number of possibilities for triangles of mutually opposite simplices is too
large to be practical or useful. The following result provides an alternative
to 8.1. The condition that $J$ is a self-opposite type is not essential, but
convenient, and we will only need it in that case.

######  Lemma 8.17.

Let $\Delta$ be a spherical building over the type set $I$ and let $J\subseteq
I$ be a self-opposite type. Suppose that for each pair of simplices
$F,F^{\prime}$ of type $J$, the subgraph $\Gamma_{J}^{\\{F,F^{\prime}\\}}$ of
$\Gamma_{J}$ induced by the vertices opposite both $F$ and $F^{\prime}$ is
connected. Suppose also that there is a simplex of type $J$ opposite any given
set of three simplices of type $J$. Let $F$ be a given simplex of type $J$.
Denote by $\Pi_{4}(F)$ the set of all self-projectivities
$F\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}F_{4}\text{\,\footnotesize$\overline{\land}$\,}F$
of $F$ of length $4$ with $F\sim F_{3}$, $F_{2}\sim F_{4}$. Suppose that
$\Pi_{4}(F)$ is geometric. Then $\Pi^{+}(F)=\langle\Pi_{4}(F)\rangle$.

###### Proof.

We first prove the following property for four simplices
$F_{1},F_{2},F_{3},F_{4}$, where
$\operatorname{\mathsf{typ}}(F_{1})=\operatorname{\mathsf{typ}}(F_{3})=J$ and
both $F_{2}$ and $F_{4}$ are opposite both $F_{1}$ and $F_{3}$.

  * (*)

_The projectivity $\rho\colon
F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}F_{4}$
can be written as a product of a perspectivity
$F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{4}$ and conjugates of
members of $\Pi_{4}(F_{1})$._

Indeed, let $F_{1}=F_{1}^{\prime}\sim F_{2}^{\prime}\sim\cdots
F_{n}^{\prime}=F_{3}$ be a path in $\Gamma_{J}^{\\{F_{2},F_{4}\\}}$. Define
$\rho_{i}\colon
F_{4}\text{\,\footnotesize$\overline{\land}$\,}F_{i}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{i+1}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F_{4}$,
$i\in\\{1,2,\ldots,n-1\\}$. Denote by $\rho_{0}$ the perspectivity
$F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{4}$. Then it is elementary
to see that $\rho=\rho_{0}\rho_{1}\rho_{2}\cdots\rho_{n-1}$. So, since
$\Pi_{4}(F)$ is geometric, it suffices to show that each $\rho_{i}$ can be
written as the product of conjugates of members of $\Pi_{4}(F_{1})$. It
follows from letting $(F_{4},F_{i}^{\prime},F_{2},F_{i+1}^{\prime})$ play the
role of $(F_{1},F_{2},F_{3},F_{4})$ in the previous argument that $\rho_{i}$
is a product of conjugates of members of $\Pi_{4}(F_{i+1}^{\prime})$. Hence
(*) is proved.

Now let $\rho\colon
F\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}\cdots\text{\,\footnotesize$\overline{\land}$\,}F_{2\ell-1}\text{\,\footnotesize$\overline{\land}$\,}F_{2\ell}\text{\,\footnotesize$\overline{\land}$\,}F$
be an arbitrary even projectivity. We prove by induction on
$\ell\in\\{1,2,\ldots\\}$ that $\rho$ is the product of conjugates of members
of $\Pi_{4}(F)$. This is trivial for $\ell=1$ and it is equivalent to (*) for
$\ell=2$. So let $\ell\geq 3$. Select a simplex $F_{2}^{\prime}$ opposite each
of $F,F_{3}$ and $F_{5}$ (since these all have the same type, this is still
possible if $J$ is not self-opposite). Setting

$\begin{cases}\rho^{*}_{1}\colon
F\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}F_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F,\\\
\rho^{*}_{2}\colon
F_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}F_{4}\text{\,\footnotesize$\overline{\land}$\,}F_{5}\text{\,\footnotesize$\overline{\land}$\,}F_{2}^{\prime},\\\
\rho^{\prime}\colon
F\text{\,\footnotesize$\overline{\land}$\,}F_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F_{5}\text{\,\footnotesize$\overline{\land}$\,}F_{6}\cdots\text{\,\footnotesize$\overline{\land}$\,}F_{2\ell}\text{\,\footnotesize$\overline{\land}$\,}F,\end{cases}$

we see that, if $\rho_{0}\colon
F\text{\,\footnotesize$\overline{\land}$\,}F_{2}^{\prime}$, we have

$\rho=\rho_{1}^{*}\cdot(\rho_{0}\rho_{2}^{*}\rho_{0}^{-1})\cdot\rho^{\prime},$

where we know by the induction hypothesis that all factors are products of
members of $\Pi_{4}(F)$, using the fact that $\Pi_{4}(F)$ is geometric (and
hence closed under conjugation). ∎

In order to be able to apply 8.17, we have to check the conditions in the
various cases. It turns out we will use 8.17 in exactly three different cases,
for which we now prove the condition on the corresponding graph $\Gamma_{J}$.

######  Lemma 8.18.

Let $\Delta$ be the spherical building over the field $\mathbb{K}$,
$|\mathbb{K}|>2$, of type either $\mathsf{E_{6}}$ or $\mathsf{E_{7}}$. Let
$J=\\{2\\}$ if $\operatorname{\mathsf{typ}}\Delta=\mathsf{E_{6}}$, and
$J\in\\{\\{1\\},\\{3\\}\\}$ if
$\operatorname{\mathsf{typ}}\Delta=\mathsf{E_{7}}$. Let $v$ and $v^{\prime}$
be vertices of type $J$. Then the subgraph $\Gamma_{J}^{\prime}$ of
$\Gamma_{J}$ induced on the vertices opposite both $v$ and $v^{\prime}$ is
connected. If $J\neq\\{3\\}$, then, more generally, the complement of the
union of two geometric hyperplanes of the corresponding geometries of type
$\mathsf{E}_{n,J}$, $n=6,7$, is connected.

###### Proof.

The claims we refer to in this proof are those of Section 8.3.

We start by proving the cases $J=\\{2\\}$ if
$\operatorname{\mathsf{typ}}\Delta=\mathsf{E_{6}}$ and for $J=\\{1\\}$ if
$\operatorname{\mathsf{typ}}\Delta=\mathsf{E_{7}}$. Indeed, let $C$ be the
complement of two proper geometric hyperplanes $H_{1},H_{2}$ of the long root
geometry $\Gamma$ of type $\mathsf{E_{6}}$ or $\mathsf{E_{7}}$ over the field
$\mathbb{K}$, with $|\mathbb{K}|>2$. Let $p$ and $q$ be two points of $C$. A
moment’s thought reveals that we may assume that $p$ and $q$ are opposite in
$\Gamma$. The equator geometry $E(p,q)$ is not the union of two geometric
hyperplanes, because lines contain at least $3$ points and hence, if $E(p,q)$
is contained in neither $H_{1}$ nor $H_{2}$, then there is some point $r\in
E(p,q)\cap C$. Claim 3 yields paths in $C$ connecting $p$ with $r$ and $r$
with $q$. So, we may assume that $E(p,q)\subseteq H_{1}$. Since $E(p,q)$
contains planes, we find a point $x\in E(p,q)\cap H_{1}\cap H_{2}$. Let $\xi$
be the symp through $p$ and $x$, and let $\zeta$ be a symp opposite $\xi$
through $q$. Set $H_{\xi,1}:=H_{1}\cap\xi$, $H_{\xi,2}:=H_{2}\cap\xi$, and let
$H_{\xi,3}$ be the (pointwise) projection of $\zeta\cap H_{1}$ onto $\xi$;
likewise let $H_{\xi,4}$ be the pointwise projection of $\zeta\cap H_{2}$ onto
$\xi$ (projecting here means mapping to the symplectic points). Since the
projection of $q$ is $x$, we see that $H_{\xi,1}\cap H_{\xi,2}\ni x\notin
H_{\xi,3}\cap H_{\xi,4}$. Hence Claim 2 yields a point $z\in C\cap\xi$ such
that the unique point $z^{\prime}$ of $\zeta$ symplectic to $z$ also belongs
to $C$. Now Claim 3 yields paths $p\perp\perp z\perp\perp z^{\prime}\perp\perp
q$ and the proof for these two cases is finished.

Now assume $J=\\{3\\}$ and $\Delta$ has type $\mathsf{E_{7}}$. Consider the
parapolar space of type $\mathsf{E_{7,7}}$ corresponding to $\Delta$. Then $v$
and $v^{\prime}$ correspond to two $5$-spaces $V$ and $V^{\prime}$. Adjacency
in $\Gamma_{\\{3\\}}$ coincides with the $\sim$ relation of the proof of Claim
6 in any symp containing the two $5$-spaces.

