#  $2$-generated axial algebras of Monster type

Clara Franchi, Mario Mainardis, Sergey Shpectorov

###### Abstract.

In this paper we provide the basic setup for a project, initiated by Felix
Rehren in [16], aiming at classifying all $2$-generated primitive axial
agebras of Monster type $(\alpha,\beta)$. We first revise Rehren’s
construction of an initial object in the cathegory of primitive $n$-generated
axial algebras giving a formal one, filling some gaps and, though correcting
some inaccuracies, confirm Rehren’s results. Then we focus on $2$-generated
algebras which naturally part into three cases: the two critical cases
$\alpha=2\beta$ and $\alpha=4\beta$, and the generic case (i.e. all the rest).
About these cases, which will be dealt in detail in subsequent papers, we give
bounds on the dimensions (the generic case already treated by Rehen) and
classify all 2-generated primitive axial algebras of Monster type
$(\alpha,\beta)$ over ${\mathbb{Q}}(\alpha,\beta)$ for $\alpha$ and $\beta$
algebraically independent indeterminates over ${\mathbb{Q}}$. Finally we
restrict to the $2$-generated Majorana algebras (i.e. when
$\alpha=\frac{1}{4}$ and $\beta=\frac{1}{32}$), showing that these fall
precisely into the nine isomorphism types of the Norton-Sakuma algebras.

##  1\. Introduction

Axial algebras constitute a class of commutative non-associative algebras
generated by idempotent elements called axes such that their adjoint action is
semisimple and the relative eigenvectors satisfy a prescribed fusion law. Let
$R$ be a ring, $\\{\alpha,\beta\\}\subseteq R\setminus\\{0,1\\}$ and
$\alpha\neq\beta$. An axial algebra over $R$ is called of Monster type
$(\alpha,\beta)$ if it satisfies the fusion law $\mathcal{M}(\alpha,\beta)$
given in Table 1.

$\begin{array}[]{|c||c|c|c|c|}\hline\cr\star&1&0&\alpha&\beta\\\
\hline\cr\hline\cr 1&1&\emptyset&\alpha&\beta\\\ \hline\cr
0&\emptyset&0&\alpha&\beta\\\ \hline\cr\alpha&\alpha&\alpha&1,0&\beta\\\
\hline\cr\beta&\beta&\beta&\beta&1,0,\alpha\\\ \hline\cr\end{array}$ Table 1.
Fusion table $\mathcal{M}(\alpha,\beta)$

This means that the adjoint action of every axis has spectrum
$\\{1,0,\alpha,\beta\\}$ and, for any two eigenvectors $v_{\gamma}$,
$v_{\delta}$ with relative eigenvalues
$\gamma,\delta\in\\{1,0,\alpha,\beta\\}$, the product $v_{\gamma}\cdot
v_{\delta}$ is a sum of eigenvectors relative to eigenvalues contained in
$\gamma\star\delta$. This class was introduced by J. Hall, F. Rehren and S.
Shpectorov [7] in order to axiomatise some key features of important classes
of algebras, including the weight-2 components of OZ-type vertex operator
algebras, Jordan algebras and Matsuo algebras (see the introductions of [7],
[16] and [5]). They are also of particular interest for finite group theorists
as most of the finite simple groups, or their automorphism groups, can be
faithfully and effectively represented as automorphism groups of these
algebras.

The motivating example is the Griess algebra $V_{2}^{\sharp}$ which is a real
axial algebra of Monster type $(\frac{1}{4},\frac{1}{32})$ and coincides with
the weight-2 component of the Monster vertex operator algebra $V^{\sharp}$.
Here axes are associated to the involutions of type $2A$ in the Monster (i.e.
those having the double cover of the Baby Monster as centraliser). The
subalgebras of the Griess algebra which are generated by two axes were first
classified by S. Norton in [14] who showed that there are nine isomorphism
classes of such algebras, corresponding to the $9$ conjugacy classes in the
Monster group $M$ of the dihedral subgroups generated by the pairs of $2A$
involutions associated to the two generating axes. These algebras are labelled
$1A$, $2A$, $3A$, $4A$, $5A$, $6A$, $4B$, $2B$.

On the basis of an earlier work by M. Miyamoto [12], who observed that Ising
vectors in a vertex operator algebra of CFT-type satisfy the Monster fusion
law, in [17] S. Sakuma classified all $OZ$-type vertex operator algebras
generated by a pair of two Ising conformal vectors showing that, up to
rescaling, the isomorphism types of their weight-$2$ subspaces match precisely
the $9$ classes of Norton, now often called Norton-Sakuma algebras.

By extracting the relevant properties of these weight-$2$ subspaces, A. A.
Ivanov introduced in 2009 the concept of Majorana algebras [10], which are
real axial algebras of Monster type $(\frac{1}{4},\frac{1}{32})$ satisfying
some additional properties, in particular they are endowed with a positive
definite associative bilinear form. In 2010 A. Ivanov, D. Pasechnik, A. Seress
and S. Shpectorov obtained Norton’s classification within the axiomatic
context of Majorana algebras (see [11]). A further development was achieved by
J. Hall, F. Rehren and S. Shpectorov [7] who constructed a universal object
for primitive Frobenius axial algebras with a prescribed fusion law and
extended Norton-Sakuma theorem to 2-generated primitive Frobenius axial
algebras of Monster type $(\frac{1}{4},\frac{1}{32})$. Subsequently F. Rehren
[16, 15] addressed the general case dropping the assumption on the existence
of the bilinear form (which characterises Frobenius axial algebras) and
described a universal object for primitive axial algebras with a prescribed
fusion law. In the particular case of $2$-generated algebras of Monster type
$(\alpha,\beta)$ with $\alpha\not\in\\{2\beta,4\beta\\}$ and in characteristic
other than $2$, he produced a spanning set of $8$ elements and computed the
structure constants with respect to these elements. Finally he produced new
examples of $2$-generated primitive Frobenius axial algebras of Monster type.
This paper is part of a project of the authors aiming to classifying
$2$-generated primitive axial algebras of Monster type over fields. We start
by giving, for any positive integer $n$, a formal construction of a universal
$n$-generated primitive axial algebra mapping epimorphically onto every
$n$-generated axial algebra with a prescribed fusion law. We then focus on
$2$-generated primitive axial algebras of Monster type. We say that such an
algebra is symmetric if the map that swaps the two generating axes extends to
an automorphism of the entire algebra. All the algebras considered by Rehren
in [15, 16] are symmetric. In Section 4, we re-prove Rehren’s result on the
number of generators and get the following bound for symmetric algebras in the
case $\alpha=4\beta$.

######  Theorem 1.1.

Every $2$-generated primitive axial algebra of Monster type $(\alpha,\beta)$
over a field ${\mathbb{F}}$ of characteristic other than $2$ has dimension at
most $8$, provided $\alpha\not\in\\{2\beta,4\beta\\}$.

######  Theorem 1.2.

Every $2$-generated symmetric primitive axial algebra of Monster type
$(4\beta,\beta)$ over a field ${\mathbb{F}}$ of characteristic other than $2$
has dimension at most $8$, except possibly when
$(\alpha,\beta)=(2,\frac{1}{2})$.

The case $(\alpha,\beta)=(2,\frac{1}{2})$ is truly exceptional, as the
infinite dimensional $2$-generated symmetric primitive axial algebra of Moster
type $(2,\frac{1}{2})$ constructed in [4] shows. On the other hand, the same
bound $8$ holds for $2$-generated primitive axial algebras of Monster type
$(2\beta,\beta)$ over any ring in which $2$ and $\beta$ are invertible and it
is the best possible (see [3]).

In Section 5 we consider in more details the case when $\alpha-2\beta$ and
$\alpha-4\beta$ are invertible in the field ${\mathbb{F}}$, which we call the
generic case. Denote by ${\mathbb{F}}_{0}$ the prime subfield of
${\mathbb{F}}$ and let ${\mathbb{F}}_{0}(\alpha,\beta)[x,y,z,t]$ be the
polynomial ring in $4$ variables over ${\mathbb{F}}_{0}(\alpha,\beta)$. We
prove the following result.

######  Theorem 1.3.

There exists a subset $T\subseteq{\mathbb{F}}_{0}(\alpha,\beta)[x,y,z,t]$ of
size $4$, depending only on ${\mathbb{F}}_{0}$, $\alpha$, and $\beta$, such
that every $2$-generated primitive axial algebra of Monster type
$(\alpha,\beta)$ over a field ${\mathbb{F}}$ of characteristic other than $2$,
with $\alpha,\beta\in{\mathbb{F}}$ and $\alpha\not\in\\{2\beta,4\beta\\}$ is
completely determined, up to homomorphic images, by a quadruple
$(x_{0},y_{0},z_{0},t_{0})\in{\mathbb{F}}^{4}$ which is a common zero of all
the elements of $T$.

Using Theorem 1.3, we classify the algebras defined over the field
${\mathbb{Q}}(\alpha,\beta)$ with $\alpha$ and $\beta$ independent
indeterminates over ${\mathbb{Q}}$. We refer to [8] for the definition of the
algebras of type $1A$, $2B$, $3C(\eta)$, $\eta\in{\mathbb{F}}$. We denote by
$3A(\alpha,\beta)$ the algebra of dimension $4$ defined in [16, Table 9] for
$\alpha\neq\frac{1}{2}$.

######  Theorem 1.4.

Let $V$ be a $2$-generated primitive axial algebra of Monster type
$(\alpha,\beta)$ over the field ${\mathbb{Q}}(\alpha,\beta)$, with $\alpha$
and $\beta$ algebraically independent indeterminates over ${\mathbb{Q}}$. Then
we have one of the following:

  1. (1)

$V$ is the trivial algebra ${\mathbb{Q}}(\alpha,\beta)$ of type $1A$;

  2. (2)

$V$ is an algebra of type $2B$;

  3. (3)

$V$ is an algebra of Jordan type $\alpha$ of type $3C(\alpha)$;

  4. (4)

$V$ is an algebra of Jordan type $\beta$ of type $3C(\beta)$;

  5. (5)

$V$ is an algebra of dimension $4$ of type $3A(\alpha,\beta)$.

In particular, $V$ is symmetric.

Recall that the fusion law $\mathcal{M}(\alpha,\beta)$ admits a
${\mathbb{Z}}_{2}$-grading $\mathcal{M}(\alpha,\beta)_{+}=\\{1,0,\alpha\\}$
and $\mathcal{M}(\alpha,\beta)_{-}=\\{\beta\\}$ and this implies that every
axis induces an automorphism of the algebra called Miyamoto involution. The
Miyamoto group is the group generated by all Miyamoto involutions (see [9]).

######  Corollary 1.5.

Let $V$ be a primitive finitely generated axial algebra of Monster type
$(\alpha,\beta)$ over the field ${\mathbb{Q}}(\alpha,\beta)$, with $\alpha$
and $\beta$ independent indeterminates over ${\mathbb{Q}}$. Then the Miyamoto
group of $V$ is a group of $3$-transpositions.

Finally, as a consequence of Theorem 1.3, we get that the Norton-Sakuma
Theorem holds in general for $2$-generated primitive axial algebra of Monster
type $(\frac{1}{4},\frac{1}{32})$ over a field of characteristic zero, without
any assumption on the existence of a Frobenius form. This fact has been also
checked computationally in [19].

######  Theorem 1.6.

Every $2$-generated primitive axial algebra of Monster type
$(\frac{1}{4},\frac{1}{32})$ over a field of characteristic zero is a Norton-
Sakuma algebra.

Throughout this paper $R$ is a commutative associative ring with $1$.

While this paper was in preparation, Yabe posted a preprint in arXiv giving an
almost complete classification of the primitive symmetric $2$-generated axial
algebras of Monster type $(\alpha,\beta)$ [18].

##  2\. Primitive axial algebras

We begin with some basic results about endomorphisms of $R$-modules, which are
well known for vector spaces. The main difference is that, when considering
$R$-modules instead of vector spaces, it is no longer true in general that
eigenvectors relative to different eigenvalues are linearly independent.

Let $V$ be an $R$-module. For $\xi\in End(V)$, $\lambda\in R$, and
$\Gamma\subseteq R$, define

$V_{\lambda}^{\xi}:=\\{v\in V|\xi(v)=\lambda v\\}\mbox{ and
}V_{\Gamma}^{\xi}:=\sum_{\lambda\in\Gamma}V_{\lambda}^{\xi}.$

If $V$ is also an $R$-algebra and $a\in V$, denote by ${\rm ad}_{a}$ the
endomorphism of $V$ induced by multiplication by $a$:

$\begin{array}[]{rccc}{\rm ad}_{a}:&V&\to&V\\\ &x&\mapsto&ax\end{array}.$

In this case, we’ll write simply $V_{\lambda}^{a}$ and $V_{\Gamma}^{a}$
instead of $V_{\lambda}^{{\rm ad}_{a}}$ and $V_{\Gamma}^{{\rm ad}_{a}}$,
respectively.

