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# Matsuo algebras in characteristic 2

Tom De Medts  Ghent University
Department of Mathematics: Algebra and Geometry
Krijgslaan 281–S25
9000 Gent
Belgium<EMAIL_ADDRESS>and  Mathias Stout  KU Leuven
Department of Mathematics
Celestijnenlaan 200B bus 2400
B-3001 Leuven
Belgium<EMAIL_ADDRESS>
(Date: August 22, 2023)

###### Abstract.

We extend the theory of Matsuo algebras, which are certain non-associative
algebras related to 3-transposition groups, to characteristic 2. Instead of
idempotent elements associated to points in the corresponding Fischer space,
our algebras are now generated by nilpotent elements associated to lines. For
many 3-transposition groups, this still gives rise to a
$\mathbb{Z}/2\mathbb{Z}$-graded fusion law, and we provide a complete
classification of when this occurs. In one particular small case, arising from
the 3-transposition group $\operatorname{Sym}(4)$, the fusion law is even
stronger, and the resulting Miyamoto group is an algebraic group
$\mathbb{G}_{a}^{2}\rtimes\mathbb{G}_{m}$.

###### Key words and phrases:

Non-associative algebras, Matsuo algebras, axial algebras, decomposition
algebras, $3$-transposition groups, Fischer spaces

###### 2020 Mathematics Subject Classification:

20B25, 20F29, 17A99, 51E30

##  1\. Introduction

_Axial algebras_ have been introduced by Jon Hall, Felix Rehren and Sergey
Shpectorov in [HRS15]. They form a class of non-associative algebras generated
by _idempotents_ , such that each idempotent gives rise to a decomposition of
the algebra respecting a certain _fusion law_.

One of the early examples that has been studied in more detail, are the so-
called _Matsuo algebras_ , related to $3$-transposition groups. More
precisely, if $(G,D)$ is a $3$-transposition group (where $G$ is the group and
$D$ is the distinguished generating set of involutions), then for each
commutative field $k$ with $\operatorname{char}(k)\neq 2$ and each $\eta\in
k\setminus\\{0,1\\}$, there is a $k$-algebra $A$ _spanned_ by idempotents
corresponding to the elements of $D$, with a very precise multiplication table
(see Definition 3.1 below). Each of these generating idempotents $e$ then
gives rise to a decomposition of $A$ into eigenspaces for the left
multiplication by $e$, and the multiplication of these eigenspaces satisfies
the _Jordan fusion law_ , making it very similar in structure as the Peirce
decomposition for Jordan algebras. (We refer to [DMR17] for a deeper
connection between Matsuo algebras and Jordan algebras.)

Extending the theory to characteristic $2$ is not obvious. First, the
decomposition into eigenspaces for an idempotent can never have a
$\mathbb{Z}/2\mathbb{Z}$-graded fusion law, as we observe in Proposition 4.1
below. Therefore, it seems more natural to use a spanning set of _nilpotent_
elements rather than idempotents. (In this case, the role of the parameter
$\eta$ becomes irrelevant.) Even so, just using nilpotents associated to the
elements of $D$ gives rise to highly uninteresting decompositions, because the
multiplication by such a nilpotent turns out to be a nilpotent operator (and
hence has $0$ as its only eigenvalue).

We solve this by using nilpotent elements associated to _lines_ in the Fischer
space associated to the $3$-transposition group, or put differently, to
triples of involutions contained in a common $\operatorname{Sym}(3)$-subgroup
of $G$. Somewhat surprisingly, this does indeed give rise to interesting
decompositions in many cases.

Perhaps the most interesting case is the smallest non-trivial case arising
from the $3$-transposition group $\operatorname{Sym}(4)$ with as generating
set $D$ the set of all $6$ transpositions. In this case, we have only $4$
lines, but the decompositions satisfy a fusion law stronger than a
$\mathbb{Z}/2\mathbb{Z}$-grading and allows for the action of a multiplicative
torus $\mathbb{G}_{m}$ on each decomposition. Together, these four tori
generate a group of the form $\mathbb{G}_{a}^{2}\rtimes\mathbb{G}_{m}$; see
Theorem 6.11. We also compute the full automorphism group of this algebra in
Proposition 6.14.

For Fischer spaces of _symplectic type_ , we get (ordinary)
$\mathbb{Z}/2\mathbb{Z}$-graded decompositions, as we show in Theorem 5.24. If
the Fischer space is not of symplectic type, then additional problems arise.
In the smallest case, which is the affine plane of order $3$, associated to
the $3$-transposition group $3^{2}:2$, we have to pass to the _generalized
eigenspaces_ (rather than the eigenspaces), which does not fit with the
original definition of axial algebras from [HRS15] but still complies with the
more general framework of _decomposition algebras_ developed in [DMPSVC20]. In
this specific case, the resulting decomposition algebra is again
$\mathbb{Z}/2\mathbb{Z}$-graded; see Theorem 5.17.

However, as soon as the Fischer space is not of symplectic type and has rank
$\geq 4$, then things break down and the resulting nilpotent Matsuo algebra is
no longer $\mathbb{Z}/2\mathbb{Z}$-graded, as we show in Theorem 5.27 below.

When we had almost finished writing this paper, we became aware of the
surprising fact that in [CHPS12], the authors study exactly the same algebras
as we do, but only over the field $\mathbb{F}_{2}$. They had different goals
than we do, so our results are almost completely disjoint from each other.

##  2\. Axial decomposition algebras

We first recall the definition of (commutative) axial decomposition algebras
from [DMPSVC20]. All algebras will be _commutative_ , but _not necessarily
associative nor unital_ algebras over a commutative base ring $R$.

######  Definition 2.1.

  1. (1)

A _fusion law_ is a pair $(X,*)$ where $X$ is a (usually finite) set and $*$
is a map from $X\times X$ to $2^{X}$, where $2^{X}$ denotes the power set of
$X$. A fusion law $(X,*)$ is called _symmetric_ if $x*y=y*x$ for all $x,y\in
X$. We call $e\in X$ a _unit_ if $e*x\subseteq\\{x\\}$ and
$x*e\subseteq\\{x\\}$ for all $x\in X$.

  2. (2)

The _Jordan fusion law_ is the symmetric fusion law $(X,*)$ with
$X=\\{e,z,h\\}$ and with $*$ given by the following table:

$*$ | $e$ | $z$ | $h$
---|---|---|---
$e$ | $\\{e\\}$ | $\emptyset$ | $\\{h\\}$
$z$ | $\emptyset$ | $\\{z\\}$ | $\\{h\\}$
$h$ | $\\{h\\}$ | $\\{h\\}$ | $\\{e,z\\}$

######  Definition 2.2.

Let $\Phi=(X,*)$ be a fusion law.

  1. (i)

A _$\Phi$ -decomposition_ of an $R$-algebra $A$ is a direct sum decomposition
$A=\bigoplus_{x\in X}A_{x}$ (as $R$-modules) such that $A_{x}A_{y}\subseteq
A_{x*y}$ for all $x,y\in X$, where $A_{Y}:=\bigoplus_{y\in Y}A_{y}$ for all
$Y\subseteq X$.

  2. (ii)

A _$\Phi$ -decomposition algebra_ is a triple $(A,\mathcal{I},\Omega)$ where
$A$ is an $R$-algebra, $\mathcal{I}$ is an index set and $\Omega$ is a tuple
of $\Phi$-decompositions of $A$ indexed by $\mathcal{I}$. We will usually
write the corresponding decompositions as $A=\bigoplus_{x\in X}A_{x}^{i}$, so

$\Omega=\bigl{(}(A_{x}^{i})_{x\in X}\mid i\in\mathcal{I}\bigr{)}.$

######  Definition 2.3.

Let $\Phi=(X,*)$ be a fusion law with a distinguished unit $e\in X$. For each
$x\in X$, let $\lambda_{x}\in R$. A $\Phi$-decomposition algebra
$(A,\mathcal{I},\Omega)$ will be called _axial_ (with _parameters_
$\lambda_{x}$) if for each $i\in\mathcal{I}$, there is some non-zero $a_{i}\in
A_{e}^{i}$ (called an _axis_) such that:

(2.4) $a_{i}\cdot b=\lambda_{x}b\quad\text{for all }x\in X\text{ and for all
}b\in A_{x}^{i}.$

In particular, $A$ decomposes into eigenspaces for the left multiplication
operator

$\operatorname{ad}_{a_{i}}\colon A\to A\colon b\mapsto a_{i}\cdot b.$

(Notice, however, that some of the parameters $\lambda_{x}$ might coincide, so
the summands $A_{x}^{i}$ are not necessarily eigenspaces.)

The most interesting examples of decomposition algebras arise from fusion laws
with a _grading_.

######  Definition 2.5.

  1. (1)

Let $\Gamma$ be a group. Then the map

$*\colon\Gamma\times\Gamma\to 2^{\Gamma}\colon(g,h)\mapsto\\{gh\\}$

makes $(\Gamma,*)$ into a fusion law, which we call a _group fusion law_. The
identity element of $\Gamma$ is the unique unit of the fusion law
$(\Gamma,*)$.

  2. (2)

Let $(X,*)$ and $(Y,*)$ be two fusion laws. A _morphism_ from $(X,*)$ to
$(Y,*)$ is a map $\xi\colon X\to Y$ such that

$\xi(x_{1}*x_{2})\subseteq\xi(x_{1})*\xi(x_{2})$

for all $x_{1},x_{2}\in X$, where we have denoted the obvious extension of
$\xi$ to a map $2^{X}\to 2^{Y}$ also by $\xi$.

  3. (3)

Let $(X,*)$ be a fusion law and let $(\Gamma,*)$ be a group fusion law. A
_$\Gamma$ -grading_ of $(X,*)$ is a morphism $\xi\colon(X,*)\to(\Gamma,*)$. In
this paper, the grading group $\Gamma$ will always be abelian.

The Jordan fusion law from Definition 2.12 is $\mathbb{Z}/2\mathbb{Z}$-graded.
Indeed, the map $\xi\colon X\to\mathbb{Z}/2\mathbb{Z}$ mapping $e$ and $z$ to
$0$ and $h$ to $1$ is a fusion law morphism. If the fusion law is graded by a
non-trivial group, then we can construct an interesting subgroup of the
automorphism group $\operatorname{Aut}(A)$ of the algebra $A$.

######  Definition 2.6.

  1. (1)

Let $R^{\times}$ be the group of invertible elements of the base ring $R$. An
$R$-character of $\Gamma$ is a group homomorphism $\chi\colon\Gamma\to
R^{\times}$. The $R$-character group of $\Gamma$ is the group $\mathcal{X}$
consisting of all $R$-characters of $\Gamma$, with group operation induced by
multiplication in $R^{\times}$.

  2. (2)

Let $(A,\mathcal{I},\Omega)$ be a $\Gamma$-decomposition algebra. Let
$\chi\in\mathcal{X}$. For each decomposition
$(A^{i}_{g})_{g\in\Gamma}\in\Omega$, we define a linear map

$\tau_{i,\chi}\colon A\to A\colon a\mapsto\chi(g)a\quad\text{for all }a\in
A^{i}_{g};$

we call this a _Miyamoto map_. It follows immediately from the definitions
that each $\tau_{i,\chi}$ is an automorphism of the $R$-algebra $A$.

