# Primitive 4-generated axial algebras of Jordan type

Tom De Medts Louis Rowen Yoav Segev  Tom De Medts
Department of Mathematics: Algebra and Geometry
Ghent University
Krijgslaan 281 – S25, 9000 Gent, Belgium<EMAIL_ADDRESS>Louis Rowen
Department of Mathematics
Bar-Ilan University
Ramat Gan
Israel<EMAIL_ADDRESS>Yoav Segev
Department of Mathematics
Ben-Gurion University
Beer-Sheva 84105
Israel<EMAIL_ADDRESS>
###### Abstract.

We show that primitive $4$-generated axial algebras of Jordan type are at most
81-dimensional.

##  1\. Introduction

Axial algebras were introduced in 2015 by Jonathan Hall, Felix Rehren and
Sergey Shpectorov [HRS15b]. They are non-associative commutative algebras
generated by _axes_ , i.e., idempotents for which the left multiplication
operator is semisimple and such that the resulting eigenspaces multiply
according to a given _fusion law_ (see §2 for precise definitions).

In the easiest interesting case, these multiplication operators admit
precisely $3$ eigenvalues $0$, $1$ and $\eta$. A typical example is provided
by _Jordan algebras_ , where each idempotent gives rise to a _Peirce
decomposition_ of the algebra. In this case, we have $\eta=\tfrac{1}{2}$, and
the fusion law is the following.

$\begin{array}[]{c|ccc}*&1&0&\eta\\\ \hline\cr
1&\\{1\\}&\emptyset&\\{\eta\\}\\\ 0&\emptyset&\\{0\\}&\\{\eta\\}\\\
\eta&\\{\eta\\}&\\{\eta\\}&\\{1,0\\}\end{array}$ Table 1. The Jordan fusion
law $\Phi(\eta)$

We call the axial algebras with a fusion law $\Phi(\eta)$ _axial algebras of
Jordan type $\eta$_. Other than Jordan algebras themselves, there are other
interesting examples of axial algebras of Jordan type (for arbitrary values of
$\eta\neq 0,1$), namely the _Matsuo algebras_ arising from $3$-transposition
groups. In this case, the dimension of the algebra is equal to the size of the
normal generating set of $3$-transpositions of the group. (See Example 2.5
below for details.)

The classification of $3$-transposition groups has a long history (see [CH95,
Hal22] and the references therein). It is a highly non-trivial fact that
finitely generated $3$-transposition groups are finite. In fact, this is a
consequence of the classification of finite simple groups, and a direct proof
of this fact would be very valuable. (See [CH95, Theorem (1.3), p. 153].)

One possible approach for such a direct proof is precisely via the
corresponding Matsuo algebras. More generally, we ask the following question.
(We refer to Definition 2.3 below for the precise meaning.)

######  Question.

Let $A$ be a primitive axial algebra of Jordan type. Assume that $A$ is
generated by a finite set of axes. Can we conclude that $A$ is finite-
dimensional?

Notice that, by Corollary 2.8 below, a positive answer to this question would
show, in particular, that finitely generated $3$-transposition groups are
finite. In fact, for $\eta\neq\tfrac{1}{2}$, it is equivalent.

It is natural to try to answer this question for an increasing number of axes.
For $2$-generated primitive axial algebras of Jordan type, this is almost
trivial: such algebras are at most $3$-dimensional. (In fact, much more can be
said: [HRS15a, Theorem 1.1] gives a complete classification of such algebras.)

For $3$-generated algebras, this question was answered affirmatively in the
recent paper [GS20]: such algebras are at most $9$-dimensional.

Our main result is the following.

######  Main Theorem.

Primitive $4$-generated axial algebras of Jordan type $\eta$ are at most
$81$-dimensional, for any $\eta.$ Moreover, this result is best possible.

To go from $3$-generated to $4$-generated primitive axial algebras of Jordan
type is a large step that required substantial new ideas. In fact, in our new
setup, it is almost a triviality to recover the earlier result from [GS20]
that such $3$-generated algebras are at most $9$-dimensional. One of the key
ideas is that we will almost never use the actual multiplication in the
algebra, but instead, we use _sequences of Miyamoto involutions_ (see
Definition 3.3 below). These sequences will allow us to formulate many
“rewriting rules” that we can use to systematically deal with larger and
larger expressions, until we eventually “wrap up” so that we can reduce every
possible expression of length larger than $6$. The precise meaning of this
will be explained below and can be seen in Theorem 5.1, which is a more
detailed version of our Main Theorem.

It is worth pointing out that going to the next step, primitive $5$-generated
axial algebras of Jordan type, is expected to be increasingly more difficult,
because the upper bound of the dimension will be at least $3^{12}=531441$. (In
fact, this is our conjectured upper bound.) In addition, one of the examples
(of dimension $306936$) arises from the largest sporadic Fischer group
$\mathrm{Fi}_{24}$.

##  2\. Primitive axial algebras of Jordan type

Throughout the paper, $\mathbb{F}$ will be a commutative field with
$\operatorname{char}\mathbb{F}\neq 2$. All our algebras will be commutative
but non-associative111As usual, non-associative means “not necessarily
associative”. $\mathbb{F}$-algebras.

For the definition of fusion laws and axial algebras, we rely on [DMPSVC20].

######  Definition 2.1.

  1. (1)

A _fusion law_ is a pair $(X,*),$ where $X$ is a 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 $x,y\in X$.

  2. (2)

The _Jordan fusion law_ is the fusion law with $X=\\{0,1,\eta\\}$ (where
$\eta$ is just a symbol) and with $*$ given by Table 1 above.

######  Definition 2.2.

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

  1. (1)

A _$\Phi$ -decomposition_ of an algebra $A$ is a direct sum decomposition
$A=\bigoplus_{x\in X}A_{x}$ (as vector spaces) 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. (2)

A _$\Phi$ -decomposition algebra_ is a triple $(A,\mathcal{I},\Omega)$ where
$A$ is an $\mathbb{F}$-algebra, $\mathcal{I}$ is an index set and $\Omega$ is
a tuple of $\Phi$-decompositions of $A$ indexed by $\mathcal{I}$. In other
words, for each $i\in\mathcal{I}$, we have a corresponding
$\Phi$-decomposition $A=\bigoplus_{x\in X}A^{(i)}_{x}$ of the algebra $A$.

######  Definition 2.3.

Let $\Phi=(X,*)$ be a fusion law with $1\in X\subseteq\mathbb{F}$.

  1. (1)

For each $a\in A$, we write $\operatorname{ad}_{a}$ for the left
multiplication by $a$, i.e., $\operatorname{ad}_{a}\colon A\to A\colon
x\mapsto ax$.

  2. (2)

An element $a\in A$ is called a _$\Phi$ -axis_ if it is idempotent (i.e.,
$a^{2}=a$) and the decomposition of $A$ into the eigenspaces for
$\operatorname{ad}_{a}$ is a $\Phi$-decomposition.

  3. (3)

The algebra $A$ is a _$\Phi$ -axial algebra_ if it is generated by a set of
$\Phi$-axes. This makes $A$ into a $\Phi$-decomposition algebra (with
$\mathcal{I}$ identified with the given set of axes).

  4. (4)

A $\Phi$-axial algebra $A$ is _primitive_ if for each axis $a$ of the
generating $\Phi$-axes of $A,$ the $1$-eigenspace $A^{(a)}_{1}$ is
$1$-dimensional, i.e., is equal to $\mathbb{F}a$.

