{Res}}}_Y(A)\) is an ovoid in \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(A)\), each point of which is collinear to the point corresponding to \(x\). This contradicts [15, Lemma 3.1.3]. Note that this argument in fact shows that no point of \(\Gamma\) is collinear to all points of \(Y\).

Let \(U\) be a singular subspace of \(\Gamma\) complementary to \(S\) (that is, of dimension \(i-1\)), which exists by the definition of subspace of corank \(i\). Set \(\Pi’=\left\langle T,U\right\rangle\). Then \(\Pi’\) intersects \(X\) in the point set \(X’\) of an induced polar space \(\Gamma’\) of rank at least \(i\) (because it contains \(U\)) and at most \(i\) (since \(T\) does not contain points of \(X\)), hence rank exactly \(i\) (note for the moment we do not know whether \(\Gamma’\) is non-degenerate, that is, whether it contains no point collinear to all others; we will prove this below). We claim that \(X’\cap S=\varnothing\). Indeed, suppose for a contradiction that \(x’\in X’\cap S\). Then \({x’}^{\perp}\) contains a singular subspace of \(\Pi’\) of dimension \(i-1\), necessarily disjoint from \(T\) (as \(T\cap X=\varnothing\)), and \({x’}^\perp\) also contains \(T\), as \(T=S^\rho\) and \(x’\in S\). Hence \({x’}^\rho\) contains \(\Pi’\), and so \(\left\langle x’,U\right\rangle\) is singular. But since \(\mathop{\mathrm{\mathsf{dim}}}\left\langle x’,U\right\rangle=i\), it intersects \(T\) in a point, a contradiction as \(T\cap X=\varnothing\). We can now also prove that \(\Gamma’\) is non-degenerate. Indeed, any point \(x’\) collinear to all points satisfies \(T\subseteq\Pi’\subseteq {x’}^\rho\), which implies \(x\in S\), contradicting the previous claim.

Hence \(\Gamma’\) is a true non-degenerate polar space and we can pick a maximal singular subspace \(U’\) of it disjoint from \(U\). Set \(U_0=\left\langle S,T\right\rangle\cap U\) and \(U_0’=\left\langle S,T\right\rangle\cap U’\). We claim dat \(\left\langle U_0,U_0’\right\rangle\) is disjoint from \(S\cap T\). This is equivalent with \(\left\langle S\cap T,U_0,U_0’\right\rangle=\left\langle S,T\right\rangle\). This, in turn, is equivalent to \[(S\cap T)^\rho\cap U_0^\perp\cap U_{0′}^\perp=\varnothing.\]

Set \(\mathop{\mathrm{\mathsf{dim}}}(S\cap T)=j-1\). Suppose first for a contradiction that \(U_0^\perp\cap U_0’\neq\varnothing\), say \(u_0’\in U_0^\perp\cap U_0’\). Since \(T\) is complementary to \(U_0\) in \(\left\langle T,S\right\rangle\), we find a line \(L\) containing \(u_0’\), a point of \(U_0\) and a point \(t\) of \(T\). Clearly \(L\) is singular, contradicting \(T\cap X=\varnothing\). Hence \(U_0^\perp\cap U_0’=\varnothing\). Similarly, \(U_{0′}^\perp U_0=\varnothing\). Then \(U_0^\perp\cap U’\) is a subspace of dimension \(j\) disjoint from \(\left\langle S,T\right\rangle\). It follows that \(\left\langle S,T\right\rangle\cap U_0^\perp=U_0\). Since \(U_{0′}^\perp\cap U_0=\varnothing\), the claim follows. Now Lemma 4.14 yields a collineation \(\varphi\) fixing \(S\) pointwise, stabilising all subspaces through \(T\) and mapping \(U\) to \(U’\). Since \(S\) is the axis of \(\varphi\), the latter also pointwise fixes \(Y\) and no other point of \(X\). We now show that \(\varphi\) preserves \(\Gamma\), that is, maps \(X\) into \(X\). It will then also follow that \(X\) is mapped onto \(X\) by considering the inverse of \(\varphi\). This will prove the proposition.

We first claim that each point of \(X\) in \(U^\perp\) is mapped into \(X\). Indeed, if \(u\in U^\perp\setminus U\), then \(\left\langle u,U\right\rangle\) has dimension \(i\) and hence contains a (unique) point \(y\) of \(Y\). Then \(\left\langle y,U\right\rangle=\left\langle u,U\right\rangle\) gets mapped onto \(\left\langle y,U’\right\rangle\). But \(y^\perp\) contains \(U\) and \(y^\rho\) contains \(T\), hence \(y^\rho\), which equals \(y^\perp\), contains \(\left\langle T,U\right\rangle\supseteq U’\). It follows that \(\left\langle y,U’\right\rangle\) is singular and so the image of \(u\) under \(\varphi\) belongs to \(X\).

Now let \(w\in X\) be arbitrary, but not contained in \(\left\langle U,U’\right\rangle\). We may suppose it is not collinear to all points of \(U\) and it is not contained in \(Y\). By the latter assumption, and the first paragraph of this proof, there exists a point \(y\in Y\) not collinear to \(w\). We claim that we can choose \(y\) such that \(y\perp U\). Indeed, suppose for a contradiction that \(U^\perp\cap S\subseteq w^\perp\cap T\). Then \[w\in (w^\perp\cap T)^\perp\subseteq (U^\perp\cap S)^\perp=\left\langle U,T\right\rangle=\left\langle U,U’\right\rangle,\] contrary to our assumption. Hence \(w^\perp\cap\left\langle y,U\right\rangle\) is a singular subspace \(U_1\) of dimension \(i-1\) complementary to \(S\). Since it is contained in \(\left\langle y,U\right\rangle\), we know that its image \(U_1’\) lies in \(X\) and naturally is contained in \(\left\langle T,U_1\right\rangle\) as \(\varphi\) has centre \(T\). Hence we can interchange the roles of \((U,U’)\) and \((U_1,U_1′)\). Since \(w\in U_1^\perp\), our previous claim proves that the image of \(w\) under \(\varphi\) belongs to \(X\). Now, if \(w\in\left\langle U,U’\right\rangle\), then we can find a line \(L_w\) of \(\Gamma\) not contained in \(\left\langle U,U’\right\rangle\) and intersecting it in \(w\). All points of \(L^\varphi_w\setminus\{w^\varphi\}\) belong to \(X\), by the foregoing, hence also \(w^\varphi\).

Noting that hyperbolic quadrics in any characteristic have unique embeddings, the proposition is proved. \(\square\)

Proof. In order to prove (1), we first assume that \(n\in\{3,4\}\). Let \(L\in\mathcal{L}\) be arbitrary. Let \(\pi\) be any plane through \(L\) and pick \(x\in\pi\setminus L\). Then the line \(M\in\mathcal{L}\) through \(x\) is collinear to \(L\), hence \(n=4\) so that \(L\) and \(M\) are contained in a (singular) solid \(S\), which they span. Let \(y\) be an arbitrary point of \(S\) not on \(L\cup M\), and let \(K\in\mathcal{L}\) contain \(y\). Then \(K\) is collinear to \(L\) and \(M\), and hence to \(S\), implying \(K\subseteq S\). Therefore, in this case, (1) is proved.

Now assume \(n\geq 5\). Let \(S\) be a singular solid spanned by two members \(L_1,L_2\in\mathcal{L}\). Let \(p\) be a point collinear to \(L_1\) but not to \(L_2\). Then the member \(L_3\in\mathcal{L}\) through \(p\) is collinear to \(L_1\) but opposite to \(L_2\). Then we consider the polar space \(\Gamma’\) induced in \(L_2^\perp\cap L_3^\perp\). Any member of \(\mathcal{L}\) having a point in \(\Gamma’\) is completely contained in \(\Gamma’\) (since \(\mathcal{L}\) is a composition spread it is collinear to both \(L_2\) and \(L_3\)). Hence \(\mathcal{L}\) induces in \(\Gamma’\) a spread \(\mathcal{S}_{\Gamma’}\). The fact that \(\mathcal{S}\) does not contain special pairs is inherited by \(\mathcal{S}_{\Gamma’}\). Consequently, we can apply the induction hypothesis and obtain that \(\mathcal{S}_{\Gamma’}\) satisfies (1). Since \(\Gamma’\) has rank at least \(3\), and hence at least \(4\) by the previous paragraph, we can find two lines \(L_4,L_5\in\mathcal{L}_{\Gamma’}\) collinear to \(L_1\), but mutually opposite. Hence also the polar space with point set \(L_4^\perp\cap L_5^\perp\) satisfies \((1)\). But this polar space contains both \(L_1\) and \(L_2\), so \((1)\) follows.

Note that the previous argument also shows that \(n\) is even. We now show \((2)\). We again use induction on \(n\). Pick \(L\subseteq U\) with \(L\in\mathcal{L}\). The argument in the previous paragraph easily shows that the singular solids through \(L\) in which \(\mathcal{L}\) induces a line spread form a composition line spread in \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(L)\). Applying induction to that spread in \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(L)\) proves (2). \(\square\)

Proof. Let \(p\) be a point and let \(M\) be a maximal singular subspace belonging to \(\Sigma\). Then, by Proposition 4.1, \(\Sigma\) induces a composition line spread in \(M\), and by dualizing, also in the dual of \(M\). If \(p\in M\), then the assertion is trivial. So we may assume \(p\notin M\). Let \(H\) be the subspace of \(M\), all points of which are collinear to \(p\). Since \(\Sigma\) defines a composition line spread in the dual of \(M\), there is a subspace \(U\) of dimension \(n-3\) of \(H\) contained in \(\Sigma\). Since \(\Sigma\) induces in \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(U)\) a spread (by definition of ovoidal subcomplex), the singular subspace generated by \(U\) and \(p\) is contained in a maximal singular subspace belonging to \(\Sigma\). The assertion is now clear. \(\square\)

Proof. We show the implications \((i)\implies(ii)\implies(iii)\implies(iv)\implies (i)\).

That (ii) follows from (i) is immediate as otherwise a point belongs to the convex closure of a line and a maximal singular subspace belongs to \(\Sigma\), contradicting condition (WCC).

Now let \(\Sigma\) be as in (ii) and suppose some pair of lines \(L_1,L_2\in\Sigma\) is special. Then there exists a plane \(\pi\) of \(\Gamma\) containing \(L_1\) and intersecting \(L_2\) in a point \(x\). Applying Lemma 4.18 in \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(L_1)\), we deduce the existence of a maximal singular subspace \(M\) of \(\Gamma\) containing \(\pi\) and belonging to \(\Sigma\). Since \(L_2\) has points not collinear to points of \(L_1\), the line \(L_2\) is not contained in \(M\), but intersects it in a point, a contradiction.

Now let \(\Sigma\) be as in (iii). In view of Proposition 4.1 and Lemma 4.18 it is sufficient to prove that no point is contained in two distinct lines belonging to \(\Sigma\). Suppose for a contradiction that the distinct lines \(L_1\) and \(L_2\) of \(\Sigma\) have a point \(p\) in common. Select a solid \(S\in\Sigma\) containing \(L_1\). If some point \(x\) of \(S\) is not collinear to \(L_2\), then the line of the spread induced in \(S\) by \(\Sigma\) is special to \(L_2\), a contradiction. Hence there is a \(4\)-space containing \(S\) and \(L_2\). Applying Lemma 4.18 to \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(S)\) yields a \(5\)-space \(U\in\Sigma\) containing \(S\) and \(L_2\), hence containing \(L_1\) and \(L_2\), contradicting the fact that \(\Sigma\) induces a line spread in \(U\) by Proposition 4.1.

Finally, suppose that \(\Sigma\) is a subcomplex defining a composition line spread of \(\Gamma\) such that the line spread in each solid belonging to \(\Sigma\) is a bi-spread. We show (i). In view of Proposition 4.1 we only need to show that the maximal singular subspaces of \(\Sigma\) containing a given singular subspace \(U\) of dimension \(n-3\) form a spread in \(\mathop{\mathrm{\mathsf{Res}}}_\Gamma(U)\) (in type \(\mathsf{D}_{2n}\) this amounts to all maximal singular subspaces through \(U\) of a given natural system). But this follows immediately from Lemma 4.17(2). \(\square\)

Proof. We may assume that \(\theta\) is non-trivial and not anisotropic. Suppose it is uniclass. Then, by Proposition 4.9 and Proposition 4.16, its fix structure is either an ideal subspace, or a composition line spread. We already noted that an ideal subspace has a corank, and hence is a convex ovoidal subcomplex by definition. Also, a composition spread fixed by a non-trivial automorphism induces a bi-spread in each of the \(3\)-spaces generated by two members of it, as follows from the arguments in the proof of Theorem 4.5. Then Proposition 4.19 implies that it is an ovoidal subspace satisfying (WCC).

Conversely, let \(\theta\) fix an ovoidal subcomplex satisfying (WCC). Then by Proposition 4.13 and Proposition 4.19, it is either an ideal subspace, or a composition line spread. Then the assertion follows from Proposition 4.9 and Proposition 4.16. \(\square\)

Proof. Let \(\Sigma\) be the ovoidal subcomplex defined by \(\mathcal{S}\). We prove three claims.

Claim 1. We claim that, if \(L,L’\in\mathcal{L}\) with \(L\) and \(L’\) opposite, then \(\mathcal{R}(L,L’)\subseteq\mathcal{S}\).