  3. (3)

We define the _(full) Miyamoto group_ of $(A,\mathcal{I},\Omega)$ by

$\operatorname{Miy}(A,\mathcal{I},\Omega):=\langle\tau_{i,\chi}\mid
i\in\mathcal{I},\chi\in\mathcal{X}\rangle\leq\operatorname{Aut}(A).$

##  3\. Matsuo algebras

One of the first classes of examples of axial decomposition algebras with a
Jordan fusion law that has been systematically studied, is the class of
_Matsuo algebras_ arising from $3$-transposition groups. We recall their
definition.

######  Definition 3.1.

  1. (1)

A _3-transposition group_ is a pair $(G,D)$ where $G$ is a group and $D\subset
G$ is a conjugacy class of involutions in $G$ such that $G=\langle D\rangle$
and such that for all $d,e\in D$, the order $o(de)$ of the element $de\in G$
is at most $3$.

  2. (2)

Assume that $2$ is invertible in $R$, and let $\eta\in R$ be arbitrary. Let
$A$ be a free $R$-module with basis $D$. We make $A$ into an $R$-algebra by
declaring

$d\cdot e:=\begin{cases}d&\text{if }d=e,\\\ 0&\text{if }o(de)=2,\\\
\tfrac{\eta}{2}(d+e-f)&\text{if }o(de)=3,\text{ where
$f=ded=ede$}.\end{cases}$

This algebra will be denoted by $M_{R}(G,D,\eta)$ and will be called the
_Matsuo algebra_ (over $R$) of the $3$-transposition group $(G,D)$ with
parameter $\eta$.

The 3-transposition groups have a geometric counterpart known as _Fischer
spaces_.

######  Definition 3.2.

  1. (1)

A partial linear space $\mathcal{G}$ is a pair
$\mathcal{G}=(\mathcal{P},\mathcal{L})$, where $\mathcal{P}$ is a set of
points and $\mathcal{L}\subseteq 2^{\mathcal{P}}$ is a set of lines such that
any $\ell\in\mathcal{L}$ has size at least $2$ and such that any two distinct
lines intersect in at most one point. Two distinct points $x,y\in\mathcal{P}$
are called _collinear_ if they are contained in a (necessarily unique) common
line, and we denote this by $x\sim y$.

  2. (2)

A partial linear space is called a partial triple system if every line has
exactly $3$ points. In this case, for any two collinear points
$x,y\in\mathcal{P}$, the unique third point on the line through $x$ and $y$
will be denoted by $x\wedge y$.

  3. (3)

A non-empty subset $S$ of $\mathcal{P}$ is called a subspace of $\mathcal{G}$
if it is closed under the operation $\wedge$, and if $T$ is any subset of
$\mathcal{P}$, then we write $\langle T\rangle$ for the subspace generated by
$T$, i.e., the smallest subspace of $\mathcal{G}$ containing $T$.

  4. (4)

Assume now that $(G,D)$ is a $3$-transposition group. Then we can associate a
partial triple system to $(G,D)$ with point set $D$, where a triple
$\\{d,e,f\\}\subseteq D$ forms a line if and only if $d,e,f$ are the three
involutions in a $\operatorname{Sym}(3)$-subgroup of $G$. In particular, two
distinct points $d$ and $e$ are collinear if and only if $o(de)=3$ in $G$.
Such a partial triple system is called a _Fischer space_.

######  Remark 3.3.

Sometimes, the defining set $D$ of a $3$-transposition group $(G,D)$ is
allowed to be a union of several conjugacy classes, but our assumption is not
a severe one and is common. It has as a consequence that the resulting Fischer
space is _connected_. See, e.g., [CH95, Theorem (3.1)].

######  Definition 3.4.

The _rank_ of a Fischer space $\mathcal{G}$ is the maximal length $r$ of a
chain of connected subspaces

$\mathcal{G}_{0}<\mathcal{G}_{1}<\dots<\mathcal{G}_{r}=\mathcal{G}.$

In particular, a Fischer space of rank $1$ is a point and a Fischer space of
rank $2$ is a line (with three points).

The following two examples play an important role, as they are the only two
Fischer spaces of rank $3$; see Theorem 3.6.

######  Examples 3.5.

  1. (1)

The $3$-transposition group $G=\operatorname{Sym}(4)$ with generating set of
involutions $D$ consisting of the six transpositions in
$\operatorname{Sym}(4)$ has as associated Fischer space the dual affine plane
of order $2$, also known as the _complete quadrilateral_ :

$(23)$$(13)$$(12)$$(14)$$(24)$$(34)$

  2. (2)

The $3$-transposition group $G=3^{2}\colon 2$ (where the action of $C_{2}$ on
$C_{3}\times C_{3}$ is given by inversion) with generating set $D$ consisting
of all $9$ involutions in $G$, has as associated Fischer space the affine
plane of order $3$. We will refer to this example of a Fischer space simply as
_the affine plane_.

$1$$2$$3$$6$$5$$4$$7$$8$$9$

######  Theorem 3.6.

Let $\mathcal{G}$ be a connected Fischer space of rank $3$. Then $\mathcal{G}$
is isomorphic to one of the two examples from 3.5.

In particular, if $\mathcal{G}$ is a Fischer space of arbitrary rank, then the
subspace generated by two distinct intersecting lines of $\mathcal{G}$ is
isomorphic to one of these two examples.

###### Proof.

See, e.g., [CH95, Section 3.2]. In fact, this property is often used as an
alternative equivalent _definition_ of Fischer spaces, as in [CH92, p. 343]. ∎

######  Definition 3.7.

A Fischer space $\mathcal{G}$ is said to be of _symplectic type_ if all
subspaces generated by two distinct intersecting lines are complete
quadrilaterals.

It is often useful to go directly from the Fischer space to the Matsuo
algebra.

######  Definition 3.8.

Let $\mathcal{G}=(\mathcal{P},\mathcal{L})$ be a Fischer space. Then we define
the Matsuo algebra $M_{R}(\mathcal{G},\eta)$ as the $R$-algebra with
underlying free $R$-module with basis $\mathcal{P}$, with multiplication given
by

$x\cdot y:=\begin{cases}x&\text{if }x=y,\\\ 0&\text{if }x\neq y\text{ and
}x\not\sim y,\\\ \tfrac{\eta}{2}(x+y-x\wedge y)&\text{if }x\sim y.\end{cases}$

For the rest of the paper, this is the point of view we will adopt, so all our
algebras will be defined starting from a Fischer space rather than from the
underlying $3$-transposition group.

Matsuo algebras are examples of axial decomposition algebras, with all
elements of the basis $\mathcal{P}$ as axes. More precisely, for each
$x\in\mathcal{P}$, there is a corresponding decomposition of
$A=M_{R}(\mathcal{G},\eta)$ into three parts, with parameters $0$, $1$ and
$\eta$. When $\eta\not\in\\{0,1\\}$, these summands are precisely equal to the
eigenspaces of the multiplication operator $\operatorname{ad}_{x}$.

##  4\. Axial algebras in characteristic 2

In the original definition of axial algebras as in [HRS15], the axes were
always assumed to be _idempotent_ elements, and this is also how the Matsuo
algebras had originally been studied. In terms of the defining parameters in
2.4, this corresponds to the assumption that $\lambda_{e}=1$. Often, however,
this is an unnecessary and sometimes unwanted restriction; in fact, if $2=0$
in $R$, this essentially excludes interesting decomposition algebras with a
$\mathbb{Z}/2\mathbb{Z}$-grading, as the following observation shows. Recall
that all algebras are assumed to be commutative.

######  Proposition 4.1.

Assume that $2=0$ in $R$ and let $A$ be an axial decomposition algebra over
$R$ with respect to a $\mathbb{Z}/2\mathbb{Z}$-graded fusion law $(X,*)$ such
that all axes are idempotents. Then all axes are contained in the _zero-core_
, i.e., the intersection of the $0$-graded part of all of the decompositions.

In particular, the algebra $A$ is not generated by its axes unless all its
$\mathbb{Z}/2\mathbb{Z}$-decompositions are trivial.

###### Proof.

Let $a,b\in A$ be two axes, and decompose $b$ with respect to the
$\mathbb{Z}/2\mathbb{Z}$-decomposition corresponding to $a$, as
$b=b_{0}+b_{1}$, with $b_{0}\in A_{0}^{a}$ and $b_{1}\in A_{1}^{a}$. Then,
since $2=0$, we have

$b=b^{2}=(b_{0}+b_{1})^{2}=b_{0}^{2}+b_{1}^{2}\in A_{0}^{a}$

because both $A_{0}^{a}\cdot A_{0}^{a}\subseteq A_{0}^{a}$ and $A_{1}^{a}\cdot
A_{1}^{a}\subseteq A_{0}^{a}$. ∎

Because of Proposition 4.1, we will, instead, assume that our axes are
_nilpotent_ elements. We point out that this also makes sense for the deeper
reason that the origin of the theory of axial algebras (and Matsuo algebras in
particular) is found in the theory of VOAs, where an important role is played
by elements $e$ for which $e^{2}=2e$ (rather than $e^{2}=e$). This seems like
an innocent rescaling but further illustrates that over arbitrary rings, the
assumption $e^{2}=2e$ is more natural than $e^{2}=e$ and naturally leads to
nilpotent elements in the case that $2=0$.

##  5\. Nilpotent Matsuo algebras

Assume, from now on, that $2=0$ in $R$. In view of the previous section, the
following definition is natural.

######  Definition 5.1.

Let $\mathcal{G}=(\mathcal{P},\mathcal{L})$ be a Fischer space. Assume that
$2=0$ in $R$ and let $\eta\in R$ with $\eta\neq 0$. Then we define the
_nilpotent Matsuo algebra_ $M_{R}(\mathcal{G},\eta)$ as the $R$-algebra with
underlying free $R$-module with basis $\mathcal{P}$, with multiplication given
by

$x\cdot y:=\begin{cases}0&\text{if }x=y\text{ or }x\not\sim y,\\\
\eta(x+y+x\wedge y)&\text{if }x\sim y.\end{cases}$

######  Remark 5.2.

  1. (1)

For each line $\ell\in\mathcal{L}$, the product of two distinct points on
$\ell$ is independent of the choice of these points; it is always equal to
$\eta$ times the sum of these three points.

  2. (2)

The role of the parameter $\eta$ is negligible. In fact, if $R$ is a field,
then we can simply replace each point $x$ by $\eta^{-1}x$ and get the same
multiplication on $A$ with $\eta$ replaced by $1$.

From now on, we keep the assumptions that
$\mathcal{G}=(\mathcal{P},\mathcal{L})$ is a Fischer space, that $2=0$ in $R$
and that $A=M_{R}(\mathcal{G},1)$ is the corresponding nilpotent Matsuo
algebra (so $\eta=1$).

A natural attempt now is to construct decompositions corresponding to the
elements of the basis $\mathcal{P}$. The algebra $A$ no longer decomposes into
eigenspaces for $\operatorname{ad}_{x}$, but even passing to generalized
eigenspaces does not work, because of the following observation.