Let $W_{1}$ and $W_{2}$ be two arbitrary $5$-spaces both opposite both $V$ and
$V^{\prime}$. We show that they belong to the same connected component of the
graph $\Gamma^{\prime}_{\\{3\\}}$. As before, we may assume that they are
opposite in $\Delta$. Consider opposite symps $\xi_{1}\supseteq W_{1}$ and
$\xi_{2}\supseteq W_{2}$. Let $U_{i}$ and $U_{i}^{\prime}$ be the projection
of $V$ and $V^{\prime}$, respectively, onto $\xi_{i}$, $i=1,2$. Then, by 2.1,
a $5$-space in $\xi_{i}$ is opposite $V$ or $V^{\prime}$ if, and only if, it
is opposite $W_{i}$ or $W_{i}^{\prime}$, respectively, $i=1,2$. Let $Z_{1}$
and $Z_{1}^{\prime}$ be the projections of $U_{2}$ and $U_{2}^{\prime}$,
respectively, onto $\xi_{1}$. Let $L_{1}$ be a line of $\xi_{1}$ disjoint from
$U_{1}\cup U_{1}^{\prime}\cup Z_{1}\cup Z_{1}^{\prime}$, guaranteed to exist
by Claim 4. Let $L_{2}$ be its projection onto $\xi_{2}$. Then $L_{2}$ is
disjoint from $U_{2}\cup U_{2}^{\prime}$. Claim $5$ now yields a $5$-space
$W_{i}^{*}$ through $L_{i}$ disjoint from $U_{i}\cup U_{i}^{\prime}$, $i=1,2$,
and Claim 6 for $d=6$ implies that it now suffices to prove that $W_{1}^{*}$
and $W_{2}^{*}$ are in the same connected component of
$\Gamma^{\prime}_{\\{3\\}}$.

Let $\zeta$ be the symp through $L_{1}$ and $L_{2}$. Let $U$ and $U^{\prime}$
be the projections of $V$ and $V^{\prime}$, respectively, onto $\zeta$. Since
no point of $L_{i}$, $i=1,2$, is collinear to a point of $V\cup V^{\prime}$,
the $5$-spaces $U$ and $U^{\prime}$ are disjoint from $L_{1}\cup L_{2}$. Claim
5 yields $5$-spaces $Y_{1}$ and $Y_{2}$ containing $L_{1}$ and $L_{2}$,
respectively, disjoint from $U\cup U^{\prime}$. Claim 6 for $d=6$ implies that
$Y_{1}$ and $Y_{2}$ are in the same connected component of
$\Gamma^{\prime}_{\\{3\\}}$. Hence, in order to prove the lemma, it suffices
to show that $W_{i}^{*}$ and $Y_{i}$ are in the same connected component, for
$i=1,2$. But this now follows from Claim 7 for $d=5$ applied to the residue of
$L_{i}$.

The lemma is proved. ∎

######  Theorem 8.19.

Let $\Delta$ be a building of type $\mathsf{E_{6}}$, $\mathsf{E_{7}}$ or
$\mathsf{E_{8}}$ over the field $\mathbb{K}$. Let $F$ be a simplex of $\Delta$
such that $\mathsf{Res}_{\Delta}(F)$ is irreducible. Then $\Pi(F)$ and
$\Pi^{+}(F)$ are given as in _Table 2_, where the last column contains a
checkmark if $\operatorname{\mathsf{typ}}(F)$ is polar closed. Again, a long
hyphen in the table in the column of $\Pi(F)$ means that
$\operatorname{\mathsf{typ}}(F)$ is not self-opposite and so $\Pi(F)$ is
trivially isomorphic to $\Pi^{+}(F)$ — it must again be read as a “bysame”
symbol. Grey rows correspond to projectivity groups which are not necessarily
full linear groups.

Reference | $\operatorname{\mathsf{typ}}(\Delta)$ | $\mathsf{Res}_{\Delta}(F)$ | $\operatorname{\mathsf{cotyp}}(F)$ | $\Pi^{+}(F)$ | $\Pi(F)$ |
---|---|---|---|---|---|---
(A1) |  | $\mathsf{A_{1}}$ |  | $\mathsf{PGL}_{2}(\mathbb{K})$ | $\mathsf{PGL}_{2}(\mathbb{K})$ |
(A2) | $\mathsf{E_{6}}$ | $\mathsf{A_{2}}$ | $\\{2,4\\}$ | $\mathsf{PGL}_{3}(\mathbb{K})$ | $\mathsf{PGL}_{3}(\mathbb{K}).2$ |
$\mathsf{E_{6}}$ | $\mathsf{A_{2}}$ | $\neq\\{2,4\\}$ | $\mathsf{PGL}_{3}(\mathbb{K})$ | —— |
$\mathsf{E_{7},E_{8}}$ | $\mathsf{A_{2}}$ |  | $\mathsf{PGL}_{3}(\mathbb{K})$ | $\mathsf{PGL}_{3}(\mathbb{K}).2$ |
(A3) | $\mathsf{E_{6}}$ | $\mathsf{A_{3}}$ | $\\{3,4,5\\}$ | $\mathsf{PGL}_{4}(\mathbb{K})$ | $\mathsf{PGL}_{4}(\mathbb{K})$ | $\checkmark$
$\mathsf{E_{6}}$ | $\mathsf{A_{3}}$ | $\neq\\{3,4,5\\}$ | $\mathsf{PGL}_{4}(\mathbb{K})$ | —— |
$\mathsf{E_{7},E_{8}}$ | $\mathsf{A_{3}}$ |  | $\mathsf{PGL}_{4}(\mathbb{K})$ | $\mathsf{PGL}_{4}(\mathbb{K}).2$ |
(A4) | $\mathsf{E_{6}}$ | $\mathsf{A_{4}}$ |  | $\mathsf{PGL}_{5}(\mathbb{K})$ | —— |
$\mathsf{E_{7},E_{8}}$ | $\mathsf{A_{4}}$ |  | $\mathsf{PGL}_{5}(\mathbb{K})$ | $\mathsf{PGL}_{5}(\mathbb{K}).2$ |
(A5) | $\mathsf{E_{6}}$ | $\mathsf{A_{5}}$ |  | $\mathsf{PSL}_{6}(\mathbb{K},3)$ | $\mathsf{PSL}_{6}(\mathbb{K},3)$ | $\checkmark$
$\mathsf{E_{7}}$ | $\mathsf{A_{5}}$ | $\\{2,4,5,6,7\\}$ | $\mathsf{PSL}_{6}(\mathbb{K},2)$ | $\mathsf{PSL}_{6}(\mathbb{K},2).2$ |
$\mathsf{E_{7}}$ | $\mathsf{A_{5}}$ | $2\notin\operatorname{\mathsf{cotyp}}(F)$ | $\mathsf{PGL}_{6}(\mathbb{K})$ | $\mathsf{PGL}_{6}(\mathbb{K}).2$ |
$\mathsf{E_{8}}$ | $\mathsf{A_{5}}$ |  | $\mathsf{PGL}_{6}(\mathbb{K})$ | $\mathsf{PGL}_{6}(\mathbb{K}).2$ |
(A6) | $\mathsf{E_{7},E_{8}}$ | $\mathsf{A_{6}}$ |  | $\mathsf{PGL}_{7}(\mathbb{K})$ | $\mathsf{PGL}_{7}(\mathbb{K}).2$ |
(A7) | $\mathsf{E_{8}}$ | $\mathsf{A_{7}}$ |  | $\mathsf{PGL}_{8}(\mathbb{K})$ | $\mathsf{PGL}_{8}(\mathbb{K}).2$ |
(D4) | $\mathsf{E_{6}}$ | $\mathsf{D_{4}}$ |  | $\overline{\mathsf{PGO}}_{8}^{\circ}(\mathbb{K})$ | $\overline{\mathsf{PGO}}_{8}(\mathbb{K})$ |
$\mathsf{E_{7},E_{8}}$ | $\mathsf{D_{4}}$ |  | $\overline{\mathsf{PGO}}^{\circ}_{8}(\mathbb{K})$ | $\overline{\mathsf{PGO}}^{\circ}_{8}(\mathbb{K})$ | $\checkmark$
(D5) | $\mathsf{E_{6}}$ | $\mathsf{D_{5}}$ |  | $\overline{\mathsf{PGO}}_{10}^{\circ}(\mathbb{K})$ | —— |
$\mathsf{E_{7},E_{8}}$ | $\mathsf{D_{5}}$ |  | $\overline{\mathsf{PGO}}_{10}^{\circ}(\mathbb{K})$ | $\overline{\mathsf{PGO}}_{10}(\mathbb{K})$ |
(D6) | $\mathsf{E_{7}}$ | $\mathsf{D_{6}}$ |  | $\overline{\mathsf{P\Omega}}_{12}(\mathbb{K})$ | $\overline{\mathsf{P\Omega}}_{12}(\mathbb{K})$ | $\checkmark$
$\mathsf{E_{8}}$ | $\mathsf{D_{6}}$ |  | $\overline{\mathsf{PGO}}^{\circ}_{12}(\mathbb{K})$ | $\overline{\mathsf{PGO}}^{\circ}_{12}(\mathbb{K})$ | $\checkmark$
(D7) | $\mathsf{E_{8}}$ | $\mathsf{D_{7}}$ |  | $\overline{\mathsf{PGO}}_{14}^{\circ}(\mathbb{K})$ | $\overline{\mathsf{PGO}}_{14}(\mathbb{K})$ |
(E6) | $\mathsf{E_{7},E_{8}}$ | $\mathsf{E_{6}}$ |  | $\mathsf{PGE_{6}}(\mathbb{K})$ | $\mathsf{PGE_{6}}(\mathbb{K}).2$ |
(E7) | $\mathsf{E_{8}}$ | $\mathsf{E_{7}}$ |  | $\mathsf{PGE}_{7}(\mathbb{K})$ | $\mathsf{PGE}_{7}(\mathbb{K})$ | $\checkmark$
Table 2. Projectivity groups in the exceptional cases
$\mathsf{E_{6},E_{7},E_{8}}$