Two elements $\alpha$ and $\beta$ of $R$ are called distinguishable if
$\alpha-\beta$ is a unit in $R$. In the remainder of this section we assume
that $\Gamma$ is a finite set of pairwise distinguishable elements of $R$.
Note that every nontrivial ring homomorphism maps sets of pairwise
distinguishable elements into sets of pairwise distinguishable elements. Let
$R[x]$ be the ring of polynomials over $R$ with indeterminate $x$.

######  Lemma 2.1.

If $g\in R[x]$ satisfies $g(\lambda)=0$ for every $\lambda\in\Gamma$ and $\deg
g<|\Gamma|$, then $g=0$.

###### Proof.

We proceed by induction on $|\Gamma|$. If $|\Gamma|=1$, then $g$ is a constant
and since the value of $g$ is zero in at least one element (the element of
$\Gamma$), it must be $g=0$. Suppose $|\Gamma|>1$. Assume by contradiction
that $g$ is not the zero polynomial and set $k:=\deg g$. Then $k<|\Gamma|$.
Let $\lambda\in\Gamma$, then, since $x-\lambda$ is monic, we have
$g=q\cdot(x-\lambda)+r$, with $r=g(\lambda)=0$, so $g=q\cdot(x-\lambda)$. Then
clearly $\deg q=k-1<|\Gamma^{\prime}|$, where
$\Gamma^{\prime}=\Gamma\setminus\\{\lambda\\}$. Also for
$\mu\in\Gamma^{\prime}$, $0=g(\mu)=q(\mu)\cdot(\mu-\lambda)$. Since $\mu$ and
$\lambda$ are distinguishable, $(\mu-\lambda)$ is a unit and so $q(\mu)=0$. By
induction this means that $q=0$, whence also $g=0$, a contradiction.∎

For $\mu\in\Gamma$, define

(1) $f_{\mu}:=\prod_{\lambda\in{\Gamma}\setminus\\{\mu\\}}(x-\lambda),$

and

(2) $f:=\prod_{\lambda\in{\Gamma}}(x-\lambda),$

clearly

$f=(x-\mu)f_{\mu}$

for every $\mu\in\Gamma$. Note that, since elements of $\Gamma$ are pairwise
distinguishable, $f_{\mu}(\mu)$ is a unit in $R$.

######  Corollary 2.2.

$\sum_{\mu\in\Gamma}\frac{1}{f_{\mu}(\mu)}f_{\mu}=1.$

###### Proof.

Define

$g:=\sum_{\mu\in\Gamma}\frac{1}{f_{\mu}(\mu)}f_{\mu}-1,$

then $\deg g<|\Gamma|$ and clearly $g(\lambda)=0$ for every
$\lambda\in\Gamma$. Hence, by Lemma 2.1, $g=0$. ∎

######  Lemma 2.3.

For every $\xi\in End(V)$, the following statements are equivalent:

  1. (1)

$V=\bigoplus_{\lambda\in\Gamma}V_{\lambda}^{\xi}$;

  2. (2)

$V=V_{\Gamma}^{\xi}$;

  3. (3)

$f(\xi)V=0$.

###### Proof.

Clearly (1) implies (2) and (2) implies (3). Suppose (3) is satisfied. Then,
for every $\mu\in\Gamma$, and $v\in V$,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(f(\xi))(v)=\left(\prod_{\lambda\in{\Gamma}}(\xi-\lambda)\right)(v)$
$\displaystyle=$
$\displaystyle(\xi-\mu)\left(\prod_{\lambda\in{\Gamma\setminus\\{\mu\\}}}(\xi-\lambda)\right)(v)$
$\displaystyle=$ $\displaystyle(\xi-\mu)((f_{\mu}(\xi))(v)),$

whence $(f_{\mu}(\xi))(v)\in V_{\mu}^{\xi}$. Set

$v_{\mu}:=\frac{1}{f_{\mu}(\mu)}(f_{\mu}(\xi))(v).$

By Corollary 2.2,

$id_{V}=\sum_{\mu\in\Gamma}\frac{1}{f_{\mu}(\mu)}f_{\mu}(\xi)$

and so

$v=\sum_{\mu\in\Gamma}\frac{1}{f_{\mu}(\mu)}(f_{\mu}(\xi))(v)=\sum_{\mu\in\Gamma}v_{\mu}.$

Now, assume $v=\sum_{\mu\in\Gamma}w_{\mu}$ for some $w_{\mu}\in
V_{\mu}^{\xi}$. Since

$\frac{1}{f_{\mu}(\mu)}(f_{\mu}(\xi))(w_{\lambda})=\delta_{\lambda\mu}w_{\mu}$

(where $\delta_{\lambda\mu}$ is the Kronecker delta), we get

$v_{\mu}=\frac{1}{f_{\mu}(\mu)}(f_{\mu}(\xi))(v)=\sum_{\mu\in\Gamma}\frac{1}{f_{\mu}(\mu)}(f_{\mu}(\xi))(w_{\lambda})=w_{\mu},$

giving (1). ∎

Recall [2] that a fusion law is a pair $(\mathcal{S},\ast)$ such that
$\mathcal{S}$ is a set and $\ast$ is a map from the cartesian product
${\mathcal{S}}\times{\mathcal{S}}$ to the power set $2^{\mathcal{S}}$. A
morphism between two fusion laws $(\mathcal{S}_{1},\ast_{1})$ and
$(\mathcal{S}_{2},\ast_{2})$ is a map

$\phi\colon{\mathcal{S}_{1}}\to{\mathcal{S}_{2}}$

such that, for $\alpha,\beta\in{\mathcal{S}_{1}}$,

$\phi(\alpha\ast_{1}\beta)\subseteq\phi(\alpha)\ast_{2}\phi(\beta).$

An isomorphism of fusion laws is a bijective morphism such that its inverse is
also a morphism. A fusion law $(\mathcal{S},\ast)$ is said to be finite if
${\mathcal{S}}$ is a finite set. In this paper we deal with fusion laws
$(\mathcal{S},\ast)$ where $\mathcal{S}$ is a finite set containing the
spectrum of the adjoint action of an idempotent element in an $R$-algebra.
Therefore, we assume $1_{R}\in\mathcal{S}\subseteq R$. Without loss of
generality, we may also assume that $0_{R}\in\mathcal{S}$. Further, for every
morphism $\phi$ of fusion laws, we’ll assume that $1^{\phi}=1$ and
$0^{\phi}=0$. More generally, when considering morphisms between fusion laws,
one may want to preserve some possible algebraic relations between the
elements of the set $\mathcal{S}$. To this aim, we call a morphism $\phi$ of
fusion laws an algebraic morphism if it is a ${\mathbb{Z}}$-linear map. An
axial algebra over $R$ with generating set $\mathcal{A}$ and fusion law
$(\mathcal{S},\star)$ is a quadruple

$(R,V,\mathcal{A},(\mathcal{S},\star))$

such that

  1. (1)

$R$ is an associative commutative ring with identity $1$;

  2. (2)

$\mathcal{S}$ is a subset of $R$ containing $1$ and $0$;

  3. (3)

$V$ is a commutative non associative $R$-algebra;

  4. (4)

$\mathcal{A}$ is a set of idempotent elements (called axes) of $V$ that
generate $V$ as an $R$-algebra and such that

  1. Ax1

$V=V_{\mathcal{S}}^{a}$ for every $a\in{\mathcal{A}}$;

  2. Ax2

$V_{\lambda}^{a}V_{\mu}^{a}\subseteq V_{\lambda\star\mu}^{a}$ for every
$\lambda,\mu\in\mathcal{S}$ and $a\in\mathcal{A}$.

Further, $V$ is called primitive if,

  1. Ax3

$V_{1}^{a}=Ra$ for every $a\in{\mathcal{A}}$.

A Frobenius axial algebra is an axial algebra
$(R,V,\mathcal{A},(\mathcal{S},\star))$ endowed with an associative bilinear
form $\kappa:V\times V\to R$ such that the map $a\mapsto\kappa(a,a)$ is
constant on the set of axes.

Let $(R,V,\mathcal{A},(\mathcal{S},\star))$ be an axial algebra and assume the
elements of $\mathcal{S}$ are pairwise distinguishable. As in the proof of
Lemma 2.3, for every $v\in V$, denote by $v_{1}$ the projection of $v$ into
$V_{1}^{a}$ with respect to the decomposition of $V$ into
$ad_{a}$-eigenspaces. If $V$ is primitive, we have

$v_{1}=\lambda_{a}(v)a$

for some $\lambda_{a}(v)\in R$ which is generally not unique. On the other
hand, if the annihilator ideal

$Ann_{R}(a):=\\{r\in R|ra=0\\}$

of $a$ in $R$ is trivial, then $\lambda_{a}(v)$ is unique, and we say that $a$
is a free axis. Clearly this condition is satisfied when $R$ is a field. As an
immediate consequence we have the main result of this section.

######  Proposition 2.4.

Let $(R,V,\mathcal{A},(\mathcal{S},\star))$ be a primitive axial algebra and
assume that the elements of $\mathcal{S}$ are pairwise distinguishable and the
axes in $\mathcal{A}$ are free. Then, for every $a\in\mathcal{A}$, there is a
well defined $R$-linear map

(3) $\begin{array}[]{rcccc}\lambda_{a}&\colon&V&\to&R\\\
&&v&\mapsto&\lambda_{a}(v)\end{array}$

such that every $v\in V$ decomposes uniquely as

(4) $v=\lambda_{a}(v)a+\sum_{\mu\in\mathcal{S}\setminus\\{1\\}}v_{\mu},$

with $v_{\mu}\in V_{\mu}^{a}$.

We say that $V$ is weak primitive if, for every $a\in\mathcal{A}$ and every
element $v\in V$, $v$ can be decomposed as

(5) $v=l_{a}(v)a+\sum_{\mu\in\mathcal{S}\setminus\\{1\\}}v_{\mu}$

where $l_{a}(v)\in R$ depends on $v$ and $a$, and $v_{\mu}\in V^{a}_{\mu}$.
Note that in general the decomposition in (5) and $l_{a}(v)$ are not uniquely
determined by $v$.

######  Lemma 2.5.

If the elements of $\mathcal{S}$ are pairwise distinguishable, in particular,
if $R$ is a field, then weak primitivity is equivalent to primitivity.

###### Proof.

Trivially, primitivity implies weak primitivity. Conversely, suppose that $V$
is weak primitive, fix $a\in\mathcal{A}$ and let $v_{1}\in V^{a}_{1}$. Then,
by weak primitivity, there exist $l\in R$, $v_{\mu}\in V^{a}_{\mu}$, such that

$v_{1}=la+\sum_{\mu\in\mathcal{S}\setminus\\{1\\}}v_{\mu}.$

Hence

$\sum_{\mu\in\mathcal{S}\setminus\\{1\\}}v_{\mu}=v_{1}-la\in(\sum_{\mu\in\mathcal{S}\setminus\\{1\\}}V^{a}_{\mu})\cap
V^{a}_{1},$

and the last intersection is trivial by Lemma 2.3. Thus $v_{1}=la\in Ra$. ∎

We conclude this section with the following straightforward observation.

######  Lemma 2.6.

Let $(R,V,\mathcal{A},(\mathcal{S},\star))$ be a primitive axial algebra and
assume that the elements of $\mathcal{S}$ are pairwise distinguishable, and
the axes in $\mathcal{A}$ are free. Then, for every $a\in\mathcal{A}$,
$\gamma,\delta\in\mathcal{S}$, and $v,w\in V$, we have

$(f_{1}({\rm ad}_{a}))(w-\lambda_{a}(w)a)=0$

and

$\left(\prod_{\eta\in\gamma\star\delta}({\rm
ad}_{c}-\eta)\right)\left((f_{\gamma}(ad_{c}))(v)\cdot(f_{\delta}(ad_{c}))(w)\right)=0.$

##  3\. Universal primitive axial algebras

In this section we fix a positive integer $k$, a fusion law
$(\mathcal{S}_{0},\ast_{0})$, and denote by ${\mathcal{O}}$ the class whose
objects are the primitive axial algebras

$(R,V,{\mathcal{A}},(\mathcal{S},\ast))$

such that

  * O1

${\mathcal{A}}$ has size at most $k$ and its elements are free axes,

  * O2

$(\mathcal{S},\ast)$ is isomorphic to $(\mathcal{S}_{0},\ast_{0})$, and

  * O3

the elements of $\mathcal{S}$ are pairwise distinguishable in $R$.

For two elements

(6)
${\mathcal{V}_{1}}:=(R_{1},V_{1},{\mathcal{A}}_{1},(\mathcal{S}_{1},\ast_{1}))\mbox{
and
}{\mathcal{V}}_{2}:=(R_{2},V_{2},{\mathcal{A}}_{2},(\mathcal{S}_{2},\ast_{2}))$

in ${\mathcal{O}}$, let ${\mathcal{H}om}(\mathcal{V}_{1},{\mathcal{V}}_{2})$
be the set of maps

$\phi\colon R_{1}\cup V_{1}\to R_{2}\cup V_{2}$

satisfying the following conditions:

  1. H1

$\phi_{|{R_{1}}}$ is an homomorphism of rings with identity between $R_{1}$
and $R_{2}$ that induces by restriction an isomorphism of fusion laws between
$({\mathcal{S}}_{1},\ast_{1})$ and $({\mathcal{S}}_{2},\ast_{2})$;

  2. H2

$\phi_{|{V_{1}}}$ is a (non-associative) ring homomorphism between $V_{1}$ and
$V_{2}$ such that $\phi({\mathcal{A}}_{1})\subseteq{\mathcal{A}}_{2}$;

  3. H3

$(\gamma v)^{\phi}=\gamma^{\phi}v^{\phi}$, for every $\gamma\in R_{1}$ and
$v\in V_{1}$.