######  Proposition 5.3.

For each $x\in\mathcal{P}$, the operator $\operatorname{ad}_{x}$ is nilpotent.
More precisely, we have $\operatorname{ad}_{x}^{2}=0$.

###### Proof.

It suffices to show that $x\cdot xy=0$ for each $y\in\mathcal{P}$. If $x=y$ or
$x\not\sim y$, this is obvious, so assume that $x\sim y$. Then $x\cdot
xy=x(x+y+x\wedge y)=xy+x(x\wedge y)$. Since $xy=x(x\wedge y)$ by Remark 5.21,
the result follows. ∎

However, not only the points $x\in\mathcal{P}$ are nilpotent, but in fact,
every single element in $A$ has square $0$. It turns out that the _line
nilpotents_ are a better candidate to obtain decompositions of the algebra
$A$.

######  Definition 5.4.

For each line $\ell\in\mathcal{L}$, we let $s_{\ell}\in A$ be the sum of the
three points on $\ell$, and we call this a _line nilpotent_ of $A$. In fact,
we will simply write $\ell$ for $s_{\ell}$; this will not cause confusion.

We first collect some easy computations that we will use repeatedly. Recall
the two examples from 3.5.

######  Lemma 5.5.

Let $\ell\in\mathcal{L}$ and $x\in\mathcal{P}$.

  1. (1)

If $x\in\ell$ or if $x$ is not collinear to any point of $\ell$, then
$x\ell=0$ in $A$.

  2. (2)

If the subspace spanned by $x$ and $\ell$ is a complete quadrilateral, then
there are precisely two lines $m,n$ joining $x$ with $\ell$; in this case,
$x\ell=m+n$ in $A$. We also have $x\ell=\ell+\ell^{\prime}$ where
$\ell^{\prime}$ is the unique line in this subspace different from $\ell$ and
not containing $x$.

  3. (3)

If the subspace spanned by $x$ and $\ell$ is an affine plane, then there is a
unique line $m\neq\ell$ in this subspace parallel to $\ell$ and not containing
$x$; in this case, $x\ell=x+\ell+m$ in $A$.

###### Proof.

This is straightforward. ∎

######  Lemma 5.6.

Let $\ell,m\in\mathcal{L}$.

  1. (1)

If $\ell=m$, then $\ell m=0$ in $A$.

  2. (2)

If $\ell\cap m=\\{x\\}$ and they span a complete quadrilateral, then $\ell
m=\ell+m$ in $A$.

  3. (3)

If $\ell\cap m=\\{x\\}$ and they span an affine plane, then $\ell m=a+b+c+d$
in $A$, where $a,b,c,d$ are the four points in this subspace not on $\ell$ nor
on $m$.

  4. (4)

If $\ell\cap m=\emptyset$ and they span an affine plane, then $\ell m=s$ in
$A$, where $s$ is the sum of all $9$ points in this subspace.

###### Proof.

This is straightforward. ∎

Nilpotent Matsuo algebras are always somewhat degenerate, as they contain
annihilating elements.

######  Definition 5.7.

An element $a\in A$ is _annihilating_ if $ab=0$ for all $b\in A$. We write

$\operatorname{Ann}(A):=\\{a\in A\mid a\text{ is annihilating}\\}.$

######  Lemma 5.8.

Let $s$ be the sum of all points in $\mathcal{P}$. Then
$\operatorname{Ann}(A)=\langle s\rangle$.

###### Proof.

To show that $s\in\operatorname{Ann}(A)$, it suffices to show that $sx=0$ for
all $x\in\mathcal{P}$. Let $\mathcal{L}_{x}$ be the set of lines containing
$x$. As $yx=0$ whenever $y=x$ or $y\not\sim x$, the product $sx$ is equal to
the sum $\sum_{\ell\in\mathcal{L}_{x}}\ell x$, which is $0$ by Lemma 5.51.

Conversely, assume that $a\in\operatorname{Ann}(A)$ and write
$a=\sum_{p\in\mathcal{P}}a_{p}p$ with $a_{p}\in R$. Let $p\sim q$ be any pair
of distinct collinear points in $\mathcal{G}$ and let $x=p\wedge q$. Then the
fact that $ax=0$ implies that $a_{p}=a_{q}$. Because $\mathcal{G}$ is
connected by Remark 3.3, we conclude that all $a_{p}$ are equal and hence $a$
is a multiple of $s$. ∎

######  Definition 5.9.

We define the _reduced nilpotent Matsuo algebra_ to be the quotient
$A^{\prime}:=A/\operatorname{Ann}(A)$.

We will now first analyze the two Fischer spaces from 3.5.

###  5.1. The complete quadrilateral

$c$$b$$a$$z$$y$$x$$\ell$$m$$n$ Figure 1. The complete quadrilateral.

Let $\mathcal{G}$ be the complete quadrilateral labeled as in Fig. 1 and let
$\ell$ be the line $\\{a,b,c\\}$. We will analyze the decomposition of $A$
arising from the operator $\operatorname{ad}_{\ell}$. For our computations, we
will also need the lines $m=\\{a,y,z\\}$ and $n=\\{b,x,z\\}$ as in Fig. 1.

We begin with a lemma that is useful also for Fischer spaces of symplectic
type of higher rank; see Definition 3.7.

######  Lemma 5.10.

Let $\mathcal{G}$ be a Fischer space of symplectic type and let
$\ell\in\mathcal{L}$. Then the operator $\operatorname{ad}_{\ell}$ is
idempotent, i.e., $\ell\cdot\ell a=\ell a$ for all $a\in A$.

###### Proof.

It suffices to show that $\ell\cdot\ell x=\ell x$ for all $x\in\mathcal{P}$,
and this is obvious if $x\in\ell$ or if $x$ is not collinear with any point of
$\ell$. Assume now that $x$ is collinear with a point of $\ell$ and consider
the subspace generated by $\ell$ and $x$. By assumption, this subspace is a
complete quadrilateral, so we may assume that $\ell$ and $x$ are as in Fig. 1.
Then $\ell x=\ell+m$ and hence, by Lemma 5.52 and Lemma 5.62,

$\ell\cdot\ell x=\ell(\ell+m)=\ell m=\ell+m=\ell x.\qed$

######  Remark 5.11.

The statement of Lemma 5.10 is false for the affine plane, so it is false for
any Fischer space that is _not_ of symplectic type.

We now assume again that $\mathcal{G}$ is a complete quadrilateral. As before,
let $s=a+b+c+x+y+z\in A$ be the sum of all points.

######  Proposition 5.12.

We have a decomposition $A=A^{\ell}_{0}\oplus A^{\ell}_{1}$ into eigenspaces
for the operator $\operatorname{ad}_{\ell}$ with eigenvalues $0$ and $1$,
respectively, where

$\displaystyle A^{\ell}_{0}$ $\displaystyle=\langle a,b,c,x+y+z\rangle=\langle
a,b,c,s\rangle,$ $\displaystyle A^{\ell}_{1}$ $\displaystyle=\langle\ell
x,\ell y\rangle.$

###### Proof.

We have $\ell a=\ell b=\ell c=0$ by Lemma 5.51 and $\ell s=0$ by Lemma 5.8.
Hence

$\langle a,b,c,s\rangle\leq A^{\ell}_{0}.$

By Lemma 5.10, we already know that

$\langle\ell x,\ell y\rangle\leq A^{\ell}_{1}.$

Notice that $\ell x$ and $\ell y$ are linearly independent because $\ell
x=b+c+y+z$ and $\ell y=a+c+x+z$. By dimension count, the result follows. ∎

######  Remark 5.13.

The subspace $A_{1}^{\ell}$ turns out to be independent of the choice of
$\ell$. Indeed, it is equal to $\langle s+a+x,s+b+y,s+c+z\rangle$ (notice that
the sum of these three vectors is $0$) and this is independent of the choice
of $\ell$.

######  Theorem 5.14.

Let $\mathcal{G}$ be the complete quadrilateral. Then the nilpotent Matsuo
algebra $A=M_{R}(\mathcal{G},1)$ is a $6$-dimensional decomposition algebra
with fusion law

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\emptyset$

,

where for each of the $4$ lines $\ell\in\mathcal{L}$, there is a decomposition
$A=A^{\ell}_{0}\oplus A^{\ell}_{1}$ into eigenspaces for
$\operatorname{ad}_{\ell}$ with eigenvalues $0$ and $1$, respectively. The
subspaces $A^{\ell}_{1}$ coincide for all $4$ choices of $\ell\in\mathcal{L}$.

###### Proof.

We keep the notations from Proposition 5.12. It remains to show that
$A^{\ell}_{0}$ is a subalgebra of $A$, that $A^{\ell}_{0}\cdot
A^{\ell}_{1}\subseteq A^{\ell}_{1}$ and that $A^{\ell}_{1}\cdot
A^{\ell}_{1}=0$.

To show that $A^{\ell}_{0}$ is a subalgebra of $A$, we first observe that
$\langle a,b,c\rangle$ is a subalgebra of $A$ because the line
$\ell=\\{a,b,c\\}$ is a subspace of $\mathcal{G}$. Since
$s\in\operatorname{Ann}(A)$ by Lemma 5.8, this implies immediately that
$\langle a,b,c,s\rangle$ is indeed a subalgebra of $A$.

Next, we compute that

$\displaystyle a\cdot\ell x$
$\displaystyle=a(\ell+m)=a\ell+am=0\quad\text{and}$ $\displaystyle a\cdot\ell
y$ $\displaystyle=a(\ell+n)=an=\ell+m=\ell x\in A^{\ell}_{1}.$

Finally, we have

$\ell x\cdot\ell y=(\ell+m)(\ell+n)=\ell m+\ell
n+mn=(\ell+m)+(\ell+n)+(m+n)=0.\qed$

######  Corollary 5.15.

Let $\mathcal{G}$ be the complete quadrilateral and let
$A=M_{R}(\mathcal{G},1)$. Then the reduced nilpotent Matsuo algebra
$A^{\prime}=A/\operatorname{Ann}(A)$ is a $5$-dimensional decomposition
algebra with fusion law

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\emptyset$

,

where for each of the $4$ lines $\ell\in\mathcal{L}$, there is a decomposition
$A^{\prime}=A^{\prime\ell}_{0}\oplus A^{\prime\ell}_{1}$ into eigenspaces for
$\operatorname{ad}_{\ell}$ with eigenvalues $0$ and $1$, respectively. The
subspaces $A^{\prime\ell}_{1}$ coincide for all $4$ choices of
$\ell\in\mathcal{L}$.

###  5.2. The affine plane of order 3

The situation for the affine plane is more complicated. Again, the only
eigenvalues for $\operatorname{ad}_{\ell}$ will be $0$ and $1$, but the
operator $\operatorname{ad}_{\ell}$ is not semisimple, so we need to pass to
the generalized eigenspaces instead.

$a_{1}$$a_{2}$$a_{3}$$b_{3}$$b_{2}$$b_{1}$$c_{1}$$c_{2}$$c_{3}$

Let $\mathcal{G}$ be the affine plane labeled as above and let $\ell$ be the
line $\\{a_{1},a_{2},a_{3}\\}$. We also set $m=\\{b_{1},b_{2},b_{3}\\}$ and
$n=\\{c_{1},c_{2},c_{3}\\}$. Let $s$ be the sum of all $9$ points of
$\mathcal{G}$.