###### Proof.

The case (A1) was handled in D. We now handle the other cases. Note that we
may again assume that $|\mathbb{K}|\geq 3$ as otherwise the linear groups are
unique.

Cases (A2) and (A3) Every subdiagram of type $\mathsf{A_{2}}$ or
$\mathsf{A_{3}}$ of $\mathsf{E}_{r}$, $r=6,7,8$, is contained in a subdiagram
of type $\mathsf{A_{3}}$ or $\mathsf{A_{4}}$, respectively. Then the
assertions all follow from 7.2 and Theorem 8.4.

Case (A4) If $2\notin\operatorname{\mathsf{cotyp}}(F)$ for $\mathsf{E_{6}}$,
or if $\operatorname{\mathsf{cotyp}}(F)\neq\\{1,2,3,4\\}$ for
$\mathsf{E_{7},E_{8}}$, then we can again embed the diagram of
$\mathsf{Res}_{\Delta}(F)$ in diagram of type $\mathsf{A_{5}}$ and use 7.2 and
Theorem 8.4.

Now suppose $2\in\operatorname{\mathsf{cotyp}}(F)$ for $\mathsf{E_{6}}$ and
$\operatorname{\mathsf{cotyp}}(F)=\\{1,2,3,4\\}$ for $\mathsf{E_{7}}$ and
$\mathsf{E_{8}}$. Then the assertion follows from 7.2 and Case (A∗∗) for $r=5$
of Theorem 8.8.

Case (A5) In the Coxeter diagram of type $\mathsf{E_{8}}$, every subdiagram of
type $\mathsf{A_{5}}$ is contained in one of type $\mathsf{A_{6}}$ and hence
the assertion for $\mathsf{E_{8}}$ follows from 7.2. The same thing holds in
the Coxeter diagram of type $\mathsf{E_{7}}$ if the subdiagram of type
$\mathsf{A_{5}}$ does not contain the vertex of type $2$.

Now suppose $\Delta$ is the building of type $\mathsf{E_{6}}$ over the field
$\mathbb{K}$, and $F$ is a vertex of type $2$.

We argue in the corresponding Lie incidence geometry of type
$\mathsf{E_{6,1}}$. There, $F$ is a $5$-space. Let $F_{1},F_{2},F_{3}$ be
three $5$-spaces, with both $F_{1}$ and $F_{3}$ opposite both $F$ and $F_{2}$,
and with $F$ adjacent to $F_{2}$, and $F_{1}$ adjacent to $F_{3}$, that is,
$\pi_{0}:=F\cap F_{2}$ and $\pi_{1}:=F_{1}\cap F_{3}$ are planes. We also
initially assume that $\pi_{0}$ and $\pi_{1}$ are opposite. Consider the
projectivity $\rho\colon
F\text{\,\footnotesize$\overline{\land}$\,}F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}F$.
We claim that $\rho$ fixes each point of $\pi_{0}$. Indeed, let
$p_{0}\in\pi_{0}$ be such a point. Then clearly, since $F\cap F_{2}$ contains
$p_{0}$, the projectivities ${F\perp
F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{2}}$ and ${F_{2}\perp
F_{3}\text{\,\footnotesize$\overline{\land}$\,}F}$ fix $p_{0}$, hence
$p_{0}^{\rho}=p_{0}$ and the claim is proved. Likewise, $\rho$ fixes each
point of $F$ collinear with a point of $\pi_{1}$. The set of such points forms
a plane $\pi_{0}^{\prime}$ of $F$, disjoint from $\pi_{0}$. Choosing a basis
of $F$ in $\pi_{0}\cup\pi_{0}^{\prime}$, a matrix of $\rho$ is a diagonal
matrix with diagonal elements three times $1$ and three times some scalar
$k\in\mathbb{K}$. We now show that $k$ can be arbitrary. This is equivalent to
showing that,

  * (*)

given $F_{1},F_{2}$ and $F_{3}$ as above, given a line $L_{0}$ in $F$
containing points $x_{0}\in\pi_{0}$ and $x_{0}^{\prime}\in\pi_{0}^{\prime}$,
and given two points $p,q\in L_{0}\setminus\\{x_{0},x_{0}^{\prime}\\}$, we can
re-choose $F_{3}$ through $\pi_{1}$ such that $\rho$ maps $p$ to $q$.

We now prove (*). Let $p_{1}$ be the projection of $p$ onto $F_{1}$ and let
$p_{2}$ be the projection of $p_{1}$ onto $F_{2}$. If $p$ and $p_{2}$ were not
collinear, then the symp $\xi(p,p_{2})$ would contain $p_{1}$ and $\pi_{0}$,
leading to additional points in $\pi_{0}$ collinear to $p_{1}$ inside
$\xi(p,p_{2})$, contradicting the fact that $F$ and $F_{1}$ are opposite and
hence $p_{1}$ is far from $F$. Hence there is some singular $4$-space $U$
containing $\pi_{0},L$ and $p_{2}$. (Note that, since $U$ intersects $F$ in a
$3$-space, 2.4$(ii)$ implies that $U$ is really a $4$-space and not a
$4^{\prime}$-space.) Set $\xi:=\xi(x_{0},p_{1})$. Then $\xi$ contains
$p,q,p_{1},p_{2}$ and the unique point $x_{1}\in\pi_{1}$ collinear to
$x_{0}^{\prime}$. It is clear that $\pi_{1}$ intersects $\xi$ in only $x_{1}$,
as otherwise there would be a point of $\pi_{1}$ collinear to $x_{0}$,
contradicting the fact that $\pi_{0}$ and $\pi_{1}$ are opposite. So, 2.5
yields a plane $\alpha\subseteq\xi$ collinear to $\pi_{1}$. 2.4 implies that
$\alpha$ and $\pi_{1}$ are contained in a unique $4^{\prime}$-space $U_{2}$,
which is itself contained in a unique $5$-space $F_{3}$. Now both $q$ and
$p_{2}$ are (inside $\xi$) collinear to all points of respective lines of
$\alpha$, implying that they are collinear to a common point $p_{3}\in F_{3}$.
Now (*) follows.

It now also follows that the set of such projectivities $\rho$ (as above with
$\pi_{0}$ and $\pi_{1}$ opposite) is geometric (they are the homologies with
two disjoint planes as centres). Now we drop the assumption of $\pi_{0}$ being
opposite $\pi_{1}$. We claim that in this more general case, the projectivity
$\rho$, as defined above, is the product of homologies with disjoint planes as
centres. Indeed, set $\pi_{0}^{\prime}:=\mathsf{proj}^{F_{1}}_{F}(\pi_{1})$ as
above. If $\pi_{0}^{\prime}$ is disjoint from $\pi_{0}$, then by 2.1,
$\pi_{0}$ and $\pi_{1}$ are opposite. Now we treat the other cases. Set
$d=\dim(\pi_{0}\cap\pi_{0}^{\prime})$ and note that $d=-1$ is precisely the
case we already proved.

