An infinite-dimensional 2-generated primitive axial algebra of Monster type

Abstract

Rehren proved in Axial algebras. Ph.D. thesis, University of Birmingham (2015), Trans Am Math Soc 369:6953–6986 (2017) that a primitive 2-generated axial algebra of Monster type $(\alpha ,\beta )$, over a field of characteristic other than 2, has dimension at most 8 if $\alpha \notin \{2\beta ,4\beta \}$. In this note, we show that Rehren’s bound does not hold in the case $\alpha =4\beta$ by providing an example (essentially the unique one) of an infinite-dimensional 2-generated primitive axial algebra of Monster type $(2,\frac{1}{2})$ over an arbitrary field ${{\mathbb {F}}}$ of characteristic other than 2 and 3. We further determine its group of automorphisms and describe some of its relevant features.

1 Introduction

This note is part of a project of the authors aimed at classifying all 2-generated primitive axial algebras of Monster type. Axial algebras were introduced by Hall, Rehren and Shpectorov [6, 7] in order to axiomatize some key features, that are relevant for the study of finite simple groups, of certain classes of algebras, such as the weight-2 components of OZ-type vertex operator algebras, Jordan algebras, Matsuo algebras, and Majorana algebras (see [8] and the introductions of [6] and [16]). In this paper, we assume that the underlying field ${{\mathbb {F}}}$ has characteristic different from 2, being the case of characteristic 2 somehow degenerate because the algebra turns out to be associative (Lemma 2.1). On the other hand, the case of characteristic other than 2 is of particular interest, since most of the finite simple groups can be faithfully represented as groups generated by special involutory automorphisms, called Miyamoto involutions [13], of such algebras. In particular, the Griess algebra (see [4]) is a real axial algebra of Monster type $(\frac{1}{4},\frac{1}{32})$ and the Miyamoto involutions of this algebra (also called Majorana involutions) are precisely the involutions of type 2A in the Monster, i.e., those whose centralizer is the double cover of the Baby Monster.

The classification of 2-generated axial algebras has a fundamental rôle in the development of the theory of axial algebras and of the representations of the finite groups on them (see [2, 3, 9,10,11,12]). In a pioneering work [14], Norton classified subalgebras of the Griess algebra generated by two axes. Norton showed that there are exactly nine isomorphism classes of such subalgebras, corresponding to the nine conjugacy classes of dihedral subgroups of the Monster generated by two involutions of type 2A. These algebras have been proven to be, up to isomorphisms, the only 2-generated primitive real axial algebras of Monster type $(\frac{1}{4},\frac{1}{32})$ and are now known as Norton–Sakuma algebras [6, 9, 17]. In the minimal non-associative case of axial algebras of Jordan type $\eta$, the classification has been obtained by Hall, Rehren and Shpectorov in [7]. Note that axial algebras of Jordan type $\eta$ are also axial algebras of Monster type $(\alpha ,\beta )$ when $\eta \in \{\alpha ,\beta \}$. In [15] and [16], Rehren started a systematic study of axial algebras of Monster type $(\alpha ,\beta )$. In particular, he showed that, when $\alpha \not \in \{2\beta ,4\beta \}$ every 2-generated primitive axial algebra of Monster type $(\alpha ,\beta )$ has dimension at most 8.

In an unpublished work (arXiv:2101.10315, Jan. 25th 2021), the authors extend Rehren’s result showing that, if $(\alpha ,\beta )\ne (2,\frac{1}{2})$, then any symmetric 2-generated primitive axial algebra of Monster type $(\alpha ,\beta )$ has dimension at most 8 (symmetric means that the algebra has an automorphism that swaps the generating axes). Still, the question of the existence of 2-generated axial algebras of infinite dimension has been around for some years. Here we give an elementary construction of such an example, proving that the case $(\alpha ,\beta )=(2,\frac{1}{2})$ is indeed an exception.

Theorem 1.1

For every field ${{\mathbb {F}}}$ of characteristic different from 2 and 3, there exists an infinite-dimensional 2-generated symmetric primitive axial algebra of Monster type $(2,\frac{1}{2})$ over ${{\mathbb {F}}}$.