Indeed, pick \(M\in\Sigma\) and \(M’\in\Sigma\) maximal singular subspaces containing \(L\) and \(L’\), respectively. Let \(K\in\mathcal{R}(L,L’)\setminus\{L,L’\}\) and pick \(p\in K\). Then \(p^\perp\cap M\) is a hyperplane of \(M\) containing a unique subspace \(U\in\Sigma\) of dimension \(2r-3\). The unique member \(L_p\) of \(\mathcal{S}\) containing \(p\) is contained in the unique maximal singular subspace \(W\) containing \(p\) and \(U\) of the same type as \(M\) (or, equivalently, disjoint from \(L\)). Likewise, \(L-p\) is contained in the unique maximal singular subspace \(W’\) containing the subspace \(U’\in\Sigma\) of \(M’\) of dimension \(2r-3\) collinear with \(p\). Now note that \(K\subseteq W\). To see this, let \(x\in L\) and \(x’\in L’\) be the unique points of \(L\) and \(L’\), respectively, collinear to \(p\) (then \(p,x,x’\) are contained in one line). Since \(x\in M\) and hence \(x\perp U\), we also have \(x’\perp U\). The definition of composition spread then implies \(L’\perp U\). Consequently, every point on every line joining a point of \(L’\) with a point of \(L\) is collinear to \(U\), and so, also \(K\) is. This implies \(K\subseteq W\) and, similarly, \(K\subseteq W’\). Now \(W\cap W’\) does not contain a plane, as this plane has to intersect both \(U\) and \(U’\) non-trivially, and so this would imply that \(p\) lies on a line together with points of \(U\) and \(U’\). But then, by uniqueness of that line (see also the comment after Proposition 6.4.2 of [36]), we would have \(x\in U\) and \(x’\in U’\), a contradiction. Hence \(K=W\cap W’=L_p\). This proves Claim 1.

Claim 2. Now we claim that \(\mathcal{S}\) is determined by two opposite maximal singular subspaces \(M,M’\in\Sigma\) and the spread \(\mathcal{S}_M\) induced by \(\mathcal{S}\) in \(M\).

Indeed, by projection, also the spread of \(M’\) induced by \(\mathcal{S}\) is determined. Now let \(W\in\Sigma\) be a maximal singular subspace intersecting \(M\) in a subspace of dimension \(2r-3\). Then \(W\cap M\in\Sigma\) and hence all such \(W\) are determined by \(\mathcal{S}_M\). Suppose first that \(W\cap M’=\varnothing\). Pick \(q\in W\setminus M\). Let \(K_q\) be the unique line through \(q\) intersecting \(M\) and \(M’\) in points \(y\) and \(y’\), respectively. If the unique member \(L_y\in\mathcal{S}\) through \(y\) (and contained in \(M\)) were collinear to the unique member of \(\mathcal{S}\) through \(y’\) (and contained in \(M’\)), then \(q\perp L_y\), implying \(L_y\in W\cap M\), hence \(K_q\in W\), a contradiction since \(W\cap M’=\varnothing\). Hence Claim 1 yields the unique line of \(\mathcal{S}\) through \(q\). If \(W\cap M’\neq\varnothing\), then we replace \(M’\) by another maximal singular subspace belong to \(\Sigma\), which intersects \(M’\) in a subspace of dimension \(2r-3\) of \(\Sigma\) disjoint from \(W\cap M’\) (which is always possible). Then the previous argument shows that also in this case, the spread induced in \(W\) by \(\mathcal{S}\) is determined. Now, by connectivity, the whole of \(\mathcal{S}\) is determined and Claim 2 is proved.

Claim 3. We finally claim that, with previous notation, the spread \(\mathcal{S}_M\) is regular.

Indeed, let \(L,L’\in\mathcal{S}_M\) be two arbitrary but distinct lines. It is easy to find a line \(T\in\mathcal{S}\) opposite both \(L\) and \(L’\). Each point \(t\in T\) defines a line \(K_t\) in \(M\) spanned by \(t^\perp\cap L\) and \(t^\perp\cap L’\). Let \(R\in\mathcal{R}(L,T)\setminus\{L,T\}\) and \(R’\in\mathcal{R}(L’,T)\setminus\{L’,T\}\). Then, by Claim 1, both \(R\) and \(R’\) belong to \(\mathcal{S}\). Let \(s\) and \(s’\) be the unique points on \(R\) and \(R’\), respectively, collinear to \(t\in T\). Then the line \(\left\langle r,r’\right\rangle\) is contained in the plane \(\left\langle t,K_t\right\rangle\) and hence intersects \(K_t\). Since this is the case for each \(t\in T\), the line \(N\) obtained as the union of all points \(\left\langle s,s’\right\rangle\cap M\) belongs to a regulus which also contains \(L\) and \(L’\). If \(R\perp R’\), then \(N\) is the intersection of the solid spanned by \(R\) and \(R’\), hence belonging to \(\Sigma\), with \(M\) and consequently belongs to \(\mathcal{S}_M\). If \(R\) is opposite \(R’\), then \(N\in\mathcal{S}\) be Claim 1. In either case (note that \(R\) and \(R’\) are not special by Proposition 4.19(iii)) we conclude that \(\mathcal{S}_M\) contains a regulus, and hence by its transitivity properties, it is a regular spread.

Now, by the last assertion of [21, Proposition 3.24], each regular spread of any maximal singular subspace embeds in a composition spread of \(\Delta\) elementwise fixed by a non-trivial (point domestic) collineation. Since the automorphism group of \(\Delta\) is transitive on pairs of opposite maximal singular subspaces, Claims 2 and 3 now complete the proof of the proposition. \(\square\)

Proof. Let \(\Sigma\) be an ovoidal subcomplex of class \(1\). According to Main Result 1.4 it has Weyl diagram \(\mathsf{E_{6;2}}\). Let \(V\) be the set of points of \(\Sigma\). We check the defining properties of an ideal Veronesean for \(V\). Suppose \(x,y\in V\) are collinear. The definition of ovoidal subcomplex yields a symp \(\xi\) of \(\Sigma\) containing \(x\) and containing the line \(xy\). But now \(\xi\) contains two collinear points of \(\Sigma\), a contradiction.

Clearly the points of no ovoidal subcomplex are contained in a single symp, as through each of its points \(x\) there are at least two symps \(\xi_1,\xi_2\) of \(\Sigma\) containing points not collinear to \(x\). Such points are not contained in one symp with \(x\) as that symp otherwise has to coincide with both \(\xi_1\) and \(\xi_2\), a contradiction.

Now let \(x,y\) be two points of \(\Sigma\), necessarily non-collinear. Let \(\xi\) be the unique symp containing \(x\) and \(y\). Suppose for a contradiction that \(\xi\notin \Sigma\). Since \(\Sigma\) is ovoidal, there exist many, and hence at least one symp \(\zeta\) of \(\Sigma\) containing \(x\) and intersecting \(\xi\) in a \(4\)-space \(U\). Since \(y\) is certainly collinear with some points of \(U\), it is close to \(\zeta\) and hence collinear to a \(4’\)-space of \(\zeta\). Since the points of \(\Sigma\) in \(\zeta\) form an ovoid, one is contained in \(U\) and hence collinear to \(y\), contradicting the first paragraph of this proof. \(\square\)

Proof. We start by noting that each ovoid \(O\) belonging to \(V\), viewed as a subset of the ambient projective space in the natural representation of the symp \(\xi\) containing \(O\) spans a subspace of dimension at least \(5\). Indeed, suppose for a contradiction that \(O\) is contained in a \(4\)-dimensional subspace \(S\). Let \(W\) be any \(4\)-dimensional singular subspace. Then \(W\cap O=\{x\}\). Let \(p\) be any point of \(\xi\) not collinear to \(x\) and let \(W_p\) be the unique singular \(4\)-space through \(p\) intersecting \(W\) in a \(3\)-space. Since \(O\) is an ovoid, \(W_p\) contains a point of \(O\) and since this point is not contained in \(W\), we find \(p\in\left\langle O,W\right\rangle\). Since every point of \(x^\perp\setminus\{x\}\) lies on a line having at least two points not collinear to \(x\), we conclude that \(\xi\) is contained in \(\left\langle O,W\right\rangle\), which is however at most \(8\)-dimensional, a contradiction.

Note that a similar argument shows that, if, with the notation of the previous paragraph, \(O\) spans a subspace \(S\) of dimension exactly \(5\), then \(Q=S\cap\xi\) (indeed, if not then \(S\) contains a line \(L\) of \(\xi\) through some point of \(O\); consider a maximal singular subspace \(W\) through \(L\), then the argument of the previous paragraph shows \(\xi\subseteq\left\langle S,W\right\rangle\), which is however only \(8\)-dimensional).

Consider the standard \(26\)-dimensional representation of \(\mathsf{E_{6,1}}(\mathbb{K})\) in a projective space \(\mathsf{PG}(26,\mathbb{K})\) over \(\mathbb{K}\). Since in this representation the subspaces spanned by symps that are not adjacent intersect in a unique point (as has been explicitly checked in [26, §7]), it follows from [11, Main Result 4.3] that \(V\) is a Veronese variety spanning a subspace \(U\) of \(\mathsf{PG}(26,\mathbb{K})\) of dimension \(14\) or \(26\) (since the ovoids of that Veronese variety span at least \(5\)-spaces by the first paragraph). Now the second paragraph implies that, if the dimension of \(U\) is \(14\), then each ovoid arises as the intersection of a symp with a subspace of the ambient projective space of the symp. Hence in this case the assertion follows from [18, Remark 5.4].

Hence we may assume that each ovoid of \(V\) spans the entire ambient subspace of the symp it is contained in. Since \(V\) is now an octonion Veronesean, it lives in a projective space \(\mathsf{PG}(26,\mathbb{K}’)\), \(\mathbb{K}’\leq\mathbb{K}\), in which every symp is a quadric. It follows that \(\mathbb{K}’\neq\mathbb{K}\), as non-degenerate quadrics cannot be contained in each other over the same field. It also follows that \(\mathsf{PG}(26,\mathbb{K}’)\) does not contain any point \(p\) of \(\Delta\setminus V\), as each such point is contained in some symp \(\xi\) containing an ovoid \(O\) of \(V\), cp. [18, Lemma 4.11]. Then \(\xi\) intersects \(\mathsf{PG}(26,\mathbb{K}’)\) in a quadric containing \(O\) and \(p\), a contradiction again.

Now we claim that \(\mathbb{K}’\) is a maximal subfield of \(\mathbb{K}\). Indeed, suppose \(\mathbb{K}’\leq\mathbb{K}”\leq\mathbb{K}\), with \(\mathbb{K}’\neq\mathbb{K}”\neq\mathbb{K}\). Let \(\xi\) be a symp containing an ovoid \(O\) of \(V\). Let \(\xi”\) be the (possibly degenerate) subquadric of \(\xi\) induced by \(\mathbb{K}”\) (note that \(\xi”\) is non-empty since it contains \(O\); also, since \(O\) is an ovoid, \(\xi”\) contains lines). Let \(M\) be a maximal singular subspace of \(\xi\) containing a maximal singular subspace \(M”\) of \(\xi”\). Since \(\mathbb{K}”\) is a proper subfield of \(\mathbb{K}\), one can find a hyperplane \(H\) of \(M\), disjoint from \(O\), and intersecting \(M”\) in a subspace \(U”\) of dimension at most \(\dim M”-2\). Let \(M^*\) be the maximal singular subspace of \(\xi\) distinct from \(M\) and containing \(H\). Let \(\{x\}=M^*\cap O\). Then \(x\in \xi”\). But \(x\) is only collinear to all points of \(U”\) of \(M”\) in \(\xi”\), contradicting the fact that \(\xi”\) is a (possibly degenerate) polar space.

Now we claim that, since \(p:=\mathop{\mathrm{\mathsf{char}}}\mathbb{K}\neq 2\), the field extension \(\mathbb{K}/\mathbb{K}’\) is separable. Indeed, suppose it is inseparable. Then, since \(\mathbb{K}’\) is maximal, we have \(\mathbb{K}’=\mathbb{K}^p\), the set of all \(p\)th powers of elements of \(\mathbb{K}\). By [36, Lemma 3.4.2], there is an anisotropic quadratic form over \(\mathbb{K}’\) in eight variables describing \(O\). Since \(p\) is odd, we may diagonalise this form and so we find elements \(k_1,\ldots,k_7\in\mathbb{K}’\) such that the form \(x_0^2+k_1x_1^2+\ldots k_7x_7^2\) in the variables \(x_0,\ldots,x_8\) has only the trivial zero over \(\mathbb{K}’\), but it completely splits over \(\mathbb{K}\) (as \(\xi\) is hyperbolic). Let \((a_0,\ldots,a_7)\) be a zero over \(\mathbb{K}\). Raising to the power \(p\) and dividing by \(k_0^{\frac{p-1}2}\), we obtain that \[k_0+k_1\left(\left(\frac{k_1}{k_0}\right)^{\frac{p-1}2}x_1^p\right)^2+\cdots+k_8\left(\left(\frac{k_8}{k_0}\right)^{\frac{p-1}2}x_8^p\right)^2=0,\] hence producing a zero over \(\mathbb{K}’\) of the above anisotropic form over \(\mathbb{K}’\), a contradiction. The claim is proved.

Consequently, there is a (field) automorphism \(\sigma\) just fixing \(V\); it indeed preserves \(\Delta\) as each symp is preserved by the fact that each ovoid uniquely determines its host symp as they share infinitely many (generating) points and tangent hyperplanes.