######  Proposition 5.16.

The operator $\operatorname{ad}_{\ell}$ has eigenvalues $0$ and $1$, with
corresponding eigenspaces

$\displaystyle A^{\ell}_{0}$ $\displaystyle=\langle
a_{1},a_{2},a_{3},s\rangle,$ $\displaystyle A^{\ell}_{1}$
$\displaystyle=\langle b_{1}+b_{2},b_{1}+b_{3},c_{1}+c_{2},c_{1}+c_{3}\rangle$

and generalized eigenspaces

$\displaystyle\widetilde{A}^{\ell}_{0}$ $\displaystyle=\langle
a_{1},a_{2},a_{3},m,n\rangle,$ $\displaystyle\widetilde{A}^{\ell}_{1}$
$\displaystyle=A^{\ell}_{1}.$

In particular, $A=\widetilde{A}^{\ell}_{0}\oplus A^{\ell}_{1}$.

###### Proof.

By Lemma 5.51 and Lemma 5.8, we already know that

$\langle a_{1},a_{2},a_{3},s\rangle\leq A^{\ell}_{0}.$

Next, we have $b_{i}\ell=b_{i}+\ell+n$ and $c_{i}\ell=c_{i}+\ell+m$ for each
$i$ by Lemma 5.53, so indeed $(b_{i}+b_{j})\ell=b_{i}+b_{j}$ and
$(c_{i}+c_{j})\ell=c_{i}+c_{j}$ for all $i\neq j$, hence

$\langle b_{1}+b_{2},b_{1}+b_{3},c_{1}+c_{2},c_{1}+c_{3}\rangle\leq
A^{\ell}_{1}.$

Finally, by Lemma 5.64, we have

$\ell m=\ell n=s\in A^{\ell}_{0},$

hence $m,n\in\widetilde{A}^{\ell}_{0}$. (Notice that
$a_{1}+a_{2}+a_{3}+m+n=s$.) By dimension count, the result follows. ∎

######  Theorem 5.17.

Let $\mathcal{G}$ be the affine plane of order $3$. Then the nilpotent Matsuo
algebra $A=M_{R}(\mathcal{G},1)$ is a $9$-dimensional decomposition algebra
with fusion law

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\\{0\\}$

,

where for each of the $12$ lines $\ell\in\mathcal{L}$, there is a
decomposition $A=\widetilde{A}^{\ell}_{0}\oplus A^{\ell}_{1}$ into generalized
eigenspaces for $\operatorname{ad}_{\ell}$ with eigenvalues $0$ and $1$,
respectively.

###### Proof.

We keep the notations from Proposition 5.16. It remains to show that
$\widetilde{A}^{\ell}_{0}$ is a subalgebra of $A$, that
$\widetilde{A}^{\ell}_{0}\cdot A^{\ell}_{1}\subseteq A^{\ell}_{1}$ and that
$A^{\ell}_{1}\cdot A^{\ell}_{1}\subseteq\widetilde{A}^{\ell}_{0}$.

We already know that $\langle a_{1},a_{2},a_{3}\rangle$ is a subalgebra of
$A$. By Lemma 5.64, we have $mn=s\in\widetilde{A}^{\ell}_{0}$. Further, we
have $a_{i}m=a_{i}n=a_{i}+m+n\in\widetilde{A}^{\ell}_{0}$ for each $i$ by
Lemma 5.53, so we conclude that $\widetilde{A}^{\ell}_{0}$ is indeed a
subalgebra of $A$.

Next, we have

$a_{1}(b_{1}+b_{2})=(b_{1}+b_{2})+(c_{1}+c_{3})\in A_{1}^{\ell},$

and similarly for all other products of any $a_{i}$ with any basis element of
$A_{1}^{\ell}$. Moreover, $m(b_{1}+b_{2})=mb_{1}+mb_{2}=0$ by Lemma 5.51 and

$m(c_{1}+c_{2})=mc_{1}+mc_{2}=(\ell+m+c_{1})+(\ell+m+c_{2})=c_{1}+c_{2}\in
A_{1}^{\ell}$

by Lemma 5.53. Similarly, $n(b_{1}+b_{2})\in A_{1}^{\ell}$ and
$n(c_{1}+c_{2})=0$, so we conclude that indeed $\widetilde{A}^{\ell}_{0}\cdot
A^{\ell}_{1}\subseteq A^{\ell}_{1}$.

Finally, we compute that

$\displaystyle(b_{1}+b_{2})(b_{1}+b_{3})$
$\displaystyle=b_{1}b_{3}+b_{1}b_{2}+b_{2}b_{3}=m+m+m=m\in\tilde{A}_{0}^{\ell}\
\text{ and}$ $\displaystyle(b_{1}+b_{2})(c_{1}+c_{2})$
$\displaystyle=b_{1}c_{1}+b_{1}c_{2}+b_{2}c_{1}+b_{2}c_{2}$
$\displaystyle=(a_{1}+b_{1}+c_{1})+(a_{3}+b_{1}+c_{2})+(a_{3}+b_{2}+c_{1})+(a_{2}+b_{2}+c_{2})$
$\displaystyle=a_{1}+a_{2}\in\tilde{A}_{0}^{\ell}$

(and similarly for the other products) so we conclude that also
$A^{\ell}_{1}\cdot A^{\ell}_{1}\subseteq\widetilde{A}^{\ell}_{0}$. ∎

For the reduced nilpotent Matsuo algebra, however, we again have a
decomposition into (proper) eigenspaces, i.e., the operators
$\operatorname{ad}_{\ell}$ are semisimple.

######  Corollary 5.18.

Let $\mathcal{G}$ be the affine plane of order $3$ and let
$A=M_{R}(\mathcal{G},1)$. Then the reduced nilpotent Matsuo algebra
$A^{\prime}=A/\operatorname{Ann}(A)$ is an $8$-dimensional decomposition
algebra with fusion law

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\\{0\\}$

,

where for each of the $12$ lines $\ell\in\mathcal{L}$, there is a
decomposition $A^{\prime}=A^{\prime\ell}_{0}\oplus A^{\prime\ell}_{1}$ into
eigenspaces for $\operatorname{ad}_{\ell}$ with eigenvalues $0$ and $1$,
respectively.

###### Proof.

By Theorem 5.17, we have, for each $\ell\in\mathcal{L}$, a decomposition into
generalized eigenspaces $A^{\prime}=\widetilde{A}^{\prime\ell}_{0}\oplus
A^{\prime\ell}_{1}$. However, since $s=0$ in $A^{\prime}$, we have $\ell
m=\ell n=0$, so that $m$ and $n$ are eigenvectors with eigenvalue $0$ in
$A^{\prime}$ and hence $\widetilde{A}^{\prime\ell}_{0}=A^{\prime\ell}_{0}$. ∎

###  5.3. Fischer spaces of symplectic type

We can now move on to Fischer spaces of higher rank. We first deal with
Fischer spaces of symplectic type (see Definition 3.7). Let $\mathcal{G}$ be a
Fischer space of symplectic type and let $\ell$ be any line of $\mathcal{G}$.
Then any point of $p\not\in\ell$ is collinear with either $0$ or $2$ points of
$\ell$. (Indeed, as soon as a point $p$ is collinear with one point of $\ell$,
the subspace spanned by $p$ and $\ell$ must be a complete quadrilateral.)

If $\pi\subset\mathcal{P}$ is the set of points of a complete quadrilateral
inside $\mathcal{G}$, then we will denote the sum of the points of $\pi$ in
the algebra $A$ also by $\pi$.

######  Proposition 5.19.

Let $\mathcal{G}$ be a Fischer space of symplectic type and let
$\ell=\\{a,b,c\\}$ be any line of $\mathcal{G}$. Let $\mathcal{P}_{0}$ be the
set of points of $\mathcal{G}$ not on $\ell$ and not collinear with $a$, $b$
nor $c$ and let $\mathcal{P}_{2}$ be the set of points collinear with exactly
$2$ points of $\ell$.

Then we have a decomposition $A=A^{\ell}_{0}\oplus A^{\ell}_{1}$ into
eigenspaces for the operator $\operatorname{ad}_{\ell}$ with eigenvalues $0$
and $1$, respectively, where

$\displaystyle A^{\ell}_{0}$ $\displaystyle=\langle
a,b,c\rangle\oplus\langle\pi\mid\pi\text{ a complete quadrilateral with
}\ell\subset\pi\rangle\oplus\langle w\mid w\in\mathcal{P}_{0}\rangle,$
$\displaystyle A^{\ell}_{1}$ $\displaystyle=\langle\ell z\mid
z\in\mathcal{P}_{2}\rangle.$

###### Proof.

Since $\operatorname{ad}_{\ell}$ is idempotent by Lemma 5.10, it is semisimple
with eigenvalues $0$ and $1$ only, so we have a decomposition
$A=A^{\ell}_{0}\oplus A^{\ell}_{1}$ into eigenspaces. Of course, $\langle
a,b,c\rangle\oplus\langle w\mid w\in\mathcal{P}_{0}\rangle\leq A^{\ell}_{0}$.
Moreover, by Proposition 5.12, we already know that for each complete
quadrilateral $\pi$ containing $\ell$, we have

$\displaystyle\langle\pi\rangle\leq A^{\ell}_{0}\quad\text{and}$
$\displaystyle\langle\ell z\mid z\in\pi\rangle\leq A^{\ell}_{1}.$

For each choice of $\pi$, this gives us a $1$-dimensional contribution to
$A_{0}^{\ell}$ and a $2$-dimensional contribution to $A_{1}^{\ell}$. Since
$\mathcal{P}$ is a basis for $A$, these contributions all intersect trivially
and together with $\langle a,b,c\rangle$ and $\langle w\mid
w\in\mathcal{P}_{0}\rangle$, they span all of $A$. The result follows. ∎

We will show in Theorem 5.24 below that these decompositions will again give
rise to a $\mathbb{Z}/2\mathbb{Z}$-graded decomposition algebra. We need a
number of auxiliary results first. To begin, we need to understand the
possible interactions between two different complete quadrilaterals through
the line $\ell$.

######  Proposition 5.20.

Let $\mathcal{G}$ be a Fischer space of symplectic type and let
$\ell=\\{a,b,c\\}$ be a line of $\mathcal{G}$. Suppose that $\ell$ is
contained in two different complete quadrilaterals $\pi$ and $\pi^{\prime}$,
labeled as follows:

$c$$b$$a$$z$$y$$x$ $c$$b$$a$$r$$q$$p$

Then precisely two cases can occur:

  1. (a)

Either

$\displaystyle p\sim x,$ $\displaystyle p\nsim y,$ $\displaystyle p\nsim z$
$\displaystyle q\nsim x,$ $\displaystyle q\sim y,$ $\displaystyle q\nsim z$
$\displaystyle r\nsim x,$ $\displaystyle r\nsim y,$ $\displaystyle r\sim z.$

In this case, $w:=p\wedge x=q\wedge y=r\wedge z$. Moreover,
$w\in\mathcal{P}_{0}$, i.e., $w\nsim a$, $w\nsim b$ and $w\nsim c$.