  5. (5)

An _axial algebra of Jordan type $\eta$_ is a $\Phi$-axial algebra for the
fusion law $\Phi=\Phi(\eta)$ as in Table 1.

As we mentioned in the introduction, the two main sources of examples of axial
algebras of Jordan type are (1) Jordan algebras, and (2) Matsuo algebras. We
give some details.

######  Example 2.4.

Let $J$ be a Jordan algebra over $\mathbb{F}$, i.e., $J$ is a unital
commutative non-associative algebra such that $a^{2}(ab)=a(a^{2}b)$ for all
$a,b\in J$. If $e\in J$ is an idempotent, then it is an axis for the Jordan
fusion law $\Phi(\tfrac{1}{2})$; this is the famous _Peirce decomposition_ for
Jordan algebras (see, e.g., [Jac68, Chapter III]). In particular, if $J$ is
generated by idempotents, then it is an axial algebra of Jordan type
$\tfrac{1}{2}$.

######  Example 2.5.

Let $(G,D)$ be a _$3$ -transposition group_, i.e., $G$ is a group and
$D\subseteq G$ is a generating set of involutions, closed under conjugation in
$G$, such that the product of any two elements in $D$ has order at most $3$.
Let $\eta\in\mathbb{F}\setminus\\{0,1\\}$ be arbitrary. Then the _Matsuo
algebra_ $M_{\eta}(G,D)$ is the algebra with basis $D$, with multiplication
given by

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

By [HRS15a, Theorem 6.5], $M_{\eta}(G,D)$ is a primitive axial algebra of
Jordan type $\eta$.

Axial algebras of Jordan type, and more generally any type of decomposition
algebras where the fusion law admits a $\mathbb{Z}/2$-grading, admit many
involutory automorphisms, the so-called _Miyamoto involutions_.

######  Definition 2.6.

  1. (1)

A _$\mathbb{Z}/2$ -grading_ of a fusion law $(X,*)$ is a map $\theta\colon
X\to\mathbb{Z}/2$ such that $x*y\subseteq\theta^{-1}(\theta(x)+\theta(y))$ for
all $x,y\in X$. For instance, the Jordan fusion law from Table 1 is
$\mathbb{Z}/2$-graded with $\theta(0)=\theta(1)=0$ and $\theta(\eta)=1$.

  2. (2)

If $(A,\mathcal{I},\Omega)$ is a $\Phi$-decomposition algebra for a
$\mathbb{Z}/2$-graded fusion law $(X,*)$, then for each $i\in\mathcal{I}$, we
define a _Miyamoto involution_

$\tau_{i}\colon A\to A\colon a_{x}\mapsto(-1)^{\theta(x)}a_{x},\quad\text{
when }a_{x}\in A^{(i)}_{x}.$

In other words, $\tau_{i}$ fixes the $0$-graded elements and negates the
$1$-graded elements with respect to the $i$-th decomposition of $A$.

Corollary 2.8 below is an important motivation for the main result of our
paper.

######  Proposition 2.7.

Let $(G,D)$ be a $3$-transposition group. The following are equivalent:

  1. (a)

$G$ is finite.

  2. (b)

$D$ is finite.

  3. (c)

$M_{\eta}(G,D)$ is finite-dimensional.

###### Proof.

Of course, (a) implies (b), and (b) and (c) are equivalent because
$M_{\eta}(G,D)$ has dimension $|D|$. In particular, the dimension of
$M_{\eta}(G,D)$ is independent of the choice of the base field $\mathbb{F}$
and of $\eta\in\mathbb{F}$, so to show that (c) implies (a), we may assume
that $\mathbb{F}$ is a finite field.

Then $A:=M_{\eta}(G,D)$ is finite. By [DMR17, p. 325] (which relies on [Asc97,
p. 92, Example (4)]), $G/Z(G)$ is embedded in $\operatorname{Aut}(A)$, so
$G/Z(G)$ is a finite group. By a theorem of Schur, [Asc00, (33.9), p. 168],
the derived subgroup $G^{\prime}$ is finite. Since $G/G^{\prime}$ is an
abelian group generated by a finite number of involutions, it is finite, so we
conclude that $G$ is finite. ∎

######  Corollary 2.8.

The following are equivalent:

  1. (a)

Every finitely generated $3$-transposition group $(G,D)$ is finite.

  2. (b)

Every primitive axial algebra $A$ of Jordan type $\eta\neq\tfrac{1}{2}$
generated by a finite set of axes $X$ is finite-dimensional.

###### Proof.

  * (a) $\Rightarrow$ (b)

Let $A$ and $X$ be as in (b). For $x\in X,$ let $\tau_{x}$ be the Miyamoto
involution associated with $x.$ By [HRS15a, Theorem (5.4), p. 105], the group
$G=\langle\tau_{x}\mid x\in X\rangle$ is a $3$-transposition group. By the
assumption, $G$ is finite. By [HRS15a, Corollary (1.2), p. 81], $A$ is spanned
by $\\{x^{g}\mid x\in X,g\in G\\},$ so $A$ is finite-dimensional.

  * (b) $\Rightarrow$ (a)

Let $(G,D)$ be a finitely generated $3$-transposition group. Then $G$ is
generated by a finite number of elements from $D$, hence the algebra
$M_{\eta}(G,D)$ is finitely generated. Thus, by the assumption, it is finite-
dimensional. Proposition 2.7 then tells us that $G$ is finite. ∎

In order to get an idea about the complexity of the primitive $4$-generated
axial algebras of Jordan type, it is useful to look at the list of
$4$-generated $3$-transposition groups first. In particular, this will provide
us with an example of such an algebra of dimension $81$, which is precisely
the upper bound that we will obtain in our main result.

######  Theorem 2.9.

Let $(G,D)$ be a $3$-transposition group generated by $4$ elements from $D$
(but not by less than $4$). Then its central type is one of the following:

  1. (1)

$W(A_{4})$, the Weyl group of type $A_{4}$ (with $|D|=10$);

  2. (2)

$W(D_{4})$, the Weyl group of type $D_{4}$ (with $|D|=12$);

  3. (3)

$3^{3}\colon\operatorname{Sym}(4)$ (with $|D|=18$);

  4. (4)

$2^{1+6}\colon\mathrm{SU}_{3}(2)^{\prime}$ (with $|D|=36$);

  5. (5)

Hall’s $3$-transposition group $[3^{10}]\colon 2$ (with $|D|=81$) or its
affine quotient $3^{3+3}\colon 2$ (with $|D|=27$).

###### Proof.

The definition of central type, and the proof of this fact (together with the
size of $D$ 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. ∎

The unique $3$-transposition group in this list attaining the upper bound
$|D|=81$ is particularly interesting because it arises as a $3$-transposition
subgroup of the sporadic Fischer groups $\mathrm{Fi}_{23}$ and
$\mathrm{Fi}_{24}$. We give an explicit construction of the resulting Matsuo
algebra, based on [LB83, §4.1]. In fact, we had implemented this example on a
computer to experiment with identities, which is how some of our ideas arose.