  1. $d=0$

Let $\pi_{2}$ be a plane in $F$ sharing a line with $\pi_{0}$ but disjoint
from $\pi_{0}^{\prime}$. Then it is easy to check that the unique $4$-space
$U$ containing the $3$-space generated by $\pi_{2}$ and $\pi_{0}$ is disjoint
from $\mathsf{proj}^{F_{1}}_{F_{2}}(\pi_{1})$. Hence there exists a $5$-space
$F_{2}^{\prime}\neq F$ containing $\pi_{2}$ and opposite both $F_{1}$ and
$F_{3}$, and we have that $\pi_{1}$ is opposite both $F\cap F_{2}^{\prime}$
and $F_{2}\cap F_{2}^{\prime}$. We can now write $\rho$ as the product of
$F\text{\,\footnotesize$\overline{\land}$\,}F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F_{3}\text{\,\footnotesize$\overline{\land}$\,}F$
and the conjugate of
$F_{3}\text{\,\footnotesize$\overline{\land}$\,}F_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}F_{1}\text{\,\footnotesize$\overline{\land}$\,}F_{2}\text{\,\footnotesize$\overline{\land}$\,}F_{3}$
by $F\text{\,\footnotesize$\overline{\land}$\,}F_{3}$, reducing this case to
the case $d=-1$, which we already proved.

  2. $d=1$

Let $\pi_{2}$ be a plane in $F$ sharing a line with $\pi_{0}$ and exactly one
point (necessarily in $\pi_{0}$) with $\pi_{0}^{\prime}$. Then, similarly as
in the case $d=0$, we can choose a $5$-space $F_{2}^{\prime}\neq F$ through
$\pi_{2}$ opposite both $F_{1}$ and $F_{3}$ and such that $\pi_{1}$ has a
unique point collinear to some point of $F\cap F_{2}^{\prime}$, and that point
is also the unique point of $\pi_{1}$ collinear to some point of $F_{2}\cap
F_{2}^{\prime}$. We have hence reduced this case to two times the case $d=0$,
which we proved above.

  3. $d=2$

This case is similarly reduced to the case $d=1$. We leave the
(straightforward) details to the reader.

The claim is proved. Hence, thanks to 8.18, we can apply 8.17 and obtain that
$\Pi^{+}(F)$ is generated by all homologies with disjoint planes as centres.
This group contains $\mathsf{PSL}_{6}(\mathbb{K})$ and then clearly
corresponds to all $6\times 6$ matrices with a determinant equal to some
nonzero 3th power. Also, $\Pi(F)=\Pi^{+}(F)$ by virtue of B.

Now suppose $\operatorname{\mathsf{cotyp}}(F)=\\{2,4,5,6,7\\}$ in case of
$\mathsf{E_{7}}$. Here we can take for $F$ a pair consisting of a $5$-space
$W$ and a symp $\xi$ containing $W$ in the Lie incidence geometry of type
$\mathsf{E_{7,7}}$ over the field $\mathbb{K}$. We employ the same method as
in the previous case (Case $\mathsf{A_{5}}$ in $\mathsf{E_{6}}$), noting that
a projectivity
$\\{W,\xi\\}\text{\,\footnotesize$\overline{\land}$\,}\\{W^{\prime},\xi^{\prime}\\}\text{\,\footnotesize$\overline{\land}$\,}\\{W^{\prime\prime},\xi^{\prime\prime}\\}$,
where the simplices $\\{W,\xi\\}$ and
$\\{W^{\prime\prime},\xi^{\prime\prime}\\}$ are adjacent, is trivial as soon
as $W=W^{\prime\prime}$, and so we may always assume that in such a
(sub)sequence $W\neq W^{\prime\prime}$ and $\xi=\xi^{\prime\prime}$. However,
since the action of the projectivity is apparently independent of the symps
$\xi$ and $\xi^{\prime}$, we may only consider projections from $5$-spaces
onto $5$-spaces. Hence let $W_{1},W_{2},W_{3}$ be three $5$-spaces with both
$W_{1}$ and $W_{3}$ opposite both $W$ and $W_{2}$, and $\Sigma_{1}:=W_{1}\cap
W_{3}$ and $\Sigma_{0}:=W\cap W_{2}$ $3$-spaces such that the symps $\xi_{0}$
and $\xi_{1}$ containing $W,W_{2}$, and $W_{1},W_{3}$, respectively, are also
opposite. Similarly as in the previous case (type $\mathsf{A_{5}}$ inside
$\mathsf{E_{6}}$), we may from the beginning assume that $\Sigma_{0}$ and
$\Sigma_{1}$ are opposite $3$-spaces. Set
$L_{0}:=\mathsf{proj}^{W_{1}}_{W}(\Sigma_{1})$. Then, by 2.1 $L_{0}$ and
$\Sigma_{0}$ are disjoint. Set
$L_{2}:=\mathsf{proj}^{W_{1}}_{W_{2}}(\Sigma_{1})$, then likewise $L_{2}$ and
$\Sigma_{0}$ are disjoint. Let $x_{0}$ be an arbitrary point on $L_{0}$. Then
inside $\xi_{0}$ one sees that there is a unique point $x_{2}$ on $L_{2}$
collinear to $x_{0}$. We claim that the projectivity $\rho_{1}\colon W\perp
W_{1}\perp W_{2}$ maps $x_{0}$ to $x_{2}$. Indeed, set
$W_{1}^{\prime}:=\mathsf{proj}^{\xi_{1}}_{\xi_{0}}(W_{1})$. Then, again by
2.1, $W_{1}^{\prime}$ is disjoint from both $W$ and $W_{2}$. Set
$U_{1}:=\mathsf{proj}^{W}_{W_{1}}(x_{0})$ and
$U_{1}^{\prime}:=\mathsf{proj}^{\xi_{1}}_{\xi_{0}}(U_{1})$ and note that
$\Sigma_{1}\subseteq U_{1}$. Then $U_{1}^{\prime}\subseteq W_{1}^{\prime}$.
Since $x_{0}$ is at distance $2$ from each point of $U_{1}$, it follows by 2.6
that $x_{0}$ is collinear to all points of $U_{1}^{\prime}$. Hence $x_{0}$ is
contained in the unique $5^{\prime}$-space $V_{0}$ of $\xi_{0}$ containing
$U_{1}^{\prime}$. Likewise, if
$x_{2}^{\prime}=\mathsf{proj}^{W_{1}}_{W_{2}}(U_{1})$, then $x_{2}^{\prime}\in
V_{0}$. Hence $x_{0}$ and $x_{2}^{\prime}$, which is contained in $L_{2}$ as
$U_{1}$ contains $\Sigma_{1}$, are collinear. Consequently,
$x_{2}^{\prime}=x_{2}$ and the claim is proved.

It now also follows that $\rho_{3}\colon
W_{2}\text{\,\footnotesize$\overline{\land}$\,}W_{3}\text{\,\footnotesize$\overline{\land}$\,}W$
maps $x_{2}$ back to $x_{0}$, since $x_{0}$ is the unique point on $L_{0}$
collinear to $x_{2}$. Consequently, the projectivity $\rho\colon
W\text{\,\footnotesize$\overline{\land}$\,}W_{1}\text{\,\footnotesize$\overline{\land}$\,}W_{2}\text{\,\footnotesize$\overline{\land}$\,}W_{3}\text{\,\footnotesize$\overline{\land}$\,}W$
fixes each point of $L_{0}$. It is easy to see that it also fixes every point
of $\Sigma_{0}$. Hence it is a homology corresponding to a diagonal matrix
with the diagonal consisting of four times a $1$ and two times a scalar
$k\in\mathbb{K}^{\times}$. If we can now show that every nonzero scalar $k$
can occur, then, similarly to the case $\mathsf{A_{5}}$ in $\mathsf{E_{6}}$,
using 8.17 and 8.18, we are done.

But it follows from the arguments in the previous paragraphs that the
projectivity $\rho_{1}$ coincides with the projectivity
$W\text{\,\footnotesize$\overline{\land}$\,}W_{1}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}W_{2}$
inside $\xi_{0}$. Likewise the projectivity $\rho_{3}$ coincides with the
projectivity
$W_{2}\text{\,\footnotesize$\overline{\land}$\,}W_{3}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}W$
inside $\xi_{0}$, with
$W_{3}^{\prime}:=\mathsf{proj}^{\xi_{1}}_{\xi_{0}}(W_{3})$. Now the assertion
follows with exactly the same arguments as Case (A**) in the proof of Theorem
8.8.

This concludes Case (A5).

Case (A6) In a Coxeter diagram of type $\mathsf{E_{8}}$ a subdiagram of type
$\mathsf{A}_{6}$ is either contained in a subdiagram of type $\mathsf{A}_{7}$,
in which case we are done by 7.2 and Theorem 8.4, or it has type
$\\{2,4,5,6,7,8\\}$, in which case it is contained in a subdiagram of type
$\mathsf{D_{7}}$ and we are done by 7.2 and Theorem 8.8(A**).

In a Coxeter diagram of type $\mathsf{E_{7}}$ a subdiagram of type
$\mathsf{A}_{6}$ necessarily has type $\\{1,3,4,5,6,7\\}$. Although a proof
using the Lie incidence geometry of type $\mathsf{E_{7,7}}$, where $F$ is a
6-space, is feasible, we prefer to consider the Lie incidence geometry of type
$\mathsf{E_{7,1}}$, where the proof is entirely similar to the case (A7) of
$\mathsf{A_{7}}$ in $\mathsf{E_{8}}$ below. So we refer to that case for the
details.