Note that, since $\phi_{|{R_{1}}}$ is a ring homomorphism, the induced
isomorphism of fusion law in (H1) is in fact an algebraic isomorphism. Denote
by $\mathcal{H}$ the class of all
$\phi\in{\mathcal{H}om}(\mathcal{V}_{1},{\mathcal{V}}_{2})$ where
${\mathcal{V}_{1}}$ and ${\mathcal{V}_{2}}$ range in $\mathcal{O}$. Then
clearly $({\mathcal{O}},{\mathcal{H}})$ is a category.

It will turn convenient to define some subcategories of
$({\mathcal{O}},{\mathcal{H}})$ in the following way. Let

  * -

$n:=|\mathcal{S}\setminus\\{0,1\\}|$;

  * -

$x_{i},y_{i},w_{i},z_{i,j},t_{h}$ be algebraically independent indeterminates
over ${\mathbb{Z}}$, for $i,j\in\\{1,\ldots,n\\}$, with $i<j$,
$h\in{\mathbb{N}}$;

  * -

$D$ be the polynomial ring

${{\mathbb{Z}}}[x_{i},y_{i},w_{i},z_{i,j},t_{h}\>|\>i,j\in\\{1,\ldots
n\\},i<j,h\in{\mathbb{N}}];$

  * -

$L$ be a proper ideal of $D$ containing the set

$\Sigma:=\\{x_{i}y_{i}-1,\>\>(1-x_{i})w_{i}-1,\mbox{ and
}\>(x_{i}-x_{j})z_{i,j}-1\mbox{ for all }1\leq i<j\leq n\\};$

  * -

$\hat{D}:=D/L$. For $d\in D$, we denote the element $L+d$ by $\hat{d}$.

The extra indeterminates $t_{h}$ in the definition of $D$ have been introduced
here in order to guarantee, when necessary, the invertibility of certain
elements in the ring $\hat{D}$.

Since $L$ is proper and, for $1\leq i<j\leq n$, the elements
$\hat{x}_{i}-\hat{x}_{j}$ are invertible in $\hat{D}$, the elements
$\hat{x}_{1},\ldots,\hat{x}_{n}$ are still pairwise distinguishable in
${\hat{D}}$. Define $({\mathcal{O}}_{L},{\mathcal{H}}_{L})$ as the full
subcategory of $({{\mathcal{O}},{\mathcal{H}}})$ whose objects are the
primitive axial algebras
$(R,V,{\mathcal{A}},(\mathcal{S},\ast))\in{\mathcal{O}}$ that satisfy the
further condition

  * O4

$R$ is a ring containing a subring isomorphic to a factor of $\hat{D}$.

Clearly, if $(\Sigma)$ is the ideal of $D$ generated by the set $\Sigma$, then
${\mathcal{O}}_{(\Sigma)}={\mathcal{O}}$.

######  Lemma 3.1.

Let $\mathcal{V}_{1},{\mathcal{V}}_{2}\in\mathcal{O}_{L}$ be as in Equation
(6) and let $\phi\in{\mathcal{H}om}(\mathcal{V}_{1},{\mathcal{V}}_{2})$. Then,
for every $a\in\mathcal{A}_{1},v\in V_{1}$, we have

$\lambda_{a^{\phi}}(v^{\phi})=(\lambda_{a}(v))^{\phi}.$

###### Proof.

Since $\mathcal{V}_{1},{\mathcal{V}}_{2}$ are primitive axial algebras, the
result follows applying $\phi$ to the decomposition of $v$ in Equation (4). ∎

We construct a universal object for each category
$(\mathcal{O}_{L},{\mathcal{H}}_{L})$ as follows. Let

  * -

$\mathcal{A}$ be a set of size $k$;

  * -

$W$ be the free commutative magma generated by the elements of $\mathcal{A}$
subject to the condition that every element of $\mathcal{A}$ is idempotent;

  * -

${\hat{R}}:={\hat{D}}[\Lambda]$ be the ring of polynomials with coefficients
in $\hat{D}$ and indeterminates set
$\Lambda:=\\{\lambda_{c,w}\>|\>c\in\mathcal{A},w\in W,c\neq w\\}$ where
$\lambda_{c,w}=\lambda_{c^{\prime},w^{\prime}}$ if and only if $c=c^{\prime}$
and $w=w^{\prime}$.

  * -

${\hat{V}}:={\hat{R}}[W]$ be the set of all formal linear combinations
$\sum_{w\in W}\gamma_{w}w$ of the elements of $W$ with coefficients in
${\hat{R}}$, with only finitely many coefficients different from zero. Endow
${\hat{V}}$ with the usual structure of a commutative non associative
${\hat{R}}$-algebra;

  * -

${\hat{\mathcal{S}}}$ be the set $\\{1,0,\hat{x}_{1},\ldots,\hat{x}_{n}\\}$.

Let $\star\colon{\hat{\mathcal{S}}}\times{\hat{\mathcal{S}}}\to
2^{{\hat{\mathcal{S}}}}$ be a map such that $({\hat{\mathcal{S}}},\star)$ is a
fusion law isomorphic to $(\mathcal{S},\ast)$. Since, obviously, a fusion law
is isomorphic to $(S,\ast)$ if and only if it is isomorphic to
$({\hat{\mathcal{S}}},\star)$, we may assume
$(\mathcal{S},\ast)=({\hat{\mathcal{S}}},\star)$. For
$\mu\in\hat{\mathcal{S}}$, let $f_{\mu}$ be the polynomial defined in Equation
(1), for every $c\in\mathcal{A}$, let $\lambda_{c,c}:=1$, and let $J$ be the
ideal of ${\hat{V}}$ generated by all the elements

(7) $(f_{1}({\rm ad}_{c}))(w-\lambda_{c,w}c)\>\>\>\>\mbox{ for all
}c\in\mathcal{A}\mbox{ and }w\in W$

and

(8) $\left(\prod_{\eta\in\gamma\star\delta}({\rm ad}_{c}-\eta\>{\rm
id}_{\hat{V}})\right)\left((f_{\gamma}(ad_{c}))(v)\cdot(f_{\delta}(ad_{c}))(w)\right)$

for all $v,w\in{\hat{V}}$, $\gamma,\delta\in{\hat{\mathcal{S}}}$,
$c\in\mathcal{A}$. Let $I_{0}$ be the ideal of ${\hat{R}}$ generated by all
the elements

(9) $\sum_{w\in W}\gamma_{w}\lambda_{c,w}\>\>\mbox{ for all
}c\in\mathcal{A},w\in W,\gamma_{w}\in\hat{R}\>\>\mbox{ such that }\sum_{w\in
W}\gamma_{w}w\in J.$

Finally, set

  * -

$\mathcal{A}/J:=\\{c+J\>|\>c\in\mathcal{A}\\}$

  * -

$J_{0}:=J+I_{0}{\hat{V}}$,

  * -

${\overline{R}}:={\hat{R}}/I_{0}$,

  * -

${\overline{V}}:={\hat{V}}/J_{0}$,

  * -

$\alpha_{i}:=x_{i}+I_{0},\mbox{ for }i\in\\{1,\ldots,n\\}$

  * -

$\bar{c}:=c+J_{0},\mbox{ for }c\in\mathcal{A}$,

  * -

${\overline{\mathcal{A}}}:=\\{\bar{c}\>|\>c\in\mathcal{A}\\}$,

  * -

$\overline{\mathcal{S}}:=\\{1+I_{0},0+I_{0},\alpha_{i}\>|\>i\in\\{1,\ldots,n\\}\\}$.

######  Remark 3.2.

Since ${\hat{D}}\cap I_{0}=\\{0\\}$ and
${\overline{\mathcal{S}}}\leq{\hat{D}}I_{0}/I_{0}$,
$(\overline{\mathcal{S}},\overline{\star})$ is a fusion law isomorphic to
$(\mathcal{S},\ast)$ and the elements of $\overline{\mathcal{S}}$ are pairwise
distinguishable.

######  Lemma 3.3.

$({\hat{R}},{\hat{V}}/J,\mathcal{A}/J,({\hat{\mathcal{S}}},\star))$ and
$({\hat{R}},{\overline{V}},{\overline{\mathcal{A}}},({\hat{\mathcal{S}}},\star))$
are primitive axial algebras such that, for $J_{\ast}\in\\{J,J_{0}\\}$,

$\lambda_{c+J_{\ast}}(w+J_{\ast})=\lambda_{c,w}$

for every $c\in\mathcal{A}$ and $w\in W$.

###### Proof.

Clearly ${\hat{V}}/J$ is generated by $\mathcal{A}/J$ and ${\overline{V}}$ is
generated by ${\overline{\mathcal{A}}}$. Let $J_{\ast}\in\\{J,J_{0}\\}$ and
let $f$ be as in Equation (2). By the definition of $J$, for every
$a\in\mathcal{A}$, $f({\rm ad}_{a}){\hat{V}}\subseteq J$, whence, by Lemma
2.3, ${\hat{V}}/J_{\ast}$ satisfies condition Ax1. By Equation (8),
${\hat{V}}/J_{\ast}$ also satisfies the fusion law
$({\hat{\mathcal{S}}},\star)$. Furthermore, for every $c\in\mathcal{A}$ and
$w\in W$, since $f_{1}({\rm ad}_{c})(w-\lambda_{c,w}c)\in J_{\ast}$, we get

$w+J_{\ast}=(\lambda_{c,w}c+\sum_{\mu\in{\hat{\mathcal{S}}}\setminus\\{1\\}}w_{\mu})+J_{\ast},$

where $w_{\mu}+J_{\ast}$ is a $\mu$-eigenvector for ${\rm ad}_{c+J_{\ast}}$,
so ${\hat{V}}/J_{\ast}$ is also weak primitive. By Remark 3.2 and Lemma 2.5,
${\hat{V}}/J_{\ast}$ is also primitive. ∎

######  Lemma 3.4.

Let ${\mathcal{V}}:=(R,V,{\mathcal{A}},(\mathcal{S},\ast))$ be an element of
$\mathcal{O}_{L}$, let $\phi\colon{\hat{R}}\cup{\hat{V}}\to R\cup V$ be a map
that satisfies conditions (H1) (with respect to $({\hat{\mathcal{S}}},\star)$
and $(\mathcal{S},\ast)$), (H2), and (H3). Then
$I_{0}\subseteq\ker\phi_{|_{{\hat{R}}}}$,
$J_{0}\subseteq\ker\phi_{|_{{\hat{V}}}}$.

###### Proof.

By Lemma 2.6, $J\subseteq\ker\phi_{|_{{\hat{V}}}}$, so $\phi$ induces an
${\hat{R}}$-algebra homomorphism

$\begin{array}[]{clcl}\phi_{{\hat{V}}/J}:&{\hat{V}}/J&\to&V,\end{array}$
$\>\>\>\>\>\>\>\>\>\>\>\>\>\begin{array}[]{clcl}&v+J&\mapsto&v^{\phi}.\end{array}$

Since, by Lemma 3.3, ${\hat{V}}/J$ is a primitive axial algebra over the ring
${\hat{R}}$, for every $c\in\mathcal{A}$ and $w\in W$, we can write

$w+J=(\lambda_{c,w}c+\sum_{\mu\in{\hat{\mathcal{S}}}\setminus\\{1\\}}w_{\mu})+J,$

where, for every $\mu\in{\hat{\mathcal{S}}}\setminus\\{1\\}$, $w_{\mu}+J$ is a
$\mu$-eigenvector for ${\rm ad}_{c+J}$. Condition (H3) implies that, for every
$\mu\in\mathcal{S}$, $a\in\mathcal{A}$, $\phi_{{\hat{V}}/J}$ maps
$\mu$-eigenvectors for ${\rm ad}_{a+J}$ to $\mu^{\phi}$-eigenvectors for ${\rm
ad}_{a^{\phi}}$. Thus, if $v=\sum_{w\in W}\gamma_{w}w\in J$, then
$v^{\phi}=0$, whence

$\displaystyle 0=\lambda_{a^{\phi}}(v^{\phi})$ $\displaystyle=$
$\displaystyle\lambda_{a^{\phi}}\left(\sum_{w\in
W}\gamma_{w}^{\phi}w^{\phi}\right)$ $\displaystyle=$ $\displaystyle\sum_{w\in
W}\gamma_{w}^{\phi}\lambda_{a^{\phi}}(w^{\phi})$ $\displaystyle=$
$\displaystyle\sum_{w\in
W}\gamma_{w}^{\phi}\lambda_{a^{\phi}}((\lambda_{a,w}a+\sum_{\mu\in{\hat{\mathcal{S}}}\setminus\\{1\\}}w_{\mu})^{\phi})$
$\displaystyle=$ $\displaystyle\sum_{w\in
W}\gamma_{w}^{\phi}\lambda_{a^{\phi}}((\lambda_{a,w})^{\phi}a^{\phi})+\sum_{w\in
W}\gamma_{w}^{\phi}\lambda_{a^{\phi}}(\sum_{\mu\in{\hat{\mathcal{S}}}\setminus\\{1\\}}w_{\mu}^{\phi}))$
$\displaystyle=$ $\displaystyle\sum_{w\in
W}\gamma_{w}^{\phi}(\lambda_{a,w})^{\phi}=(\sum_{w\in
W}\gamma_{w}\lambda_{a,w})^{\phi}.$

Thus $I_{0}\subseteq\ker\phi_{|_{{\hat{R}}}}$. Finally, by condition (H3),
$(I_{0}{\hat{V}})^{\phi}=I_{0}^{\phi}V^{\phi}=0V_{1}=0$, whence
$J_{0}\subseteq\ker\phi_{|_{{\hat{V}}}}$. ∎

######  Lemma 3.5.

We have $J_{0}\neq{\hat{V}}$, in particular $|{\overline{\mathcal{A}}}|=k$.