######  Example 2.10.

Let $D$ be the $4$-dimensional vector space over the field $\mathbb{F}_{3}$
(so $|D|=81$). We first set

$(x_{1},x_{2},x_{3},x_{4})\bullet(y_{1},y_{2},y_{3},y_{4})\\\
:=\bigl{(}x_{1}+y_{1},\ x_{2}+y_{2},\ x_{3}+y_{3},\
x_{4}+y_{4}+(x_{1}y_{2}-x_{2}y_{1})(x_{3}-y_{3})\bigr{)}$

for all $x_{i},y_{i}\in\mathbb{F}_{3}$. Next, we set

$d*e:=(d\bullet e)\bullet(d\bullet e)$

for all $d,e\in D$. For any $\eta\in\mathbb{F}\setminus\\{0,1\\}$—recall that
$\mathbb{F}$ is still our arbitrary base field of characteristic different
from $2$—we now define an $\mathbb{F}$-algebra with basis $D$, and with
multiplication given by

$de:=\begin{cases}e&\text{ if }d=e\\\ \tfrac{\eta}{2}(d+e-d*e)&\text{ if
}d\neq e.\end{cases}$

Then by combining [LB83] with Example 2.5, we see that this is precisely the
Matsuo algebra corresponding to Hall’s $3$-transposition group $[3^{10}]\colon
2$.

##  3\. Method

From now on, we assume that $A$ is a primitive axial algebra of Jordan type
$\eta$ generated by a finite set $S$ of axes.

######  Definition 3.1.

For each $i\geq 0$, we set

$S[i]:=\langle\tau_{a_{1}}\tau_{a_{2}}\dotsm\tau_{a_{\ell}}(b)\mid\ell\leq
i,a_{1},\dots,a_{\ell},b\in S\rangle.$

In particular, $S[0]=\langle S\rangle$, and the $S[i]$ form an ascending chain
of subspaces of $A$.

Our goal is to show that $S[n]=A$ for some $n$. The following proposition
tells us that we can do this by showing that the ascending chain of the $S[i]$
stabilizes.

######  Proposition 3.2.

Assume that $S[n]=S[n+1]$ for some $n$. Then $A=S[n]$.

###### Proof.

Following [HRS15a, p. 81], we define the _closure_ of the set $S$ of axes to
be the smallest set $C$ of axes of $A$ containing $S$ such that for each $a\in
C$, we have $\tau_{a}(C)\subseteq C$. In fact,
$C=\\{\tau_{a_{1}}\tau_{a_{2}}\dotsm\tau_{a_{\ell}}(b)\mid\ell\geq
0,a_{1},\dots,a_{\ell},b\in S\\}$; see, for instance, [KMS20, Lemma 3.5]. It
now suffices to observe that if $S[n]=S[n+1]$, then $S[n]=S[\ell]$ for all
$\ell\geq n$, hence $S[n]=\langle C\rangle$. By [HRS15a, Cor. (1.2), p. 81],
however, $A$ is spanned by $C$, and the result follows. ∎

From now on, when we refer to an arbitrary _axis of $A$_, we will always mean
an element of the closure $C$ of $S$ (which is indeed always an axis for the
same fusion law).

The following two definitions will play a crucial role.

######  Definition 3.3.

  1. (1)

We let

$\llbracket
a_{1},a_{2},\dots,a_{\ell}\rrbracket:=\tau_{a_{1}}\tau_{a_{2}}\dotsm\tau_{a_{\ell}}$

for all axes $a_{1},\dots,a_{\ell}\in A$.

  2. (2)

For all $x,y\in A$, we set

$x\;\equiv_{(i)}\,y\iff x-y\in S[i].$

Notice that $x\;\equiv_{(i)}\,y$ implies $x\;\equiv_{(j)}\,y$ for all $j\geq
i$, and also implies that $\llbracket a_{1},\dots,a_{\ell}\rrbracket
x\;\equiv_{(i+\ell)}\,\llbracket a_{1},\dots,a_{\ell}\rrbracket y$ for all
$a_{1},\dots,a_{\ell}\in S$.

######  Remark 3.4.

The notation $\llbracket a_{1},\dots,a_{\ell}\rrbracket$ will also be used
when the $a_{i}$ are axes that are not necessarily contained in $S$. Some care
is needed with the use of the equivalence relations $\;\equiv_{(i)}\,$ in such
a situation, as these relations are always meant with respect to the given
generating set $S$.

By [HSS18, Theorem 4.1], primitive axial algebras of Jordan type always admit
a (necessarily unique) normalized symmetric Frobenius form.

######  Definition 3.5.

  1. (1)

A bilinear form $(\cdot,\cdot)\colon A\times A\to\mathbb{F}$ is called a
_(normalized) Frobenius form_ on $A$ if $(xy,z)=(x,yz)$ for all $x,y,z\in A$
and, in addition, $(a,a)=1$ for each axis $a\in A$.

  2. (2)

It will be useful to introduce the notation

$\epsilon_{x,y}:=1-\tfrac{2}{\eta}(x,y)$

for all $x,y\in A$.

######  Proposition 3.6.

Let $a\in A$ be an axis and $x\in A$ be arbitrary. Then

$\tau_{a}(x)=x+\tfrac{2}{\eta}(a,x)a-\tfrac{2}{\eta}ax.$

###### Proof.

This is [HSS18, Lemma 3.3] combined with the statement from [HSS18, Theorem
4.1] that $(a,x)=\varphi_{a}(x)$. ∎

######  Remark 3.7.

In [HSS18], their Lemma 3.3 is used, in fact, in the proof of their Theorem
4.1 (the existence of the Frobenius form). On the other hand, if we already
_assume_ the existence of the Frobenius form to begin with, then there is an
easy direct proof of Proposition 3.6 by simply decomposing $x$ with respect to
the eigenspaces for the axis $a$.

Proposition 3.6 has the following immediate but useful consequences.

######  Corollary 3.8.

Let $a,b\in A$ be axes. Then:

  1. (1)

$\llbracket a\rrbracket b-\llbracket b\rrbracket a=\epsilon_{a,b}(b-a)$.

  2. (2)

If $a\in S$ and $x\in S[i]$, then
$(-\tfrac{2}{\eta})ax\;\equiv_{(0)}\,\llbracket a\rrbracket
x-x\;\equiv_{(i)}\,\llbracket a\rrbracket x$.

###### Proof.

  1. (1)

By Proposition 3.6, we have

$\tau_{a}(b)-\tau_{b}(a)=\bigl{(}1-\tfrac{2}{\eta}(a,b)\bigr{)}(b-a).$

  2. (2)

This follows immediately from Proposition 3.6. ∎

We recall the following important fact, which we will be using over and over
again, often without explicitly mentioning it.

######  Proposition 3.9.

We have $\llbracket a,b,a\rrbracket=\llbracket\tau_{a}(b)\rrbracket$ for all
axes $a,b\in A$.

###### Proof.

This follows from [HRS15a, Lemma 5.1, p. 103] and the fact that
$\tau_{a}\in\operatorname{Aut}(A)$. ∎

The following result is a first instance of how useful it is.

######  Proposition 3.10.

Let $a,b\in S$ and $x\in A$. Then:

  1. (1)

$\llbracket a,b,a\rrbracket x\;\equiv_{(1)}\,x-\tfrac{2}{\eta}\tau_{a}(b)x$.

  2. (2)

$\llbracket a,b,a\rrbracket x-\llbracket b,a,b\rrbracket
x\;\equiv_{(1)}\,\epsilon_{a,b}(\llbracket b\rrbracket x-\llbracket
a\rrbracket x)$. In particular, if $x\in S[i]$ for some $i\geq 0$, then
$\llbracket a,b,a\rrbracket x\;\equiv_{(i+1)}\,\llbracket b,a,b\rrbracket x$.