  2. (b)

Or

$\displaystyle p\nsim x,$ $\displaystyle p\sim y,$ $\displaystyle p\sim z$
$\displaystyle q\sim x,$ $\displaystyle q\nsim y,$ $\displaystyle q\sim z$
$\displaystyle r\sim x,$ $\displaystyle r\sim y,$ $\displaystyle r\nsim z.$

In this case, we have

$d:=r\wedge y=q\wedge z,\quad e:=r\wedge x=p\wedge z,\quad f:=q\wedge
x=p\wedge y.$

The points $a,b,c,d,e,f$ form another complete quadrilateral
$\pi^{\prime\prime}$ through $\ell$, so in particular, we have lines
$\\{a,e,f\\}$, $\\{b,d,f\\}$ and $\\{c,d,e\\}$.

###### Proof.

We repeatedly use the fact that for any point and any line, there are either
$0$ or $2$ points on that line collinear with that point.

  1. (a)

Assume first that $p\sim x$. Since also $p\sim b$, we must have $p\nsim z$.
Similarly, since $p\sim c$, we have $p\nsim y$. On the other hand, we now have
$y\sim c$ and $y\nsim p$, hence $y\sim q$. Continuing in this fashion, we find
all $9$ relations between $p,q,r$ and $x,y,z$ as stated.

Next, we look at the subspace spanned by the lines $\\{c,y,x\\}$ and
$\\{c,q,p\\}$. This, too, is a complete quadrilateral, and since we already
have $p\sim x$ and $q\sim y$, we must have $w:=p\wedge x=q\wedge y$.
Similarly, $w=r\wedge z$.

Finally, we note that $w\sim x$ and $w\sim y$ imply $w\nsim c$, and similarly
$w\nsim a$ and $w\nsim b$, so $w\in\mathcal{P}_{0}$.

  2. (b)

Assume now that $p\nsim x$. Since $p\sim b$, this implies $p\sim z$, and since
$p\sim c$, this implies $p\sim y$. Now $y\sim p$ and $y\sim c$ imply $y\nsim
q$. Again, we continue in this fashion to find all $9$ relations between
$p,q,r$ and $x,y,z$ as stated.

Next, we look at the subspace spanned by the lines $\\{a,z,y\\}$ and
$\\{a,r,q\\}$. Again using the fact that this is a complete quadrilateral, the
known collinearities between $y,z$ and $q,r$ tell us that $r\wedge y=q\wedge
z$. Similarly, $r\wedge x=p\wedge z$ and $q\wedge x=p\wedge y$.

Finally, we look at the subspace spanned by the line $\\{a,r,q\\}$ and the
point $x$. Since $x\sim r$ and $x\sim q$, this is again a complete
quadrilateral, containing $e=r\wedge x$ and $f=q\wedge x$. This forces
$\\{a,e,f\\}$ to be a line. Similarly, we get lines $\\{b,d,f\\}$ and
$\\{c,d,e\\}$. It is now clear that the points $a,b,c,d,e,f$ form another
complete quadrilateral. ∎

######  Remark 5.21.

In principle, Proposition 5.20 is nothing more than a geometric proof of the
classification of symplectic Fischer spaces of rank $4$, which is, of course,
a well known result (see also Theorem 5.25 below). The situation of
Proposition 5.20a corresponds to the Fischer space of $W(A_{4})$, while the
situation b corresponds to the Fischer space of $W(D_{4})$. However, our
geometric version will be useful in the proof of Theorem 5.24 below.

We also have a converse for this result for points in $\mathcal{P}_{0}$.

######  Proposition 5.22.

Let $\mathcal{G}$ be a Fischer space of symplectic type, let
$\ell=\\{a,b,c\\}$ be a line of $\mathcal{G}$ and let $\mathcal{P}_{0}$ be the
set of points not collinear to any of $a,b,c$. If $w\in\mathcal{P}_{0}$, then
there exist two complete quadrilaterals containing $\ell$ such that $w$ is as
in Proposition 5.20a.

###### Proof.

First, we recall that connected Fischer spaces have diameter at most $2$ by
[Hal93]*(2.6), so for any point $w\in\mathcal{P}_{0}$, there exists a complete
quadrilateral $\pi$ labeled as in Fig. 1 such that $w\sim x$. Since $w\nsim b$
and $w\nsim c$, we must have $w\sim z$ and $w\sim y$ as well (again using the
$0$-$2$-property). Now define $p:=w\wedge x$, $q:=w\wedge y$ and $r:=w\wedge
z$. Once again using the $0$-$2$-property several times, it follows that
$\ell\cup\\{p,q,r\\}$ forms a second complete quadrilateral $\pi^{\prime}$.
Now $\pi$, $\pi^{\prime}$ and $w$ are indeed as in Proposition 5.20a, as
claimed. ∎

The points in $\mathcal{P}_{0}$ form a disconnected union of subspaces of
$\mathcal{G}$, even if $\mathcal{G}$ is not of symplectic type.

######  Proposition 5.23.

Let $\mathcal{G}$ be any Fischer space, let $\ell=\\{a,b,c\\}$ be a line of
$\mathcal{G}$ and let $\mathcal{P}_{0}$ be the set of points not collinear to
any of $a,b,c$. If $v,w\in\mathcal{P}_{0}$ with $v\sim w$, then also $v\wedge
w\in\mathcal{P}_{0}$.

###### Proof.

In any Fischer space, we have the $0$-$2$-$3$-property, i.e., given a point
and a line of $\mathcal{G}$, it is impossible to have exactly one point of
that line collinear with that point. (Indeed, in a complete quadrilateral, we
have the $0$-$2$-property, and in an affine plane, all $9$ points are
collinear with each other.)

Now assume that $v,w\in\mathcal{P}_{0}$ with $v\sim w$ but with $u:=v\wedge
w\not\in\mathcal{P}_{0}$. Then without loss of generality, we have $u\sim a$,
but then $u$ is the only point of the line $\\{u,v,w\\}$ collinear with $a$, a
contradiction. ∎

We are now well prepared to show that our algebras are indeed decomposition
algebras. Unavoidably, the proof will require several case distinctions.

######  Theorem 5.24.

Let $\mathcal{G}$ be a Fischer space of symplectic type. Then the nilpotent
Matsuo algebra $A=M_{R}(\mathcal{G},1)$ is a decomposition algebra with fusion
law

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\\{0\\}$

,

where for each line $\ell\in\mathcal{L}$, there is a decomposition
$A=A^{\ell}_{0}\oplus A^{\ell}_{1}$ into eigenspaces for
$\operatorname{ad}_{\ell}$ with eigenvalues $0$ and $1$, respectively.

###### Proof.

Let $\ell\in\mathcal{L}$ and consider the decomposition $A=A^{\ell}_{0}\oplus
A^{\ell}_{1}$ into eigenspaces from Proposition 5.19, with

$\displaystyle A^{\ell}_{0}$ $\displaystyle=\langle
a,b,c\rangle\oplus\langle\pi\mid\pi\text{ a complete quadrilateral with
}\ell\subset\pi\rangle\oplus\langle w\mid w\in\mathcal{P}_{0}\rangle,$
$\displaystyle A^{\ell}_{1}$ $\displaystyle=\langle\ell z\mid
z\in\mathcal{P}_{2}\rangle.$

We first show that $A^{\ell}_{0}$ is a subalgebra of $A$. Notice that $\langle
a,b,c\rangle$ is a subalgebra and that also $\langle w\mid
w\in\mathcal{P}_{0}\rangle$ is a subalgebra by Proposition 5.23. For any $\pi$
through $\ell$, we already know that $\langle a,b,c\rangle\cdot\pi\in
A_{0}^{\ell}$ by Theorem 5.14, and obviously $\langle a,b,c\rangle\cdot w=0$
for any $w\in\mathcal{P}_{0}$. Next, we consider two distinct complete
quadrilaterals $\pi,\pi^{\prime}$ through $\ell$. Labeling the points of $\pi$
and $\pi^{\prime}$ as in Proposition 5.20, we have

$\displaystyle\pi\cdot\pi^{\prime}$ $\displaystyle=(\ell+x+y+z)(\ell+p+q+r)$
$\displaystyle=\ell(x+y+z)+\ell(p+q+r)+(x+y+z)(p+q+r)$
$\displaystyle=(x+y+z)(p+q+r).$

If $\pi$ and $\pi^{\prime}$ are as in Proposition 5.20a, then

$\displaystyle\pi\cdot\pi^{\prime}$ $\displaystyle=(x+p+w)+(y+q+w)+(z+r+w)$
$\displaystyle=\pi+\pi^{\prime}+w\in A_{0}^{\ell}.$

If $\pi$ and $\pi^{\prime}$ are as in Proposition 5.20b, then

$\pi\cdot\pi^{\prime}=(x+q+f)+(x+r+e)+(y+p+f)+(y+r+d)\\\ +(z+p+e)+(z+q+d)=0.$

Next, let $\pi$ be any complete quadrilateral through $\ell$, labeled as in
Fig. 1, and let $w\in\mathcal{P}_{0}$. Of course, $\pi\cdot w=0$ if $w$ is not
collinear with any of $x,y,z$, so we may assume that $w\sim x$. By Proposition
5.22 (and its proof), $\pi$ and $w$ are as in Proposition 5.20a. Then

$\displaystyle\pi\cdot w=(\ell+x+y+z)w$
$\displaystyle=(x+p+w)+(y+q+w)+(z+r+w)$ $\displaystyle=\pi+\pi^{\prime}+w\in
A_{0}^{\ell}.$

This covers all cases to show that $A_{0}^{\ell}$ is indeed a subalgebra of
$A$.

We now show that $A_{0}^{\ell}\cdot A_{1}^{\ell}\subseteq A_{1}^{\ell}$.
Recall that if $\pi$ is a complete quadrilateral through $\ell$ labeled as in
Fig. 1, then $\ell z=a+b+x+y$. Moreover, we already know that $\langle
a,b,c,\pi\rangle\cdot\ell x\in A_{1}^{\ell}$ by Theorem 5.14. Next, consider a
complete quadrilateral $\pi^{\prime}$ through $\ell$ different from $\pi$ and
label it as in Proposition 5.20. Then

$\pi^{\prime}\cdot\ell z=\ell\cdot\ell z+(p+q+r)(a+b+x+y).$

Notice that $\ell\cdot\ell z=\ell z$ by Lemma 5.10. Moreover, we have

$(p+q+r)(a+b)=(p+r+b)+(q+r+a)+(r+a+q)+(r+b+p)=0.$

If $\pi,\pi^{\prime}$ are as in Proposition 5.20a, then

$\displaystyle(p+q+r)(x+y)$ $\displaystyle=(p+x+w)+(q+y+w)$
$\displaystyle=(a+b+x+y)+(a+b+p+q)=\ell z+\ell r,$

so $\pi^{\prime}\cdot\ell z=\ell r\in A_{1}^{\ell}$. If $\pi,\pi^{\prime}$ are
as in Proposition 5.20b, then

$\displaystyle(p+q+r)(x+y)$ $\displaystyle=(p+y+f)+(q+x+f)+(r+x+e)+(r+y+d)$
$\displaystyle=(a+b+d+e)+(a+b+p+q)=\ell f+\ell r,$

so $\pi^{\prime}\cdot\ell z=\ell z+\ell f+\ell r\in A_{1}^{\ell}$. Next, we
let $w\in\mathcal{P}_{0}$ and $z\in\mathcal{P}_{2}$. If $w\nsim x$ and $w\nsim
y$, then of course $w\cdot\ell z=0$, so assume that $w\sim x$ (and hence also
$w\sim y$ and $w\sim z$) so that again, $w$ and $\pi$ are as in Proposition
5.20a. Then

$\displaystyle w\cdot\ell z$ $\displaystyle=w(a+b+x+y)=(w+x+p)+(w+y+q)$
$\displaystyle=(a+b+x+y)+(a+b+p+q)=\ell z+\ell r\in A_{1}^{\ell}.$

Finally, we show that $A_{1}^{\ell}\cdot A_{1}^{\ell}\subseteq A_{0}^{\ell}$.
Let $z,z^{\prime}\in\mathcal{P}_{2}$. If $z^{\prime}$ belongs to the complete
quadrilateral $\pi$ spanned by $\ell$ and $z$, then the result follows again
from Theorem 5.14. Assume now that $\ell$ and $z^{\prime}$ span a complete
quadrilateral $\pi^{\prime}$ different from $\pi$ and label it as in
Proposition 5.20. We have to distinguish between $z^{\prime}=r$ versus
$z^{\prime}=q$ (or $p$, but this amounts to interchanging $p$ and $q$ without
loss of generality).