###### Proof.

Let ${\overline{R}}^{k}$ be the direct sum of $k$ copies of ${\overline{R}}$.
Set $\mathcal{B}:=\\{e_{1},\ldots,e_{k}\\}$, where $(e_{1},\ldots,e_{k})$ is
the canonical basis of ${\overline{R}}^{k}$. Then, for every
$i\in\\{1,\ldots,k\\}$, $e_{i}$ is an idempotent and

${\overline{R}}^{k}={\overline{R}}e_{i}\oplus\ker{\rm ad}_{e_{i}}.$

Therefore, for every fusion law
$(\mathcal{S}_{\overline{R}},\ast_{\overline{R}})$, the $4$-tuple

$({\overline{R}},{\overline{R}}^{k},\mathcal{B},(\mathcal{S}_{\overline{R}},\ast_{\overline{R}}))$

is obviously a primitive (associative) axial algebra. By the construction of
${\hat{V}}$, any bijection from $\mathcal{A}$ to $\mathcal{B}$ extends
uniquely to a map $\phi_{{\hat{V}}}\colon{\hat{V}}\to{\overline{R}}^{k}$. Let

$\phi\colon{\hat{R}}\cup{\hat{V}}\to{\overline{R}}\cup{\overline{R}}^{k}$

be the map whose restrictions to ${\hat{R}}$ and ${\hat{V}}$ are the canonical
projection on ${\overline{R}}$ and $\phi_{{\hat{V}}}$, respectively. Then
$\phi$ satisfies the conditions (H1), (H2), and (H3). Therefore, by Lemma 3.4,
$J_{0}\subseteq\ker\phi_{|_{{\hat{V}}}}\neq{\hat{V}}$. Since
$k=|\mathcal{A}|\geq|{\overline{\mathcal{A}}}|\geq|\mathcal{B}|=k$, we get
$|{\overline{\mathcal{A}}}|=k$. ∎

######  Theorem 3.6.

$\overline{\mathcal{V}}:=({\overline{R}},{\overline{V}},{\overline{\mathcal{A}}},(\overline{\mathcal{S}},\overline{\star}))$
is a universal object in the category $({\mathcal{O}_{L}},{\mathcal{H}_{L}})$.

###### Proof.

Clearly ${\overline{V}}$ is an algebra over ${\overline{R}}$ generated by the
set of idempotents ${\overline{\mathcal{A}}}$. Since by Lemma 3.3,
$({\hat{R}},{\overline{V}},{\overline{\mathcal{A}}},(\hat{\mathcal{S}},\overline{\star}))$
is a primitive axial algebra, and $I_{0}\subseteq
Ann_{\hat{R}}({\overline{V}})$, we get that
$({\overline{R}},{\overline{V}},{\overline{\mathcal{A}}},(\overline{\mathcal{S}},\star))$
is a primitive axial algebra. Finally, the axes in ${\overline{\mathcal{A}}}$
are free, for, if $c\in\mathcal{A}$ and $r\in{\hat{R}}$ are such that $rc\in
J_{0}$, then there exist $j\in J,i_{0}\in I_{0}$ and $\sum_{w\in
W}\gamma_{w}w\in{\hat{V}}$ such that

$rc-i_{0}\left(\sum_{w\in W}\gamma_{w}w\right)=j,$

whence, by the definition of $I_{0}$,

$r\in i_{0}\left(\sum_{w\in W}\gamma_{w}\lambda_{c,w}\right)+I_{0}=I_{0}.$

By Lemma 3.5, the elements of ${\hat{\mathcal{S}}}$ are pairwise
distinguishable in ${\overline{R}}$, whence
$\overline{\mathcal{V}}\in\mathcal{O}_{L}$.

Now assume
$\mathcal{V}_{1}:=(R_{1},V_{1},\mathcal{A}_{1},(\mathcal{S}_{1},\ast_{1}))$ is
an object in $\mathcal{O}_{L}$ and let
$\bar{t}:{{\overline{\mathcal{A}}}}\to\mathcal{A}_{1}$ a map of sets. Since
every non-empty subset of $\mathcal{A}_{1}$ generates a primitive axial
subalgebra of $V_{1}$ with fusion law $(\mathcal{S}_{1},\ast_{1})$ and free
axes, without loss of generality we may assume that $\bar{t}$ is surjective.
Let $t$ be the composition of $\bar{t}$ with the (bijective) projection of
$\mathcal{A}$ to ${\overline{\mathcal{A}}}$. Since $W$ is the free commutative
magma generated by the set of idempotents $\mathcal{A}$ there is a unique
magma homomorphism

$\chi:W\to V_{1},$

inducing the map $t:\mathcal{A}\to\mathcal{A}_{1}$. Since the elements of
$\Lambda$ are alegbraically independent over $\hat{D}$, there is a unique
homomorphism of $\hat{D}$-algebras

$\hat{\psi}:{\hat{R}}\to R_{1}$

such that, for $c\in\mathcal{A}$ and $w\in W\setminus\\{c\\}$,

(10) $\lambda_{c,w}^{\hat{\psi}}=\lambda_{c^{t}}(w^{\chi}),$

where $\lambda_{c^{t}}$ is the function defined in Proposition 2.4. Define

$\begin{array}[]{rcccc}\hat{\chi}&:&{\hat{V}}&\to&V_{1}\\\ &&\sum_{w\in
W}\gamma_{w}w&\mapsto&\sum_{w\in
W}\gamma_{w}^{\hat{\psi}}\>w^{\chi}\end{array}.$

Then $\hat{\chi}$ is a ring homomorphism extending $t$ and such that $(\gamma
v)^{\hat{\chi}}=\gamma^{\hat{\psi}}v^{\hat{\chi}}$ for every
$\gamma\in{\hat{R}}$ and $v\in{\hat{V}}$. Since, for every $c\in\mathcal{A}$,
$v\in{\hat{V}}$, and $\gamma\in{\mathcal{S}}$,

$[({\rm ad}_{c}-\gamma\>{\rm id}_{\hat{V}})(v)]^{\hat{\chi}}=({\rm
ad}_{c^{t}}-\gamma^{\hat{\psi}}\>{\rm id}_{V_{1}})(v^{\hat{\chi}}),$

by Lemma 2.6, it follows that $J_{0}$ is contained in $\ker\hat{\chi}$ and so
$\hat{\chi}$ induces a ring homomorphism

$\begin{array}[]{rccc}\phi_{V}:&{\overline{V}}&\to&V_{1}\end{array}$

extending $\overline{t}$ and such that
$(\gamma\overline{w})^{\phi_{V}}=\gamma^{\hat{\psi}}{\overline{w}}^{\phi_{V}}$
for every $\gamma\in{\hat{R}}$ and ${\overline{w}}\in{\overline{V}}$. As in
the proof of Lemma 3.4 we get that $I_{0}\subseteq\ker\hat{\psi}$. Let
$\phi_{R}:{\overline{R}}\to R_{1}$ be the homomorphism of $\hat{D}$-algebras
induced by $\hat{\psi}$. Then
$(\phi_{R},\phi_{V})\in{\mathcal{H}om}(\overline{\mathcal{V}},\mathcal{V}_{1})$
.

Since ${\hat{R}}=\hat{D}[\Lambda]$, $\phi_{R}$ is completely determined by its
values on the elements $\lambda_{c,w}+I_{0}$. Further, by Equation (10), for
every $c\in\mathcal{A}$, $w\in W\setminus\\{c\\}$,

$(\lambda_{c,w}+I_{0})^{\phi_{R}}=\lambda_{c^{t}}((w+J_{0})^{\phi_{V}}),$

whence $\phi_{R}$ is completely determined by the images
$(w+J_{0})^{\phi_{V}}$, with $w\in W$. Since $\phi_{V}$ is a ring homomorphism
extending $t$, such images are uniquely determined, whence the uniqueness of
$(\phi_{R},\phi_{V})$. ∎

######  Corollary 3.7.

Every permutation $\sigma$ of the set ${\overline{\mathcal{A}}}$ extends to a
unique automorphism
$f_{\sigma}\in{\mathcal{H}om}({\overline{\mathcal{V}}},{\overline{\mathcal{V}}})$.

Note that, for a generic object $\mathcal{V}$, the above assertion is false
(see e.g. the algebra $Q_{2}(\eta)$ constructed in [5, Section 5.3]). We say
that $\mathcal{V}=(R,V,\mathcal{A},(\mathcal{S},\ast))\in\mathcal{O}_{L}$ is
symmetric if every permutation $\sigma$ of the set $\mathcal{A}$ extends to a
unique automorphism $f_{\sigma}\in{\mathcal{H}om}(\mathcal{V},\mathcal{V})$.

######  Corollary 3.8.

Let $\mathcal{V}:=(R,V,\mathcal{A},(\mathcal{S},\ast))\in{\mathcal{O}}_{L}$,
then

  1. (1)

${\overline{\mathcal{V}}}\otimes
R:=({\overline{R}}\otimes_{\hat{D}}R,{\overline{V}}\otimes_{\hat{D}}R,{\overline{\mathcal{A}}}\otimes_{\hat{D}}1,(\overline{\mathcal{S}}\otimes_{\hat{D}}1,\overline{\star}))\in{\mathcal{O}}_{L}$,

  2. (2)

$R$ is isomorphic to a factor of ${\overline{R}}\otimes_{\hat{D}}R$,

  3. (3)

$V$ is isomorphic to a factor of ${\overline{V}}\otimes_{\hat{D}}R$.

######  Remark 3.9.

Note that, with the notation of Corollaries 3.7 and 3.8, $f_{\sigma}\otimes
id_{R}$ is an automorphism of ${\overline{\mathcal{V}}}\otimes R$.

Questions

  1. (1)

Can we define a variety of axial algebras corresponding to the fusion law
$(\mathcal{S},\ast)$?

  2. (2)

Is it true that any ideal $I$ of ${\hat{R}}$ containing $I_{0}$ defines an
axial algebra?

##  4\. $2$-generated primitive axial algebras of Monster type
$(\alpha,\beta)$

In this section we keep the notation of Section 3, with $k=n=2$,
$(\mathcal{S},\ast)$ equal to the Monster fusion law
$\mathcal{M}(\alpha,\beta)$, and $L$ an ideal of $D$ containing $2t_{1}-1$, so
that (the class of) $2$ is invertible in $\hat{D}$. In order to simplify
notation we’ll also identify $\alpha_{1}$ with $\alpha$ and $\alpha_{2}$ with
$\beta$.

Let $\mathcal{V}=(V,R,\mathcal{A},(\mathcal{S},\ast))\in{\mathcal{O}}_{L}$ and
$a\in\mathcal{A}$. Let ${\mathcal{S}}^{+}:=\\{1,0,\alpha\\}$ and
${\mathcal{S}}^{-}:=\\{\beta\\}$. The partition
$\\{{\mathcal{S}}^{+},{\mathcal{S}^{-}}\\}$ of $\mathcal{S}$ induces a
${\mathbb{Z}}_{2}$-grading on ${\mathcal{S}}$ which, on turn, induces, a
${\mathbb{Z}}_{2}$-grading $\\{V_{+}^{a},V_{-}^{a}\\}$ on $V$ where
$V_{+}^{a}:=V_{1}^{a}+V_{0}^{a}+V_{\alpha}^{a}$ and $V_{-}^{a}=V_{\beta}^{a}$.
It follows that, if $\tau_{a}$ is the map from $R\cup V$ to $R\cup V$ such
that $\tau_{a|_{V}}$ inverts $V_{\beta}^{a}$ and leaves invariant the elements
of $V_{+}^{a}$ and $\tau_{a|_{R}}$ is the identity, then $\tau_{a}$ is an
involutory automorphism of $\mathcal{V}$ (see [7, Proposition 3.4]). The map
$\tau_{a}$ is called the Miyamoto involution associated to the axis $a$. By
definition of $\tau_{a}$, the element $av-{\beta}v$ of $V$ is
$\tau_{a}$-invariant and, since $a$ lies in $V_{+}^{a}\leq C_{V}(\tau_{a})$,
also $av-{\beta}(a+v)$ is $\tau_{a}$-invariant. In particular, by symmetry,

######  Lemma 4.1.

Let $a$ and $b$ be axes of $V$. Then $ab-\beta(a+b)$ is fixed by the
2-generated group $\langle\tau_{a},\tau_{b}\rangle$.