The theorem is proved. \(\square\)

Proof. Part I: A uniqueness claim.

Let \((x,\xi)\) be a point-symp pair belonging to \(\Sigma\). We claim first that \(x\) is unique with respect to \(\xi\). Indeed, suppose also \((x’,\xi)\in \Sigma\), with \(x’\neq x\). First assume that \(x\perp x’\). The Weyl diagram tells us that there is a maximal \(4\)-space \(U\) contained in \(\xi\) and containing the line \(xx’\) such that \((xx’,U)\in\Sigma\). Then again the Weyl diagram implies that each point of the line \(xx’\) is associated with a unique symp through \(U\) to form a pair of \(\Sigma\). But \((x,\xi)\) and \((x’,\xi)\) are two pairs with \(U\subseteq \xi\), a contradiction.

Note that the preceding argument shows that, whenever \((p,\zeta)\in\Sigma\) and \(\zeta\ni q \perp p\), then there exists a symp \(\zeta_q\) such that \((q,\zeta_q)\in\Sigma\). Moreover, \(p\in\zeta_q\).

Next suppose that \(x\) and \(x’\) are not collinear. We may choose a \(5\)-space \(W\in\Sigma\) through \(x\) intersecting \(\xi\) in a \(4’\)-space \(V\) (this is our notation to make clear the singular subspace of dimension \(4\) is not a maximal subspace). Then there is a unique \(4\)-space \(U\) through \(x’\) intersecting \(V\) in a \(3\)-space. By the Weyl diagram, there is also a unique line \(L\ni x’\) such that \((L,U)\in \Sigma\). Let \(r\perp x\) be such that \(r\in L\). Since \(W\in \Sigma\), the Weyl diagram tells us that the line \(rx\) is contained in a \(4\)-space \(U_r\) inside \(\xi\) intersecting \(V\) in a \(3\)-space, and hence \(U\) in a plane \(\pi\). Then, as before, the point \(r\) is in a symp \(\xi_r\) going through \(U_r\) and \((r,\xi_r)\in\Sigma\); likewise, it is in a symp \(\xi_r’\) containing \(U\) such that \((r,\xi_r’)\in\Sigma\). The symps \(\xi_r\) and \(\xi_r’\) have \(\pi\) in common, hence the dual of the first part of the proof of the claim implies \(\xi_r=\xi_r’\), which must then coincide with \(\xi\), a contradiction to the first part again. The claim follows, hence the map \(x\mapsto x^\theta\) with \((x,x^\theta)\in\Sigma\) is an injective and well-defined map from \(P\) into the set of symps.

Part II: A subspaces claim.

Let \(P\) be the set of points contained in a point-symp flag belonging to \(\Sigma\). We now claim that \(P\) is a subspace of \(\Delta\). Let \(x,y\) be two collinear points in \(P\) and let \((x,\xi)\in\Sigma\), with \(\xi\) a symp, and \((y,\zeta)\in \Sigma\), with \(\zeta\) a symp. Since \(y\) is collinear to at least one point of \(\xi\), it is collinear to all points of a \(4’\)-space \(W\) in \(\xi\). It now also follows from the convexity of symps that \(\zeta\) and \(\xi\) have a(t least one) point \(r\in W\) in common. In view of the Weyl diagram of \(\Sigma\), the line \(xr\) is contained in a pair \((xr,U)\in\Sigma\), with \(U\) a maximal \(4\)-space in \(\xi\) containing \(xr\). Again by that diagram, we see that all points of the line \(xr\) belong to \(P\), and the corresponding symps contain \(U\). In particular, \(r\in P\) and \(xr\) is contained in \(r^\theta\). Likewise \(yr\subseteq r^\theta\). Hence the line \(xy\) is contained in \(r^\theta\). By our remark in the second paragraph of this proof, each point of \(xy\) belongs to \(P\). The claim is proved.

Part III: No deep points in \(P\).

We resume the notation of the previous paragraph and claim that \(U\cap W\) is a \(3\)-space. Indeed, \(U\) lies in \(\xi\) and is contained in \(r^\theta\), which also contains \(y\); hence the point \(y\) is collinear to at least a \(3\)-space of \(U\). This shows the claim. This now implies that \(U\) and \(W\) are incident (adjacent vertices in the underlying building) and so are \(U\) and the \(5\)-space \(A\) generated by \(W\) and \(y\). Hence, by the diagram (looking inside the residue of \((x,\xi)\)), the \(5\)-space \(A\) belongs to \(\Sigma\). This in turn implies that there is a \(4\)-space \(V\) such that \((ry,V)\in\Sigma\), \(V=r^\theta\cap\zeta\) and \(V\cap A\) is a \(3\)-space. It follows that \(\zeta\) and \(\xi\) share a \(4\)-space.

The previous paragraph has a nice consequence: if \(B\) is a \(5\)-space through \(xr\) which does not belong to \(\Sigma\) (and such \(5\)-spaces exist), then no point of \(B\setminus\xi\) belongs to \(P\). So, with standard terminology (see Subsection 2.1), \(P\) does not admit deep points.

Part IV: \(P\) is a geometric hyperplane.

Our next claim is that \(P\) is a geometric hyperplane. Indeed, let \(L\) be an arbitrary line in \(\Delta\). First assume that some point \(p\in P\) is collinear to \(L\). If \(L\) intersects \(p^\theta\), then by the second paragraph, some point of \(L\) belongs to \(P\). So, we may assume that \(L\) and \(p^\theta\) are disjoint. Each point of \(L\) is close to \(p^\theta\), and it is easily seen that the \(5\)-spaces generated by a point \(x\in L\) and \(x^\perp\cap p^\theta\) form the set of all \(5\)-spaces through a plane in \(p^\theta\). That plane corresponds to a line in the residue of \((p,p^\theta)\), and the Weyl diagram tells us now that at least one of these \(5\)-spaces belongs to \(\Sigma\). Hence in view of the Weyl diagram and Proposition 4.3, which yields that \(\Sigma\) induces in each of its \(5\)-spaces a symplectic polar space, the points of which are all points of the \(5\)-space, there is a point of \(L\) collinear to \(p\) in that symplectic polar space and the claim is proved in this case.

Now we reduce the general situation to the situation of the previous paragraph. Let \(p\in P\) be arbitrary. Then some point \(x\in L\) is close to \(p^\theta\), and so, by the second paragraph again, there is a point \(q\in P\) collinear to \(x\). We may assume no point on \(L\) other than \(x\) is collinear to \(q\). Suppose additionally that no point of \(q^\theta\) is collinear to any point of \(L\setminus\{x\}\). Let \(q^\theta\ni r\perp q\). Then there is a \(4\)-space \(U\subseteq q^\theta\) with \((rq,U)\in\Sigma\). Exactly one symp \(\xi^*\) through \(U\) is close to an arbitrarily chosen point \(y\) of \(L\setminus\{x\}\). By the Weyl diagram, we may assume that \(r^\theta=\xi^*\). This already shows that we may drop our additional assumption. Since \(q\in\xi^*\), we see that the line \(L\) is collinear with a plane \(\pi\) in \(\xi^*\) contained in \(x^\perp\cap\xi^*\). Each point of \(\pi\) collinear to \(r\) belongs to \(P\) and is collinear to \(L\). This reduces to the previous paragraph and proves the claim.

Part V: End of the proof.

The analysis of the different kinds of geometric hyperplanes of \(\Delta\) in Section 3.2 of [9], in particular the table given there, shows that only so-called hyperplanes of type \(\mathsf{F_4}\) do not possess deep points. Then Theorem 1 of [7] implies that there exists a polarity \(\sigma\) of \(\Delta\), the absolute points of which are precisely the points of \(P\). Moreover, Lemma 4.20 of [7] implies that \(\sigma\) assigns to each point \(x\in P\) the symp \(x^\theta\). Since this defines \(\sigma\) on the lines and planes through a point \(p\in P\) inside \(p^\theta=p^\sigma\) by intersecting, the elements of types \(1,2,3\) and \(4\) of \(\Delta\) that belong to a pair contained in \(\Sigma\) (types \(1,3\)) or belong itself to \(\Sigma\) (types \(2,4\)) form a metasymplectic space isomorphic to \(\mathsf{F_{4,4}}(\mathbb{K})\) (also by Theorem 1 of [7]).

This completes the proof of the proposition. \(\square\)

Proof. As pointed out above, we may assume that \(\theta\) pointwise fixes an ovoidal subcomplex \(\Sigma\) of class \(2\). First suppose that \(\theta\) preserves types. Then the construction in Section 6 of [7] shows that all points of \(\Delta\) are fixed (indeed, a point of \(\Delta\) outside the set \(P\) of points of the metasymplectic space associated to \(\Sigma\) by Proposition 4.25 is collinear to a unique so-called tropics geometry, and this correspondence is bijective). Hence \(\theta\) is the identity. Now suppose that \(\theta\) is a duality. Let \(\sigma\) be a polarity associated to \(\Sigma\) as in Proposition 4.25. Then \(\sigma\theta\) is type preserving and fixes each point of \(P\). The previous argument implies \(\sigma\theta=\mathrm{id}\), hence \(\theta=\sigma\). Now [17, Theorem 4.20]] completes the proof. \(\square\)

Proof. We show the implications \((i)\implies(ii)\implies(iii)\implies(iv)\implies (i)\).

Recall that an ovoidal subcomplex \(\Sigma\) of class \(1\) has Weyl diagram \(\mathsf{F_{4;2}}\). Let \(\Delta\cong\mathsf{F_{4}}(\mathbb{K},\mathbb{A})\) and identify \(\Delta\) with \(\mathsf{F_{4,1}}(\mathbb{K},\mathbb{A})\). Suppose \(\Sigma\) satisfies condition (WCC) and \((p,\xi)\) is a point-symp pair not in general position, with \(p\notin\xi\). Then the set of points of \(\xi\) collinear to \(p\) forms a line, contradicting condition (WCC).

Now suppose that \(\Sigma\) satisfies (ii). Suppose \(\{p_1,p_2\}\) is a special pair of points. Let \(p_1\perp x\perp p_2\). By the definition of ovoidal subspace, there exists a unique symp \(\xi\in \Sigma\) incident with the line \(p_1x\). Then \(p_2\) and \(\xi\) are neither incident nor in general position, a contradiction.

Now suppose \(\Sigma\) satisfies (iii). We show some claims.

Claim 1. Two points of \(\Sigma\) are never collinear; dually, two symps of \(\Sigma\) are never adjacent. Indeed, if the points \(p_1\) and \(p_2\) of \(\Sigma\) were collinear, then in the unique symp \(\xi\) of \(\Sigma\) through \(p_1\) containing the line \(p_1p_2\), we would find two collinear points of \(\Sigma\), contradicting the fact that the set of points of \(\Sigma\) in \(\xi\) is an ovoid.

Claim 2. If two symps \(\xi_1,\xi_2\) of \(\Sigma\) intersect in a unique point \(x\), then \(x\in\Sigma\). Indeed, suppose not. Select a point \(p_1\in\xi_1\cap\Sigma\cap x^\perp\). Then \(p_1\) is collinear to a line \(L_2\subseteq\xi_2\). Select a point \(p_2\in\xi_2\cap\Sigma\) not collinear to \(L_2\). Then the pair \(\{p_1,p_2\}\) is special, a contradiction.

Claim 3. If two symplectic points \(p_1,p_2\) belong to \(\Sigma\), then the unique symp \(\xi\) containing \(p_1\) and \(p_2\) also belongs to \(\Sigma\). Indeed, suppose not. Let \(L_1\) be any line in \(\xi\) through \(p_1\). Then there is a unique symp \(\xi_1\) of \(\Sigma\) containing \(L\), and it is distinct from \(\xi\). Let \(L_2\) be the unique line through \(p_2\) and not disjoint from \(L_1\). Let \(\xi_2\) be the unique symp of \(\Sigma\) containing \(L_2\). Then \(\xi_1\) and \(\xi_2\) intersect just in \(L_1\cap L_2\), which, by Claim 2, belongs to \(\Sigma\), contradicting Claim 1 as \(L_1\cap L_2\) is collinear to \(p_1\).

Claim 4. If two disjoint symplecta \(\xi_1,\xi_2\) belong to \(\Sigma\), then they are opposite. Indeed, suppose not. Then there is a symp \(\xi\) intersecting \(\xi_i\) in a plane \(\pi_i\), \(i=1,2\). Since the points of \(\Sigma\) in \(\xi_i\) form an ovoid, there is a point \(p_i\) of \(\Sigma\) in \(\pi_i\), \(i=1,2\). Since by Claim 1, \(p_1\) and \(p_2\) are not collinear, they are symplectic and Claim 3 implies that \(\xi\) belongs to \(\Sigma\). Claim 1 contradicts the fact that \(\xi\) and \(\xi_1\) are adjacent.