  3. (3)

$\llbracket b,a,b,a\rrbracket x\;\equiv_{(2)}\,\llbracket a,b\rrbracket
x+\epsilon_{a,b}(x-\llbracket b,a\rrbracket x)$. In particular, if $x\in S[i]$
for some $i\geq 2$, then $\llbracket b,a,b,a\rrbracket
x\;\equiv_{(i)}\,\llbracket a,b\rrbracket x-\epsilon_{a,b}\llbracket
b,a\rrbracket x$.

###### Proof.

  1. (1)

We apply Proposition 3.9 to $x$ and use Proposition 3.6 on the right-hand side
to get

$\llbracket a,b,a\rrbracket x=x+\tfrac{2}{\eta}(\tau_{a}(b),x)\llbracket
a\rrbracket b-\tfrac{2}{\eta}\tau_{a}(b)x.$ (3.1)

Since $\llbracket a\rrbracket b\in S[1]$, the result follows.

  2. (2)

Interchanging $a$ and $b$ in 1 and subtracting gives, using Corollary 3.81,

$\llbracket a,b,a\rrbracket x-\llbracket b,a,b\rrbracket
x\;\equiv_{(1)}\,(-\tfrac{2}{\eta})\epsilon_{a,b}(b-a)x.$

By Corollary 3.82, however,

$(-\tfrac{2}{\eta})(bx-ax)\;\equiv_{(0)}\,(\llbracket b\rrbracket
x-x)-(\llbracket a\rrbracket x-x)=\llbracket b\rrbracket x-\llbracket
a\rrbracket x,$

and the result follows.

  3. (3)

This follows immediately by applying $\tau_{b}$ on 2. ∎

######  Lemma 3.11.

Let $a,b,c\in S$. Then:

  1. (1)

$\llbracket a,b\rrbracket a=\epsilon_{a,b}a+b-\epsilon_{a,b}\llbracket
a\rrbracket b\;\equiv_{(0)}\,-\epsilon_{a,b}\llbracket a\rrbracket b$.

  2. (2)

$\llbracket a,b,a\rrbracket c=\alpha c-\alpha\llbracket a\rrbracket
b+\llbracket c,a\rrbracket b$ where $\alpha=\epsilon_{\tau_{a}(b),c}\in F$.

  3. (3)

$\llbracket a,b,c\rrbracket a\;\equiv_{(0)}\,\delta\llbracket a\rrbracket
b-\epsilon_{a,c}\llbracket a,b\rrbracket c+\llbracket c,a\rrbracket b$ for
some $\delta\in F$.

###### Proof.

  1. (1)

By Corollary 3.81,

$\llbracket a,b\rrbracket a=\llbracket a\rrbracket(\llbracket b\rrbracket
a-\llbracket a\rrbracket b+\llbracket a\rrbracket b)=\epsilon_{a,b}\llbracket
a\rrbracket(a-b)+b.$

  2. (2)

Let $\alpha=\epsilon_{\tau_{a}(b),c}$. By substituting $\tau_{a}(b)$ for $a$
and $c$ for $b$ in Corollary 3.81, we get

$\llbracket\tau_{a}(b)\rrbracket c-\llbracket c,a\rrbracket
b=\alpha(c-\llbracket a\rrbracket b).$

The result now follows from Proposition 3.9.

  3. (3)

By Corollary 3.81, we have

$\displaystyle\llbracket a,b,c\rrbracket a$ $\displaystyle=\llbracket
a,b,a\rrbracket c+\llbracket a,b\rrbracket\bigl{(}\llbracket c\rrbracket
a-\llbracket a\rrbracket c\bigr{)}$ $\displaystyle=\llbracket a,b,a\rrbracket
c+\epsilon_{a,c}\llbracket a,b\rrbracket(a-c),$

so 3 follows from 1 and 2. ∎

##  4\. Rewriting rules

In this section, we will gradually build up “rewriting rules” that will allow
us to simplify certain expressions. As the length of the expressions
increases, the proofs become more and more involved.

######  Proposition 4.1.

Let $a,b,c,d\in S$. Then:

  1. (1)

$\llbracket a\rrbracket b\;\equiv_{(0)}\,\llbracket b\rrbracket a$.

  2. (2)

$\llbracket a,b\rrbracket a\in S[1]$.

  3. (3)

$\llbracket a,b,a\rrbracket c\;\equiv_{(1)}\,\llbracket c,a\rrbracket b$.

  4. (4)

$\llbracket a,b,c\rrbracket a\;\equiv_{(1)}\,\llbracket c,b\rrbracket
a-\epsilon_{a,c}\llbracket a,b\rrbracket c$

  5. (5)

$\llbracket a,b,a,c\rrbracket d\;\equiv_{(1)}\,\llbracket c,d,c,a\rrbracket
b$.

  6. (6)

$\llbracket a,b,c,d,b\rrbracket a\;\equiv_{(2)}\,\llbracket b,d,c,b\rrbracket
a-\epsilon_{a,\tau_{b}(d)}\llbracket a,b,c\rrbracket
d\;\equiv_{(3)}\,\llbracket b,d,c,b\rrbracket a$.

  7. (7)

$\llbracket a,b,c,a,b\rrbracket d\;\equiv_{(3)}\,\llbracket
b,a,d,b,a\rrbracket c$.

###### Proof.

  1. (1)

This follows from Corollary 3.81.

  2. (2)

By 1, we have $\llbracket a,b\rrbracket a\;\equiv_{(1)}\,\llbracket
a,a\rrbracket b=b$. (Of course, this also follows from Lemma 3.111.)

  3. (3)

This follows from Lemma 3.112.

  4. (4)

This follows from Lemma 3.113 and 1.

  5. (5)

We have

$\llbracket a,b,a,c\rrbracket d-\llbracket c,d,c,a\rrbracket
b=\llbracket\tau_{a}(b)\rrbracket\tau_{c}(d)-\llbracket\tau_{c}(d)\rrbracket\tau_{a}(b),$

which is contained in $\langle\tau_{a}(b)-\tau_{c}(d)\rangle\leq S[1]$ by
Corollary 3.81.

  6. (6)

We have

$\llbracket a,b,c,d,b\rrbracket a=\llbracket
a,\tau_{b}(c),\tau_{b}(d)\rrbracket a.$

Now let $S^{\prime}=\\{a,\tau_{b}(c),\tau_{b}(d)\\}$ and apply 4 with respect
to this set $S^{\prime}$ in place of $S$. Notice that $S^{\prime}[1]\leq
S[2]$, because

$\displaystyle\llbracket\tau_{b}(c)\rrbracket a=\llbracket b,c,b\rrbracket
a\in S[2]\quad\text{(by \ref{pr:rules:aba c})},$
$\displaystyle\llbracket\tau_{b}(c)\rrbracket\tau_{b}(d)=\llbracket
b,c,b,b\rrbracket d=\llbracket b,c\rrbracket d\in S[2],$

so we see that indeed $\llbracket x\rrbracket y\in S[2]$ for all $x,y\in
S^{\prime}$. Hence