We would like to mention that a few weeks after this paper was posted in arXiv (arXiv:2007.02430, Jul. 5th 2020), another independent paper by Takahiro Yabe was submitted in that repository (arXiv:2008.01871, Aug. 4th 2020), where the same algebra appears (though with no details apart from the multiplication table) within a general result on 2-generated symmetric primitive axial algebras of Monster type. From Yabe’s results, it would follow that in characteristic other than 5, any infinite-dimensional example of such algebras must be a quotient of this one.

2 The algebra ${{\mathcal {H}}}$

We first recall the definition of axial algebras of Monster type. Let ${{\mathbb {F}}}$ be a field and let ${\mathcal {S}}$ be a finite subset of ${{\mathbb {F}}}$ with $1\in {\mathcal {S}}$. A fusion law on ${\mathcal {S}}$ is a map

$$\begin{aligned} \star :{\mathcal {S}} \times {\mathcal {S}} \rightarrow 2^{{\mathcal {S}}}. \end{aligned}$$

An axial algebra over ${{\mathbb {F}}}$ with spectrum ${\mathcal {S}}$ and fusion law $\star$ is a commutative non-associative ${{\mathbb {F}}}$-algebra V generated by a set ${\mathcal {A}}$ of nonzero idempotents (called axes) such that, for each $a\in {{\mathcal {A}}}$,

(Ax1):

$ad_a:v\mapsto av$ is a semisimple endomorphism of V with spectrum contained in ${\mathcal {S}}$;

(Ax2):

for every $\lambda ,\mu \in {\mathcal {S}}$, the product of a $\lambda$-eigenvector and a $\mu$-eigenvector of $\mathrm{ad}_a$ is the sum of $\delta$-eigenvectors, for $\delta \in \lambda \star \mu$.

Furthermore, V is called primitive if, for all $a\in {\mathcal {A}}$,

(Ax3):

the 1-eigenspace of $ad_a$ is $\langle a \rangle$.

An axial algebra over ${{\mathbb {F}}}$ is said to be of Monster type $(\alpha ,\beta )$ if it satisfies the fusion law ${\mathcal {M}}(\alpha , \beta )$ given in Table 1, with $\alpha ,\beta \in {{\mathbb {F}}}\backslash \{0,1\}$, with $\alpha \ne \beta$.

Table 1 Fusion law ${\mathcal {M}}(\alpha ,\beta )$

Note that, given an axial algebra A of Monster type $(\alpha , \beta )$, every axis a induces an obvious ${{\mathbb {Z}}}_2$-grading $A=A^+_a\oplus A^-_a$, where $A^+_a$ is the sum of the 0-, 1-, $\alpha$-eigenspaces of $\mathrm{ad}_a$ and $A^-_a$ is the $\beta$-eigenspace of $\mathrm{ad}_a$. Furthermore, we have a ${{\mathbb {Z}}}_2$-grading $A^+_a=A^{++}_a\oplus A^{+-}_a$ of $A^+_a$, where $A^{++}_a$ is the sum of the 0- and 1-eigenspaces of $\mathrm{ad}_a$ and $A^{+-}_a$ is the $\alpha$-eigenspace of $\mathrm{ad}_a$. An easy argument due to Mathias Stout (see p. 59 in Modular representation theory and applications to decomposition algebras, master thesis Gent University 2021, https://algebra.ugent.be/ $\tilde{}$ tdemedts/research-students.php#), who kindly allowed us to include it here, shows that

Lemma 2.1

If A is a primitive axial algebra of Monster type $(\alpha , \beta )$ over a field of characteristic 2, then A is associative.

Proof

Let b be an axis and assume $b=b_++b_-$, with $b_+\in A^+_a$ and $b_-\in A^-_a$. Then, by the fusion law,

$$\begin{aligned} b=b^2=(b_++b_-)^2=b_+^2+b_-^2\in A^+_a. \end{aligned}$$

So all axes lie in $A^+_a$, whence $A=A^+_a$. Now, the same argument on $A^+_a$ shows that also the $\alpha$-eigenspace of $\mathrm{ad}_a$ is trivial. Thus, the adjoint of every axis has spectrum $\{1,0\}$, which is equivalent to the algebra being associative (see [6]). $\square$

Therefore, from now on we assume that ${{\mathbb {F}}}$ is a field of characteristic other than 2. A straightforward computation shows that the map that negates $A^-_a$ and induces the identity on $A^+_a$ is an involutory algebra automorphism called Miyamoto involution (see [9, 13]).