Now we show that \(\Sigma\) defines a Moufang quadrangle, the points of which are the points of \(\Sigma\) and the lines of which are determined by the symps in \(\Sigma\). Let \(p\) be a point of \(\Sigma\) and \(\xi\) a symp of \(\Sigma\) not containing \(p\). If \(p\) is collinear to a line \(L\) of \(\xi\), then each point of \(\Sigma\) in \(\xi\) not collinear to \(L\) is special to \(p\), contradicting our assumption. Hence there is a unique point \(x\in\xi\) symplectic to \(p\). If \(x\notin\Sigma\), then there is a point \(y\in\Sigma\) collinear to \(x\), and consequently special to \(p\), a contradiction. Therefore \(x\in \Sigma\), and the symp through \(p\) and \(x\) also belongs to \(\Sigma\) (by Claim 3) so that \(\Sigma\) defines a generalized quadrangle. By [28, Main Theorem 1], it is a Moufang quadrangle. \(\square\)

Proof. Let \(\Sigma\) be an ideal quadrangular Veronesean of a separable building \(\Delta\) of type \(\mathsf{F}_4\). We view \(\Delta\) as the corresponding metasymplectic space of type \(\mathsf{F_{4,4}}\). Let \(\xi\) and \(\xi’\) be two non-disjoint symps contained in \(\Sigma\) and let \(O\) and \(O’\) be the respective ovoids induced by the elements of \(\Sigma\) in \(\xi\) and \(\xi’\). Since, by [28], \(\Sigma\) defines a Moufang quadrangle, there is a (short) root elation group \(E\) with root centred at \(\xi’\) acting sharply transitively on \(O\setminus O’\). Since \(O’\) is fixed pointwise, all lines of \(\xi\) through the point \(\infty=O\cap O’\) are stabilized. We now consider the description of \(\xi\) provided in [36, Chapter 5], valid for all cases, in particular for the non-embeddable polar spaces. We will not repeat the whole construction here, but restrict ourselves to recalling that the points opposite a fixed point \(\infty\) can be given coordinates \((x_{-2},x_{-1},x_1,x_2;\ell)\), with \(x_i\in\mathbb{A}\) and \(\ell\in\mathbb{K}\), \(i\in\{-2,-1,1,2\}\), where \(\mathbb{K}\) is the underlying field and \(\mathbb{A}\) is an alternative quadratic division algebra over \(\mathbb{K}\). The fact that we are in the separable case translates to \(\mathbb{K}\) not being of characteristic \(2\) if the standard involution is trivial. Also, the complete short root elation group of \(\Delta\) whose root contains \(\xi’\) in the centre and \(\xi\) in the boundary, acts on \(\xi\) as the group of unipotent collineations fixing \(\infty\) linewise. Exactly that group is displayed on page 78 of [36]. It consists of the following transformations, given by the action on the points described above. For each \(y_i\), \(i\in\{-2,-1,1,2\}\) and each \(k\in \mathbb{K}\) we define \(\rho_{y_{-2},y_{-1},y_1,y_2;k}: (x_{-2},x_{-1},x_1,x_2;\ell)\mapsto\) \[(x_{-2}+y_{-2},x_{-1}+y_{-1},x_1+y_1,x_2+y_2;\ell+k+\overline{y}_2x_{-2}+\overline{y}_1x_{-1}+\overline{x}_{-1}y_1+\overline{x}_{-2}y_2),\] where the standard involution of \(\mathbb{A}\) is denoted by \(x\mapsto \overline{x}\). Let \(E\) be the set of all such transformations stabilising \(O\). Clearly, \((x_{-2},x_{-1},x_1,x_2;\ell)\in O\) if, and only if, \(\rho_{x_{-2},x_{-1},x_1,x_2;\ell}\in E\). Hence it suffices to prove that \(\rho_{0,0,0,0;k}\in E\), for all \(k\in\mathbb{K}\).

It follows from the description of \(\xi\) in [36] that there exist functions \(f\) and \(g\) such that \(O=\{(a,b,f(a,b,k),g(a,b,k);k)\mid a,b\in\mathbb{A}, k\in\mathbb{K}\}\). Hence we have to show that \(f(0,0,k)=g(0,0,k)=0\), for all \(k\in \mathbb{K}\). Suppose, for a contradiction, that for some element \(k\in\mathbb{K}\) this is not true, and we may assume \(f(0,0,k_0)=a_0\neq 0\). Letting \(x\in\mathbb{A}\) be arbitrary, then we calculate the commutator (denoting \(g(0,0,k_0)\) by \(b_0\)) as \[[\rho_{0,0,a_0,b_0;k_0},\rho_{0,x,f(0,x,0),g(0,x,0),0}]=\rho_{0,0,0,0;-\overline{a}_0x-\overline{x}a_0}.\]

If \(x\) runs through \(\mathbb{A}\), then also \(-\overline{a}_0x\) runs through \(\mathbb{A}\). Hence it follows from the commutator above that \((0,0,0,0;\overline{y}+y)\in O\), for all \(y\in\mathbb{A}\). Since the map \(x\mapsto \overline{x}+x\) is linear and not zero (remember that the characteristic of \(\mathbb{K}\) is not \(2\) if \(\overline{x}=x\) for all \(x\in\mathbb{A}\)), it is surjective onto \(\mathbb{K}\) and the proposition follows. \(\square\)

Proof. In [15, Section 6] it is shown that, if a separable thick metasymplectic space \(\mathsf{F_{4,1}}(\mathbb{K},\mathbb{A})\) contains a flat and linear ideal quadrangular Veronesean \(\Sigma\), then \(\mathop{\mathrm{\mathsf{dim}}}_{\mathbb{K}}\mathbb{A}\in\{2,4,8\}\) (recall that \(\mathbb{K}=\mathbb{A}\) is ruled out by the simple fact that a symplectic polar space cannot admit a linear ovoid as each point is collinear to at least one point of a given hyperbolic line). Now [15, Remark 6.4.2] shows the assertion for \(\mathop{\mathrm{\mathsf{dim}}}_{\mathbb{K}}\mathbb{A}\in\{2,4\}\). Finally, let \(\mathbb{A}\) be a Cayley division algebra over \(\mathbb{K}\). Then the same method as in [15, §6.4] shows that the rank 1 residue of the Moufang quadrangle defined by \(\Sigma\) corresponding to the residue of the vertices of type \(4\) has Tits index

However, the only Tits index of rank \(2\) containing this one is, with the notation of [30], the index \(\mathsf{^2D_{7,2}^{(1)}}\). But the other rank 1 residue is just a projective line over \(\mathbb{K}\), which can never be the case for an ovoid of a polar space of rank 3 over \(\mathbb{K}\) (the arguments in [15, Section 6.4] show that this quadrangle is actually a subquadrangle induced by an extended equator geometry, and hence can never coincide with the complete Moufang quadrangle defined by the ideal Veronesean).

This completes the proof of the proposition. \(\square\)

Proof. It is clear that (ii) follows from (i) and that (i) follows from (iii). Hence it remains to show that (iii) follows from (ii). So assume \(\Sigma\) is ovoidal of type \(\mathsf{^2F_{4;2}}\) satisfying the additional condition mentioned in (ii). Our first principal aim is to show that \(\Sigma\) defines a generalized octagon. The first subgoal is to prove that for each pair of simplices of \(\Sigma\) of type \(\{1,4\}\), the distance between the vertices of type \(1\) is the same as the distance between the vertices of type \(4\). For convenience we will work in the corresponding metasymplectic space of type \(\mathsf{F_{4,1}}\), where \(\Sigma\) is identified with a set of point-symp and line-plane pairs. We denote this space also by \(\Delta\). Note that the corresponding geometry of type \(\mathsf{F_{4,4}}\) is also a metasymplectic space, and we call it the dual of \(\Delta\), denoted by \(\Delta^*\). We will occasionally appeal to dual arguments. Also, we denote opposite object \(S,S’\) in \(\Delta\) as \(S\equiv S’\).

We prove some numbered claims so that we may easily refer to them.

Claim 1. Each pair \((\{p_1,\xi_1\},\{p_2,\xi_2\})\subseteq\Sigma\) of point-symp pairs, \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), has the property that, if \(\xi_1\) and \(\xi_2\) are opposite, then so are \(p_1\) and \(p_2\).

Indeed, choose such pairs with \(\xi_1\) and \(\xi_2\) opposite and suppose \(p_1,p_2\) are not opposite. First suppose that \(p_1\) and \(p_2\) are special. Let \(p_1’\) be the unique point of \(\xi_1\) symplectic to \(p_2\). We can find a line-plane pair \(\{L_1,\pi_1\}\in\Sigma\) incident with \(\{p_1,\xi_1\}\) and such that \(L_1\) is not collinear with \(p_1’\). Then all points of \(L_1\setminus\{p_1\}\) are opposite \(p_2\). There exists a unique symp \(\zeta_1\) through \(\pi_1\) not opposite \(\xi_2\). Since \(\Sigma\) is ovoidal, there exists a point \(x_1\in L_1\) such that \(\{x_1,\zeta_1\}\in\Sigma\). Since \(\zeta_1\neq\xi_1\), we have \(x_1\neq p_1\). Hence \(\{x_1,\zeta_1\}\) and \(\{p_2,\xi_2\}\) are not opposite while \(x_1,p_2\) are. This contradicts the assumption of (ii).

Now suppose \(p_2\) is symplectic to \(p_1\). On any line \(M_1\) through \(p_1\), which belongs to a line-plane pair \(\{M,\alpha_1\}\) in \(\Sigma\), we can find a point \(y_1\) such that the point-symp pair \(\{y_1,\eta_1\}\in\Sigma\) incident with \(\{M_1,\alpha_1\}\) has its symp \(\eta_1\) opposite \(\xi_2\). This contradicts the conclusion in the previous paragraph. We conclude that Claim 1 is proved.

Claim 1 implies that the assumptions are now self-dual. So from now on, we may appeal to the duality principle. We already do so for the next consequence and claim.

Consequence 1. If \(L,L’\) are lines, \(\pi,\pi’\) planes and \(\{L,\pi\},\{L’,\pi’\}\in\Sigma\), then \(L\) is opposite \(L’\) if, and only if, \(\pi\) is opposite \(\pi’\).

Indeed, by our previous remark it is enough to show one direction, since the other direction is the dual. Suppose \(L\equiv L’\) and suppose \(\pi\not\equiv\pi’\). Then some point \(x\in\pi\) is symplectic to some point \(x’\in\pi’\). There is a symplecton \(\zeta\) containing \(\pi\) and a line of \(\xi(x,x’)\). Clearly, \(\zeta\) is not opposite any symp through \(\pi’\), whereas there exists a point \(p\in L\) such that \(\{p,\zeta\}\in\Sigma\), and \(p\) is opposite at least one point of \(L’\), contradicting Claim 1. Consequence 1 is proved.

Claim 2. Each pair \((\{p_1,\xi_1\},\{p_2,\xi_2\})\subseteq\Sigma\) of point-symp pairs, \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), has the property that \(\xi_1\) and \(\xi_2\) are special if, and only if, \(p_1\) and \(p_2\) are special.

Indeed, suppose \(\xi_1\) and \(\xi_2\) are special, with \(\xi\) a symp sharing a plane \(\alpha_i\) with \(\xi_i\), \(i=1,2\). By assumption, \(p_1\) and \(p_2\) are not opposite. If they are special, there is nothing to prove, so suppose now first that they are symplectic. Then we may assume that \(p_1\in\alpha_1\). Suppose \(p_2\notin\alpha_2\). Then \(p_1^\perp\cap\alpha_2=p_2^\perp\cap \alpha_2=:L\). Since \(\Sigma\) is ovoidal, we find a line-plane pair \(\{L_2,\pi_2\}\in\Sigma\) with \(p_2\in L_2\) and \(\pi_2\) disjoint from \(\alpha_2\). Each symp through \(\pi_2\) distinct from \(\xi_2\) is opposite \(\xi_1\), whereas no point on \(L_2\) is opposite \(p_1\), a contradiction to Claim 1. Suppose now \(p_2\in \alpha_2\). Let \(K_i\) be the unique line through \(p_i\) in \(\alpha_i\) contained in a line-plane pair of \(\Sigma\), \(i=1,2\). Since lines contain at least three points, we can select a point \(x_1\in p_2^\perp\cap\alpha_1\) distinct from both \(K_2^\perp\cap\alpha_1\) and \(p_2^\perp\cap L_1\). Set \(M_1=p_1x_1\) and \(M_2=x_1^\perp\cap\alpha_2\). Setting \(i=1,2\), select a plane \(\beta_i\neq\alpha_i\) in \(\xi_i\) through \(M_i\), not belonging to a line-plane pair of \(\Sigma\) incident with \(\{p_i,\xi_i\}\). Then there is a unique line-plane pair \(\{N_i,\gamma_i\}\in\Sigma\) incident with \(\{p_i,\xi_i\}\) such that \(N_i\subseteq \beta_i\). Pick a point \(y_i\in N_i\setminus\{p_i\}\). Let \(\{y_i,\zeta_i\}\in\Sigma\) be incident with \(\{N_i,\gamma_i\}\). Then, since \(\zeta_1\cap\xi=p_1\) is symplectic to \(p_2=\zeta_2\cap\xi\), we find that \(\zeta_1\) is opposite \(\zeta_2\). However, \(y_1\) is collinear to \(x_1\), which is symplectic to \(y_2\), as \(M_2\subseteq x_1^\perp\cap y_2^\perp\). Hence \(y_1\) and \(y_2\) are not opposite, a contradiction to Claim 1 again.