$\displaystyle\llbracket a,\tau_{b}(c),\tau_{b}(d)\rrbracket a$
$\displaystyle\;\equiv_{(2)}\,\llbracket\tau_{b}(d),\tau_{b}(c)\rrbracket
a-\epsilon_{a,\tau_{b}(d)}\llbracket a,\tau_{b}(c)\rrbracket\tau_{b}(d)$
$\displaystyle=\llbracket b,d,c,b\rrbracket
a-\epsilon_{a,\tau_{b}(d)}\llbracket a,b,c\rrbracket d$

so we conclude that indeed

$\llbracket a,b,c,d,b\rrbracket a\;\equiv_{(2)}\,\llbracket b,d,c,b\rrbracket
a-\epsilon_{a,\tau_{b}(d)}\llbracket a,b,c\rrbracket
d\;\equiv_{(3)}\,\llbracket b,d,c,b\rrbracket a.$

  7. (7)

We start from

$\displaystyle\llbracket b,a,c,a,b\rrbracket d$
$\displaystyle=\llbracket\tau_{b}\tau_{a}(c)\rrbracket d$
$\displaystyle=\llbracket
d\rrbracket\tau_{b}\tau_{a}(c)+\epsilon_{d,\tau_{b}\tau_{a}(c)}(d-\tau_{b}\tau_{a}(c))$
$\displaystyle=\llbracket d,b,a\rrbracket
c+\epsilon_{d,\tau_{b}\tau_{a}(c)}(d-\llbracket b,a\rrbracket c)$
$\displaystyle\;\equiv_{(0)}\,\llbracket d,b,a\rrbracket
c-\epsilon_{d,\tau_{b}\tau_{a}(c)}\llbracket b,a\rrbracket c.$

In particular, $\llbracket b,a,c,a,b\rrbracket d\in S[3]$. Moreover, applying
$\llbracket b,a\rrbracket$ to this equivalence yields

$\displaystyle\llbracket b,a,b,a,c,a,b\rrbracket d$
$\displaystyle\;\equiv_{(2)}\,\llbracket b,a,d,b,a\rrbracket
c-\epsilon_{d,\tau_{b}\tau_{a}(c)}\llbracket b,a,b,a\rrbracket c$
$\displaystyle\;\equiv_{(2)}\,\llbracket b,a,d,b,a\rrbracket c,$ (4.1)

where the last equivalence holds because by 3 and 4, we have

$\llbracket b,a,b,a\rrbracket c\;\equiv_{(2)}\,\llbracket b,c,a\rrbracket b\in
S[2].$

We now apply Proposition 3.103 with $x=\llbracket c,a,b\rrbracket d\in S[3]$,
which gives

$\displaystyle\llbracket b,a,b,a,c,a,b\rrbracket d$
$\displaystyle\;\equiv_{(3)}\,\llbracket a,b,c,a,b\rrbracket
d-\epsilon_{a,b}\llbracket b,a,c,a,b\rrbracket d$
$\displaystyle\;\equiv_{(3)}\,\llbracket a,b,c,a,b\rrbracket d.$

The claim follows by combining this with (4.1). ∎

For our next rewriting rule in Proposition 4.4, we first need the following
lemma.

######  Lemma 4.2.

Let $a,b,c,d\in S$ and let $S^{\prime}=\\{a,b,c,\tau_{a}(d)\\}$. Then
$S^{\prime}[3]\subseteq S[4]$.

###### Proof.

Let $\llbracket x,y,z\rrbracket w$ be any element with $x,y,z,w\in
S^{\prime}$. Of course, if none of these four elements is equal to
$\tau_{a}(d)$, then $\llbracket x,y,z\rrbracket w\in S[3]\subseteq S[4]$, and
if all four elements are equal to $\tau_{a}(d)$, then $\llbracket
x,y,z\rrbracket w=\tau_{a}(d)\in S[1]\subseteq S[4]$.

Case 1. Suppose that only one of these four elements is equal to
$\tau_{a}(d)$.

If $w=\tau_{a}(d)$, then $\llbracket x,y,z\rrbracket w=\llbracket
x,y,z,a\rrbracket d\in S[4]$. If $z=\tau_{a}(d)$, then, by Proposition 4.13,

$\llbracket x,y,z\rrbracket w=\llbracket x,y,a,d,a\rrbracket w\in S[4].$

If $y=\tau_{a}(d)$, then, by Proposition 4.15,

$\llbracket x,y,z\rrbracket w=\llbracket x,a,d,a,z\rrbracket
w\;\equiv_{(2)}\,\llbracket x,z,w,z,a\rrbracket d.$

If $x=a$ or $z=a$, then $\llbracket x,y,z\rrbracket w\in S[4]$. If $w=a$, then
$\llbracket x,y,z\rrbracket w\in S[4]$, by Proposition 4.12. We may thus
assume that $z=b$ and $w=c$. If $x=b$, then we see that $\llbracket
x,y,z\rrbracket w\in S[4]$. If $x=c$, then

$\llbracket x,y,z\rrbracket w\;\equiv_{(2)}\,\llbracket c,b,c,b,a\rrbracket
d\in S[3],$

by Proposition 3.103.

If $x=\tau_{a}(d)$, then, assuming without loss that $y=b$,

$\llbracket x,y,z\rrbracket w=\llbracket a,d,a,b,z\rrbracket w.$

If $z=a$, then by Proposition 4.13, $\llbracket x,y,z\rrbracket w\in S[4]$.
Hence we may assume $z=c$, and by Proposition 4.12 we may assume that $w=a$.
In this case, Proposition 4.14 shows that $\llbracket x,y,z\rrbracket w\in
S[4]$.

Case 2. Suppose that three of the four elements $x,y,z,w$ are equal to
$\tau_{a}(d)$.

If $x=y=z=\tau_{a}(d)$, then of course $\llbracket x,y,z\rrbracket
w=\llbracket\tau_{a}(d)\rrbracket w=\llbracket a,d,a\rrbracket w\in S[2]$. For
the other cases, we simply observe that

$\displaystyle\llbracket\tau_{a}(d),x,\tau_{a}(d)\rrbracket\tau_{a}(d)=\llbracket\tau_{a}(d),x\rrbracket\tau_{a}(d)=\llbracket
a,d,a,x,a\rrbracket d\in S[4],$
$\displaystyle\llbracket\tau_{a}(d),\tau_{a}(d),x\rrbracket\tau_{a}(d)=\llbracket
x,a\rrbracket d\in S[2],$ $\displaystyle\llbracket
x,\tau_{a}(d),\tau_{a}(d)\rrbracket\tau_{a}(d)=\llbracket x,a\rrbracket d\in
S[2].$

Case 3. Exactly two of the four elements $x,y,z,w$ are equal to $\tau_{a}(d)$.

We have

$\displaystyle\llbracket\tau_{a}(d),\tau_{a}(d),z\rrbracket w=\llbracket
z\rrbracket w\in S[1],$ $\displaystyle\llbracket
x,\tau_{a}(d),\tau_{a}(d)\rrbracket w=\llbracket x\rrbracket w\in S[1],\text{
and}$ $\displaystyle\llbracket x,y,\tau_{a}(d)\rrbracket\tau_{a}(d)=\llbracket
x,y,a\rrbracket d\in S[3].$

If $x=w=\tau_{a}(d)$ and $y,z\in\\{a,b,c\\}$, then, by 6,

$\llbracket\tau_{a}(d),y,z\rrbracket\tau_{a}(d)=\llbracket
a,d,a,y,z,a\rrbracket d\;\equiv_{(4)}\,\llbracket a,a,z,y,a\rrbracket d\in
S[3].$

Next, if $x=w=\tau_{a}(d)$ and $y,z\in\\{a,b,c\\}$, then, by Lemma 3.11(ii)
(with $\tau_{a}(d)$ in place of $a$),
$\llbracket\tau_{a}(d),y,\tau_{a}(d)\rrbracket w\in S[4]$.