Let ${{\mathcal {H}}}$ be the infinite-dimensional ${{\mathbb {F}}}$-vector space with basis $\{a_i,\sigma _j\mid i\in {{\mathbb {Z}}},j\in {{\mathbb {Z}}}_+\}$,

$$\begin{aligned} {{\mathcal {H}}}:=\bigoplus _{i\in {{\mathbb {Z}}}}{{\mathbb {F}}}a_i\oplus \bigoplus _{j\in {{\mathbb {Z}}}_+}{{\mathbb {F}}}\sigma _j. \end{aligned}$$

Set $\sigma _0=0$. Define a commutative non-associative product on ${{\mathcal {H}}}$ by linearly extending the following values on the basis elements:

(${{\mathcal {H}}}$1):

$a_ia_j:=\frac{1}{2}(a_i+a_j)+\sigma _{|i-j|}$;

(${{\mathcal {H}}}$2):

$a_i\sigma _j:=-\frac{3}{4}a_i+\frac{3}{8}(a_{i-j}+a_{i+j})+\frac{3}{2}\sigma _j$;

(${{\mathcal {H}}}$3):

$\sigma _i\sigma _j:=\frac{3}{4}(\sigma _i+\sigma _j)-\frac{3}{8}(\sigma _{|i-j|}+\sigma _{i+j})$.

In particular, $a_i^2=\frac{1}{2}(a_i+a_i)+\sigma _0=a_i$, so each $a_i$ is an idempotent.

We call ${{\mathcal {H}}}$ the Highwater algebra because it was discovered in Venice during the disastrous floods in November 2019. In what follows, double angular brackets denote algebra generation while single brackets denote linear span.

Theorem 2.2

If $\mathrm {char}({{\mathbb {F}}})\ne 3$ then ${{\mathcal {H}}}=\langle \!\langle a_0,a_1\rangle \!\rangle$ is a symmetric primitive axial algebra of Monster type $(2,\frac{1}{2})$.

Manifestly, this result implies Theorem 1.1.

If $\mathrm {char}({{\mathbb {F}}})=3$ then $2=\frac{1}{2}$ and so the concept of an algebra of Monster type $(2,\frac{1}{2})$ is not defined. However, the four-term decomposition typical for algebras of Monster type still exists, and so ${{\mathcal {H}}}$ in characteristic 3 is an example of an axial decomposition algebra as defined in [1]. We also prove that ${{\mathcal {H}}}$ in this case is a Jordan algebra. Note, however, that because a lot of structure constants in ${{\mathcal {H}}}$ become zero in characteristic 3, ${{\mathcal {H}}}$ is no longer generated by $a_0$ and $a_1$. In fact, every pair of distinct axes $a_i, a_j$ generates a 3-dimensional subalgebra, linearly spanned by $a_i$, $a_j$, and $\sigma _{|i-j|}$, and isomorphic to the 3-dimensional Jordan algebra denoted by $Cl^{00}({{\mathbb {F}}}^2,b_2)$ in [7, Theorem (1.1)].

We start with a number of observations concerning the properties of ${{\mathcal {H}}}$. First of all, we show that it is not a simple algebra. Consider the linear map $\lambda :{{\mathcal {H}}}\rightarrow {{\mathbb {F}}}$ defined on the basis of ${{\mathcal {H}}}$ as follows: $\lambda (a_i)=1$ for all $i\in {{\mathbb {Z}}}$ and $\lambda (\sigma _j)=0$ for all $j\in {{\mathbb {Z}}}_+$.

Lemma 2.3

The map $\lambda$ is a homomorphism of algebras.