So we first show that $\ell z\cdot\ell r\in A_{0}^{\ell}$. (In fact, we will
have $\ell z\cdot\ell r=0$.) Notice that $a\wedge b=p\wedge q=x\wedge y=c$, so
by Lemma 5.62, we have

$\displaystyle(a+b)(x+y)$ $\displaystyle=a+b+x+y,$ $\displaystyle(a+b)(p+q)$
$\displaystyle=a+b+p+q,$ $\displaystyle(x+y)(p+q)$ $\displaystyle=x+y+p+q,$

hence

$\ell z\cdot\ell r=(a+b+x+y)(a+b+p+q)=0.$

We now show that $\ell z\cdot\ell q\in A_{0}^{\ell}$. Using the fact that
$(a+b)(a+c)=a+b+c$, that $(a+b)(p+r)=a+r+q$ and $(a+c)(x+y)=a+y+z$, we get

$\displaystyle\ell z\cdot\ell q$ $\displaystyle=(a+b+x+y)(a+c+x+r)$
$\displaystyle=a+b+c+q+r+y+z+(x+y)(p+r).$

If $\pi,\pi^{\prime}$ are as in Proposition 5.20a, then we get

$\ell z\cdot\ell q=(a+b+c)+\pi+\pi^{\prime}+w\in A_{0}^{\ell}.$

If $\pi,\pi^{\prime}$ are as in Proposition 5.20b, then we get

$\ell z\cdot\ell q=\pi+\pi^{\prime}+\pi^{\prime\prime}+w\in A_{0}^{\ell},$

where $\pi^{\prime\prime}$ is as in the statement of Proposition 5.20b. We
have now shown that $A_{1}^{\ell}\cdot A_{1}^{\ell}\subseteq A_{0}^{\ell}$ in
all possible cases. ∎

###  5.4. Fischer spaces not of symplectic type

Finally, we show that when the Fischer space $\mathcal{G}$ is not of
symplectic type, then the corresponding nilpotent Matsuo algebra has a
decomposition that is no longer $\mathbb{Z}/2\mathbb{Z}$-graded. It suffices
to go through the list of possible Fischer spaces of rank $4$ that are not of
symplectic type. It will turn out that in each case, the relation
$A_{1}^{\ell}\cdot A_{1}^{\ell}\subseteq\tilde{A}_{0}^{\ell}$ fails for at
least one line $\ell$. (See, however, Remark 5.28 below.)

Since this amounts to finding a single instance of a line and two elements of
$A_{1}^{\ell}$ whose product does not lie in $\tilde{A}_{0}^{\ell}$, we have
performed this task by computer, and we only present the result.

We first recall the classification of Fischer spaces of rank $4$.

######  Theorem 5.25.

Let $\mathcal{G}=(\mathcal{P},\mathcal{L})$ be a Fischer space of rank $4$.
Then $\mathcal{G}$ is the Fischer space of one of the following
$3$-transposition groups:

  1. (1)

$W(A_{4})$, the Weyl group of type $A_{4}$ (with $|\mathcal{P}|=10$), of
symplectic type;

  2. (2)

$W(D_{4})$, the Weyl group of type $D_{4}$ (with $|\mathcal{P}|=12$), of
symplectic type;

  3. (3)

$3^{3}\colon\operatorname{Sym}(4)$ (with $|\mathcal{P}|=18$), not of
symplectic type;

  4. (4)

$2^{6}\colon\mathrm{SU}_{3}(2)^{\prime}$ (with $|\mathcal{P}|=36$), not of
symplectic type;

  5. (5)

Marshall Hall’s $3$-transposition group $[3^{10}]\colon 2$ (with
$|\mathcal{P}|=81$) or its affine quotient $3^{3+3}\colon 2$ (with
$|\mathcal{P}|=27$), both of “Moufang type”, i.e., not containing any complete
quadrilaterals but only affine planes.

###### Proof.

The proof of this fact, together with the size of $\mathcal{P}$ in each case,
can be found in [HS95, Proposition (4.2)], where the authors point out that
this classification has been proven independently by Zara, Hall and Moori; the
first written source seems to be Zara’s (unpublished) thesis from 1984. ∎

######  Examples 5.26.

  1. (1)

Let $\mathcal{G}$ be of type $3^{3}\colon\operatorname{Sym}(4)$ and let
$\ell=\\{a,b,c\\}$ be a line contained in a complete quadrilateral labeled as
in Fig. 1. Let $m=\\{a^{\prime},b^{\prime},c^{\prime}\\}$ be a line parallel
to $\ell$ in some affine plane through $\ell$, labeled such that the lines
$aa^{\prime}$, $bb^{\prime}$ and $cc^{\prime}$ are parallel. Then
$a^{\prime}+b^{\prime}\in A_{1}^{\ell}$ and $\ell z\in A_{1}^{\ell}$, but
their product is not contained in $\tilde{A}_{0}^{\ell}$.

  2. (2)

Let $\mathcal{G}$ be the affine $3$-space, and label its $27$ points by
$[p,q,r]$ with $p,q,r\in\mathbb{F}_{3}$. Let $\ell$ be the line
$\\{[0,0,0],[1,0,0],[2,0,0]\\}$. Then $[0,1,0]+[1,1,0]\in A_{1}^{\ell}$ and
$[1,0,1]+[2,0,1]\in A_{1}^{\ell}$, but their product is not contained in
$\tilde{A}_{0}^{\ell}$. (In fact, this product is again an element of
$A_{1}^{\ell}$ itself.)

  3. (3)

Let $\mathcal{G}$ be the Fischer space of type
$2^{6}\colon\mathrm{SU}_{3}(2)^{\prime}$. Explicitly, this group is the
semidirect product of a $3$-dimensional vector space $V$ over
$\mathbb{F}_{4}=\\{0,1,\omega,\omega+1\\}$ (with $\omega^{2}=\omega+1$) with
the group $\mathrm{SU}_{3}(2)^{\prime}$ generated by the involutions

$d=\begin{pmatrix}1&0&0\\\ 1&1&0\\\ 1&0&1\end{pmatrix},\
e=\begin{pmatrix}1&1&0\\\ 0&1&0\\\ 0&\omega+1&1\end{pmatrix},\
f=\begin{pmatrix}1&0&1\\\ 0&1&\omega\\\ 0&0&1\end{pmatrix}.$

(See [Hal93, p. 288].) The points of $\mathcal{G}$ are now in one-to-one
correspondence with the conjugates of these $3$ involutions inside this
semidirect product, and we will represent them in the form $[v,g]$ where $v\in
V$ and $g\in\langle d,e,f\rangle$, with product given by $[v,g][w,h]=[v\cdot
h+w,gh]$.

Now let $\ell$ be the line $\\{[0,d],[0,e],[0,ded]\\}$. Then
$[0,f]+[0,defed]\in A_{1}^{\ell}$ and
$[(0,0,\omega),f]+[(\omega+1,1,\omega),defed]\in A_{1}^{\ell}$, but their
product is not contained in $\tilde{A}_{0}^{\ell}$ (and again, this is, in
fact, an element of $A_{1}^{\ell}$).

  4. (4)

Let $\mathcal{G}$ be the Fischer space of type $[3^{10}]\colon 2$, with $81$
points. We can use a similar configuration as in 2. Explicitly, label the $81$
points of $\mathcal{G}$ by $[p,q,r,s]$ with $p,q,r,s\in\mathbb{F}_{3}$ as in
[DS23, Example 2.7] (which is based on [LB83]). For $\ell$, we then take the
line $\\{[0,0,0,0],[1,0,0,0],[2,0,0,0]\\}$. Then $[0,1,0,0]+[1,1,0,0]\in
A_{1}^{\ell}$ and $[0,0,0,1]+[1,0,0,1]\in A_{1}^{\ell}$, but their product is
not contained in $\tilde{A}_{0}^{\ell}$ (and once again, this is an element of
$A_{1}^{\ell}$).

######  Theorem 5.27.

Let $\mathcal{G}$ be a Fischer space, with corresponding nilpotent Matsuo
algebra $A=M_{R}(\mathcal{G},1)$. For each line $\ell$ of $\mathcal{G}$,
consider the decomposition into the generalized eigenspaces
$\tilde{A}_{0}^{\ell}\oplus\tilde{A}_{1}^{\ell}$.

Then $A$ is a decomposition algebra with fusion law

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\\{0\\}$

,

if and only if $\mathcal{G}$ is of symplectic type or $\mathcal{G}$ is an
affine plane.

###### Proof.

This now follows from Theorems 5.17, 5.24 and 5.26. ∎

######  Remark 5.28.

Interestingly, it can happen that for _some_ lines $\ell$, the corresponding
decomposition still admits a $\mathbb{Z}/2\mathbb{Z}$-grading, but not for all
lines. This happens, for instance, for the Fischer spaces of type
$3^{n}\colon\operatorname{Sym}(n+1)$, for all $n\geq 3$. If we would restrict
to only the “good” lines in such a Fischer space, then we would still get a
decomposition algebra with a $\mathbb{Z}/2\mathbb{Z}$-grading, but with too
few decompositions to be interesting. Notice that these “good lines” are
precisely the ones that are only contained in affine planes (and not in any
complete quadrilateral). The $3$ points on such a line are
“$\theta$-equivalent” in the terminology of [CH95, p. 150].

##  6\. Miyamoto group for the complete quadrilateral

The small example of the complete quadrilateral is particularly interesting
because it admits a $\mathbb{Z}$-grading, and not just a
$\mathbb{Z}/2\mathbb{Z}$-grading. Indeed, recall from Theorem 5.14 that the
fusion law is given by

$*$ | $0$ | $1$
---|---|---
$0$ | $\\{0\\}$ | $\\{1\\}$
$1$ | $\\{1\\}$ | $\emptyset$

.