Now suppose \(p_1\perp p_2\). Let \(K_i\) be as in the previous case and suppose first that \(p_i^\perp\cap\alpha_j=K_j\), \(\{i,j\}=\{1,2\}\). Then every point of \(K_1\setminus\{p_1\}\) is symplectic to every point of \(K_2\setminus\{p_2\}\). Define \(\{K_i,\beta_i\}\in\Sigma\) to be incident with \(\{p_i,\xi_i\}\), \(i=1,2\). For a given symp through \(\beta_1\) distinct from \(\xi\) (which is only relevant if \(\beta_1=\alpha_2\)), there is at most one symp through \(\beta_2\) not special to the given symp. Hence we can find points \(z_i\in K_i\setminus\{p_i\}\), \(i=1,2\), such that the unique symps \(\zeta_i\) with \(\{z_i,\zeta_i\}\in\Sigma\) incident with \(\{K_i,\beta_i\}\) are special. Since \(z_1\) and \(z_2\) are symplectic, that contradicts the previous paragraph.

Hence we may assume that \(K_1\neq p_2^\perp\cap\alpha_1:=K_1’\). There are a unique plane \(\gamma_1\) through \(K_1’\) in \(\xi_1\) and a unique line \(M_1\subseteq\gamma_1\) through \(p_1\) such that \(\{M_1,\gamma_1\}\in\Sigma\). Note that every point of \(M_1\setminus\{p_1\}\) is symplectic to \(p_2\), but only one symp through \(\gamma_1\) is symplectic to \(\xi_2\). Hence we can again find a simplex \(\{x_1,\zeta_1\}\in\Sigma\), with \(x_1\in M_1\setminus\{p_1\}\) and \(\gamma_1\subseteq\zeta_1\), and \(x_1\) symplectic to \(p_1\), while \(\zeta_1\) special to \(\xi_2\), again contradicting a previous paragraph.

Hence we have shown that, if \(\{p_1,\xi_1\},\{p_2,\xi_2\}\in\Sigma\) with \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), and \(\xi_1\) and \(\xi_2\) are special, then \(p_1\) and \(p_2\) are special as well. Since the converse is the dual, the proof of Claim 2 is complete.

Claim 3. Each pair \((\{p_1,\xi_1\},\{p_2,\xi_2\})\subseteq\Sigma\) of point-symp pairs, \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), has the property that \(\xi_1\) and \(\xi_2\) are symplectic if, and only if, \(p_1\) and \(p_2\) are symplectic.

Indeed, let \(\xi_1\) and \(\xi_2\) be symplectic, that is, they intersect in a unique point \(p\). If \(p_1=p=p_2\), then by considering an arbitrary point \(q_1\) on a line \(L_1\) such that \(\{L_1,\alpha_1\}\in\Sigma\) and \(p_1\in L_1\subseteq \alpha_1\subseteq\xi_1\), with \(q_1\) distinct from \(p_1\) and such that the unique symp \(\zeta_1\) with \(\{q_1,\zeta_1\}\in\Sigma\) and \(\alpha_1\subseteq\zeta_1\) is not adjacent to \(\xi_2\), we reduce this situation to \(p_1\neq p=p_2\). Then there exists a line-plane pair \(\{M_1,\beta_1\}\in\Sigma\) with \(p_1\in M_1\subseteq\xi_1\) but \(p\notin\beta_1\). Selecting an arbitrary point \(p_1’\in M_1\setminus p_1\), we see that the corresponding symp through \(\beta_1\) is special to \(\xi_2\), whereas \(p_1’\) is symplectic or collinear to \(p_2\), contradicting Claim 2.

So, we may assume that \(p_1\neq p\neq p_2\). We do have \(p_1\perp p\perp p_2\perp p_1\) as otherwise we contradict one of the previous claims. So, we can still choose \(p_1’\) as in the previous paragraph, now taking care that \(p_1’\) is not collinear to \(p\) and we denote the corresponding symp as \(\xi_1’\). We now choose \(\{p_2′,\xi_2’\}\) analogously, but with respect to \(p_2\) and \(\xi_2\). Then \(p_1’\) is opposite \(p_2’\), but the symps \(\xi_1’\) and \(\xi_2’\) contain collinear points \(p_1\) and \(p_2\), respectively, contradicting Claim 1. This proves Claim 3.

Claim 4. Each pair \((\{p_1,\xi_1\},\{p_2,\xi_2\})\subseteq\Sigma\) of point-symp pairs, \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), has the property that \(\xi_1\) and \(\xi_2\) are adjacent if, and only if, \(p_1\) and \(p_2\) are collinear. Moreover, if both conditions are satisfied, then \(p_1p_2\subseteq\xi_1\cap\xi_2\) and \(\{p_1p_2,\xi_1\cap\xi_2\}\in\Sigma\).

Indeed, suppose \(\xi_1\) and \(\xi_2\) are adjacent while \(p_1\) is not collinear to \(p_2\). Then, by the foregoing claims, \(p_1=p_2\). Then there exist a plane \(\alpha_1\) through \(p_1\) inside \(\xi_1\) intersecting \(\xi_1\cap\xi_2\) in exactly \(p_1\), and a line \(L_1\subseteq \alpha_1\) with \(p_1\in L_1\) such that \(\{L_1,\alpha_1\}\in\Sigma\). Let \(\{q_1,\zeta_1\}\in\Sigma\) be such that \(q_1\in L_1\setminus\{p_1\}\) and \(\alpha_1\subseteq \zeta_1\). Then \(\zeta_1\) and \(\xi_2\) intersect in exactly \(p_1=p_2\), but \(q_1\perp p_2\), contradicting Claim 3.

Hence \(p_1\perp p_2\), \(p_1\neq p_2\). Suppose \(p_1\notin\xi_2\). Then clearly \(p_2\in\xi_1\). There exist a plane \(\alpha_1\subseteq\xi_1\) with \(p_1\in\alpha_1\), and a line \(L_1\subseteq\alpha_1\) with again \(p_1\in L_1\), such that \(\{L_1,\alpha_1\}\in\Sigma\) and \(\alpha_1\cap\xi_2\) is empty. It is also easy to see that we can choose \(L_1\) not contained in \(p_2^\perp\). As before this yields \(\{q_1,\zeta_1\}\in\Sigma\) with \(p_1\neq q_1\in L_1\subseteq\alpha_1\subseteq\xi_1\). Then \(q_1\) is symplectic to \(p_2\), but \(\zeta_1\) is special to \(\xi_2\), contradicting Claims 2 and 3. Hence \(p_1p_2=:L\subseteq \alpha:=\xi_1\cap\xi_2\). Set \(i=1,2\). There exist a unique line \(L_i\subseteq\alpha\) and a unique plane \(\beta_i\) with \(p_i\in L_i\subseteq \beta_i\subseteq\xi_i\) and \(\{L_i,\beta_i\}\in\Sigma\). Suppose there exists \(p\in L_1\cap L_2\setminus L\). By the dual of the previous paragraph, there is a unique symp \(\xi\ni p\) with \(\{p,\xi\}\in\Sigma\), and it has to contain \(\beta_1\) and \(\beta_2\). This is only possible if at least one of \(\beta_1,\beta_2\) coincides with \(\xi_1\cap\xi_2\). But then \(\xi\) coincides with either \(\xi_1\) or \(\xi_2\), a contradiction. So we may assume \(L_1=L\). Then \(p_2\in L_1\) and so \(\beta_2\subseteq\xi_1\), implying \(\beta_2=\alpha\). But now, by the Weyl diagram, there exists a unique point \(q\in L_2\setminus\{p_2\}\) with \(\{q,\xi_2\}\in\Sigma\), contradicting the dual of the first paragraph again.

Now it follows that for each point-symp pair \(\{p,\xi\}\in\Sigma\), the symp \(\xi\) is uniquely determined by \(p\) and we write \(\xi=p^\sigma\) and \(p=\xi^\sigma\). A point (symp) in the domain of \(\sigma\) will be called a Ree point (symp). Similarly we define Ree lines and Ree planes, and for a Ree line \(L\), there is a unique plane \(L^\sigma\) such that \(\{L,L^\sigma\}\in\Sigma\). We also write \(L=(L^\sigma)^\sigma\).

Similar results as Claim 4 pinning down the mutual positions of pairs of point-symp pairs of \(\Sigma\) exist. We prove them now, going our way up from close to far. We will briefly call a pair of point-symp pairs of \(\Sigma\) collinear, symplectic, special, opposite, if the points of these pairs are collinear, symplectic, special, opposite, respectively.

Claim 5. If a pair \((\{p_1,\xi_1\},\{p_2,\xi_2\})\subseteq\Sigma\) of point-symp pairs, \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), is symplectic, then, writing \(p:=\xi_1\cap\xi_2\) and letting \(\xi\) be the symp containing \(p_1\) and \(p_2\), the pair \(\{p,\xi\}\) belongs to \(\Sigma\). Moreover, \(p_1\perp p\perp p_2\).

Indeed, assume first for a contradiction that \(p=p_1\). Similar arguments as before yield a Ree point \(q\perp p\) in \(\xi_1\) such that \(q^\sigma\) intersects \(\xi_2\) in exactly \(p\). Since \(q\) is special to \(p_2\) (noting that \(p_1\) and \(p_2\) are not collinear!), this contradicts Claim 2. Hence \(p_1\perp p\perp p_2\), with \(p_1\neq p\neq p_2\). Assume next for a contradiction that the line \(K_1:=pp_1\) is not a Ree line. Set \(L_1:=p_2^\perp\cap \xi_1\). Then, since \(p_1\) is symplectic to \(p_2\), we find \(L_1\subseteq p_1^\perp\). Set \(\beta_1:=\left\langle p_1,L_1\right\rangle\). Let \(\{M_1,\alpha_1\}\in\Sigma\) be incident with \(\{p_1,\xi_1\}\) such that \(M_1\subseteq\beta_1\). Set \(x=M_1\cap L_1\). Then, by Claim 4, \(x\in\xi_2\), a contradiction. Hence \(p\) belongs to a point-symp pair of \(\Sigma\) and, dually, \(p^\sigma=\xi\). Claim 5 is proved.

Claim 6. If a pair \((\{p_1,\xi_1\},\{p_2,\xi_2\})\subseteq\Sigma\) of point-symp pairs, \(p_1,p_2\) points and \(\xi_1,\xi_2\) symps of \(\Delta\), is special, then neither \(p_1\) nor \(p_2\) is contained in the unique symp \(\xi\) adjacent to both \(\xi_1\) and \(\xi_2\) (implying that \(\xi\) is not a Ree symp). Moreover, there exist unique Ree points \(q_1\) and \(q_2\) with \(p_1\perp q_1\perp q_2\perp p_2\).

Indeed, suppose first that \(p_1\in\xi\). Then \(p_2\notin \xi\) and we find a Ree point \(q_2\perp p_2\) in \(\xi\). If \(q_2\perp p_1\), then \(q_2\in p_1^\sigma=\xi_1\) by Claim 4, a contradiction. Hence \(q_2\) is symplectic to \(p_1\). Claim 5 yields a Ree point \(q\in\xi\) with \(q^\sigma=\xi\). Claim 4 again implies \(q\perp \xi_1\cap\xi_2\), a contradiction. Hence \(p_1\) is far from \(\xi_2\) and \(p_2\) is far from \(\xi_1\). We now also see that \(\xi\) is not a Ree symp, as by Claim 4 the point \(\xi^\sigma\) would have to belong to both \(\xi_1\) and \(\xi_2\), a contradiction. Dually, the unique point in \(\Delta\) collinear to both \(p_1\) and \(p_2\) is not a Ree point.

Let \(i=1,2\). Set \(\alpha_i=:\xi_i\cap\xi\) and \(L_i=:p_i^\perp\cap\alpha_i\). Since \(p_1\) is special to \(p_2\), there exists a unique point \(r_i\in L_i\cap L_{3-i}^\perp\). We find a Ree point \(q_i\in L_i\) on the unique Ree line contained in \(\left\langle p_i,L_i\right\rangle\). If \(q_1\) is symplectic to \(q_2\), then \(\xi\) is a Ree symp by Claim 5, contradicting our previous observation. So \(q_1\perp q_2\) and we may assume without loss of generality that \(q_1=r_1\). If \(q_2\neq r_2\), then \(q_2\) is special to \(p_1\), with \(q_2\perp q_1\perp p_1\). This contradicts our earlier observation that the unique point collinear to two special Ree points is not a Ree point. Hence \(r_1\) and \(r_2\) are both Ree points. Clearly, the sequence \((p_1,r_1,r_2,p_2)\) is unique as every other Ree point collinear to \(p_1\) is either special to or opposite \(p_2\), and so only longer sequences can occur. Claim 6 is proved.

Before we prove that we actually have a generalized octagon, we make a couple of observations.

Indeed , if \(q\perp p_2\) is a Ree point not on \(p_1p_2\), then it lies in \(p_2^\sigma\) and the axioms of an ovoidal subcomplex of type 2 imply the assertion.

Indeed, Claim \(5\) proves the existence of \(p\) and the uniqueness of \((p^\sigma)^\sigma\) proves the uniqueness of \(p\) (if \(q\) would be another Ree point collinear with both \(p_1,p_2\), then \(q^\sigma\) contains \(p_1,p_2\) and hence coincides with \(p^\sigma\)). Now let \(p_3\perp p_2\) be a Ree point not on \(p_2p\). In \(p_2^\sigma\), the point \(p_3\) is not collinear to \(p\), which means that \(p_3^\perp\cap p^\sigma\) is not contained in \(p_1^\perp\). This implies that \(p_3\) is special to \(p_1\).

The first part of this observation is contained in Claim 6. The second part follows from the construction of \(q_1,q_2\) and an argument similar to the one used in the proof of Observation 4.35. The last assertion follows directly from the proof of Claim 6.

We are now ready to prove our first principal goal.

Claim 7. The set of Ree points, furnished with the Ree lines, is a generalized octagon.