Let $\mathcal{A}:=\\{a_{0},a_{1}\\}$ and, for $i\in\\{1,2\\}$, let $\tau_{i}$
be the Miyamoto involutions associated to $a_{i}$. Set
$\rho:=\tau_{0}\tau_{1}$, and for $i\in{\mathbb{Z}}$,
$a_{2i}:=a_{0}^{\rho^{i}}$ and $a_{2i+1}:=a_{1}^{\rho^{i}}$. Since $\rho$ is
an automorphism of $V$, for every $j\in{\mathbb{Z}}$, $a_{j}$ is an axis.
Denote by $\tau_{j}:=\tau_{a_{j}}$ the corresponding Miyamoto involution.

######  Lemma 4.2.

For every $n\in{\mathbb{N}}$, and $i,j\in{\mathbb{Z}}$ such that $i\equiv
j\>\bmod n$ we have

$a_{i}a_{i+n}-\beta(a_{i}+a_{i+n})=a_{j}a_{j+n}-\beta(a_{j}+a_{j+n}),$

###### Proof.

This follows immediately from Lemma 4.1. ∎

For $n\in{\mathbb{N}}$ and $r\in\\{0,\ldots,n-1\\}$ set

(11) $s_{r,n}:=a_{r}a_{r+n}-\beta(a_{r}+a_{r+n}).$

For every $a\in\mathcal{A}$, let $\lambda_{a}$ be as in Proposition 2.4.

######  Lemma 4.3.

For $i\in\\{1,2,3\\}$ we have

$\displaystyle a_{0}s_{0,i}$ $\displaystyle=$
$\displaystyle(\alpha-\beta)s_{0,i}+[(1-\alpha)\lambda_{a_{0}}(a_{i})+\beta(\alpha-\beta-1)]a_{0}+\frac{1}{2}\beta(\alpha-\beta)(a_{i}+a_{-i}).$

###### Proof.

This is [16, Lemma 3.1]. ∎

For $i\in\\{1,2,3\\}$, let

(12) $a_{i}=\lambda_{a_{0}}(a_{i})a_{0}+u_{i}+v_{i}+w_{i}$

be the decomposition of $a_{i}$ into $ad_{a_{0}}$-eigenvectors, where $u_{i}$
is a $0$-eigenvector, $v_{i}$ is an $\alpha$-eigenvector and $w_{i}$ is a
$\beta$-eigenvector.

######  Lemma 4.4.

With the above notation,

  1. (1)

$u_{i}=\frac{1}{\alpha}((\lambda_{a_{0}}(a_{i})-\beta-\alpha\lambda_{a_{0}}(a_{i}))a_{0}+\frac{1}{2}(\alpha-\beta)(a_{i}+a_{-i})-s_{0,i})$;

  2. (2)

$v_{i}=\frac{1}{\alpha}((\beta-\lambda_{a_{0}}(a_{i}))a_{0}+\frac{\beta}{2}(a_{i}+a_{-i})+s_{0,i})$;

  3. (3)

$w_{i}=\frac{1}{2}(a_{i}-a_{-i})$.

###### Proof.

(3) follows from the definitions of $\tau_{0}$ and $a_{i}$, (2) is just a
rearranging of $a_{0}a_{i}=\lambda_{a_{0}}(a_{i})a_{0}+\alpha v_{i}+\beta
w_{i}$ using Equation (11), and (1) follows rearranging Equation (12). ∎

For $i,j\in\\{1,2,3\\}$, set

$P_{ij}:=u_{i}u_{j}+u_{i}v_{j}\>\>\mbox{ and
}\>\>Q_{ij}:=u_{i}v_{j}-\frac{1}{\alpha^{2}}s_{0,i}s_{0,j}.$

######  Lemma 4.5.

For $i,j\in\\{1,2,3\\}$ we have

(13) $s_{0,i}\cdot s_{0,j}=\alpha(a_{0}P_{ij}-\alpha Q_{ij}).$

###### Proof.

Since $u_{i}$ and $v_{j}$ are a $0$-eigenvector and an $\alpha$-eigenvector
for ${\rm ad}_{a_{0}}$, respectively, by the fusion rule, we have
$a_{0}P_{ij}=\alpha(u_{i}\cdot v_{j})$ and the result follows. ∎

From now on we assume ${\mathcal{V}}={{\overline{\mathcal{V}}}}$. Let $f$ be
the automorphism of ${\overline{\mathcal{V}}}$ induced by the permutation that
swaps $a_{0}$ with $a_{1}$ as defined in Corollary 3.7. For $i\in{\mathbb{N}}$
define

(14) $\lambda_{i}:=\lambda_{a_{0}}(a_{i}).$

Note that, by Lemma 3.1, for every $i\in{\mathbb{N}}$, we have

$\lambda_{a_{0}}(a_{-i})=\lambda_{i},\>\>\lambda_{a_{1}}(a_{0})=\lambda_{1}^{f},\mbox{
and }\>\>\lambda_{a_{1}}(a_{-1})=\lambda_{2}^{f}.$

Set $T_{0}:=\langle\tau_{0},\tau_{1}\rangle$ and
$T:=\langle\tau_{0},f\rangle$.

######  Lemma 4.6.

The groups $T_{0}$ and $T$ are dihedral groups, $T_{0}$ is a normal subgroup
of $T$ such that $|T:T_{0}|\leq 2$. For every $n\in{\mathbb{N}}$, the set
$\\{s_{0,n},\ldots,s_{n-1,n}\\}$ is invariant under the action of $T$. In
particular, if $K_{n}$ is the kernel of this action, we have

  1. (1)

$K_{1}=T$;

  2. (2)

$K_{2}=T_{0}$, in particular $s_{0,2}^{f}=s_{1,2}$;

  3. (3)

$T/K_{3}$ induces the full permutation group on the set
$\\{s_{0,3},s_{1,3},s_{2,3}\\}$ with point stabilisers generated by
$\tau_{0}K_{3}$, $\tau_{1}K_{3}$ and $fK_{3}$, respectively. In particular
$s_{0,3}^{f}=s_{1,3}$ and $s_{0,3}^{\tau_{1}}=s_{2,3}$.

###### Proof.

This follows immediately from the definitions. ∎

######  Lemma 4.7.

In the algebra $\overline{\mathcal{V}}$, the following equalities hold:

$\displaystyle(\alpha-2\beta)a_{0}s_{1,2}=$
$\displaystyle\beta^{2}(\alpha-\beta)(a_{-2}+a_{2})$
$\displaystyle+\left[-2\alpha\beta\lambda_{1}+2\beta(1-\alpha)\lambda_{1}^{f}+\frac{\beta}{2}(4\alpha^{2}-2\alpha\beta-\alpha+4\beta^{2}-2\beta)\right](a_{1}+a_{-1})$
$\displaystyle+\frac{1}{(\alpha-\beta)}\left[(6\alpha^{2}-8\alpha\beta-2\alpha+4\beta)\lambda_{1}^{2}+(2\alpha^{2}-2\alpha)\lambda_{1}\lambda_{1}^{f}\right.$
$\displaystyle+2(-2\alpha^{2}-2\alpha\beta+\alpha)(\alpha-\beta)\lambda_{1}-4\beta(\alpha-1)(\alpha-\beta)\lambda_{1}^{f}-\alpha\beta(\alpha-\beta)\lambda_{2}$
$\displaystyle\left.+(4\alpha^{2}\beta-2\alpha\beta+2\beta^{3})(\alpha-\beta)\right]a_{0}$
$\displaystyle+\left[-4\alpha\lambda_{1}-4(\alpha-1)\lambda_{1}^{f}+(4\alpha^{2}-2\alpha\beta-\alpha+4\beta^{2}-2\beta)\right]s_{0,1}$
$\displaystyle+2\beta(\alpha-\beta)s_{0,2}$

and

$\displaystyle 4(\alpha-2\beta)s_{0,1}\cdot s_{0,1}=$
$\displaystyle\beta(\alpha-\beta)^{2}(\alpha-4\beta)(a_{-2}+a_{2})$
$\displaystyle+\left[4\alpha\beta(\alpha-\beta)\lambda_{1}+2(-\alpha^{3}+5\alpha^{2}\beta+\alpha^{2}-4\alpha\beta^{2}-5\alpha\beta+4\beta^{2})\lambda_{1}^{f}\right.$
$\displaystyle\>\>\;\;\;\left.\beta(-10\alpha^{2}\beta-\alpha^{2}+14\alpha\beta^{2}+7\alpha\beta-4\beta^{3}-6\beta^{2})\right](a_{-1}+a_{1})$
$\displaystyle+2\left[2(-3\alpha^{2}+4\alpha\beta+\alpha-2\beta)\lambda_{1}^{2}+2\alpha(1-\alpha)\lambda_{1}\lambda_{1}^{f}\right.$
$\displaystyle\>\>\>\>\>\left.+2(\alpha^{3}+4\alpha^{2}\beta-6\alpha\beta^{2}-3\alpha\beta+4\beta^{2})\lambda_{1}+2\alpha\beta(\alpha-1)\lambda_{1}^{f}+\alpha\beta(\alpha-\beta)\lambda_{2}\right.$
$\displaystyle\>\>\>\>\>\left.+\beta(-\alpha^{3}-8\alpha^{2}\beta+13\alpha\beta^{2}+4\alpha\beta-4\beta^{3}-4\beta^{2})\right]a_{0}$
$\displaystyle+4\left[2\alpha(\alpha-\beta)\lambda_{1}+\alpha(\alpha-1)\lambda_{1}^{f}+(-6\alpha^{2}\beta+10\alpha\beta^{2}+\alpha\beta-4\beta^{3})\right]s_{0,1}$
$\displaystyle-2\alpha\beta(\alpha-\beta)s_{0,2}+2\beta(\alpha-\beta)(\alpha-2\beta)s_{1,2}.$

###### Proof.

By the fusion law,

(15) $a_{0}(u_{1}\cdot u_{1}-v_{1}\cdot v_{1}+\lambda_{a_{0}}(v_{1}\cdot
v_{1})a_{0})=0.$

Substituting in the left side of (15) the values for $u_{1}$ and $v_{1}$ given
in Lemma 4.4 we get the first equality. The expression for
$(\alpha-2\beta)a_{)}s_{1,2}$ allows us to write explicitly as a linear
combination of $a_{-2},a_{-1},a_{0},a_{1},a_{2},s_{0,1},s_{0,2},s_{1,2}$ the
vector

$(\alpha-2\beta)(a_{0}P_{11}-\alpha Q_{11}).$

Thus, the second equality then follows from Equation (13) in Lemma 4.5. ∎

######  Lemma 4.8.

In the algebra $\overline{\mathcal{V}}$, the following equalities hold:

  1. (1)

$\beta(\alpha-\beta)^{2}(\alpha-4\beta)(a_{3}-a_{-2})=c$,

  2. (2)

$\beta(\alpha-\beta)^{2}(\alpha-4\beta)(a_{4}-a_{-1})=-c^{\tau_{1}},$

where

$\displaystyle c=$
$\displaystyle(\alpha-\beta)\left[4\alpha\beta\lambda_{1}-2(\alpha-1)(\alpha-4\beta)\lambda_{1}^{f}+\beta(-\alpha^{2}-5\alpha\beta-\alpha+6\beta)\right]a_{-1}$
$\displaystyle+\left[4(-3\alpha^{2}+4\alpha\beta+\alpha-2\beta)\lambda_{1}^{2}-4\alpha(\alpha-1)\lambda_{1}\lambda_{1}^{f}\right.$
$\displaystyle+(6\alpha^{3}+6\alpha^{2}\beta-2\alpha^{2}-16\alpha\beta^{2}-2\alpha\beta+8\beta^{2})\lambda_{1}+4\alpha\beta(\alpha-1)\lambda_{1}^{f}+2\alpha\beta(\alpha-\beta)\lambda_{2}$
$\displaystyle\left.+\beta(\alpha-\beta)(-2\alpha^{2}-8\alpha\beta+\alpha+4\beta^{2}+2\beta)\right]a_{0}$
$\displaystyle+\left[4\alpha(\alpha-1)\lambda_{1}\lambda_{1}^{f}+4(13\alpha^{2}-4\alpha\beta-\alpha+2\beta){\lambda_{1}^{f}}^{2}-4\alpha\beta(\beta-1)\lambda_{1}\right.$
$\displaystyle+(-6\alpha^{3}-6\alpha^{2}\beta+2\alpha^{2}+16\alpha\beta^{2}+2\alpha\beta-8\beta^{2})\lambda_{1}^{f}-2\alpha\beta(\alpha-\beta)\lambda_{2}^{f}$
$\displaystyle\left.+\beta(\alpha-\beta)(2\alpha^{2}+8\alpha\beta-\alpha-4\beta^{2}-2\beta)\right]a_{1}$
$\displaystyle+(\alpha-\beta)\left[2(\alpha-1)(\alpha-4\beta)\lambda_{1}-4\alpha\beta\lambda_{1}^{f}+\beta(\alpha^{2}+5\alpha\beta+\alpha-6\beta)\right]a_{2}$
$\displaystyle+(\alpha-\beta)\left[4\alpha(\alpha-2\beta+1)(\lambda_{1}-\lambda_{1}^{f})\right]s_{0,1}$
$\displaystyle-4\beta(\alpha-\beta)^{2}(s_{0,2}-s_{1,2}).$