Notice that the map from $X=\\{0,1\\}$ to $\mathbb{Z}$ sending $0$ to $0$ and
$1$ to $1$ determines a fusion law morphism (see Definition 2.5). In
particular, the algebras of Theorem 5.14 and Corollary 5.15 can be viewed as
$\mathbb{Z}$-decomposition algebras. Observe that the $R$-character group of
$\Gamma=\mathbb{Z}$ is naturally isomorphic to $R^{\times}$.

######  Definition 6.1.

Let $(A,\mathcal{I},\Omega)$ be a $\mathbb{Z}$-decomposition algebra. For
$i\in\mathcal{I}$ and $\lambda\in R^{\times}$, write $\tau_{i,\lambda}$ for
the Miyamoto map given by

$\tau_{i,\lambda}(a)=\lambda^{k}a\quad\text{for all }a\in A^{i}_{k}.$

######  Remark 6.2.

If $A=M_{R}(\mathcal{G},1)$ and $\ell$ is a line of $\mathcal{G}$ such that
$A$ decomposes into eigenspaces $A=A_{0}^{\ell}\oplus A_{1}^{\ell}$, then we
can write the corresponding Miyamoto map as

$\tau_{\ell,\lambda}=\mathrm{id}+(1+\lambda)\operatorname{ad}_{\ell}.$

Let $(A,\mathcal{I},\Omega)$ be a $\mathbb{Z}$-decomposition algebra. For each
fixed $i$, we have a multiplicative group

$T_{i}(R):=\langle\tau_{i,\lambda}\mid\lambda\in R^{\times}\rangle\cong
R^{\times}$

of automorphisms, so it makes sense to investigate the corresponding Miyamoto
groups as _algebraic groups_ , rather than as abstract ones (at least when $A$
is finite-dimensional over a field). Following [Mil17], we will view algebraic
groups as functors from the category of commutative $k$-algebras to the
category of groups.

######  Definition 6.3.

Fix a base field $k$ and let $(A,\mathcal{I},\Omega)$ be a
$\mathbb{Z}$-decomposition algebra, with $A$ finite-dimensional over $k$.
Define the _(full) Miyamoto group_ of $(A,\mathcal{I},\Omega)$ as the
algebraic group over $k$ generated by the tori $T_{i}$:

$\operatorname{Miy}(A,\mathcal{I},\Omega):=\langle T_{i}\mid
i\in\mathcal{I}\rangle\leq\operatorname{Aut}(A).$

######  Remark 6.4.

Recall that, by definition, this is the smallest _closed_ subgroup of
$\operatorname{Aut}(A)$ containing these tori $T_{i}$. In particular, it will
_not_ be true in general that the group of $R$-points
$\operatorname{Miy}(A,\mathcal{I},\Omega)(R)$ is generated by the groups
$T_{i}(R)$ for all $k$-algebras $R$. We will see this explicitly in Remark
6.13 below.

From now on, let $A$ be the $6$-dimensional algebra from Theorem 5.14 and let
$A^{\prime}:=A/\operatorname{Ann}(A)$ be its $5$-dimensional quotient as in
Corollary 5.15. We investigate their Miyamoto groups as algebraic groups over
$\mathbb{F}_{2}$. We keep the notations from Section 5.1, so in particular,
the points $a,b,c,x,y,z$ are labeled as in Fig. 1.

######  Lemma 6.5.

Let $\ell=\\{a,b,c\\}$ as in Fig. 1 and let $\lambda\in R^{\times}$. Then the
Miyamoto maps $\tau_{\ell,\lambda}$ of $A$ fix $a$, $b$ and $c$ and

$\displaystyle\tau_{\ell,\lambda}(x)$ $\displaystyle=x+(1+\lambda)\ell x,$
$\displaystyle\tau_{\ell,\lambda}(y)$ $\displaystyle=y+(1+\lambda)\ell y,$
$\displaystyle\tau_{\ell,\lambda}(z)$ $\displaystyle=z+(1+\lambda)\ell z.$

###### Proof.

This follows immediately from Remark 6.2. ∎

The previous lemma suggests to use the basis

(6.6) $\mathcal{B}=(a,\ b,\ \ell,\ \underbrace{b+c+y+z}_{\ell x},\
\underbrace{a+c+x+z}_{\ell y},\ s)$

for $A$, and that is what we will be doing. In fact, the multiplication table
of $A$ with respect to this basis $\mathcal{B}$ looks very simple:

$\begin{array}[]{c|cccccc}\cdot&a&b&\ell&\ell x&\ell y&s\\\ \hline\cr
a&0&\ell&0&0&\ell x&0\\\ b&\ell&0&0&\ell y&0&0\\\ \ell&0&0&0&\ell x&\ell
y&0\\\ \ell x&0&\ell y&\ell x&0&0&0\\\ \ell y&\ell x&0&\ell y&0&0&0\\\
s&0&0&0&0&0&0\end{array}$

######  Remark 6.7.

From the multiplication table, we see that $A\cdot A=\langle\ell,\ell x,\ell
y\rangle$ and $A\cdot(A\cdot A)=\langle\ell x,\ell y\rangle$, which gives us
two interesting subalgebras that are invariant under arbitrary automorphisms
of $A$. Recall that $\operatorname{Ann}(A)=\langle s\rangle$, which gives us a
third subalgebra invariant under $\operatorname{Aut}(A)$.

The following notation will turn out to be useful.

######  Definition 6.8.

  1. (1)

For each $\alpha,\beta\in R$, we define a $3$ by $3$ matrix

$M_{\alpha,\beta}:=\begin{pmatrix}\alpha&0&\beta\\\ 0&\beta&\alpha\\\
0&0&0\end{pmatrix}.$

  2. (2)

For each $3$ by $3$ matrix $M$ over $R$ and each $\lambda\in R^{\times}$, we
define a $6$ by $6$ matrix

$S_{M,\lambda}:=\begin{pmatrix}I_{3}&O_{3}\\\ M&\begin{pmatrix}\lambda&&\\\
&\lambda&\\\ &&1\end{pmatrix}\end{pmatrix},$

where $I_{3}$ is the $3$ by $3$ identity matrix and $O_{3}$ is the $3$ by $3$
zero matrix. Moreover, we define

$S_{\alpha,\beta,\lambda}:=S_{M_{\alpha,\beta},\lambda}$

for all $\alpha,\beta\in R$ and $\lambda\in R^{\times}$.

######  Lemma 6.9.

Consider the algebraic group $G=\mathbb{G}_{a}^{2}\rtimes\mathbb{G}_{m}$,
where the (left) action of $\mathbb{G}_{m}$ on $\mathbb{G}_{a}^{2}$ is given
by multiplication. Then the set of matrices

$\\{S_{\alpha,\beta,\lambda}\mid\alpha,\beta\in R,\lambda\in R^{\times}\\}$

forms a group under matrix multiplication, isomorphic to $G(R)$.

###### Proof.

It is straightforward to check that

$S_{\alpha,\beta,\lambda}\cdot
S_{\gamma,\delta,\mu}=S_{\alpha+\lambda\gamma,\,\beta+\lambda\delta,\,\lambda\mu}$

for all $\alpha,\beta,\gamma,\delta\in R$ and $\lambda,\mu\in R^{\times}$. In
particular,

$\displaystyle\\{S_{\alpha,\beta,1}\mid\alpha,\beta\in R\\}\cong R^{2},$
$\displaystyle\\{S_{0,0,\lambda}\mid\lambda\in R^{\times}\\}\cong R^{\times},$

and the conjugation action of the second subgroup on the first is given by

$S_{0,0,\lambda}\cdot S_{\alpha,\beta,1}\cdot
S_{0,0,\lambda}^{-1}=S_{\lambda\alpha,\lambda\beta,1}.\qed$

######  Lemma 6.10.

Consider the four lines $\ell_{1}=\\{a,b,c\\}$, $\ell_{2}=\\{b,x,z\\}$,
$\ell_{3}=\\{a,y,z\\}$ and $\ell_{4}=\\{c,x,y\\}$. Then with respect to the
basis $\mathcal{B}$ as in 6.6, the Miyamoto maps with respect to these four
lines have matrix representations

$\displaystyle\tau_{\ell_{1},\lambda}$ $\displaystyle=S_{0,\,0,\,\lambda},$
$\displaystyle\tau_{\ell_{2},\lambda}$
$\displaystyle=S_{1+\lambda,\,0,\,\lambda},$
$\displaystyle\tau_{\ell_{3},\lambda}$
$\displaystyle=S_{0,\,1+\lambda,\,\lambda},$
$\displaystyle\tau_{\ell_{4},\lambda}$
$\displaystyle=S_{1+\lambda,\,1+\lambda,\,\lambda}.$

###### Proof.

This follows from Lemma 6.5 and a little bit of computation. ∎

######  Theorem 6.11.

We have
$\operatorname{Miy}(A,\mathcal{I},\Omega)\cong\operatorname{Miy}(A^{\prime},\mathcal{I},\Omega)\cong\mathbb{G}_{a}^{2}\rtimes\mathbb{G}_{m}$,
where the action of $\mathbb{G}_{m}$ on $\mathbb{G}_{a}^{2}$ is given by
multiplication.

###### Proof.

Let $G=\mathbb{G}_{a}^{2}\rtimes\mathbb{G}_{m}$. For each $k$-algebra $R$, let

$H(R)=\\{S_{\alpha,\beta,\lambda}\mid\alpha,\beta\in R,\lambda\in
R^{\times}\\}\cong G(R)$

as in Lemma 6.9. By Lemma 6.10, we already have
$\operatorname{Miy}(A,\mathcal{I},\Omega)\leq H$, and to show equality, it
only remains to show that the algebraic group $T$ generated by the four tori

$\displaystyle T_{1}(R)$ $\displaystyle:=\langle
S_{0,\,0,\,\lambda}\mid\lambda\in R^{\times}\rangle,$ $\displaystyle T_{2}(R)$
$\displaystyle:=\langle S_{1+\lambda,\,0,\,\lambda}\mid\lambda\in
R^{\times}\rangle,$ $\displaystyle T_{3}(R)$ $\displaystyle:=\langle
S_{0,\,1+\lambda,\,\lambda}\mid\lambda\in R^{\times}\rangle,$ $\displaystyle
T_{4}(R)$ $\displaystyle:=\langle
S_{1+\lambda,\,1+\lambda,\,\lambda}\mid\lambda\in R^{\times}\rangle,$

coincides with the whole group $H$. Observe that
$S_{\alpha,\beta,\lambda}\cdot S_{0,0,\lambda^{-1}}=S_{\alpha,\beta,1}$ for
all $\alpha,\beta\in R,\lambda\in R^{\times}$, so we already have
$S_{1+\lambda,0,1}\in T(R)$ and $S_{0,1+\lambda,1}\in T(R)$ for all
$\lambda\in R^{\times}$. For $R=k^{s}$, the separable closure of $k$, these
matrices clearly span all matrices of the form $S_{\alpha,\beta,1}$, so
together with the $S_{0,0,\lambda}$ for all $\lambda\in(k^{s})^{\times}$, we
get the whole group $\\{S_{\alpha,\beta,\lambda}\mid\alpha,\beta\in
k^{s},\lambda\in(k^{s})^{\times}\\}=H(k^{s})$, so $T(k^{s})=H(k^{s})$. Since
$T\leq H$ and $H\cong G$ is smooth, this implies that $T=H$ as algebraic
groups.