Finally, if $y=w=\tau_{a}(d)$ and $x,z\in\\{a,b,c\\}$, then, by Lemma 3.11(i)

$\llbracket x,\tau_{a}(d),z\rrbracket\tau_{a}(d)\in S[4].\qed$

The following corollary will play an important role in the proof of
Proposition 4.5.

######  Corollary 4.3.

Let $a,b,c,d\in S$ and let $T=\\{a,\tau_{a}(b),\tau_{a}(c),d\\}$. Then
$T[4]\subseteq S[6]$.

###### Proof.

Let $S^{\prime}=\\{a,b,c,\tau_{a}(d)\\}$ as in Lemma 4.2 and notice that
$T=\tau_{a}(S^{\prime})$, i.e., $T$ is obtained from $S^{\prime}$ by applying
$\tau_{a}$ on each element. By Proposition 3.9, for all
$x_{1},\dots,x_{k},y\in S^{\prime}$ we have

$\llbracket\tau_{a}(x_{1}),\dots,\tau_{a}(x_{k})\rrbracket\tau_{a}(y)=\llbracket
a,x_{1},\dots,x_{k},a\rrbracket\tau_{a}(y)=\tau_{a}(\llbracket
x_{1},\dots,x_{k}\rrbracket y),$

so $T[i]=\tau_{a}(S^{\prime}[i])$ for all $i$.

By Lemma 4.2, we have $S^{\prime}[3]\subseteq S[4]$. Now

$\displaystyle S^{\prime}[4]$ $\displaystyle=\llbracket a\rrbracket
S^{\prime}[3]\cup\llbracket b\rrbracket S^{\prime}[3]\cup\llbracket
c\rrbracket S^{\prime}[3]\cup\llbracket\tau_{a}(d)\rrbracket S^{\prime}[3]$
$\displaystyle\subseteq S[5]\cup\llbracket a,d,a\rrbracket S[4],$

and hence

$T[4]=\llbracket a\rrbracket S^{\prime}[4]\subseteq\llbracket a\rrbracket
S[5]\cup\llbracket d,a\rrbracket S[4]\subseteq S[6].\qed$

######  Proposition 4.4.

Let $a,b,c,d\in S$. Then

$\llbracket a,b,c,a,b,c\rrbracket d\;\equiv_{(4)}\,\llbracket
b,c,a,b,c,a\rrbracket d\;\equiv_{(4)}\,\llbracket c,a,b,c,a,b\rrbracket d.$

###### Proof.

Let $S^{\prime}=\\{a,b,c,\tau_{a}(d)\\}$. By Lemma 4.2, we have
$S^{\prime}[3]\subseteq S[4]$. We can thus apply Proposition 4.17 with respect
to $S^{\prime}$ to get

$\llbracket c,b,\tau_{a}(d),c,b\rrbracket a\;\equiv_{(4)}\,\llbracket
b,c,a,b,c\rrbracket\tau_{a}(d),$

hence

$\llbracket c,b,a,d,a,c,b\rrbracket a\;\equiv_{(4)}\,\llbracket
b,c,a,b,c,a\rrbracket d.$ (4.2)

On the other hand, we apply $\llbracket c,b,a,d\rrbracket$ to the equivalence
in Lemma 3.113 (with $b$ and $c$ interchanged) to get

$\llbracket c,b,a,d,a,c,b\rrbracket a\\\ \;\equiv_{(4)}\,\delta\llbracket
c,b,a,d,a\rrbracket c-\epsilon_{a,b}\llbracket c,b,a,d,a,c\rrbracket
b+\llbracket c,b,a,d,b,a\rrbracket c$

for some $\delta\in F$. Now $\llbracket c,b,a,d,a\rrbracket c\in S[4]$ by
Proposition 4.13. Also, by Proposition 4.15 and Proposition 3.103, we have

$\llbracket c,b,a,d,a,c\rrbracket b\;\equiv_{(3)}\,\llbracket
c,b,c,b,c,a\rrbracket d\in S[4].$

Thus, by Proposition 4.17,

$\llbracket c,b,a,d,a,c,b\rrbracket a\;\equiv_{(4)}\,\llbracket
c,b,a,d,b,a\rrbracket c\;\equiv_{(4)}\,\llbracket c,a,b,c,a,b\rrbracket d.$
(4.3)

Combining (4.2) and (4.3), we see that

$\llbracket b,c,a,b,c,a\rrbracket d\;\equiv_{(4)}\,\llbracket
c,a,b,c,a,b\rrbracket d.$

It now suffices to cyclically permute $a$$,b$$,c$ to also get the other
equivalence. ∎

We now come to the final and most challenging rewriting rule, which will
effectively put a bound on the dimension of $4$-generated primitive axial
algebras of Jordan type.

######  Proposition 4.5.

Let $a,b,c,d\in S$. Then $\llbracket d,a,b,c,a,b,c\rrbracket d\in S[6]$.

###### Proof.

Let

$T=\\{\tau_{d}(a),\tau_{d}(b),c,d\\}.$

By Corollary 4.3, we have $T[4]\subseteq S[6]$. By Proposition 4.4 applied to
$T$, this implies that

$\llbracket\tau_{d}(a),\tau_{d}(b),c,\tau_{d}(a),\tau_{d}(b),c\rrbracket
d\;\equiv_{(6)}\,\llbracket
c,\tau_{d}(a),\tau_{d}(b),c,\tau_{d}(a),\tau_{d}(b)\rrbracket d.$ (4.4)

We will proceed in two steps: We first show that

$\llbracket\tau_{d}(a),\tau_{d}(b),c,\tau_{d}(a),\tau_{d}(b),c\rrbracket
d\;\equiv_{(6)}\,\llbracket d,a,c,b,a,c,b\rrbracket d,$ (4.5)

and then we show that

$\llbracket c,\tau_{d}(a),\tau_{d}(b),c,\tau_{d}(a),\tau_{d}(b)\rrbracket d\in
S[6].$ (4.6)

Interchanging the role of $b$ and $c$, it will then follow from (4.4), (4.5)
and (4.6) that $\llbracket d,a,b,c,a,b,c\rrbracket d\in S[6]$.

Step 1. Proof of (4.5).

By Lemma 3.111 applied on $\llbracket d,c\rrbracket d$, we have

$\llbracket\tau_{d}(a),\tau_{d}(b),c,\tau_{d}(a),\tau_{d}(b),c\rrbracket d\\\
\begin{aligned} &=\llbracket d,a,b,d,c,d,a,b,d,c\rrbracket d\\\ &=\llbracket
d,a,b,d,c,d,a,b\rrbracket c+\epsilon_{c,d}\llbracket d,a,b,d,c,d,a,b\rrbracket
d\end{aligned}\\\ -\epsilon_{c,d}\llbracket d,a,b,d,c,d,a,b,d\rrbracket c.$

Now let $\gamma=-\epsilon_{c,\tau_{d}(b)}$; then by Proposition 4.16 and
Proposition 4.15, we have

$\displaystyle\llbracket d,a,b,d,c,d,a,b,d\rrbracket c$
$\displaystyle\;\equiv_{(6)}\,\llbracket d,a,b,d,d,b,a,d\rrbracket
c+\gamma\llbracket d,a,b,d,c,d,a\rrbracket b$
$\displaystyle\;\equiv_{(0)}\,\gamma\llbracket d,a,b,d,c,d,a\rrbracket b$
$\displaystyle\;\equiv_{(4)}\,\gamma\llbracket d,a,b,a,b,a,d\rrbracket c\in
S[6],$

by Proposition 3.10(iii).