###### Proof.

Since $s_{0,1}s_{0,1}$ is invariant under $f$, we have
$4(\alpha-2\beta)[s_{0,1}s_{0,1}-(s_{0,1}s_{0,1})^{f}]=0$. Then, equality
$(1)$ follows from the expression for $4(\alpha-2\beta)s_{0,1}s_{0,1}$ given
in Lemma 4.7. By applying $\tau_{1}$ to equation in $(1)$ we get (2). ∎

From the formulae in Lemmas 4.7 and 4.8, it is clear that we have different
pictures according whether $\alpha-2\beta$ and $\alpha-4\beta$ are invertible
in ${\overline{R}}$ or not. Since we are most concerned with algebras over a
field, later we will also assume that $\alpha-2\beta$ and $\alpha-4\beta$ are
either invertible or zero. Thus we deal separately with the following three
cases:

  1. (1)

The generic case: the ideal $L$ contains the elements $t_{h}$ for
$h\in{\mathbb{N}}$, and $h\geq 3$, $2t_{1}-1$, $(x_{1}-2x_{2})t_{2}-1$, and
$(x_{1}-4x_{2})t_{3}-1$. In this case we set
$\mathcal{O}_{g}:=\mathcal{O}_{L}$.

  2. (2)

The $\alpha=2\beta$ case: the ideal $L$ contains the elements $t_{h}$ for
$h\in{\mathbb{N}}$, and $h\geq 1$, $2t_{1}-1$, and $x_{1}-2x_{2}$. In this
case we set $\mathcal{O}_{2\beta}:=\mathcal{O}_{L}$.

  3. (3)

The $\alpha=4\beta$ case: the ideal $L$ contains the elements $t_{h}$ for
$h\in{\mathbb{N}}$, and $h\geq 1$, $2t_{1}-1$, and $x_{1}-4x_{2}$. In this
case we set $\mathcal{O}_{4\beta}:=\mathcal{O}_{L}$.

In [3] the case $\alpha=2\beta$ is considered in details: in particular it is
shown that every $2$-generated primitive axial algebra in
$\mathcal{O}_{2\beta}$ is at most $8$ dimensional and this bound is attained.
The following result, which can be compared to Theorem 3.7 in [16], is a
consequence of the resurrection principle [11, Lemma 1.7].

######  Proposition 4.9.

Let
$\overline{\mathcal{V}}=({\overline{R}},{\overline{V}},{\overline{\mathcal{A}}},(\overline{\mathcal{S}},\overline{\star}))$
be the universal object in the cathegory $\mathcal{O}_{g}$. Then
${\overline{V}}$ is linearly spanned by the set
$\\{a_{-2},a_{-1},a_{0},a_{1},a_{2},s_{0,1},s_{0,2},s_{1,2}\\}$.

###### Proof.

Let $U$ be the linear span in ${\overline{V}}$ of the set

$B:=\\{a_{-2},a_{-1},a_{0},a_{1},a_{2},s_{0,1},s_{0,2},s_{1,2}\\}$

with coefficients in ${\overline{R}}$. From Lemmas 4.7 and 4.8, since
$\alpha-2\beta$ and $\alpha-4\beta$ are invertible in ${\overline{R}}$, we get
that

$a_{0}\cdot s_{1,2},a_{3}\in U.$

The set $B$ is clearly invariant under the action of $\tau_{0}$ and since
$a_{-2}^{f}=a_{3}$, $U$ is also invariant under $f$. By Equation (11),
$a_{0}a_{1}$ and $a_{0}a_{2}$ are contained in $U$; by applying alternatively
$\tau_{0}$ and $f$ we get that $U$ contains all the products $a_{i}a_{j}$ for
$i,j\in{\mathbb{Z}}$. Similarly, since by Lemma 4.3, for $i\in\\{1,2\\}$,
$a_{0}s_{0,i}\in U$, and by Lemma 4.7, $a_{0}s_{1,2}\in U$, we get that $U$
contains all the products $a_{j}s_{0,i}$ and $a_{j}s_{1,2}$ for
$j\in{\mathbb{Z}},i\in\\{1,2\\}$.

It follows that, for $i,j\in\\{1,2\\}$, the expression on the righthand side
of Equation (13) is contained in $U$, whence $s_{0,i}\cdot s_{0,j}$ is
contained in $U$. As $U$ is invariant under $f$, we have also $s_{1,2}\cdot
s_{1,2}\in U$. Finally, with a similar argument, using the identity

$a_{0}\cdot(u_{i}\cdot u_{-1}+u_{i}\cdot v_{-1})=\alpha(u_{i}\cdot v_{-1}),$

we can express $s_{0,i}\cdot s_{1,2}$ as linear combination of elements of
$B$. Hence $U$ is a subalgebra of ${\overline{V}}$, and since ${\overline{V}}$
is generated by $a_{0}$ and $a_{1}$, we get that $U={\overline{V}}$. ∎

######  Remark 4.10.

Note that the above proof gives an explicit way to compute the structure
constants of the algebra ${\overline{V}}$ relative to the generating set $B$.
This has been done with the use of GAP [6]. The explicit expressions however
are far too long to be written explicitly here.

######  Corollary 4.11.

Let
$\overline{\mathcal{V}}=({\overline{R}},{\overline{V}},{\overline{\mathcal{A}}},(\overline{\mathcal{S}},\overline{\star}))$
be the universal object in the cathegory $\mathcal{O}_{g}$. Then,
${\overline{R}}$ is generated as a $\hat{D}$-algebra by $\lambda_{1}$,
$\lambda_{2}$, $\lambda_{1}^{f}$, and $\lambda_{2}^{f}$.

###### Proof.

Since, for every $v\in{\overline{V}}$,
$\lambda_{a_{1}}(v)=(\lambda_{a_{0}}(v^{f}))^{f}$, $\lambda_{a_{0}}$ is a
linear function, and ${\overline{R}}={\overline{R}}^{f}$, by Proposition 4.9,
we just need to show that, for every
$v\in\\{a_{-2},a_{-1},a_{0},a_{1},a_{2},s_{0,1},s_{0,2},s_{1,2}\\}$,
$\lambda_{a_{0}}(v)$ can be written as a linear combination, with coefficients
in $\hat{D}$, of products of $\lambda_{1}$, $\lambda_{2}$, $\lambda_{1}^{f}$,
and $\lambda_{2}^{f}$. By definition we have

$\lambda_{a_{0}}(a_{0})=1,\>\>\lambda_{a_{0}}(a_{1})=\lambda_{1},\mbox{ and
}\>\>\lambda_{a_{0}}(a_{2})=\lambda_{2}.$

Since $\tau_{0}$ is an ${\overline{R}}$-automorphism of ${\overline{V}}$
fixing $a_{0}$, we get

$\lambda_{a_{0}}(a_{-1})=\lambda_{a_{0}}((a_{1})^{\tau_{0}})=\lambda_{1},$
$\lambda_{a_{0}}(a_{-2})=\lambda_{a_{0}}((a_{2})^{\tau_{0}})=\lambda_{2},$

and

$\lambda_{a_{0}}(s_{0,1})=\lambda_{a_{0}}(a_{0}a_{1}-\beta a_{0}-\beta
a_{1})=\lambda_{1}-\beta-\beta\lambda_{1},$
$\lambda_{a_{0}}(s_{0,2})=\lambda_{a_{0}}(aa_{2}-\beta a-\beta
a_{2})=\lambda_{2}-\beta-\beta\lambda_{2}.$

Finally, by the fusion law, $u_{1}u_{1}+u_{1}v_{1}$ is a sum of a $0$ and an
$\alpha$-eigenvector for ${\rm ad}_{a_{0}}$, whence
$\lambda_{a_{0}}(u_{1}u_{1}+u_{1}v_{1})=0$. By Lemma 4.3, we can compute
$u_{1}u_{1}+u_{1}v_{1}$ and find

$u_{1}u_{1}+u_{1}v_{1}=w+\frac{(\alpha-\beta)}{2\alpha}s_{1,2},$

with $w\in\langle a_{-2},a_{-1},a_{0},a_{1},a_{2},s_{0,1}\rangle$. So, we can
express $\lambda_{a_{0}}(s_{1,2})$ and we obtain

$\lambda_{a_{0}}(s_{1,2})=\frac{2(\alpha-1)}{\alpha-\beta}\lambda_{1}^{2}-\frac{2(\alpha-1)}{\alpha-\beta}\lambda_{1}\lambda_{1}^{f}+(1-2\beta)\lambda_{1}+\beta\lambda_{2}-\beta.$

∎

We conclude this section with a similar result for symmetric algebras over a
field in the case $\alpha=4\beta$. Note that, in this case, since we are
assuming $\alpha\neq\beta$, we also assume that the field is of characteristic
other than $3$.

######  Proposition 4.12.

Let $V$ be a primitive symmetric axial algebra of Monster type
$(4\beta,\beta)$ over a field ${\mathbb{F}}$ of characteristic greater than
$3$, generated by two axes $\overline{a}_{0}$ and $\overline{a}_{1}$. Then $V$
has dimension at most $8$, unless $2\beta-1=0$ and

$\lambda_{\overline{a}_{0}}(\overline{a}_{1})=\lambda_{\overline{a}_{1}}(\overline{a}_{0})=\lambda_{\overline{a}_{0}}(\overline{a}_{0}^{\tau_{\overline{a}_{1}}})=\lambda_{\overline{a}_{1}}(\overline{a}_{1}^{\tau_{\overline{a}_{0}}})=1.$

###### Proof.

Let
$\overline{\mathcal{V}}=({\overline{R}},{\overline{V}},{\overline{\mathcal{A}}},(\overline{\mathcal{S}},\overline{\star}))$
be the universal object in the cathegory $\mathcal{O}_{4\beta}$. Since
$\alpha-2\beta=2\beta$ is invertible in ${\overline{R}}$, Lemma 4.7 yields
that, $a_{0}\cdot s_{1,2}$ is contained in $\langle
a_{-2},a_{-1},a_{0},a_{1},a_{2},s_{0,1},s_{0,2},s_{1,2}\rangle$. Since
$\alpha=4\beta$, Equation $(1)$ in Lemma 4.8 becomes

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\left[(48\beta^{3})\lambda_{1}-(108\beta^{4}-6\beta^{3})\right]a_{-1}$
$\displaystyle+\left[(-128\beta^{2}+8\beta)\lambda_{1}^{2}+(-64\beta^{2}+16\beta)\lambda_{1}\lambda_{1}^{f}+(416\beta^{3}-32\beta^{2})\lambda_{1}\right.$
$\displaystyle\left.+(16\beta^{3}-16\beta^{2})\lambda_{1}^{f}+(24\beta^{3})\lambda_{2}+(-180\beta^{4}+18\beta^{3})\right]a_{0}$
$\displaystyle+\left[(64\beta^{2}-16\beta)\lambda_{1}\lambda_{1}^{f}+(128\beta^{2}-8\beta){\lambda_{1}^{f}}^{2}+(-16\beta^{3}+16\beta^{2})\lambda_{1}\right.$
$\displaystyle\left.+(-416\beta^{3}+32\beta^{2})\lambda_{1}^{f}+(-24\beta^{3})\lambda_{2}^{f}+(180\beta^{4}-18\beta^{3})\right]a_{1}$
$\displaystyle+\left[(-48\beta^{3})\lambda_{1}^{f}+(108\beta^{4}-6\beta^{3})\right]a_{2}$
$\displaystyle+48\beta^{2}(2\beta+1)(\lambda_{1}-\lambda_{1}^{f})s_{0,1}$
$\displaystyle-36\beta^{3}(s_{0,2}-s_{1,2}).$