Case (A7) In a Coxeter diagram of type $\mathsf{E_{8}}$ a subdiagram of type
$\mathsf{A}_{7}$ has cotype $2$, which corresponds to a 7-space. Let $\Delta$
be the building of type $\mathsf{E_{8}}$ over the field $\mathbb{K}$.

Let $W_{0}$ and $W_{2}$ be two $7$-spaces in $\Delta$ intersecting in a
$4$-space that we denote by $U$. Let $W_{1}$ be a $7$-space opposite to both
$W_{0}$ and $W_{2}$. Then $U$ projects to a plane $\alpha$ in $W_{1}$.
Projection here means that $U$ and $\alpha$ are special; for every pair of
points $(u,a)$ with $u\in U$ and $a\in\alpha$, there exists a point $p_{u,a}$
that is collinear to $u$ and $a$. Let $U^{\prime}$ be a $4$-space in $W_{1}$
that has no intersection with $\alpha$. Let $W_{3}$ be a $7$-space that
intersects $W_{1}$ in $U^{\prime}$ and is opposite both $W_{0}$ and $W_{3}$.
Set

$\rho\colon
W_{0}\text{\,\footnotesize$\overline{\land}$\,}W_{1}\text{\,\footnotesize$\overline{\land}$\,}W_{2}\text{\,\footnotesize$\overline{\land}$\,}W_{3}\text{\,\footnotesize$\overline{\land}$\,}W_{0}.$

We claim that points of $U$ are fixed under $\rho$. Indeed, a point of
$W_{0}\cap W_{2}$ first maps to a hyperplane of $W_{1}$, then back to itself,
then to a hyperplane of $W_{3}$ and again back to itself. That means $U$ is
fixed pointwise under $\rho$. The claim is proved.

The projection of $U^{\prime}$ onto $W_{0}$ is a plane that we will denote by
$\beta$. A point $p$ of $\beta$ maps to a hyperplane of $W_{1}$ that contains
$U^{\prime}$ and intersects $\alpha$ in a line. We will call this hyperplane
$H$. The projection of $H$ onto $W_{2}$ is a point that we will denote by
$p^{\prime}$. We claim that the point $p$ is collinear to $p^{\prime}$.

Indeed, suppose for a contradiction that $p$ were not collinear to
$p^{\prime}$. The points $p$ and $p^{\prime}$ cannot have distance $3$ or be a
special pair, since they are both collinear to each point of $U$. That means
$p$ and $p^{\prime}$ are symplectic. The symp $\xi(p,p^{\prime})$ intersects
$W_{0}$ and $W_{2}$ each in a $6^{\prime}$ space that contains $U$. The
projection of the $6^{\prime}$-space $W_{0}\cap\xi(p,p^{\prime})$ onto $W_{1}$
is a point $w$. The point $w$ is in $\alpha$, since $\alpha$ was the
projection of $W_{0}$ onto $W_{1}$. Let $F$ be a $6^{\prime}$ space in $W_{1}$
not through $w$. Then $F$ is opposite $W_{0}\cap\xi(p,p^{\prime})$. Let $\xi$
be the unique symp through $F$. Then $\xi$ is opposite $\xi(p,p^{\prime})$.

The projections of $p$ and $q$ onto $W_{1}$ define the two $6^{\prime}$-spaces
of $W_{1}$, for which it is not the case, that all points are opposite $p$ and
$q$ and the intersection of them is a $5$-space that we will denote by
$V_{1}$. Let $V_{0}$ be the $5$-space that is the projection of $V_{1}$ onto
$\xi(p,p^{\prime})$. The $5$-space spanned by $U$ and $p$ in
$\xi(p,p^{\prime})$ is disjoint from $V_{0}$, but $p$ and $p^{\prime}$ are
both collinear to $V_{0}$. Let $x$ be an arbitrary point in $V_{1}$ and
$x^{\prime}$ be a point at distance $2$ from $x$ in $V_{0}$. The point
$x^{\prime}$ is collinear to both $p$ and $p^{\prime}$. There exists a unique
symp $\xi^{\prime}$ that intersects $\xi(p,p^{\prime})$ exactly in
$x^{\prime}$ and $\xi$ exactly in $x$.

The $6^{\prime}$-spaces $\langle V_{0},p\rangle$ and $\langle
V_{0},p^{\prime}\rangle$ in $\xi(p,p^{\prime})$ are of the same type as the
$6^{\prime}$ spaces $W_{0}\cap\xi(p,p^{\prime})$ and
$W_{2}\cap\xi(p,p^{\prime})$. But the intersections $\langle
V_{0},p\rangle\cap(W_{0}\cap\xi(p,p^{\prime}))$ and $\langle
V_{0},p^{\prime}\rangle\cap(W_{2}\cap\xi(p,p^{\prime}))$ only contain a unique
point. That contradicts the fact that subspaces of the same type in polar
spaces of type $D_{7}$ have to intersect in even codimension. We conclude that
$p$ and $p^{\prime}$ have to be collinear and the claim is proved.

Now we claim that all points of $\beta$ are fixed under $\rho$. Indeed, the
point $p$ projects to the hyperplane $H$ in $W_{1}$. This hyperplane projects
to $p^{\prime}$ in $W_{2}$. This already implied that $p$ and $p^{\prime}$ are
collinear. If we project a point of $W_{2}$ to $W_{3}$ and to $W_{0}$, we also
get, that this point and the point in $W_{0}$ have to be collinear. The point
$p$ is the only point of $\beta$ that $p^{\prime}$ is collinear to. Hence
$p^{\prime}$ projects to a hyperplane in $W_{3}$ and then back to $p$.

Let $xy$ be a line in $W_{0}$ between a point $x\in U$ and $y\in\beta$. Let
$a$ and $b$ be two distinct points on $xy$ not equal to either $x$ or $y$. We
claim that we can re-define $W_{3}$ such that $\rho$ maps $a$ to $b$. Let
$a^{\prime}:=W_{0}\text{\,\footnotesize$\overline{\land}$\,}W_{1}\text{\,\footnotesize$\overline{\land}$\,}W_{2}(a)$,
$b^{\prime}:=W_{0}\text{\,\footnotesize$\overline{\land}$\,}W_{1}\text{\,\footnotesize$\overline{\land}$\,}W_{2}(b)$.
Since $a$, $b$, $a^{\prime}$, $y$, $y^{\prime}$ and $U$ form a $6$-space,
$\langle a,b,a^{\prime}\rangle$ forms a plane, $ba^{\prime}$ and $yy^{\prime}$
intersect in a point $s$.

Let $W^{\prime}$ be a $7$-space through $U$ and $s$. The projection of
$W^{\prime}$ onto $U^{\prime}$ is a $7$-space that is not opposite
$W^{\prime}$ and that we will denote by $W_{3}$. Since $W_{3}$ and
$W^{\prime}$ are not opposite, there exists, by 2.14, a plane $\gamma_{3}$ in
$W_{3}$ such that no point of $\gamma_{3}$ is opposite any point of
$W^{\prime}$ and there exists a plane $\gamma^{\prime}$ in $W^{\prime}$ such
that no point of $\gamma^{\prime}$ is opposite any point of $W_{3}$.

Let $y^{\prime}$ be the point in $\gamma^{\prime}$ on a line with the point
$y$ of $\beta$. The line $yy^{\prime}$ is the unique line between
$\gamma^{\prime}$ and $\beta$. The points $y$ and $y^{\prime}$ are not
opposite all points of $U^{\prime}$. Hence the point $s$ is not opposite all
points of $U^{\prime}$. The point $s$ is also not opposite any of the points
of $\gamma_{3}$ since $s\in W^{\prime}$. That means $s$ is not opposite any
point of $W_{3}$ and hence has to be contained in $\gamma^{\prime}$.

The line $a^{\prime}b$ contains $s$. That means the points $a^{\prime}$ and
$b$ project to the same hyperplane of $W_{3}$. Then the point $a$ maps to a
hyperplane of $W_{1}$, then to $a^{\prime}$ in $W_{2}$ and the point
$a^{\prime}$ projects to a hyperplane of $W_{3}$ and then to $b$. Our claim is
proved.

By A, $\Pi^{+}(W_{0})$ contains $\mathsf{PSL}_{8}(\mathbb{K})$. By the above,
it also contains all diagonal matrices with diagonal $(1,1,1,1,1,k,k,k)$,
$k\in\mathbb{K}^{\times}$ arbitrarily, and the entries $k$ can be anywhere.
This readily implies that $\Pi^{+}(W_{0})$ contains the diagonal matrices with
diagonal $(1,1,1,1,1,1,1,k)$, hence all homologies, hence
$\Pi^{+}(W_{0})=\mathsf{PGL}_{8}(\mathbb{K})$ and
$\Pi(W_{0})=\mathsf{PGL}_{8}(\mathbb{K}).2$.