Also, by Proposition 4.1(iv), we have

$\llbracket d,a,b,d,c,d,a,b\rrbracket d\\\ \begin{aligned}
&\;\equiv_{(6)}\,\llbracket d,a,b,d,c,b,a\rrbracket d-\epsilon_{b,d}\llbracket
d,a,b,d,c,d,a\rrbracket b\\\ &\;\equiv_{(6)}\,\llbracket
d,a,b,d,c,b,d\rrbracket a-\epsilon_{b,d}\llbracket d,a,b,a,b,a,d\rrbracket
c\quad\text{(by \ref{pr:rules}\ref{pr:rules:a b} and
\ref{pr:rules}\ref{pr:rules:abac d})}\\\ &\;\equiv_{(6)}\,\llbracket
d,a,d,b,a,d,b\rrbracket c-\epsilon_{b,d}\llbracket d,a,b,a,b,a,c\rrbracket
d\quad\text{(by \ref{pr:rules}\ref{pr:rules:abcab d} and
\ref{pr:rules}\ref{pr:rules:a b})}\\\ &\ \in S[6],\end{aligned}$

by Proposition 4.4 and Proposition 3.103.

Finally, by Proposition 3.102,

$\llbracket d,a,b,d,c,d,a,b\rrbracket c\\\ \begin{aligned}
&\;\equiv_{(6)}\,\llbracket d,a,b,c,d,c,a,b\rrbracket c\\\
&\;\equiv_{(6)}\,\llbracket d,a,b,c,d,b,a\rrbracket c-\epsilon_{b,c}\llbracket
d,a,b,c,d,c,a\rrbracket b\quad\text{(by \ref{pr:rules}\ref{pr:rules:abc
a})}\\\ &\;\equiv_{(6)}\,\llbracket d,a,b,c,d,b,c\rrbracket
a-\epsilon_{b,c}\llbracket d,a,b,a,b,a,c\rrbracket d\quad\text{(by
\ref{pr:rules}\ref{pr:rules:a b} and \ref{pr:rules}\ref{pr:rules:abac d})}\\\
&\;\equiv_{(5)}\,\llbracket d,a,c,b,a,c,b\rrbracket d\quad\text{(by
\ref{pr:rules}\ref{pr:rules:abcab d} and
\ref{pr:aba}\ref{pr:aba:baba}).}\end{aligned}$

This proves (4.5).

Step 2. Proof of (4.6).

By Lemma 3.11(iii), there exists $\delta\in F$ with

$\llbracket c,\tau_{d}(a),\tau_{d}(b),c,\tau_{d}(a),\tau_{d}(b)\rrbracket d\\\
\begin{aligned} &=\llbracket c,d,a,b,d,c,d,a,b\rrbracket d\\\
&\;\equiv_{(6)}\,\delta\llbracket c,d,a,b,d,c,d\rrbracket
a-\epsilon_{b,d}\llbracket c,d,a,b,d,c,d,a\rrbracket b\end{aligned}\\\
+\llbracket c,d,a,b,d,c,b,d\rrbracket a.$

Now by Proposition 4.13, $\llbracket c,d,a,b,d,c,d\rrbracket a\in S[6]$. By
Proposition 4.15 and Proposition 3.103, we have

$\llbracket c,d,a,b,d,c,d,a\rrbracket b\;\equiv_{(5)}\,\llbracket
c,d,a,b,a,b,a,d\rrbracket c\in S[6].$

Finally, by Proposition 4.17 and Proposition 4.4, we also have

$\displaystyle\llbracket c,d,a,b,d,c,b,d\rrbracket a$
$\displaystyle\;\equiv_{(6)}\,\llbracket c,d,a,d,b,a,d,b\rrbracket c$
$\displaystyle\;\equiv_{(6)}\,\llbracket c,d,d,b,a,d,b,a\rrbracket
c=\llbracket c,b,a,d,b,a\rrbracket c\in S[6].$

This proves (4.6) and thus finishes the proof of this proposition. ∎

##  5\. $4$-generated primitive axial algebras of Jordan type

We are now ready to prove our main result. Although it requires some care to
write down the proof, the hard work has already been done in Propositions 4.1,
4.4 and 4.5.

######  Theorem 5.1.

Assume that $A$ is generated by a set $S=\\{a,b,c,d\\}$ of $4$ axes. Then
$A=S[6]$ and $A$ is at most $81$-dimensional.

More precisely, let $G$ be the group $\operatorname{Sym}(S)$ of all
permutations of $S$. Define222There is some obvious abuse of notation here: a
priori, the group $G$ does not act on $A$, so when we write an expression like
$\\{\llbracket a,b,c\rrbracket d\\}^{G}$, we really mean $\\{\llbracket
a^{\rho},b^{\rho},c^{\rho}\rrbracket(d^{\rho})\mid\rho\in G\\}$.

$\displaystyle\Gamma_{0}$ $\displaystyle=\\{a\\}^{G},$
$\displaystyle\Gamma_{1}$ $\displaystyle=\\{\llbracket a\rrbracket b\\}^{G},$
$\displaystyle\Gamma_{2}$ $\displaystyle=\\{\llbracket a,b\rrbracket
c\\}^{G},$ $\displaystyle\Gamma_{3}$ $\displaystyle=\\{\llbracket
a,b,c\rrbracket d\\}^{G},$ $\displaystyle\Gamma_{4}$
$\displaystyle=\\{\llbracket a,b,a,c\rrbracket d,\ \llbracket
a,b,c,a\rrbracket d\\}^{G},$ $\displaystyle\Gamma_{5}$
$\displaystyle=\\{\llbracket a,b,c,a,b\rrbracket d\\}^{G},$
$\displaystyle\Gamma_{6}$ $\displaystyle=\\{\llbracket a,b,c,a,b,c\rrbracket
d\\}^{G}.$

Then for each $i\in\\{1,\dots,6\\}$, we have
$S[i]=\langle\Gamma_{0},\dots,\Gamma_{i}\rangle$. In particular,
$A=\langle\Gamma_{0},\dots,\Gamma_{6}\rangle$.

Moreover, there is some redundancy in these spanning sets: The dimension of
each of the $S[i]$ is at most $4$, $10$, $22$, $34$, $61$, $73$ and $81$,
respectively.

###### Proof.

For each $i\leq 6$, let $T[i]$ be the subspace of $A$ spanned by
$\Gamma_{0},\dots,\Gamma_{i}$. Obviously, we have $T[i]\leq S[i]$ for each
$i$. We will show recursively that for each $i\leq 6$, $S[i]=T[i]$, and that
$S[7]=S[6]$. We will, at the same time, compute the maximal possible dimension
of each $T[i]$. Notice that for each $i$, the subspace $S[i+1]$ is spanned by
$S[i]$ and all elements obtained by applying the four operations $\llbracket
a\rrbracket$, $\llbracket b\rrbracket$, $\llbracket c\rrbracket$ and
$\llbracket d\rrbracket$ on the elements of $S[i]$. In order to go from
$S[i]=T[i]$ to the next step $S[i+1]$, it will suffice, by $G$-symmetry, to
apply these four operations on the given representative of the set
$\Gamma_{i}$.