By Corollary 3.8, $V$ is a homomorphic image of
${\overline{V}}\otimes_{\hat{D}}{\mathbb{F}}$, via a homomorphism $\phi_{V}$
mapping $a_{i}$ to $\overline{a}_{i}$, for $i\in\\{0,1\\}$ and ${\mathbb{F}}$
is a homomorphic image of ${\overline{R}}\otimes_{\hat{D}}{\mathbb{F}}$ via
$\phi_{R}$. We use the bar notation to denote the images of the elements of
${\overline{V}}\otimes_{\hat{D}}{\mathbb{F}}$ via $\phi_{V}$, while we
identify the image under $\phi_{R}$ of an element of
${\overline{R}}\otimes_{\hat{D}}{\mathbb{F}}$ with the element itself. When we
apply $\phi_{V}$ to the relation (4) we get a similar relation in $V$. If the
coefficient of $a_{2}$ is not zero in ${\mathbb{F}}$, then we get
$\overline{a}_{2}\in
U_{0}:=\langle\overline{a}_{-1},\overline{a}_{0},\overline{a}_{1},\overline{s}_{0,1},\overline{s}_{0,2},\overline{s}_{1,2}\rangle$.
Since $V$ is symmetric, $f$ induces an automorphism $\bar{f}$ of $V$ and
$U_{0}$ is $\bar{f}$ invariant. Since $U_{0}$ is also
$\tau_{\bar{a}_{0}}$-invariant, we get also $\overline{a}_{-2}\in U_{0}$. More
generally, by applying alternatively $\tau_{\bar{a}_{0}}$ and $\bar{f}$ we get
that $\bar{a}_{i}\in U_{0}$ for every $i\in{\mathbb{Z}}$. The argument used in
the proof of Proposition 4.9 yields $V=U_{0}$. If the coefficient of
$\overline{a}_{2}$ in the Equation (4) is zero in ${\mathbb{F}}$, then we may
consider the coefficient of $\overline{a}_{-1}$ and if it is not zero we
deduce as above that
$\overline{a}_{-1}\in\langle\overline{a}_{0},\overline{a}_{1},\overline{s}_{0,1},\overline{s}_{0,2},\overline{s}_{1,2}\rangle$.
By proceeding as in the previous case, we get
$V=\langle\overline{a}_{0},\overline{a}_{1},\overline{s}_{0,1},\overline{s}_{0,2},\overline{s}_{1,2}\rangle$.
If the coefficients of $\overline{a}_{2}$ and $\overline{a}_{-1}$ are both
zero, then we get

$\lambda_{\bar{a}_{0}}(\bar{a}_{1})=\lambda_{\bar{a}_{1}}(\bar{a}_{0})=\frac{18\beta-1}{8}.$

As above, if the coefficient of $\overline{a}_{0}$ (or the coefficient of
$\overline{a}_{1}$) is not zero, we can express $\overline{a}_{0}$ (or
$\overline{a}_{1}$ respectively) as a linear combination of
$\overline{a}_{-1},\overline{s}_{0,1},\overline{s}_{0,2}-\overline{s}_{1,2}$
($\overline{a}_{0},\overline{s}_{0,1},\overline{s}_{0,2}-\overline{s}_{1,2}$
respectively). In both cases, it follows that
$V=\langle\overline{a}_{-1},\overline{a}_{0},\overline{a}_{1},\overline{s}_{0,1},\overline{s}_{0,2},\overline{s}_{1,2}\rangle$.
If also the coefficients of $\overline{a}_{0}$ and $\overline{a}_{1}$ are both
zero, then we get

$\lambda_{\bar{a}_{0}}(\bar{a}_{2})=\lambda_{\bar{a}_{1}}(\bar{a}_{-1})=\frac{480\beta^{3}-228\beta^{2}+28\beta-1}{64\beta^{2}}$

and Equation (4) becomes

$0=36\beta^{2}(\overline{s}_{0,2}-\overline{s}_{1,2}).$

Hence, since ${\mathbb{F}}$ has caracteristic greater than $3$,
$\overline{s}_{0,2}=\overline{s}_{1,2}$ and the identity
$\lambda_{\overline{a}_{0}}(\overline{s}_{0,2})=\lambda_{\overline{a}_{0}}(\overline{s}_{1,2})$
gives that $\beta$ satisfies the relation

(17) $(2\beta-1)^{2}(12\beta-1)(14\beta-1)=0.$

From now on assume
$\beta\in\\{\frac{1}{12},\frac{1}{14}\\}\setminus\\{\frac{1}{2}\\}$, in
particular ${\mathbb{F}}$ has characteristic other than $5$. Set
$U_{1}:=\langle\overline{a}_{-3},\overline{a}_{-2},\overline{a}_{-1},\overline{a}_{0},\overline{a}_{1},\overline{a}_{2},\overline{a}_{3},\overline{s}_{0,1},\overline{s}_{0,2},\overline{s}_{0,3}\rangle$.
From the identity
$\overline{a}_{0}(\overline{u}_{1}\overline{u}_{2}-\overline{v}_{1}\overline{v}_{2}+\lambda_{\overline{a}_{0}}(\overline{v}_{1}\overline{v}_{2})\overline{a}_{0})=0$
we can express $\overline{a}_{0}(\overline{s}_{1,3}+\overline{s}_{3,2})$ as a
linear combination of $\overline{a}_{-3}$, $\overline{a}_{-2}$,
$\overline{a}_{-1}$, $\overline{a}_{0}$, $\overline{a}_{1}$,
$\overline{a}_{2}$, $\overline{a}_{3}$, $\overline{s}_{0,1}$, and
$\overline{s}_{0,2}$ and then, by Lemma 4.5, we get that
$\overline{s}_{0,1}\overline{s}_{0,2}\in U_{1}$. Then, from the identity
$\overline{s}_{0,1}\overline{s}_{0,2}-(\overline{s}_{0,1}\overline{s}_{0,2})^{f}=0$,
we derive that $\overline{s}_{1,3}\in U_{1}$, whence also
$\overline{s}_{2,3}=(\overline{s}_{1,3})^{\tau_{\overline{a}_{0}}}\in U_{1}$.
From the identity $\overline{s}_{2,3}-(\overline{s}_{2,3})^{f}=0$ we then get
$\overline{a}_{4}\in U_{1}$. It follows that $U_{1}$ is invariant under $f$
and $\tau_{\overline{a}_{0}}$, hence $\overline{a}_{\pm i}\in U_{1}$ for
$i\geq 4$. Since $U_{1}$ is also ${\rm ad}_{\overline{a}_{0}}$-invariant, it
follows that $U_{1}$ contains $\overline{s}_{r,n}$ for every $n\geq 1$,
$r\in\\{0,\ldots,n-1\\}$. Thus $U_{1}$ is a subalgebra of $V$, whence
$V=U_{1}$. From the identity
$\overline{a}_{0}(\overline{v}_{1}\overline{v}_{2}-\lambda_{\overline{a}_{0}}(\overline{v}_{1}\overline{v}_{2})\overline{a}_{0})=0$
we get
$\overline{s}_{0,3}\in\langle\overline{a}_{-3},\overline{a}_{-2},\overline{a}_{-1},\overline{a}_{0},\overline{a}_{1},\overline{a}_{2},\overline{a}_{3},\overline{s}_{0,1},\overline{s}_{0,2}\rangle$.
Finally, from the identity $\overline{a}_{5}^{2}-\overline{a}_{5}=0$ we get
$\overline{a}_{-3}\in\langle\overline{a}_{-2},\overline{a}_{-1},\overline{a}_{0},\overline{a}_{1},\overline{a}_{2},\overline{a}_{3},\overline{s}_{0,1},\overline{s}_{0,2}\rangle$,
that is $V$ has dimension at most $8$. ∎

##  5\. The generic case

Let
$\overline{\mathcal{V}}=({\overline{R}},{\overline{V}},{\overline{\mathcal{A}}},(\overline{\mathcal{S}},\overline{\star}))$
be the universal object in the cathegory $\mathcal{O}_{g}$. Note that in this
case we have

$\hat{D}={\mathbb{Z}}[1/2,x_{1},x_{2},x_{1}^{-1},x_{2}^{-1},(x_{1}-x_{2})^{-1},(x_{1}-2x_{2})^{-1},(x_{1}-4x_{2})^{-1}].$

By Corollary 4.11,
${\overline{R}}=\hat{D}[\lambda_{1},\lambda_{1}^{f},\lambda_{2},\lambda_{2}^{f}]$.
The elements $\lambda_{1},\lambda_{1}^{f},\lambda_{2},\lambda_{2}^{f}$ are not
necessarily indeterminates on $\hat{D}$, as they have to satisfy various
relations imposed by the definition of ${\overline{R}}$. In particular, since
by Lemma 4.6, $s_{2,3}-s_{2,3}^{f}=0$, in the ring ${\overline{R}}$ the
following relations hold

  1. (1)

$\lambda_{a_{0}}(s_{2,3}-(s_{2,3})^{f})=0$,

  2. (2)

$\lambda_{a_{0}}((s_{2,3}-(s_{2,3})^{f})^{\tau_{1}})=0$,

  3. (3)

$\lambda_{a_{0}}(a_{3}a_{3}-a_{3})=0$,

  4. (4)

$\lambda_{a_{1}}(s_{2,3}-(s_{2,3})^{f})=0$.

By Remark 4.10, the four expressions on the left hand side of the above
identities can be computed explicitly and produce respectively four
polynomials $p_{i}(x,y,z,t)$ for $i\in\\{1,\ldots,4\\}$ in $\hat{D}[x,y,z,t]$
(with $x,y,z,t$ indeterminates on $\hat{D}$), that simultaneously annihilate
on the quadruple $(\lambda_{1},\lambda_{1}^{f},\lambda_{2},\lambda_{2}^{f})$.
Define also, for $i\in\\{1,2\\}$, $q_{i}(x,z):=p_{i}(x,x,z,z)$. The
polynomials $p_{i}$’s and $q_{i}$’s are too long to be displayed here but can
be computed using [1] or [6].

Suppose $V$ is a primitive axial algebra of Monster type $(\alpha,\beta)$ over
a field ${\mathbb{F}}$ of odd characteristic , with
$\alpha,\beta\in{\mathbb{F}}$ and $\alpha\not\in\\{2\beta,4\beta\\}$,
generated by two axes $\bar{a}_{0}$ and $\bar{a}_{1}$. Then, by Corollary 3.8,
$V$ is a homomorphic image of ${\overline{V}}\otimes_{\hat{D}}{\mathbb{F}}$
and ${\mathbb{F}}$ is a homomorphic image of
${\overline{R}}\otimes_{\hat{D}}{\mathbb{F}}$. We denote the images of an
element $\delta$ of ${\overline{R}}\otimes_{\hat{D}}{\mathbb{F}}$ in
${\mathbb{F}}$ by $\bar{\delta}$ and by $\overline{p}_{i}$ and
$\overline{q}_{i}$ the polynomials in ${\mathbb{F}}[x,y,z,t]$ and
${\mathbb{F}}[x,z]$ corresponding to $p_{i}$ and $q_{i}$, respectively. Set

$T:=\\{\overline{p}_{1},\overline{p}_{2},\overline{p}_{3},\overline{p}_{4}\\}$

and

$T_{s}:=\\{\overline{p}_{1}(x,z),\overline{p}_{2}(x,z)\\}.$

Moreover, for $P\in\\{T,T_{s}\\}$, denote by $Z(P)$ the set of common zeroes
of all the elements of $P$ in ${\mathbb{F}}^{4}$ and ${\mathbb{F}}^{2}$
respectively. It is clear from the definition that the $\overline{p}_{i}$’s
have the coefficients in the field ${\mathbb{F}}_{0}(\alpha,\beta)$. By
Proposition 4.9 and Corollary 4.11, the algebra $V$ is completely determined,
up to homomorphic images, by the quadruple

$(\lambda_{\bar{a}_{0}}(\bar{a}_{1}),\lambda_{\bar{a}_{1}}(\bar{a}_{0}),\lambda_{\bar{a}_{0}}(\bar{a}_{2}),\lambda_{\bar{a}_{1}}(\bar{a}_{-1})).$

Furthermore, this quadruple is the homomorphic image in ${\mathbb{F}}^{4}$ of
the quadruple $(\lambda_{1},\lambda_{1}^{f},\lambda_{2},\lambda_{2}^{f})$ and
so it is a common zero of the elements of $T$.

If, in addition, the algebra $V$ is symmetric, then

$\lambda_{\bar{a}_{0}}(\bar{a}_{1})=\lambda_{\bar{a}_{1}}(\bar{a}_{0})\mbox{
and }\lambda_{\bar{a}_{0}}(\bar{a}_{2}),=\lambda_{\bar{a}_{1}}(\bar{a}_{-1}))$

and the pair
$(\lambda_{\bar{a}_{0}}(\bar{a}_{1}),\lambda_{\bar{a}_{0}}(\bar{a}_{2}))$ is a
common zero of the elements of the set $T_{s}$.

We thus have proved Theorem 1.3.