Case (D5) We first consider the case of a Coxeter diagram of type
$\mathsf{E_{6}}$. Without loss of generality, we may assume that $F$ has type
$6$. Hence we consider $F$ as a symp in a geometry of type $\mathsf{E_{6,1}}$
over the field $\mathbb{K}$.

Let $p_{1}$ be a point in $\Delta$ and $\xi_{0}$ a symp opposite $p_{1}$ in
$\Delta$. Let $U$ be a maximal singular subspace in $\xi_{0}$. Then $U$ is a
$4$-space. Let $\xi_{2}$ be another symp through $U$ opposite $p_{1}$.
Opposite a $4$-space are lines. Let $L$ be a line through $p_{1}$ opposite $U$
and $V:=\mathsf{proj}_{\xi_{0}}(L)$. Let $p_{3}$ be any point on $L$ opposite
both $\xi_{0}$ and $\xi_{2}$, so that we have a projectivity
$\rho:\xi_{0}\text{\,\footnotesize$\overline{\land}$\,}p_{1}\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}\text{\,\footnotesize$\overline{\land}$\,}p_{3}\text{\,\footnotesize$\overline{\land}$\,}\xi_{0}$.
We will show that $\rho$ fixes $U$ and $V$ pointwise.

First let $x$ be a point in $U$. Then $x$ projects to a symp $\xi(x,p_{1})$,
then back to $x$, since $x\in\xi_{0}\cap\xi_{2}$, then to a symp
$\xi(x,p_{3})$ and then again back to $x$.

Now let $y$ be a point in $V$. The point $y=y_{0}$ projects to a symp
$\xi(y_{0},p_{1})=\xi_{y}$ and then to a point $y_{2}\in\xi_{2}$. Suppose
$y_{0}$ and $y_{2}$ were not collinear. The symp $\xi_{y}$ has to contain the
closure of $y_{0}$ and $y_{2}$. Both $y_{0}$ and $y_{2}$ are collinear to a
$3$-space of $U$. The intersection of these $3$-spaces contains a plane. That
means that the closure has to contain a plane of $U$ that then had to be
contained in $\xi_{y}$. But that contradicts the fact that $U$ and $p_{1}$ are
opposite, because $p_{1}$ would have to be collinear to elements of that
plane. It follows that $y_{0}\perp y_{2}$. Now, since
$V=\mathsf{proj}_{\xi_{0}}(L)$, we see that $L\subseteq\xi_{y}$. So $y_{2}$
continues mapping to $\xi_{y}$ and then back to $y_{0}$. Hence points of $V$
are fixed.

Next we want to show that we can always define $p_{3}$ on $L$ in a way, such
that the projectivity $\rho$ defined above maps an arbitrary point $p$ on a
line $xy$, with $x\in U$ and $y\in V$, to another arbitrary point $q$ on $xy$
for $p\notin U,V$ and $q\notin U,V$. Given $U,V,L$ and $p_{1}$ as before and a
line $xy$ as described above, let $p$ be an arbitrary point on $xy$ not in $U$
or $V$. Then projecting $p$ to $p_{1}$ yields a symp $\xi(p,p_{1})$ that
projects to a point $p_{2}$ onto $\xi_{2}$. Let
$y_{2}:=\mathsf{proj}_{\xi_{2}}(\mathsf{proj}_{p_{1}}(y))$. By the previous
paragraph, the points $x,y,y_{2}$ generate a singular plane, which contains
$p,q$ and $p_{2}$. Let $a:=p_{2}q\cap yy_{2}$. Suppose $a$ were collinear to
$p_{1}$. Then $a$ would be in $\xi(p,p_{1})$ and $\xi(p,p_{1})$ would contain
the plane $\langle x,y,y_{2}\rangle$ and in particular the line $xy$. But that
contradicts the fact that $\xi(p,p_{1})$ intersects $\xi_{0}$ only in $p$. It
follows that $a$ is not collinear to $p_{1}$. That means $a$ is collinear to a
different point of $L$ that we will define as $p_{3}$. This point $p_{3}$ is
not collinear to $p_{2}$ as otherwise $\xi(p,p_{1}$ would contain $L$, forcing
$p\in V$, a contradiction. Since $a$ and $p_{2}$ are in $\xi(p_{3},p_{2})$,
$\xi(p_{3},p_{2})$ contains the whole line $ap_{2}$ and hence the point $q$.
With that it follows that $p$ maps to $\xi(p,p_{1})$ to $p_{2}$ to
$\xi(p_{3},p_{2})=\xi(p_{3},q)$, and finally to $q$.

Now 8.11$(i)$ proves the assertion.

In a Coxeter diagram of type $\mathsf{E_{7}}$ or $\mathsf{E_{8}}$, a
subdiagram of type $\mathsf{D_{5}}$ is always contained in a subdiagram of
type $\mathsf{E_{6}}$, and so we can apply 7.2, the previous paragraphs, and
B.

Case (D4) Each subdiagram of type $\mathsf{D_{4}}$ in a diagram of type
$\mathsf{E}_{n}$, $n=6,7,8$, is contained in a subdiagram of type
$\mathsf{D_{5}}$. It follows that, if $F$ is a simplex of cotype
$\mathsf{D_{4}}$ in a building $\Delta$ of type $\mathsf{E}_{n}$, $n=6,7,8$,
then there is a subsimplex $F^{\prime}\subseteq F$ of cotype $\mathsf{D_{5}}$.
By the previous case and A, the stabiliser of $F^{\prime}$ in the little
projective group $\operatorname{\mathsf{Aut}}^{\dagger}(\Delta)$ of $\Delta$
acts on $\mathsf{Res}_{\Delta}(F^{\prime})$ as the complete linear type
preserving group of automorphisms. Hence the stabiliser of $F$ in
$\operatorname{\mathsf{Aut}}^{\dagger}(\Delta)$, acting on
$\mathsf{Res}_{\Delta}(F)$ contains the stabiliser in the full linear type
preserving group of $\mathsf{Res}_{\Delta}(F^{\prime})$ of the vertex
$F\setminus F^{\prime}$. This is clearly also the full linear type preserving
group of $\mathsf{Res}_{\Delta}(F)$.

Now, in case of $\operatorname{\mathsf{typ}}(\Delta)=\mathsf{E_{6}}$, it
follows from B that $\Pi^{+}(F)$ has index $2$ in $\Pi(F)$, and so $\Pi(F)$ is
the full linear group of the corresponding polar space of
$\mathsf{Res}_{\Delta}(F)$. In case of $\mathsf{E_{7}}$ or $\mathsf{E_{8}}$, B
implies that $\Pi(F)=\Pi^{+}(F)$.

Case (D7) In a Coxeter diagram of type $\mathsf{E_{8}}$ a subdiagram of type
$\mathsf{D_{7}}$ is missing vertex $1$. So $F$ is a symp in the geometry
$\Gamma$ of type $\mathsf{E_{8,8}}$ over the field $\mathbb{K}$.

Let $\xi_{0}$ and $\xi_{2}$ be two symps in $\Gamma$ intersecting in a
$6$-space that we will denote by $U$. Let $\xi_{1}$ be a symp opposite
$\xi_{0}$ and $\xi_{2}$, and $U^{\prime}$ the $6^{\prime}$-space that is the
projection of $U$ onto $\xi_{1}$. Opposite $U^{\prime}$ in $\xi_{1}$ is a
$6$-space $V$. Let $V^{\prime}$ be the $6^{\prime}$-space that is the
projection of $V$ onto $\xi_{0}$. Suppose there is a symp $\xi_{3}\supseteq V$
opposite both $\xi_{0}$ and $\xi_{2}$, such that we have a projectivity
$\rho\colon\xi_{0}\text{\,\footnotesize$\overline{\land}$\,}\xi_{1}\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}\text{\,\footnotesize$\overline{\land}$\,}\xi_{3}\text{\,\footnotesize$\overline{\land}$\,}\xi_{0}$.
We first show that $\rho$ fixes every point of $U$ and $V^{\prime}$. For
points of $U$ the assertion follows immediately. A point $v_{0}$ of
$V^{\prime}$ first projects to a point $v_{1}$ in $V$, then to some point in
$\xi_{2}$ and then back to $v_{1}$. The point $v_{1}$ maps back to $v_{0}$.

Next we want to show that there always exists a $\xi_{3}$ through $V$ opposite
both $\xi_{0}$ and $\xi_{2}$ and that we can define it in a way, such that we
can map an arbitrary point $p$ on a line $xy$, with $x\in U$ and $y\in
V^{\prime}$, to a point $q$ on $xy$ for $p\notin U,V^{\prime}$ and $q\notin
U,V^{\prime}$.