We denote the geometry of Ree points and Ree lines as \(O_R\). Observation \(1\) implies that there are no triangles in \(O_R\), and that every pair of points in \(O_R\) at mutual distance 2 are symplectic in \(\Delta\). Observation 4.35 implies that there are no quadrangles in \(O_R\) and, combined with Claim 6, it implies that there are no pentagons in \(O_R\). If also follows that points in \(O_R\) are at mutual distance \(3\) if, and only if, they are special in \(\Delta\). Claim 6 then implies that there are no hexagons in \(O_R\). Observation 4.36 now implies that there are no heptagons in \(O_R\) (as two points at distance 3 in such a heptagon should at the same time be special — considering the path of length 3 connecting them — and opposite — considering the path of length 4 connecting them — in \(\Delta\), a contradiction). It suffices to show that every line contains a point at distance at most \(3\) from a given point. Let \(p\) be a Ree point and \(L\) be a Ree line. Then \(L\) contains at least one (Ree) point not opposite \(p\) in \(\Delta\). Such a point is at distance at most \(3\) from \(p\) by our previous findings. Claim 7 is proved.

Next we want to prove our second principal aim, that is, \(O_R\) is a Ree-Tits octagon. For that, we have to prove the Moufang property for \(O_R\). To do so, it is convenient to first show that \(\Sigma\) is a convex chamber complex of \(\Delta\).

Claim 8. The complex \(\Sigma\) is convex in \(\Delta\).

This is equivalent to showing that \(\Sigma\) is closed under projection. Let \(p_1\) and \(p_2\) be two Ree points and let \(\{p_i,\xi_i\}\), \(i=1,2\), be the corresponding simplices of \(\Sigma\). First suppose that \(p_1\) and \(p_2\) are collinear. Then both those simplices are contained in a chamber of \(\Delta\) together with the simplex \(\{p_1p_2,\xi_1\cap\xi_2\}\), which belongs to \(\Sigma\). Hence the projection of \(\{p_1,\xi_1\}\) onto \(\{p_2,\xi_2\}\) is the chamber \(\{p_1,p_1p_2,\xi_1\cap\xi_2,\xi_2\}\), which also defines a chamber of \(\Sigma\).

Now suppose that \(p_1\) is symplectic to \(p_2\). By Claim 5, the simplex \(\{\xi_1\cap\xi_2,\xi(p_1,p_2)\}\) belongs to \(\Sigma\) and defines a Ree point collinear to both \(p_1,p_2\). Setting \(p:=\xi_1\cap\xi_2\) and \(\xi:=\xi(p_1,p_2)\), we see that the individual projections of \(p_2\) and \(\xi_2\) onto \(\{p_1,\xi_1\}\) in \(\Delta\) are \(\{p_1,\xi_1\cap\xi,\xi_1\}\) and \(\{p_1,p_1p,\xi_1\}\), respectively. Hence the projection of \(\{p_2,\xi_2\}\) onto \(\{p_1,\xi_1\}\) equals \(\{p_1,p_1p,\xi_1\cap\xi,\xi_1\}\). Since \(\{p_1p,\xi_1\cap\xi\}\in\Sigma\), the claim follows in this case.

Now assume \(p_1\Join p_2\). Let \(r_1,r_2\) be the unique Ree points with \(p_1\perp r_1\perp r_2\perp p_2\). The second paragraph of the proof of Claim 6 implies that the projection of \(\xi_2\) onto \(\{p_1,\xi_1\}\) is \(\{p_1,\left\langle p_1,r_2^\perp\cap\xi_1\right\rangle,\xi_1\}\). Similarly, the projection of \(p_2\) onto \(\{p_1,\xi_1\}\) is \(\{p_1,p_1r_1,\xi_1\}\). The claim now again follows.

Now let \(\{L_i,\pi_i\}\in \Sigma\), with \(L_i\) Ree lines and \(L_i^\sigma=\pi_i\), \(i=1,2\). If \(L_1\) and \(L_2\) meet, then \(p:=L_1\cap L_2\) is a Ree point and clearly, by uniqueness of \(p^\sigma\), the latter is the symp containing \(L_1\) and \(L_2\), which also contains \(\pi_1\) and \(\pi_2\). It fiollows that the projection of \(\{L_2,\pi_2\}\) onto \(\{L_1,\pi_1\}\) is equal to \(\{p,L_1,\pi_1,p^\sigma\}\) and the claim follows.

Now suppose \(L_1\) and \(L_2\) do not meet, but there are collinear Ree points \(p_i\in L_i\), \(i=1,2\). Observation 4.34 implies that every point on \(L_1\setminus\{p_1\}\) is symplectic to \(p_2\) and Observation 4.35 then implies that every such point is special to every point of \(L_2\setminus\{p_2\}\). Hence \(p_1\) is the projection of \(L_2\) onto \(\{L_1,\pi_1\}\). Likewise, \(p_1^\sigma\) is the projection of \(\Pi_2\) onto \(\{L_1,\pi_1\}\) and the claim again follows.

The proof for \(L_1\) and \(L_2\) not meeting and not containing collinear points, but a pair of symplectic points is completely similar.

The case where we project a point-symp pair \(\{p,\xi\}\in\Sigma\) onto a line-plane pair \(\{L,\pi\}\) is easy: there is always a unique point \(q\in L\) nearest to \(p\) and, dually, a unique symp \(\zeta\supseteq\pi\) nearest to \(\xi\). It follows from earlier observations that \(\{q,\zeta\}\) belongs to \(\Sigma\), which shows the claim once again.

Finally, if we project a line-plane pair \(\{L,\pi\}\in \Sigma\) onto a point-symp pair \(\{p,\xi\}\in\Sigma\), then that is equivalent to projecting the projection \(\{q,L,\pi,\zeta\}\) of \(\{p,\xi\}\) onto \(\{L,\pi\}\), onto \(\{p,\xi\}\), which coincides with the projection of \(\{q,\zeta\}\) onto \(\{p,\xi\}\) because this projection is a chamber in \(\Delta\), and which was proved to belong to \(\Sigma\) before.

This completes the proof of Claim 8. In the next claim we freely use the terminology of [16].

Claim 9. The generalised octagon \(O_R\) is a Moufang octagons and all elations are induced by unipotent elements of \(\mathsf{Aut}(\Delta)\).

By Claim 8, \(\Sigma\) is a thick twisting of \(\Delta\) in the sense of [16, §3]. Since each incident point-line pair in \(O_R\) defines a unique chamber of \(\Delta\), it is also a thick folding of Type 3 in the sense of [16, §2], which can be seen by considering an apartment through two opposite chambers of \(O_R\). Then by either Proposition 2.8 or 3.1 of [16], \(O_R\) is a Moufang octagon. Also, [16, Lemma 2.7] now shows that elations are induced by unipotent elements of \(\Delta\). Note that Lemma 2.7 of [16] holds for both thick foldings and thick twistings, but it is only explicitly proved in [16] for foldings; the proof for twistings is completely similar, as noted in [16, §3]. This completes the proof of Claim 9.

Our final principal aim is to show that \(\Sigma\) is the fixed complex of a polarity of \(\Delta\). Let \(P:=\{p,\xi\}\in\Sigma\), with \(p\) a Ree point. Then \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(P)\), which is equal to \(\mathop{\mathrm{\mathsf{Res}}}_\xi(p)\), is a generalised quadrangle, which we denote as \(\Gamma\). The simplices of \(\Sigma\) incident with \(P\) define an ovoid-spread pairing \(\beta:\mathcal{O}\to\mathcal{S}\), with \(\mathcal{O}\) an ovoid of \(\Gamma\), \(\mathcal{S}\) a spread, and \(\beta\) a bijection.

Claim 10. Let \(x_i\in\mathcal{O}\), \(i=1,2,3\). Then the unique points on \(x_1^\beta\) collinear to \(x_2\) and \(x_3\), respectively, coincide, if, and only if, the unique lines through \(x_1\) concurrent with \(x_2^\beta\) and \(x_3^\beta\), respectively, coincide.

Let \((x_i,x_i^\beta\}\) be arbitrary, with \(x_i\in\mathcal{O}\), \(i=1,2,3\). Then Claim 9 implies that there exists a unipotent element \(\varphi\) of \(\Gamma\) fixing \(x_1\) and \(x_1^\beta\) and mapping \((x_2,x_2^\beta)\) to \((x_3,x_3^\beta)\). Moreover, varying \(x_3\), the set of all such unipotent elements constitutes a root group \(U\) of a (generalised) Suzuki group \(G\) (see [34]). Note that, if \(\mathbb{K}\) is finite, then the fact that no ovoids exist in generalised quadrangles with \(q+1\) points per line and \(q^2+1\) lines per line pencil, implies that \(\Delta\) is split. We assume that, if \(\Delta\) is finite, it is not defined over \(\mathbb{F}_2\), that is, \(\Gamma\) is not isomorphic to the smallest (thick) generalised quadrangle. Then \(U\) is nilpotent of class 2, and the centre \(Z(U)\) consists precisely of the involutions of \(U\) (plus the identity, of course), and is generated by all commutators (see [34]). Let \(U\) fix the point \(x\) and the line \(L\ni x\), with \(x\in\mathcal{O}\) and \(L=x^\beta\). By the classification of buildings of type \(\mathsf{F_4}\) (see Chapter 9 of [31]), the action of \(U\) induced on \(L\) is isomorphic to the action of the additive group of an alternative division algebra, and the same holds for the action of \(U\) induced on the line pencil of \(\Gamma\) at \(x\). Note that these actions are that of an elementary abelian \(2\)-group. Hence each commutator in \(U\) pointwise fixes \(L\) and linewise fixes \(x\). It follows that each member of \(Z(G)\) pointwise fixes \(L\) and linewise fixes \(x\). Suppose now some member \(u\in U\) linewise fixes \(x\). Then \(u\) induces on each of these lines an involution (for the same reason as before) and so \(u^2\) pointwise fixes \(x^\perp\). Since every member of \(\mathcal{S}\) distinct from \(L\) contains a unique point of \(x^\perp\), each member of \(\mathcal{S}\) is fixed by \(u^2\) and hence \(u^2\) is the identity, showing \(u\in Z(U)\), and hence \(u\) pointwise fixes \(L\). Conversely one shows that, dually, if \(u\in U\) pointwise fixes \(L\), then it linewise fixes \(x\).

Now let \(\{x_2,x_2^\beta\}\) and \(\{x_3,x_3^\beta\}\) be two members of the ovoid-spread pairing, distinct from \(\{x,L\}\) (with \(x_2,x_3\in\mathcal{O}\)). Suppose \(x_2\) and \(x_3\) are collinear to the same point \(p\) on \(L\). Let \(u\in U\) be the unique unipotent element mapping \(\{x_2,x_2^\beta\}\) onto \(\{x_3,x_3^\beta\}\). Then \(u\) fixes \(x\) and \(p\) on \(L\) and hence fixes \(L\) pointwise. It follows that \(u\) fixes \(x\) linewise, and consequently \(x_2^\beta\) intersects the same line through \(x\) as \(x_3^\beta\). The converse is shown dually.

Now suppose that \(\Delta\) is defined over \(\mathbb{F}_2\). Then a straight forward check reveals that the previous property is true for every possible ovoid-spread pairing in the smallest generalised quadrangle (use the fact that three points of an ovoid never occur as the intersection of two point perps and, dually, no three lines of any spread form one regulus of a \(3\times3\) grid. This completes the proof of Claim 10.

Now we define the following map \(\rho\) on \(\Gamma\). Each point \(x\in\mathcal{O}\) is mapped onto \(x^\beta\) and each line\(L\) of \(\mathcal{S}\) is mapped onto the point \(x\) for which \(x^\beta=L\). Now let \(p\) be some point with \(p\notin\mathcal{O}\). Then \(p\) is contained in a unique spread line \(x^\beta\in\mathcal{S}\). Let \(y\in \mathcal{O}\) be collinear to \(p\), but distinct from \(x\). Then there is a unique line through \(x\) intersecting \(y^\beta\). That line is by definition \(p^\rho\). By Claim 10, this is independent of \(y\perp p\), \(y\in\mathcal{O}\). Dually, one defines the image \(L^\rho\) of any line \(L\) of \(\Gamma\).

Claim 11. The map \(\rho\) is a polarity of \(\Gamma\).

Indeed, a moment’s thought reveals that, with the foregoing notation and convention, the image of \(p^\rho\) is precisely \(p\). Hence \(\rho\) is an involution. If \(L\) is an arbitrary line through \(p\), then we can choose the point \(y\in\mathcal{O}\) on \(L\), and we see that \(L^\rho\) is the projection of \(x\) onto \(y^\beta\), which belongs to \(p^\rho\). We conclude that \(\rho\) preserves incidence, and since it is an involution, it also preserves non-incidence. This means that \(\rho\) is a polarity, as we had to show. Claim 11 is proved.

To emphasise the dependence of \(\rho\) from \(x\), we sometimes also denote \(\rho\) as \(\rho_x\). Also, we sometimes view \(\rho_x\) as a map between the lines and planes through \(x\), in the obvious way.

Claim 12. Let \(\{L,\pi\}\in\Sigma\) consist of a line \(L\) and a plane \(\pi\). Then there exists a unique isomorphism \(\psi\) from the projective plane \(\pi\) to the residue \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(L)\) such that, for each line \(K\subseteq \pi\), we have \(K^\psi=K^{\rho_x}\), with \(x=K\cap L\), and for each \(x\in L\), we have \(x^\psi=x^\sigma\).