Computing the resultant of the polynomials $\overline{p}_{1}(x,z)$ and
$\overline{p}_{2}(x,z)$ with respect to $z$ one obtains a polynomial in $x$ of
degree 10, which is the product of the five linear factors

$x,\>\>x-1\>\>,2x-\alpha,\>\>2x-\beta,\>\>4(2\alpha-1)x-(3\alpha^{2}+3\alpha\beta-\alpha-2\beta)$

and a factor of degree at most $5$. This last factor has degree $5$ and is
irreducible in ${\mathbb{Q}}(\alpha,\beta)$, if $\alpha$ and $\beta$ are
indeterminates over ${\mathbb{Q}}$. On the other hand, for certain values of
$\alpha$ and $\beta$, this factor can be reducible: for example, it even
completely splits in ${\mathbb{Q}}(\alpha,\beta)[x]$ when $\alpha=2\beta$ (see
[3]), or in the Norton-Sakuma case, when $(\alpha,\beta)=(1/4,1/32)$ (see the
proof of Theorem 1.6 below).

Fixed a filed ${\mathbb{F}}$, in order to classify primitive generic axial
algebras of Monster type $(\alpha,\beta)$ over ${\mathbb{F}}$ generated by two
axes $\bar{a}_{0}$ and $\bar{a}_{1}$ we can proceed as follows. We first find
all the zeroes of the set $T_{s}$ and classify all symmetric algebras. Then we
observe that, the even subalgebra
$\langle\langle\bar{a}_{0},\bar{a}_{2}\rangle\rangle$ and the odd subalgebra
$\langle\langle\bar{a}_{-1},\bar{a}_{1}\rangle\rangle$ are symmetric, since
the automorphisms $\tau_{\bar{a}_{0}}$ and $\tau_{\bar{a}_{1}}$ respectively,
swap the generating axes. Hence, from the classification of the symmetric
case, we know all possible values for the pairs

$(\lambda_{\bar{a}_{0}}(\bar{a}_{2}),\lambda_{\bar{a}_{1}}(\bar{a}_{-1}))$

and we can look for common zeros $(x_{0},y_{0},z_{0},t_{0})$ of the set $T$
with those prescribed values for $(x_{0},z_{0})$.

Using this method, we now classify $2$-generated primitive axial algebras of
Monster type $(\alpha,\beta)$ over the field ${\mathbb{Q}}(\alpha,\beta)$,
with $\alpha$ and $\beta$ independent indeterminates over ${\mathbb{Q}}$.

######  Lemma 5.1.

If ${\mathbb{F}}={\mathbb{Q}}(\alpha,\beta)$, with $\alpha$ and $\beta$
independent indeterminates over ${\mathbb{Q}}$, the set $Z(T_{s})$ consists
exactly of the $5$ points

$(1,1),\>\>(0,1),\>\>\left(\frac{\beta}{2},\frac{\beta}{2}\right),\>\>\left(\frac{\alpha}{2},1\right),$

and

$(q(\alpha,\beta),q(\alpha,\beta)),\mbox{ with
}q(\alpha,\beta)=\frac{(3\alpha^{2}+3\alpha\beta-\alpha-2\beta)}{4(2\alpha-1)}.$

###### Proof.

The system can be solved in ${\mathbb{Q}}(\alpha,\beta)$ using [1] giving the
five solutions of the statement. ∎

######  Lemma 5.2.

Let ${\mathbb{F}}={\mathbb{Q}}(\alpha,\beta)$, with $\alpha$ and $\beta$
independent indeterminates over ${\mathbb{Q}}$ and let $(x_{0},z_{0})\in
Z(T_{s})$. Then $(x_{0},y_{0},z_{0},t_{0})\in Z(T)$ if and only if

$(y_{0},t_{0})=(x_{0},z_{0}).$

###### Proof.

This have been checked using [1]. ∎

######  Lemma 5.3.

The algebras $3C(\alpha)$, $3C(\beta)$, and $3A(\alpha,\beta)$ over the field
${\mathbb{Q}}(\alpha,\beta)$ are simple.

###### Proof.

The claim follows from [9, Theorem 4.11 and Corollary 4.6]. For algebras
$3C(\alpha)$ and $3C(\beta)$ it is proved in [8, Example 3.4]. The algebra
$3A(\alpha,\beta)$ is the same as the algebra $3A^{\prime}_{\alpha,\beta}$
defined by Reheren in [16]. By [16, Lemma 8.2], it admits a Frobenius form
wich is non degenerate over the field ${\mathbb{Q}}(\alpha,\beta)$ and such
that all the generating axes are non-singular with respect to this form.
Hence, by Theorem 4.11 in [9], every non trivial ideal contains at least one
of the generating axes. Then, Corollary 4.6 in [9] yields that the algebra is
simple. ∎

Proof of Theorem 1.4. It is straightforward to check that the algebras $1A$,
$2B$, $3C(\alpha)$, $3C(\beta)$, and $3A(\alpha,\beta)$ are $2$-generated
symmetric axial algebras of Monster type $(\alpha,\beta)$ over the field
${\mathbb{Q}}(\alpha,\beta)$ and their corresponding values of
$(\lambda_{\bar{a}_{0}}(\bar{a}_{1}),\lambda_{\bar{a}_{0}}(\bar{a}_{2}))$ are
respectively

$(1,1),\>\>(0,1),\>\>\left(\frac{\beta}{2},\frac{\beta}{2}\right),\>\>\left(\frac{\alpha}{2},1\right),\mbox{
and }(q(\alpha,\beta),q(\alpha,\beta)),$

where $q(\alpha,\beta)$ is defined in Lemma 5.1. Let $V$ be an axial algebra
of Monster type $(\alpha,\beta)$ over the field ${\mathbb{Q}}(\alpha,\beta)$
generated by the two axes $\bar{a}_{0}$ and $\bar{a}_{1}$. Set

$\bar{\lambda}_{1}:=\lambda_{\bar{a}_{0}}(\bar{a}_{1}),\>\bar{\lambda}_{1}^{\prime}:=\lambda_{\bar{a}_{1}}(\bar{a}_{0}),\>\bar{\lambda}_{2}:=\lambda_{a_{0}}(\bar{a}_{2}),\>\mbox{
and }\bar{\lambda}_{2}^{\prime}:=\lambda_{\bar{a}_{1}}(\bar{a}_{-1}).$

By Theorem 1.3, $V$ is determined, up to homomorphic images, by the quadruple
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})$,
which must be in $Z(T)$. By Lemma 5.2, we get the five quadruples

$(1,1,1,1),\>\>(0,1,0,1),\>\>\left(\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2}\right),\>\>\left(\frac{\alpha}{2},1,\frac{\alpha}{2},1\right),$

and

$(q(\alpha,\beta),q(\alpha,\beta),q(\alpha,\beta),q(\alpha,\beta)).$

By Corollary 3.8 and Proposition 4.9, $V$ is linearly generated on
${\mathbb{Q}}(\alpha,\beta)$ by the set $\bar{a}_{-2}$, $\bar{a}_{-1}$,
$\bar{a}_{0}$, $\bar{a}_{1}$, $\bar{a}_{2}$, $\bar{s}_{0,1}$, $\bar{s}_{0,2}$,
and $\bar{s}_{1,2}$. Define

$\bar{d}_{0}:=\bar{s}_{2,3}-\bar{s}_{1,3}^{\tau_{0}},\>\>\bar{d}_{1}:=\bar{d}_{0}^{f},\>\;\bar{d}_{2}:={\bar{d}_{0}}^{\tau_{1}},$

and, for $i\in\\{0,1,2\\}$,

$\bar{D}_{i}:={\bar{d}_{i}}^{\tau_{0}}-\bar{d}_{i}.$

By Lemma 4.6, all vectors $\bar{d}_{i},\bar{D}_{i}$ for $i\in\\{0,1,2\\}$ are
zero. For all the admissible values of
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})$,
the coefficient of $\bar{a}_{-2}$ in $D_{0}$ is non zero, hence we can express
$\bar{a}_{-2}$ as a linear combination of $\bar{a}_{-1}$, $\bar{a}_{0}$,
$\bar{a}_{1}$,$\bar{a}_{2}$, $\bar{s}_{0,1}$, $\bar{s}_{0,2}$, and
$\bar{s}_{1,2}$. Similarly, from identity $\bar{d}_{0}=0$ we can express
$\bar{s}_{1,2}$ as a linear combination of $\bar{a}_{-1}$, $\bar{a}_{0}$,
$\bar{a}_{1}$,$\bar{a}_{2}$, $\bar{s}_{0,1}$, and $\bar{s}_{0,2}$.

For
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})$
in

$\left\\{\left(\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2}\right),\left(q(\alpha,\beta),q(\alpha,\beta),q(\alpha,\beta),q(\alpha,\beta)\right)\right\\},$

from identity $\bar{d}_{2}=0$ we get $\bar{a}_{-1}=\bar{a}_{2}$ and
consequently $\bar{s}_{0,2}=\bar{s}_{0,1}$. Thus in this two cases the
dimension of $V$ is at most $4$. Moreover, if
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})=(\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2})$,
then from the identity
$\bar{s}_{0,2}\bar{s}_{0,1}-\bar{s}_{0,1}\bar{s}_{0,1}=0$ we get
$\bar{s}_{0,1}=-\frac{\beta}{2}(\bar{a}_{0}+\bar{a}_{1}+\bar{a}_{2})$ and
hence the dimension is at most $3$.

For
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})$
in

$\left\\{(1,1,1,1),(0,1,0,1),\left(\frac{\alpha}{2},1,\frac{\alpha}{2},1\right)\right\\},$

from the identity $\bar{D}_{2}=0$ we get $\bar{a}_{-1}=\bar{a}_{1}$. Then,
from the identity $\bar{d}_{2}=0$ we deduce $\bar{a}_{2}=\bar{a}_{0}$ and
hence $\bar{s}_{0,2}=(1-2\beta)\bar{a}_{0}$. Hence in this cases $V$ has
dimension at most $3$. Suppose
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})=(0,1,0,1)$.
Then, from the identity $\bar{s}_{0,1}\bar{s}_{0,2}+(2\beta-1)\bar{a}_{0}$ we
get $\bar{s}_{0,1}=-\beta(\bar{a}_{0}+\bar{a}_{1})$ and so
$\bar{a}_{0}\bar{a}_{1}=0$. Hence in this case $V$ is isomorphic to the
algebra $2B$. Finally, suppose
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})=(1,1,1,1)$.
From the identity $\bar{s}_{0,2}-\bar{s}_{1,2}^{f}=0$ we get
$\bar{a}_{0}=\bar{a}_{1}$, that is $V$ is the algebra $1A$.

Thus, for each
$(\bar{\lambda}_{1},\bar{\lambda}_{1}^{\prime},\bar{\lambda}_{2},\bar{\lambda}_{2}^{\prime})$
in the set

$\left\\{\left(\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2},\frac{\beta}{2}\right),\left(q(\alpha,\beta),q(\alpha,\beta),q(\alpha,\beta),q(\alpha,\beta)\right),\left(\frac{\alpha}{2},1,\frac{\alpha}{2},1\right)\right\\},$

we get that $V$ satisfies the same multiplication table as the algebras
$3C(\beta)$, $3C(\alpha)$ and $3A(\alpha,\beta)$ respectively and has at most
the same dimension. Therefore, to conclude the proof, we need only to show
that the algebras $3C(\alpha)$, $3C(\beta)$ and $3A(\alpha,\beta)$ are simple.
This follows from Lemma 5.3. $\square$

As a corollary of Theorem 1.3, we can prove now Theorem 1.6.

Proof of Theorem 1.6. Let ${\mathbb{F}}$ be a field of characteristic zero.
Then ${\mathbb{F}}$ contains ${\mathbb{Q}}$. The resultant with respect to $z$
of the polynomials in $T_{s}$ has degree $9$ and splits in ${\mathbb{Q}}[x]$
as the product of a constant and the linear factors

$x,\>x-1,\>x-\frac{1}{8},\>\left(x-\frac{1}{64}\right)^{2},\>x-\frac{13}{2^{8}},\>x-\frac{1}{32},\>x-\frac{3}{2^{7}},\>x-\frac{5}{2^{8}}.$

In ${\mathbb{Q}}^{4}$, the set $Z(T)$ consists of the $9$ points

$(1,1,1,1),\>\>(0,0,1,1),\>\>(\frac{1}{8},\frac{1}{8},1,1),\>\>(\frac{1}{64},\frac{1}{64},\frac{1}{64},\frac{1}{64}),\>\>(\frac{13}{2^{8}},\frac{13}{2^{8}},\frac{13}{2^{8}},\frac{13}{2^{8}}),$
$(\frac{1}{32},\frac{1}{32},0,0),\>\>(\frac{1}{64},\frac{1}{64},\frac{1}{8},\frac{1}{8}),\>\>(\frac{3}{2^{7}},\frac{3}{2^{7}},\frac{3}{2^{7}},\frac{3}{2^{7}}),\>\>(\frac{5}{2^{8}},\frac{5}{2^{8}},\frac{13}{2^{8}},\frac{13}{2^{8}}).$

By [11, 7], each quadruple of the above list corresponds to a Norton-Sakuma
algebra. By Corollary 4.13 in [9], every Norton-Sakuma algebra is simple,
provieded it is not of type $2B$. Hence, the thesis follows from Theorem 1.3,
once we prove that in each case the dimension of $V$ is at most equal to the
dimension of the corresponding Norton-Sakuma algebra. This can be done, by
Remark 4.10, using [6]. $\square$

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