  * $i=0$.

Obviously, $T[0]=\langle a,b,c,d\rangle=S[0]$, and $\dim S[0]\leq 4$.

  * $i=1$.

We have $\llbracket a\rrbracket a=a\in S[0]$, whereas applying any of the
other three operations $\llbracket b\rrbracket$, $\llbracket c\rrbracket$,
$\llbracket d\rrbracket$ on the representative $a\in\Gamma_{0}$ results in an
element of $\Gamma_{1}$, so $S[1]\leq T[1]$.

By Proposition 4.11, we have $\llbracket b\rrbracket
a\;\equiv_{(0)}\,\llbracket a\rrbracket b$, so the $12$ possible elements of
$\Gamma_{1}$ come in pairs that are linearly dependent modulo $S[0]$. Hence
$\dim S[1]\leq 4+12/2=10$.

  * $i=2$.

We have $\llbracket a,a\rrbracket b=b\in S[0]$ and $\llbracket b,a\rrbracket
b\in S[1]$ by Proposition 4.12. On the other hand, $\llbracket c,a\rrbracket
b$ and $\llbracket d,a\rrbracket b$ belong to $\Gamma_{2}$ and hence to
$T[2]$. Hence $S[2]\leq T[2]$.

By Proposition 4.11, we have $\llbracket a,b\rrbracket
c\;\equiv_{(1)}\,\llbracket a,c\rrbracket b$, so the $24$ possible elements of
$\Gamma_{2}$ come in pairs that are linearly dependent modulo $S[1]$. Hence
$\dim S[2]\leq 10+24/2=22$.

  * $i=3$.

We have $\llbracket a,a,b\rrbracket c=\llbracket b\rrbracket c\in S[1]$, and
we have $\llbracket b,a,b\rrbracket c\in S[2]$ by Proposition 4.13 and
$\llbracket c,a,b\rrbracket c\in S[2]$ by Proposition 4.14. On the other hand,
$\llbracket d,a,b\rrbracket c$ belongs to $\Gamma_{3}$ and hence to $T[3]$.
Hence $S[3]\leq T[3]$.

By Proposition 4.11, we have $\llbracket a,b,c\rrbracket
d\;\equiv_{(2)}\,\llbracket a,b,d\rrbracket c$, so the $24$ possible elements
of $\Gamma_{3}$ come in pairs that are linearly dependent modulo $S[2]$. Hence
$\dim S[3]\leq 22+24/2=34$.

  * $i=4$.

We have $\llbracket a,a,b,c\rrbracket d=\llbracket b,c\rrbracket d\in S[2]$.
On the other hand, $\llbracket b,a,b,c\rrbracket d$ and $\llbracket
c,a,b,c\rrbracket d$ belong to $\Gamma_{4}$ and hence to $T[4]$. Finally,
$\llbracket d,a,b,c\rrbracket d\;\equiv_{(3)}\,\llbracket d,a,b,d\rrbracket
c\in\Gamma_{4}$, so $\llbracket d,a,b,c\rrbracket d$ belongs to $\langle
S[3],\Gamma_{4}\rangle\leq T[4]$. Hence $S[4]\leq T[4]$.

By Proposition 4.15, the $24$ possible elements of $\\{\llbracket
a,b,a,c\rrbracket d\\}^{G}$ come in $8$-tuples that are pairwise linearly
dependent modulo $S[3]$:

$\llbracket a,b,a,c\rrbracket d\;\equiv_{(1)}\,\llbracket c,d,c,a\rrbracket
b\;\equiv_{(3)}\,\llbracket c,d,c,b\rrbracket a\;\equiv_{(1)}\,\llbracket
b,a,b,c\rrbracket d\\\ \;\equiv_{(3)}\,\llbracket b,a,b,d\rrbracket
c\;\equiv_{(1)}\,\llbracket d,c,d,b\rrbracket a\;\equiv_{(3)}\,\llbracket
d,c,d,a\rrbracket b\;\equiv_{(1)}\,\llbracket a,b,a,d\rrbracket c.$

On the other hand, there are no such equivalences between the $24$ possible
elements of $\\{\llbracket a,b,c,a\rrbracket d\\}^{G}$. Hence $\dim S[4]\leq
34+24/8+24=61$.

  * $i=5$.

First, because $\llbracket a,b,a,c\rrbracket d$ is $3$-equivalent to an
element beginning with any of the generators $a,b,c,d$, we see that applying
any of the four operators $\llbracket a\rrbracket$, $\llbracket b\rrbracket$,
$\llbracket c\rrbracket$, $\llbracket d\rrbracket$ on this element will result
in an element already contained in $S[3]$.

Next, we apply these operators on $\llbracket a,b,c,a\rrbracket d$. Of course,
we again have $\llbracket a,a,b,c,a\rrbracket d\in S[3]$. Next, by Proposition
3.102 and Proposition 4.13, we have $\llbracket b,a,b,c,a\rrbracket
d\;\equiv_{(3)}\,\llbracket a,b,a,c,a\rrbracket d\in S[4]$, and by Proposition
4.16, we have $\llbracket d,a,b,c,a\rrbracket d\in S[4]$. Finally, $\llbracket
c,a,b,c,a\rrbracket d\in\Gamma_{5}$. Hence $S[5]\leq T[5]$.

By Proposition 4.17, the $24$ possible elements of $\Gamma_{5}$ come in pairs
that are linearly dependent modulo $S[4]$. Hence $\dim S[5]\leq 61+24/2=73$.

  * $i=6$.

We have $\llbracket a,a,b,c,a,b\rrbracket d=\llbracket b,c,a,b\rrbracket d\in
S[4]$. By Proposition 3.102 and Proposition 4.15, we have

$\llbracket b,a,b,c,a,b\rrbracket d\;\equiv_{(4)}\,\llbracket
a,b,a,c,a,b\rrbracket d\;\equiv_{(3)}\,\llbracket a,b,b,d,b,a\rrbracket c\in
S[4].$

Next, $\llbracket c,a,b,c,a,b\rrbracket d\in\Gamma_{6}$, and finally, by
Proposition 4.17, we also have $\llbracket d,a,b,c,a,b\rrbracket
d\;\equiv_{(4)}\,\llbracket d,b,a,d,b,a\rrbracket c\in\Gamma_{6}$. Hence
$S[6]\leq T[6]$.

By Proposition 4.4, the $24$ possible elements of $\Gamma_{6}$ come in triples
that are pairwise linearly dependent module $S[5]$. Hence $\dim S[6]\leq
73+24/3=81$.

  * $i=7$.

We have $\llbracket a,a,b,c,a,b,c\rrbracket d\in S[5]$, and by Proposition
4.4, it follows that also $\llbracket b,a,b,c,a,b,c\rrbracket d$ and
$\llbracket c,a,b,c,a,b,c\rrbracket d$ belong to $S[5]$. Finally, by
Proposition 4.5, we also have $\llbracket d,a,b,c,a,b,c\rrbracket d\in S[6]$.
We conclude that $S[7]=S[6]$, and therefore $A=S[6]$. ∎

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