Indeed, we can define the image \(K^\psi\) of any line \(K\) in \(\pi\) as \(K^\sigma\) if \(K=L\), and as \(K^{\rho_x}\) if \(K\neq L\) and \(x=K\cap L\). We have to show that triples of lines intersecting at a unique point \(p\) are mapped onto triples of planes contained in a common symp \(\xi_p\). If one of these lines is \(L\), then all images are contained in \(p^\sigma\). Assume now that \(p\in\pi\setminus L\) and let \(K_1,K_2,K_3\) be three lines in \(\pi\) through \(p\). Denote their respective intersection points with \(L\) as \(x_1,x_2,x_3\). Select an arbitrary plane \(\alpha_i\) through \(K_i\) in \(x_i^\sigma\), not containing \(L\), and let \(L_i\) be the Ree line in \(\alpha_i\), \(i=1,2\). Let \(y_4\) be a Ree point opposite \(x_2\) but symplectic to some point \(y_2\in L_1\) and let \(x_1\perp y_2\perp y_3\perp y_4\) be the unique path of Ree points joining \(x_1\) and \(y_4\) (cp. Claim 6). Let \(u\) be the automorphism of \(\Delta\) inducing the root elation in \(\Gamma\) that pointwise fixes the lines \(x_1y_2\), \(y_2y_3\) and \(y_3y_4\), that fixes all Ree lines through the points \(x_1, y_2, y_3\) and \(y_4\), and that maps \(\{x_2,x_2^\sigma\}\) to \(\{x_3,x_3^\sigma\}\).

We show that \(p\) is collinear with \(L_2^u\). Equivalently, we show that \(p\) is fixed by \(u\). Since \(\pi\) is stabilised by \(u\), it suffices to show that the line \(y_2p\) (which is indeed a line as \(y_2\in L_1\subseteq\alpha_1\) and \(p\in K_1\subseteq\alpha_1\)) is fixed. Therefore, we determine the action of \(u\) on the residue \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(y_2)\). We view this residue as a polar space \(\Omega\) with point set the set of symps through \(y_2\), line set the set of planes through \(y_2\) and plane set the set of lines through \(y_2\), with natural incidence. The symp \(y_2^\sigma\) is a fixed point \(a\) of \(\Omega\). Moreover, \(u\) elementwise fixes the ovoid-spread pairing of \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(\{y_2,y_2^\sigma\})\) induced by \(\Sigma\), which easily implies that \(u\) acts as the identity on the lines of \(\Omega\) through \(a\). The Ree planes through \(y_2\) correspond to an ovoid \(\mathcal{O}_a\) in \(\mathop{\mathrm{\mathsf{Res}}}_\Omega(a)\) of lines through \(a\). Now, the commutation relation in Ree-Tits octagons numbered (8M7) in Section 5.4.5 of [35] immediately implies that \(u\) fixes all points on each line of \(O_R\) through either \(y_2\) or \(y_3\). This, in turn, implies that each point of each line belonging to \(\mathcal{O}_a\) is fixed. Now, since \(x_1\) and \(y_4\) are fixed, the special pair of symps \(\{x_1^\sigma,y_4^\sigma\}\) is fixed, and hence so is the unique symp \(\zeta\) adjacent to both of these symps. Noticing that \(\zeta\) is not a Ree symp by Claim 6, this induces an additional fixed point in \(\Omega\) collinear to \(a\), but not on one of the pointwise fixed ovoid lines. This implies that all points of \(\Omega\) collinear to \(a\) are fixed. Finally, since all Ree lines through \(y_4\) and \(y_3\) are stabilised, the plane \((y_3y_4)^\sigma\) is pointwise fixed (project the aforementioned Ree lines onto it to see this quickly). Hence every line through \(y_2\) meeting that plane is stabilised, implying in \(\Omega\) that some planes not through \(a\) are stabilized. Consequently \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(y_2)\) is elementwise fixed, implying that the line \(y_2p\) is fixed, and so, \(p\) is fixed, which we wanted to show.

Noticing that \(p\) is the unique point collinear to both \(y_2\) and some arbitrarily chosen \(z_2\in L_2\setminus\{x_2\}\), the dual token shows that the unique symp \(\xi^*\) adjacent to both \(y^2\) and \(z_2\) is stabilised by \(u\). But \(\xi^*\) contains \(K_1^\psi\) and \(K_2^\psi\). Hence it also contains \((K_2^\psi)^u=K_3^\psi\).

Now Claim 12 follows.

We are finally in a position to prove that there exists a polarity with fixed structure \(\Sigma\). To that aim, consider two opposite flags of \(O_R\), that is, two chambers \(C:=\{x,L,\pi,\xi\}\) and \(C’:=\{x’,L’,\pi’,\xi’\}\) with \(\{x,\xi\},\{x’,\xi’\},\{L,\pi\},\{L’,\pi’\}\in\Sigma\) and \(x\) opposite \(x’\) and \(L\) opposite \(L’\). Then \(C\) is opposite \(C’\) by Claim 1 and Consequence 1. Hence they determine a unique apartment \(\mathcal{A}\). The map \(\rho_x\) is a duality of \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(\{x,\xi\}\). Furthermore, by Claim 12, there exists an isomorphism \(\psi\) from \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(\{\pi,\xi\}\) to \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(\{p,L\})\) compatible with \(\rho_x\) and such that elements of \(\mathcal{A}\) are mapped onto elements of \(\mathcal{A}\). The inverse of that isomorphism has the dual properties. The fact that elements of \(\mathcal{A}\) are mapped onto elements of \(\mathcal{A}\) means that both \(\rho_x\) and \(\psi\) (and also \(\psi^{-1}\)) are compatible with the duality of \(\mathcal{A}\) induced by the involution \[p\leftrightarrow\xi, \ \ \ p’\leftrightarrow\xi’, \ \ \ L\leftrightarrow \pi, \ \ \ L’\leftrightarrow\pi’.\]

We can now apply Tits’ extension theorem [31, Theorem 4.16] to obtain a polarity \(\theta\) of \(\Delta\) fixing all members of \(\Sigma\) incident with either \(\{p,\xi\}\) or \(\{L,\pi\}\), and fixing \(\{p’,\xi’\}\) and \(\{L’,\pi’\}\). The convexity of \(\Sigma\) readily implies that \(\Sigma\) is fixed elementwise. Since the fix structure of a polarity, hence also of \(\theta\), defines a generalised octagon \(O_\theta\), and clearly \(O_R\) is a subpolygon of \(O_\theta\), which is full by considering \(L\), and ideal by considering \(p\), [35, Theorem 1.8.2] yields \(O_R=O_\theta\).

The theorem is proved. \(\square\)

Proof. We begin with the last assertion. Let \(\theta\) be a collineation of \(\mathsf{F_{4,1}}(\mathbb{K},\mathbb{K}’)\) fixing \(A\) pointwise. Let \(x\) be an arbitrary point of \(A\). Then the symp \(\xi=x^\rho\) intersects \(A\) in the lines through \(x\) of an ovoid \(O\) of \(\mathop{\mathrm{\mathsf{Res}}}_\xi(x)\), a so-called Suzuki-Tits ovoid (see Lemma 2 of Section 2.5 in [35]). We now claim that \(\theta\) pointwise fixes \(\mathop{\mathrm{\mathsf{Res}}}_\xi(x)\). If \(|\mathbb{K}|=|\mathbb{K}’|=2\), this is immediate. Otherwise [35, Theorem 7.6.10] implies that \(\theta\) fixes all lines of the corresponding spread \(\mathcal{S}\) of \(\mathop{\mathrm{\mathsf{Res}}}_\xi(x)\). Since each point of each member of \(\mathcal{S}\) is collinear to some member of \(O\), each member of \(\mathcal{S}\) is fixed pointwise by \(\theta\). Since \(\mathcal{S}\) is a spread, the claim follows.

Now select a point \(y\in A\) opposite \(x\); then the unique point \(y_0\in\xi\) symplectic to \(y\) is fixed by \(\theta\). As \(y_0\) is opposite \(x\) in \(\xi\), the previous claim then shows that \(x^\perp\cap y_0^\perp\) is fixed pointwise, and hence every line through \(y_0\) in \(\xi\) is stabilised. Since each line of \(\xi\) through \(y_0\) is opposite (opposition in the polar space \(\xi\)) at least one member of \(O\) in \(\xi\), the set \(y_0^\perp\cap\xi\) is fixed pointwise. Likewise, \(x^\perp\cap\xi\) is fixed pointwise and this readily implies that \(\xi\) is fixed pointwise. Dually, all lines through \(x\) are fixed. Similarly, all lines through \(y\) are fixed. This, in turn, implies that the equator geometry \(E(x,y)\) is fixed pointwise. It is easy to see that \(E(x,y)\) contains two opposite points \(a,b\in A\). Since also \(E(a,b)\) is fixed pointwise, [15, Lemma 5.1.5(iii)] concludes the proof of the last assertion of the statement.

Now let \(\rho’\) be a duality with the property that each point of \(A\) is absolute. We first claim that, if \(x,x’\in A\) are collinear, then \(\zeta:=x^{\rho’}\) or \(\zeta’:={x’}^{\rho’}\) contains the line \(xx’\). Indeed, suppose not, then, since \(\zeta\cap\zeta’\) is a plane, the point \(x\) is collinear to a line of that plane and to \(x’\), a contradiction. The claim follows.

Now let \(L\) be a line, all points of which belong to \(A\). We may assume \(x,x’,x”\in L\). If \(x^{\rho’}\) does not contain \(L\), then both \({x’}^{\rho’}\) and \({x”}^{\rho’}\) do by the previous claim. Since \(\rho’\) is a duality, \(x^{\rho’}\) also contains \(L\).

Now it follows easily that \(x^{\rho’}=x^\rho\), since both are apparently determined by the lines contained in \(A\) and containing \(x\). Since \(x\in A\) was arbitrary, it follows that \(a^{\rho’}=a^\rho\) for all \(a\in A\) so that \(a^{\rho’\rho}=a\) for all \(a’ \in A\). From the last assertion, proved above, we now deduce \(\rho’\rho=\mathrm{id}\), implying \(\rho=\rho’\). The proposition is proved. \(\square\)

Proof. We show the implications \((i)\implies(ii)\implies(iii)\implies(iv)\implies (i)\), the latter being immediate.

If a pair of vertices \(v_1,v_6\) of \(\Sigma\) of type \(1\) and \(6\), respectively, that are not incident with a common vertex of type \(7\), are not in general position, then there is either a type 2 or a type 3 vertex in the convex closure, a contradiction. This shows that (ii) follows from (i).

Now suppose the subcomplex \(\Sigma\) is ovoidal of class \(1\) with Weyl diagram \(\mathsf{E_{7;3}}\) and some pair of vertices \(p,q\) of type \(1\) is special. We argue in \(\mathsf{E_{7,1}}(\mathbb{K})\). Let \(L\) be the unique line through \(q\) containing a point collinear to \(p\). Lemma 4.10 yields a symp \(\xi\in\Sigma\) containing \(L\). Since \(\{p,q\}\) is special, \(\xi\) and \(p\) are not contained in a common para; since \(p\) is collinear to some point of \(L\), and hence of \(\xi\), the pair \(\{p,\xi\}\) is not in general position. This shows that (iii) follows from (ii).

Now let \(\Sigma\) be an ovoidal subcomplex with Weyl diagram \(\mathsf{E_{7;3}}\) such that no pair of vertices of type \(1\) of \(\Sigma\) is special. We again reason in the long root subgroup geometry \(\Delta:=\mathsf{E_{7,1}}(\mathbb{K})\).

We first claim that, for each point \(p\in\Sigma\), the residue \(\mathop{\mathrm{\mathsf{Res}}}_{\Sigma}(p)\) is an ideal subspace of corank 4 of \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(p)\). Suppose this is not the case for some point \(p\). Then, by the weaker condition in Proposition 4.13, we find two symps \(\xi\) and \(\zeta\) through \(p\) belonging to \(\Sigma\), not contained in a common para, and having a plane \(\pi\) in common. Then points \(x\in\xi\) and \(y\in\zeta\) with \(|x^\perp\cap\pi\cap y^\perp|=1\) form a special pair, a contradiction. The claim is proved.

Next we claim that two points of \(\Sigma\) are never collinear. Indeed, suppose \(p\perp q\), with \(p,q\in\Sigma\). In \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(p)\), the line \(pq\) represents a maximal singular subspace of the polar space with point set the paras through \(p\). Lemma 4.10 yields a para \(\Pi\in\Sigma\) containing \(p\) and \(q\). This contradicts the fact that the elements of \(\Sigma\) in \(\Pi\) constitute an ideal Veronesean. The claim is proved. In particular, the foregoing ideal Veronesean is quaternion since, looking in the residue of a point, we recover the elements of \(\Sigma\) as a subspace, hence as the intersection of the corresponding polar space with a subspace of the ambient projective space.

In view of Proposition 4.23 and the definition above, it remains to show that \(\Sigma\) defines a polar space, with point set the points of \(\Sigma\), and line set the point sets of \(\Sigma\) determined by the symps in \(\Sigma\). The non-degeneracy axioms of a polar space can easily be proved by the reader; we content ourselves with proving the main axiom, namely, the one-or-all axiom. Let \(p,\xi\in\Sigma\) be a point and symp, respectively. If \(p,\xi\) are contained in a common para, then every point \(q\in\Sigma\) contained in \(\xi\) is symplectic to \(p\) (remember it cannot be collinear). We claim that the symp \(\zeta\) defined by \(p\) and \(q\) belongs to \(\Sigma\). We prove this claim for arbitrary symplectic points \(p,q\in\Sigma\).