It remains to show that
$\operatorname{Miy}(A,\mathcal{I},\Omega)\cong\operatorname{Miy}(A^{\prime},\mathcal{I},\Omega)$.
Notice that each Miyamoto map of $(A,\mathcal{I},\Omega)$ fixes
$\operatorname{Ann}(A)$ elementwise. Since
$A^{\prime}=A/\operatorname{Ann}(A)$, we get an epimorphism

$\rho\colon\operatorname{Miy}(A,\mathcal{I},\Omega)\twoheadrightarrow\operatorname{Miy}(A^{\prime},\mathcal{I},\Omega).$

With respect to the basis $\mathcal{B}$ for $A$ as in 6.6 and the
corresponding basis

(6.12) $\mathcal{B}^{\prime}=(a,\ b,\ \ell,\ \ell x,\ \ell y)$

for $A^{\prime}$, this map is simply given by canceling the last row and the
last column of the matrix. Since all matrices $S_{\alpha,\beta,\lambda}$ have
the same last row and last column, this implies that $\rho$ is injective, and
hence an isomorphism. ∎

######  Remark 6.13.

As indicated in Remark 6.4, it is not true that $\langle
T_{i}(R)\rangle_{i=1,2,3,4}\cong(\mathbb{G}_{a}^{2}\rtimes\mathbb{G}_{m})(R)$
for all $k$-algebras $R$. For instance, take $R=k[x]$, the algebra of
polynomials over $k$. Then $T_{i}(R)=R^{\times}=k^{\times}$, so that the group
generated by the $T_{i}(R)$ is $k^{2}\rtimes k^{\times}\not\cong R^{2}\rtimes
R^{\times}$.

To finish this section, we point out that the full automorphism groups
$\operatorname{Aut}(A)$ and $\operatorname{Aut}(A^{\prime})$ are substantially
larger than the Miyamoto groups.

######  Proposition 6.14.

We have isomorphisms as algebraic groups

  1. (1)

$\operatorname{Aut}(A^{\prime})\cong\mathbb{G}_{a}^{2}\rtimes\operatorname{GL}_{2}$
and

  2. (2)

$\operatorname{Aut}(A)\cong\mathbb{G}_{a}^{2}\rtimes(\operatorname{Aut}(A^{\prime})\times\mathbb{G}_{m})$,
where $\operatorname{GL}_{2}\leq\operatorname{Aut}(A^{\prime})$ acts
“quadratically” and $\mathbb{G}_{a}^{2}\leq\operatorname{Aut}(A^{\prime})$
acts trivially on the first copy of $\mathbb{G}_{a}^{2}$, see below.

###### Proof.

We will continue to use the bases $\mathcal{B}$ for $A$ and
$\mathcal{B}^{\prime}$ for $A^{\prime}$ given in (6.6) and (6.12). It will be
convenient to use the multiplication table from page 6.

  1. (1)

Let $\theta\in\operatorname{Aut}(A^{\prime})$ be arbitrary. By Remark 6.7, its
matrix representation with respect to $\mathcal{B}^{\prime}$ must be of the
form

$\begin{pmatrix}\alpha_{1}&\beta_{1}&0&0&0\\\ \alpha_{2}&\beta_{2}&0&0&0\\\
\alpha_{3}&\beta_{3}&\gamma_{3}&0&0\\\
\alpha_{4}&\beta_{4}&\gamma_{4}&\delta_{4}&\eta_{4}\\\
\alpha_{5}&\beta_{5}&\gamma_{5}&\delta_{5}&\eta_{5}\end{pmatrix}.$

Since this matrix is invertible, so is its submatrix
$\begin{pmatrix}\delta_{4}&\eta_{4}\\\ \delta_{5}&\eta_{5}\end{pmatrix}$. The
equality $\theta(\ell)\cdot\theta(\ell x)=\theta(\ell x)\neq 0$ then implies
$\gamma_{3}=1$. From $\theta(a)\theta(\ell x)=0$ and $\theta(a)\theta(\ell
y)=\theta(\ell x)$, we get

$\begin{pmatrix}\delta_{4}&\delta_{5}\\\
\eta_{4}&\eta_{5}\end{pmatrix}\cdot\begin{pmatrix}\alpha_{3}\\\
\alpha_{1}\end{pmatrix}=\begin{pmatrix}0\\\
\delta_{4}\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}\delta_{4}&\delta_{5}\\\
\eta_{4}&\eta_{5}\end{pmatrix}\cdot\begin{pmatrix}\alpha_{2}\\\
\alpha_{3}\end{pmatrix}=\begin{pmatrix}0\\\ \delta_{5}\end{pmatrix}.$

In particular, $\alpha_{1},\alpha_{2},\alpha_{3}$ are uniquely determined by
$\delta_{4},\delta_{5},\eta_{4},\eta_{5}$, and a similar argument holds for
$\beta_{1},\beta_{2},\beta_{3}$. Next, using $\theta(a)\theta(\ell)=0$, we get

$\alpha_{4}=\alpha_{1}\gamma_{5}+\alpha_{3}\gamma_{4}\quad\text{and}\quad\alpha_{5}=\alpha_{2}\gamma_{4}+\alpha_{3}\gamma_{5},$

so $\alpha_{4}$ and $\alpha_{5}$ are uniquely determined by the $\gamma$’s,
$\delta$’s and $\epsilon$’s. The same holds for $\beta_{4}$ and $\beta_{5}$,
so we conclude that the restriction morphism

$\rho\colon\operatorname{Aut}(A^{\prime})\to\operatorname{Aut}(\langle\ell,\ell
x,\ell y\rangle)\leq\operatorname{GL}_{3}$

is injective. Moreover, by checking all remaining relations
$\theta(e_{1})\cdot\theta(e_{2})=\theta(e_{1}\cdot e_{2})$, for
$e_{1},e_{2}\in\\{a,b,\ell,\ell x,\ell y\\}$, we see that
$\operatorname{im}(\rho)\leq\operatorname{GL}_{3}$ consists of _all_
invertible matrices of the form

$\begin{pmatrix}1&0&0\\\ \gamma_{4}&\delta_{4}&\eta_{4}\\\
\gamma_{5}&\delta_{5}&\eta_{5}\end{pmatrix}.$

It is now straightforward to check that
$\operatorname{im}(\rho)\cong\mathbb{G}_{a}^{2}\rtimes\operatorname{GL}_{2}$.

  2. (2)

Notice from the multiplication table that $A_{1}=\langle a,b,\ell,\ell x,\ell
y\rangle$ is a subalgebra of $A$, isomorphic to the quotient
$A^{\prime}=A/\langle s\rangle$. Moreover, each automorphism of $A_{1}$
extends to a unique automorphism of $A$ fixing $s$, so
$\operatorname{Aut}(A^{\prime})$ is isomorphic to a subgroup of
$\operatorname{Aut}(A)$. Keeping in mind that the subalgebra $\langle\ell,\ell
x,\ell y\rangle$ is stabilized by each automorphism, it follows that an
arbitrary $\theta\in\operatorname{Aut}(A)$ has matrix form

$\left(\begin{array}[]{ccccc|c}&&&&&0\\\ &&&&&0\\\ &&\Theta&&&0\\\ &&&&&0\\\
&&&&&0\\\
\hline\cr\kappa&\lambda&0&0&0&\nu\end{array}\right)=\left(\begin{array}[]{ccccc|c}\alpha_{1}&\beta_{1}&0&0&0&0\\\
\alpha_{2}&\beta_{2}&0&0&0&0\\\ \alpha_{3}&\beta_{3}&1&0&0&0\\\
\alpha_{4}&\beta_{4}&\gamma_{4}&\delta_{4}&\eta_{4}&0\\\
\alpha_{5}&\beta_{5}&\gamma_{5}&\delta_{5}&\eta_{5}&0\\\
\hline\cr\kappa&\lambda&0&0&0&\nu\end{array}\right)$

for some $\Theta\in\operatorname{Aut}(A_{1})$ and some $\kappa,\lambda\in R$
and $\nu\in R^{\times}$. Hence
$\operatorname{Aut}(A)\cong\mathbb{G}_{a}^{2}\rtimes(\operatorname{Aut}(A_{1})\times\mathbb{G}_{m})$,
where the action can be determined by multiplying two such matrices. Clearly,
the $\mathbb{G}_{m}$ acts on $\mathbb{G}_{a}^{2}$ by multiplication. To
determine the action of $\operatorname{Aut}(A_{1})$ on $\mathbb{G}_{a}^{2}$,
we see that the action is given by right multiplication with the upper left
$(2\times 2)$-block $\begin{pmatrix}\alpha_{1}&\beta_{1}\\\
\alpha_{2}&\beta_{2}\end{pmatrix}$. Using the equations from part (i), we see
that this block can be rewritten as

$\begin{pmatrix}\alpha_{1}&\beta_{1}\\\
\alpha_{2}&\beta_{2}\end{pmatrix}=r\begin{pmatrix}\delta_{4}^{2}&\eta_{4}^{2}\\\
\delta_{5}^{2}&\eta_{5}^{2}\end{pmatrix},$

where $r$ is precisely the inverse of the determinant of the matrix
$\begin{pmatrix}\delta_{4}&\eta_{4}\\\ \delta_{5}&\eta_{5}\end{pmatrix}$, so
that this block $\begin{pmatrix}\alpha_{1}&\beta_{1}\\\
\alpha_{2}&\beta_{2}\end{pmatrix}$ has determinant $1$. In other words, the
action of $\operatorname{GL}_{2}$ on $\mathbb{G}_{a}^{2}$ is given by first
applying the group homomorphism

$\operatorname{GL}_{2}\to\operatorname{SL}_{2}\colon
M=\begin{pmatrix}\alpha&\beta\\\ \gamma&\delta\end{pmatrix}\mapsto(\det
M)^{-1}\begin{pmatrix}\alpha^{2}&\beta^{2}\\\
\gamma^{2}&\delta^{2}\end{pmatrix}$

and then applying right multiplication on $\mathbb{G}_{a}^{2}$. ∎

######  Remark 6.15.

It is natural to ask whether we can use the $\mathbb{Z}/2\mathbb{Z}$-gradings
of the other examples to produce interesting Miyamoto groups. The usual
definition of the Miyamoto maps as in Definition 2.6 seems less useful (for
instance, they are trivial if the base ring $R$ has no nilpotent elements). A
possible solution is to consider the automorphism group
$\operatorname{Aut}(A)$ as an algebraic group in characteristic $2$ and define
Miyamoto maps as elements of a copy of the algebraic group $\mu_{2}$, which is
not smooth in characteristic $2$. The groups generated by these Miyamoto maps
is then an algebraic subgroup of $\operatorname{Aut}(A)$. We have tried to
perform this computation for the symplectic Fischer space associated to
$\operatorname{Sym}(5)$, but we have not been able to find a meaningful
description of the resulting Miyamoto group.

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