We define:

$\displaystyle
y=:y_{0},~{}\mathsf{proj}_{\xi_{1}}(y_{0})=:y_{1},~{}\mathsf{proj}_{\xi_{2}}(y_{1})=:y_{2},$
$\displaystyle
p=:p_{0},~{}\mathsf{proj}_{\xi_{1}}(p_{0})=:p_{1},~{}\mathsf{proj}_{\xi_{2}}(p_{1})=:p_{2}.$

First we want to show that $p_{2}\perp q$. Suppose $p_{2}$ and $q$ were not
collinear. Then there exists a symp $\xi(p_{2},q)=\xi(p_{2},p)$ and a point
$u\in U\cap\xi(p_{2},p)$ such that $p_{1}$ is opposite $u$. But by 2.9$(iii)$
the point $p_{1}$ is supposed to be symplectic to only one point of
$\xi_{p_{2},p}$, and it is already symplectic to $p$ and $p_{2}$. It follows
that $p_{2}\perp q$ and there exists a plane $\langle p,q,p_{2}\rangle$.
Define $a:=y_{0}y_{2}\cap p_{2}q$.

Let $\xi(U,a)$ be the unique symp through $U$ containing $a$, according to
2.12. Define $\xi_{3}:=\mathsf{proj}_{V}(\xi(U,a))$. Then $\xi(U,a)$ is the
unique symp through $U$ not opposite $\xi_{3}$

In $\xi(U,a)$ let, according to 2.13, $W$ be the maximal singular subspace,
such that no point of $W$ is opposite (at distance $3$ from) any point of
$\xi_{3}$. Let $W_{3}$ be the maximal singular subspace of $\xi_{3}$, such
that no point of $W_{3}$ is opposite any point of $\xi(U,a)$. The points $y$
and $y_{2}$ are not opposite any point of $V$. Hence $a$ is not opposite any
point of $V$. But the point $a$ is also not opposite any point of $W_{3}$ by
the definition of $W_{3}$. Hence $a\in W$.

The same points of $\xi_{3}$ are not opposite $q$ and $p_{2}$, since, if we
take an arbitrary point $s$ in $\xi_{3}$ which is not opposite $q$, then $s$
is not opposite $a$ and $q$, so also not opposite $p_{2}$ and the same if we
switch the roles of $q$ and $p_{2}$. Let $q_{3}$ be the unique point of
$\xi_{3}$ symplectic to $q$. Then $q$ is opposite $\xi_{3}\setminus
q_{3}^{\perp}$. But then $\xi_{3}\setminus q_{3}^{\perp}=\xi_{3}\setminus
p_{3}^{\perp}$. The perps of points $s$ and $t$ in a polar space are the same,
if $s=t$. Hence it follows that $q_{3}=p_{3}$. That means $q$ is the unique
point of $\xi_{0}$ that $p_{3}$ is symplectic to. With that we get that
$\rho(p)=q$.

Now, as before, 8.11$(i)$ proves the assertion.

Case (D6) We first treat the case of type $\mathsf{D_{6}}$ inside type
$\mathsf{E_{7}}$. Let $\xi$ be a symp of the geometry of type
$\mathsf{E_{7,7}}$ over the field $\mathbb{K}$. We first claim that
$\Pi(\xi)$, which is equal to $\Pi^{+}(\xi)$ by B, contains all homologies
pointwise fixing two $\xi$-opposite maximal singular $5$-spaces. Let $M_{13}$
and $M$ be two such subspaces of $\xi$. Let $\xi_{3}$ be an arbitrary symp
distinct from $\xi$ and containing $M_{13}$. Let $\xi_{2}$ be a symp opposite
both $\xi$ and $\xi_{3}$ (and note that this implies that each point of
$\xi_{2}$ is opposite some point of $\xi$). There is a unique maximal singular
subspace $M_{24}$ contained in $\xi_{2}$ each point of which is collinear to
some point of $M$, that is, $M_{24}=\mathsf{proj}^{\xi}_{\xi_{2}}(M)$. Let $L$
be any given line in $\xi$ joining a point $p_{13}\in M_{13}$ and $p\in M$.
Choose two points $q,q^{\prime}\in L\setminus\\{p_{13},p\\}$. Set
$q_{2}=\mathsf{proj}_{\xi_{2}}(q)$ and $q_{3}=\mathsf{proj}_{\xi_{3}}(q_{2})$.

If $q$ were not collinear to $q_{3}$, then the symp containing them would
contain a $3$-dimensional subspace of $M_{13}$ and $q_{2}$; this would imply
that $q_{2}$ is close to $\xi$, contradicting 2.6 in view of our remark in the
previous paragraph that says that $q_{2}$ is opposite some point of $\xi$.
Hence $\langle q,q_{3},q^{\prime}\rangle$ is a plane $\pi$, contained in the
symp $\zeta$ containing $p_{13}$ and $q_{2}$. Let $\xi^{\prime}$ be any symp
containing, $M_{24}$, but distinct from $\xi_{2}$. Let $p_{24}$ be the unique
point of $\xi_{2}$ collinear to $p$, and note $p_{24}\in M_{24}$, and that
$p_{24}$ and $q_{2}$ are collinear. Hence $p_{24}\in\zeta$. This implies that
$\zeta\cap\xi^{\prime}$ is either a line or a $5$-space through $p_{24}$. In
the latter case $p_{13}$, being collinear with more than one point of that
intersection, is close to $\xi^{\prime}$, contradicting 2.6 and the fact that
$M_{24}$ is opposite $M_{13}$, and hence $p_{13}$ is opposite points of
$\xi^{\prime}$. Hence $\zeta\cap\xi^{\prime}$ is a line $K\ni p_{24}$. If
$q_{2}$ were not collinear to $K$, then $\zeta$ would contain a $3$-space of
$M_{24}$, again a similar contradiction (since $\zeta$ contains $p_{13}$). The
planes $\pi$ and $\langle q_{2},K\rangle$ are easily seen to be opposite in
$\zeta$, hence there is a unique point $q_{4}\in\langle q_{2},K\rangle$
collinear to both $q_{3}$ and $q^{\prime}$. Now let $\xi_{4}$ be the symp
containing $M_{24}$ and $q_{4}$, whose existence follows from 2.7. Then one
checks that $\xi_{4}$ is opposite both $\xi$ and $\xi_{3}$, and the
projectivity
$\xi\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}\text{\,\footnotesize$\overline{\land}$\,}\xi_{3}\text{\,\footnotesize$\overline{\land}$\,}\xi_{4}\text{\,\footnotesize$\overline{\land}$\,}\xi$
pointwise fixed both $M_{13}$ and $M$, and maps $q$ to $q^{\prime}$. This
proves the claim.

Now, if we want to apply 8.17, then we have to show that every projectivity

$\rho\colon\xi_{0}\text{\,\footnotesize$\overline{\land}$\,}\xi_{1}\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}\text{\,\footnotesize$\overline{\land}$\,}\xi_{3}\text{\,\footnotesize$\overline{\land}$\,}\xi_{0},$

with $M_{0}:=\xi_{0}\cap\xi_{2}$ and $M_{1}=\xi_{1}\cap\xi_{3}$ singular
$5$-spaces, is the product of similar projectivities, but with $M_{0}$
opposite $M_{1}$. So suppose $M_{0}$ and $M_{1}$ are not opposite. As for the
case of type $\mathsf{A_{5}}$ in type $\mathsf{E_{6}}$, there are $3$ cases to
consider, and they are again all quite similar to each other, so we consider
for example the case where the set of points of $M_{0}$ collinear to a point
of $M_{1}$ is a line $L$ (the other possibilities are a $3$-space and the
whole space $M_{1}$). Then we consider an appropriate $5$-space $M_{2}$ in
$\xi_{0}$ intersecting $M_{0}$ in a $3$-space contained in $M_{0}$, and
disjoint from $L$. Then we find a symp $\xi_{2}^{\prime}$ containing $M_{2}$,
opposite both $\xi_{1}$ and $\xi_{3}$, and intersecting $\xi_{2}$ in a
$5$-space opposite $M_{1}$. As in the case of type $\mathsf{A_{5}}$ in type
$\mathsf{E_{6}}$, we can now write $\rho$ as the product of
$\xi_{0}\text{\,\footnotesize$\overline{\land}$\,}\xi_{1}\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}\xi_{3}\text{\,\footnotesize$\overline{\land}$\,}\xi_{0}$
and the conjugate of
$\xi_{3}\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}^{\prime}\text{\,\footnotesize$\overline{\land}$\,}\xi_{1}\text{\,\footnotesize$\overline{\land}$\,}\xi_{2}\text{\,\footnotesize$\overline{\land}$\,}\xi_{3}$
by $\xi_{0}\text{\,\footnotesize$\overline{\land}$\,}\xi_{3}$.