Suppose for a contradiction that \(\zeta\) does not belong to \(\Sigma\). Then Lemma 4.10 yields a para \(\Pi\) through \(p\) intersecting \(\zeta\) in a \(4\)-space. By the point-para relations, \(q\) is collinear to all points of a \(5\)-space \(W\) of \(\Pi\), and all points of \(\Sigma\) in \(\Pi\) are collinear to a \(3\)-space of \(W\). Since no point of \(W\) belongs to \(\Sigma\) (it would otherwise be collinear to \(q\)), no symp of \(\Sigma\) in \(\Pi\) intersects \(W\) in a \(4’\)-space. If a symp of \(\Sigma\) in \(\Pi\) were disjoint from \(W\), then it would be incident with a \(5\)-space opposite \(W\) in \(\Pi\), yielding a point of \(\Sigma\) in that symp collinear to only one point of \(W\), a contradiction. Hence there is a symp \(\xi^*\in\Sigma\) in \(\Pi\) intersecting \(W\) in a line \(L\). A point of \(\Sigma\) contained in a \(4\)-space of \(\xi^*\) that intersects \(L\) in just one point is only collinear to the latter point of \(W\), again a contradiction. We conclude that the claim holds. It follows that the one-or-all axiom holds for \(p\) and \(\xi\) contained in a common para.

Next, suppose \(p\) and \(\xi\) are not contained in a common para. Then \(\xi\) contains points special to \(x\) and there are three possibilities (point-symp relations):

1. (1) The point \(p\) is collinear to a line \(M\) of \(\xi\). As before, the points of \(\Sigma\) contained in a \(4\)-space of \(\xi\) intersecting \(M\) in just a point are special to \(p\), a contradiction.

2. (2) The point \(p\) is symplectic to the points of a unique \(4\)-space of \(\xi\) and special to all other points of \(\xi\). Then \(p\) is special to te point of \(\Sigma\) contained in a \(4’\)-space of \(\xi\) disjoint from the aforementioned \(4\)-space. This is again a contradiction.

3. (3) The point \(p\) is symplectic to a unique point \(x\in\xi\), special to all points of \(\xi\) collinear to \(x\) and opposite the other points of \(\xi\). If \(x\) was not contained in \(\Sigma\), then \(p\) is special to the unique point of \(\Sigma\) contained in an arbitrary \(4\)-space of \(\xi\) through \(x\), a contradiction. Hence \(x\in\Sigma\) and one of our previous claims implies that \(op\) and \(x\) are contained in a common symp of \(\Sigma\).

This shows the one-or-all axiom and we conclude that \(\Sigma\) defines a quaternion ideal dual polar Veronesean. \(\square\)

Proof. We again show the implications \((i)\implies(ii)\implies(iii)\implies (i)\), the first implication being immediate.

Suppose \(\Sigma\) is as in (ii) and assume for a contradiction that, in \(\mathsf{E_{7,1}}(\mathbb{K})\), two symps \(\xi_1\) and \(\xi_2\) of \(\Sigma\) intersect in a point \(p\in\Sigma\) and there exists a symp \(\zeta\) intersecting \(\xi_i\) in a \(4\)-space \(U_i\), \(i=1,2\). Note that \(\{p\}=U_1\cap U_2\). We claim that we can find a point \(x\in\Sigma\) in \(\xi_1\) symplectic to \(p\). Indeed, since \(\Sigma\) is a chamber complex, we can find a plane \(\pi_1\) through \(p\) in \(\xi_1\) belonging to \(\Sigma\). By the Weyl diagram, all lines of \(\pi_1\) through \(p\), all points of these lines, and finally all lines of \(\pi_1\) belong to \(\Sigma\). let \(K_1\) be a line of \(\pi_1\) not through \(x\). Pick \(u_1\in K_1\). Then \(\mathop{\mathrm{\mathsf{Res}}}_\Sigma(\{u_1,K_1,\xi_1\})\) is an ovoid in \(\mathop{\mathrm{\mathsf{Res}}}_{\xi_1}(K_1)\). Hence we find a plane \(\pi_1’\) through \(K_1\) not collinear to \(\pi_1\). We infer that each point \(x\) of \(\pi_1’\setminus K_1\) belongs to \(\Sigma\) and is symplectic to \(p\). The claim follows. If \(x\) were collinear to a point \(x_2\in\xi_2\), then it would be collinear to at least all points of a line, which would yield a point \(y_2\) collinear to \(p\), implying by convexity of symps that \(y_2\in\xi_1\), contradictory to the fact that then \(y_2=p\) and so \(p\perp x\). Hence the condition in (ii) now says that \(x\) and \(\xi_2\) are in general position, implying that there is a point \(z_2\in\xi_2\) opposite \(x\). But any point \(u_1\) of \(U_1\) collinear to \(x\) is collinear (in \(\zeta\)) to a \(3\)-space in \(U_2\), which certainly contains a line collinear to \(z_2\). This implies that \(z_2\) is symplectic to \(u_1\perp x\), contradicting the opposition of \(x\) and \(z_2\). This shows that no pair of vertices of \(\Sigma\) of type \(6\) is special in a residue of a vertex of \(\Sigma\) of type \(1\). Proposition 4.19 now implies that the elements of \(\Sigma\) incident with a given point \(p\) of \(\Sigma\) form a Weyl substructure, in other words, they form a composition spread in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(p)\), viewed as a polar space (hence with point set the set of paras containing \(p\)).

Let \(X\) be the set of points in \(\Sigma\) and let \(\mathcal{L}\) be the set of lines in \(\Sigma\). We use Definition 2.1 to show that \(\Gamma=(X,\mathcal{L})\) is a metasymplectic space with thick symps. First we have to show that \(\Gamma\) is a parapolar space. To that aim, it is convenient to show the claim that, if two points of \(X\) are symplectic in \(\Delta\), then the unique symp they are contained in belongs to \(\Sigma\).

So let \(p,q\in X\) be symplectic in \(\Delta\). Suppose for a contradiction that the symp \(\xi\) through \(p\) and \(q\) does not belong to \(\Sigma\). Then, in the residue of \(q\), viewed as polar space \(\mathsf{D_{6,1}}(\mathbb{K})\), \(\xi\) is a line not belonging to the Weyl complex induced by \(\Sigma\) in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(q)\). Hence we can select a symp \(\zeta\) through \(q\) belonging to \(\Sigma\) and which, viewed as a line in \(\mathsf{D_{6,1}}(\mathbb{K})\), is special to \(\xi\). It then follows that \(p\) and \(\zeta\) are not in general position, and that \(p^\perp\cap\zeta\) is empty, a contradiction to (ii). Hence \(\xi\in\Sigma\). The claim is proved.

We now show that \(\Gamma\) is connected. To that aim, let \(x,y\) be two arbitrary points of \(\Sigma\). If \(x\) and \(y\) are opposite in \(\Delta\), then we can find a point on any line of \(\Sigma\) through \(y\) special to \(x\). Hence we may assume that \(x\) and \(y\) are not opposite. If they are collinear, then likewise we can find a line of \(\Sigma\) through \(y\) locally opposite the line \(xy\), and hence this line contains a point of \(\Sigma\) special to \(x\). So we may assume that \(x\) and \(y\) are special (if they are symplectic, they are connected by the unique symp through them, which belongs to \(\Sigma\) as shown above). Looking in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(y)\), it is easy to see now that there exists a symp \(\xi\in\Sigma\) through \(y\) not in general position with \(x\). Hence, by our assumption in (ii), \(x\) is collinear to a line \(L\) of \(\xi\), and it is symplectic to all points of \(\Sigma\) contained in \(\xi\) and in a maximal singular subspace containing \(L\). The connectivity of \(\Gamma\) is proved.

Now let \(x,y\) be two points of \(X\) at distance \(2\) from each other, joined by at least two paths of length \(2\) contained in \(\Sigma\). Consider one of these paths and let \(p\) be its middle point. Since \(\Sigma\) induces in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(p)\) a Weyl complex, which is convex, it follows that, if \(x\) and \(y\) were collinear in \(\Delta\), then they would also be collinear in \(\Gamma\). Hence they are symplectic and the unique symp \(\xi\) containing both belongs to \(\Sigma\), by a previous claim we proved. Since \(\xi\) is convex in \(\Delta\), the convex closure of \(x,y\) in \(\Gamma\) is contained in \(\xi\). Since the points and lines of \(\Gamma\) form a polar space \(\Gamma_\xi\) of rank 3 in \(\xi\), that polar space is the convex closure. Note that \(\Gamma_\xi\) is thick. This verifies (i) of Definition 2.1. Now assume that two symps of \(\Sigma\) have at least one point \(x\) in common. In \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(x)\), viewed as a half-spin geometry, the two symps are either disjoint, or intersect in a line, which belongs to \(\Sigma\). Hence (ii) holds. Now the residue of a point of \(\Sigma\) is a Weyl complex of type \(\mathsf{D_{6;3}^2}\). This defines a thick polar space of rank 3, completing the proof that \(\Gamma\) is a metasymplectic space. Since a Weyl complex of type \(\mathsf{D_{6;3}^2}\) in \(\mathsf{D_{6}}(\mathbb{K})\) defines a polar space with planes over a quadratic extension \(\mathbb{L}\) of \(\mathbb{K}\), \(\Sigma\) defines a metasymplectic space isomorphic to \(\mathsf{F_{4,1}}(\mathbb{K},\mathbb{L})\), fully embedded in \(\mathsf{E_{7,1}}(\mathbb{K})\).

Now (iii) follows straight from [8, Main Result]

Also the implication \((iii)\implies(i)\) follows as \(\Sigma\) is, under the assumptions in (iii), the fixed point structure of a non-trivial collineation by [8, Main Result]. The proof is complete. \(\square\)

Proof. We again show the implications \((i)\implies(ii)\implies(iii)\implies (i)\), the first implication being immediate.

The proof of the implication \((ii)\implies(iii)\) is very similar to that in Proposition 4.40. We first claim that \(\Sigma\) defines, in \(\mathsf{E_{8,8}}(\mathbb{K})\), an embedded metasymplectic space isomorphic to \(\mathsf{F_{4,1}}(\mathbb{K},\mathbb{H})\), with \(\mathbb{H}\) some quaternion division algebra over \(\mathbb{K}\), possibly inseparable (meaning, \(\mathbb{H}\) is an inseparable quadratic extension of degree 4 of \(\mathbb{K}\) in characteristic \(2\), so \(\mathbb{H}^2\leq\mathbb{K}\)). We argue in \(\mathsf{E_{8,8}}(\mathbb{K})\).

First we note that (ii) implies that no two symps of \(\Sigma\) containing a common point \(p\) of \(\Sigma\) are special to each other in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(p)\). The proof of this is the same as in Proposition 4.40. Now Proposition 4.39 implies that for each point \(x\in\Sigma\), \(\mathop{\mathrm{\mathsf{Res}}}_\Sigma(x)\) is a quaternion polar space of rank 3.

Furthermore, the same proof as above for Proposition 4.40 shows that, if \(x,y\in\Sigma\) are symplectic points, then the unique symp containing \(x,y\) belongs to \(\Sigma\).

Now let \(\xi\in\Sigma\) be a symp. Suppose for a contradiction that two planes \(\pi_1,\pi_2\) of \(\Sigma\) are contained in the same maximal singular subspace of \(\xi\). Select a line \(L_1\subseteq\pi_1\). In \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(L_1)\), the complex \(\Sigma\) induces an ovoidal subcomplex of type \(\mathsf{E_{6;2}}\), and so we can find a symp \(\zeta_1\in\Sigma\) containing \(\pi_1\) and intersecting \(\xi\) precisely in \(\pi_1\). There are now two ways to see a contradiction: (1) defining \(\zeta_2\) analogously to \(\zeta_1\), but now with respect to \(\pi_2\), we see that each point of \(\Sigma\) in \(\zeta_1\), not collinear to \(\pi_1\), is symplectic to each point of \(\pi_2\), but not collinear to any point of \(\zeta_2\); hence no such point is in general position with \(\zeta_2\), contradictory to (ii). More directly, (2) let \(x_1\in\Sigma\) be a point of \(\zeta_1\) such that \(x_1^\perp\cap\pi_1=L_1\). Pick \(x_2\in\pi_2\) arbitrarily. Then \(x_1\) and \(x_2\) are not collinear, and so the symp \(\zeta\) containing them belongs to \(\Sigma\). But in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(L_1)\) the symps \(\zeta\) and \(\zeta_1\) intersect \(\xi\) in collinear points, a contradiction to the fact that \(\Sigma\) defines a Weyl complex of type \(\mathsf{E_{6;2}}\) in \(\mathop{\mathrm{\mathsf{Res}}}_\Delta(L_1)\).

Now it follows from Proposition 4.13 that \(\Sigma\) induces in each of its symps an ovoidal subspace of corank 4, which is a thick polar space of rank 3.

The rest of the proof of the claim is now the same as in the proof of Proposition 4.40. Also, the implication \((iii)\implies(i)\) follows straight from [22, Theorem 4.1]. \(\square\)

Proof. This follows from Table 3. We explain below the notation and how Table 3 is constructed. \(\square\)