Now we can use 8.11 and, thanks to 8.18, also 8.17 to conclude that
$\Pi^{+}(F)=\overline{\mathsf{P\Omega}}_{12}(\mathbb{K})$.

Now consider the case of type $\mathsf{D_{6}}$ inside type $\mathsf{E_{8}}$
and let $F$ be a simplex of cotype $\mathsf{D_{6}}$. On the one hand, since
the $\mathsf{D_{6}}$ subdiagram is contained in a $\mathsf{E_{7}}$ subdiagram,
$\Pi^{+}(F)$ contains $\overline{\mathsf{P\Omega}}_{12}(\mathbb{K})$. On the
other hand, since the $\mathsf{D_{6}}$ subdiagram is contained in a
$\mathsf{D_{7}}$ subdiagram, $\Pi^{+}(F)$ contains
$\mathsf{PGO}_{12}^{\circ}(\mathbb{K})$, as follows from Theorem 8.8(D′). Now
B and 8.11$(iii)$ yield
$\Pi(F)=\Pi^{+}(F)=\overline{\mathsf{PGO}}_{12}^{\circ}(\mathbb{K})$.

Case (E6) Let $\Gamma$ be the parapolar space of type $\mathsf{E_{7,7}}$ over
the field $\mathbb{K}$. Let $p_{1},p_{2},p_{3}$ be three mutually opposite
points of $\Gamma$. If we show that the self-projectivity $\rho\colon
p_{1}\text{\,\footnotesize$\overline{\land}$\,}p_{2}\text{\,\footnotesize$\overline{\land}$\,}p_{3}\perp
p_{1}$ is always a symplectic polarity, then 8.1 and 8.9 implies that $\Pi(p)$
is generated by all the symplectic polarities. By Proposition 6.8$(i)$ of
[14], $\rho$ pointwise fixes a subbuilding of type $\mathsf{F_{4}}$. More
exactly, if $\mathsf{Res}_{\Delta}(p_{1})$ is viewed as a parapolar space
$\Gamma_{p_{1}}$ of type $\mathsf{E_{6,1}}$ with the lines through $p_{1}$ as
points, then $\rho$ pointwise fixes a geometric hyperplane inducing in
$\Gamma_{p_{1}}$ a geometry of type $\mathsf{F_{4,4}}$ over the field
$\mathbb{K}$. It follows from [13] that $\rho$ is a symplectic polarity. Now
8.13 shows that $\Pi^{+}(p)$ is $\mathsf{PGE_{6}}(\mathbb{K})$ and $\Pi(p)$ is
$\mathsf{PGE_{6}}(\mathbb{K}).2$. Since these groups are the respective full
linear type preserving and full not necessarily type preserving groups, 7.2
implies the assertion for type $\mathsf{E_{6}}$ in type $\mathsf{E_{8}}$.

Case (E7) In a Coxeter diagram of type $\mathsf{E_{8}}$ a subdiagram of type
$\mathsf{E_{7}}$ is missing vertex $8$. Hence, given a building $\Delta$ of
type $\mathsf{E_{8}}$ over the field $\mathbb{K}$, a simplex $F=\\{p\\}$ with
residue of type $\mathsf{E_{7}}$ can be thought of as being a point in the Lie
incidence geometry $\Gamma$ of type $\mathsf{E_{8,8}}$ over the field
$\mathbb{K}$. Let $p_{1}$ and $p_{3}$ be collinear points of $\Gamma$ opposite
$p$ and let $p_{2}$ be a point of $\Gamma$ collinear to $p$ and opposite both
$p_{1}$ and $p_{3}$. Note that the lines through $p$ form the point set of a
Lie incidence geometry of type $\mathsf{E_{7,7}}$, where a line consists of
all lines of $\Gamma$ through $p$ contained in a given plane of $\Gamma$. Set
$\rho\colon
p\text{\,\footnotesize$\overline{\land}$\,}p_{1}\text{\,\footnotesize$\overline{\land}$\,}p_{2}\text{\,\footnotesize$\overline{\land}$\,}p_{3}\text{\,\footnotesize$\overline{\land}$\,}p$.
Obviously, if $\pi$ is a plane through the line $pp_{2}$, then
$\mathsf{proj}^{p_{1}}_{p_{2}}\mathsf{proj}^{p}_{p_{1}}(\pi)=\pi$ and
$\mathsf{proj}^{p_{3}}_{p}\mathsf{proj}^{p_{2}}_{p_{3}}(\pi)=\pi$. Hence
$\rho$ fixes $\pi$. Likewise, if $L$ is the projection onto $p$ of the line
$L_{1}:=p_{1}p_{3}$, all planes through $L$ are fixed by $\rho$. By the basic
properties of Lie incidence geometries of type $\mathsf{E_{7,7}}$, there is a
unique line $M$ through $p$ no point of which is special with any point of $L$
(it is the projection of $L$ on $\pi$ in the residue
$\mathsf{Res}_{\Delta}(p)$). Let $K_{1}$ and $K_{2}$ be two distinct lines in
$\pi$ containing $p$ and distinct from both $pp_{2}$ and $M$. Let $M_{1}$ be
the line of $\pi$ containing all points of $\pi$ not opposite $p_{1}$. Set
$\\{x_{1}\\}=K_{1}\cap M_{1}$. Then there exists a unique point
$x_{1}^{\prime}$ with $x_{1}\perp x_{1}^{\prime}\perp p_{1}$. So the line
$p_{1}x_{1}^{\prime}$ is the projection onto $p_{1}$ of the line $K_{1}$.
Since $p_{2}\perp x_{1}$, we see that the line $p_{2}x_{1}$ is the projection
of the line $p_{1}x_{1}^{\prime}$ onto $p_{2}$.

Set $\\{x_{2}\\}=p_{2}x_{1}\cap K_{2}$ and redefine $p_{3}$ as the projection
of $x_{2}$ onto the line $L_{1}$. If $x_{2}\perp x_{2}^{\prime}\perp p_{3}$,
then it is clear that $p_{2}x_{1}$ is projected onto $p_{3}x_{2}^{\prime}$
from $p_{2}$ onto $p_{3}$, and $p_{3}p_{2}^{\prime}$ is projected back onto
$px_{2}=K_{2}$ from $p_{3}$ onto $p$. Hence $\rho$ maps $K_{1}$ to $K_{2}$ and
the result now follows from 8.16. ∎

This concludes the proofs of all our main results. We conclude the paper with
some remarks.

######  Remark 8.20.

It now follows from Theorem 8.19 that
$\overline{\mathsf{P\Omega}}_{12}(\mathbb{K})$ does not always coincide with
$\overline{\mathsf{PGO}}_{12}(\mathbb{K})$. Indeed, if it did, then the
special projectivity groups in the buildings of type $\mathsf{E_{7}}$ of all
irreducible residues of types contained in $\mathsf{D_{6}}$ would be the full
linear groups. This contradicts the second grey row of Table 2 for fields
containing non-square elements.

######  Remark 8.21.

The argument for case $\mathsf{E_{7}}$ in $\mathsf{E_{8}}$ of the proof of
Theorem 8.19 could also be used for the cases of $\mathsf{D_{6}}$ in
$\mathsf{E_{7}}$ and $\mathsf{A_{5}}$ in $\mathsf{E_{6}}$, if we would use the
corresponding long root geometries. We chose to use the simpler and more
accessible Lie incidence geometries of types $\mathsf{E_{7,7}}$ and
$\mathsf{E_{6,1}}$, respectively, instead, also as a warm-up for the more
complicated cases such as $\mathsf{A_{5}}$ in $\mathsf{E_{7}}$ and
$\mathsf{D_{7}}$ in $\mathsf{E_{8}}$.

######  Remark 8.22.

In the course of the proof of Theorem 8.19 we do not really need the full
strength of Lemmas 8.11$(i)$, 8.13 and 8.16, since we know by A that also the
little projective group is already contained in the group we want to generate.
This knowledge would simplify the proof, since we would only have to prove
that the little projective group together with the said homologies generate
the full linear group.

######  Remark 8.23.

One could ask what to expect of the case where the diagram is not simply
laced. For starters, the description of all spherical buildings is more
complicated. Secondly, D will not hold anymore in full generality. Indeed,
there are polar spaces of rank $n$ where $\Pi^{+}(F)$ is not permutation
equivalent to $\mathsf{PGL}_{2}(\mathbb{K})$, for $F$ of cotype $n$, even if
the set of maximal singular subspaces through a submaximal singular subspace
carries in a natural way the structure of a projective line over $\mathbb{K}$
(like a symplectic polar space). However, analogues, appropriately phrased, of
Theorems B and C should still hold. Also, A remains through across all types.

Acknowledment. The authors are grateful to Gernot Stroth for an illuminating
discussion concerning the structure and action of Levi complements in
Chevalley groups.

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