Proof

It suffices to show that $\lambda$ is multiplicative, i.e., $\lambda (uv)=\lambda (u)\lambda (v)$ for all $u,v\in {{\mathcal {H}}}$. Since this equality is linear in both u and v, it suffices to check it for the elements of the basis. If $u=a_i$ and $v=a_j$ then $\lambda (a_ia_j)=\lambda (\frac{1}{2}(a_i+a_j)+\sigma _{|i-j|})= \frac{1}{2}+\frac{1}{2}=1=1\cdot 1=\lambda (a_i)\lambda (a_j)$. If $u=a_i$ and $v=\sigma _j$ then $\lambda (a_i\sigma _j)=\lambda (-\frac{3}{4}a_i+\frac{3}{8}(a_{i-j}+a_{i+j})+\frac{3}{2}\sigma _j)= -\frac{3}{4}+\frac{3}{8}+\frac{3}{8}=0=1\cdot 0=\lambda (a_i)\lambda (\sigma _j)$. Finally, if $u=\sigma _i$ and $v=\sigma _j$ then $\lambda (\sigma _i\sigma _j)=\lambda (\frac{3}{4}(\sigma _i+\sigma _j)-\frac{3}{8}(\sigma _{|i-j|}+ \sigma _{i+j}))=0=\lambda (\sigma _i)\lambda (\sigma _j)$. So the equality holds in all cases and so $\lambda$ is indeed an algebra homomorphism. $\square$

An ${{\mathbb {F}}}$-algebra homomorphism from an ${{\mathbb {F}}}$-algebra A to ${{\mathbb {F}}}$ is called a weight function and an algebra admitting a weight function is called a baric (or weighted) algebra. Note that, if A is an axial algebra with a weight function $\omega$, then the axes, being idempotents, have weights 0 or 1 and, since the axes generate the whole algebra, at least one axis has weight 1. We say that A is a baric axial algebra if all axes have weight 1. So ${{\mathcal {H}}}$ is a baric axial algebra with weight function $\lambda$. Clearly its kernel J is an ideal of codimension 1.

Using $\lambda$, we can also define a bilinear form on ${{\mathcal {H}}}$. Namely, for $u,v\in {{\mathcal {H}}}$, we set $(u,v):=\lambda (u)\lambda (v)$. It is immediate that this is a bilinear form; furthermore, it associates with the algebra product. Indeed, for $u,v,w\in {{\mathcal {H}}}$, we have $(uv,w)=\lambda (uv)\lambda (w)= \lambda (u)\lambda (v)\lambda (w)=\lambda (u)\lambda (vw)=(u,vw)$. In the theory of axial algebras such forms are called Frobenius forms. The form $(\cdot ,\cdot )$ further satisfies the property that $(a_i,a_i)= \lambda (a_i)\lambda (a_i)=1\cdot 1=1$, which is often required in the definition of a Frobenius form.

Lemma 2.4

If $\mathrm {char}({{\mathbb {F}}})\ne 3$, the only nonzero finite-dimensional subalgebras of ${{\mathcal {H}}}$ are the 1-dimensional ones generated by axes.

Proof

Let $L\ne 0$ be a finite-dimensional subalgebra of ${{\mathcal {H}}}$. Every element e of L can be written uniquely as

$$\begin{aligned} e=\sum _{i\in {{\mathbb {Z}}}}r_i(e) a_i+\sum _{j\in {{\mathbb {Z}}}_+}s_j(e)\sigma _j \end{aligned}$$

with the coefficients $r_i(e)$ and the $s_j(e)$ in ${{\mathbb {F}}}$ and only finitely many of them being nonzero. Let

$$\begin{aligned} R:=\{i\in {{\mathbb {Z}}}\mid r_i(e)\ne 0 \text{ for } \text{ some } e\in L\} \end{aligned}$$

and

$$\begin{aligned} S:=\{j\in {{\mathbb {Z}}}_+\mid s_j(e)\ne 0 \text{ for } \text{ some } e\in L\}. \end{aligned}$$

Since L is finite dimensional, both R and S are finite sets. Also, since $L\ne 0$, at least one of the two sets is nonempty. If both R and S are nonempty, let $n:=\max (R)$ and $m:=\max (S)$. Select $e\in L$ such that $r_n(e)\ne 0$. If also $s_m(e)\ne 0$ then observe that, by (${{\mathcal {H}}}$2), $r_{n+m}(e^2)=2\cdot \frac{3}{8}r_n(e)s_m(e)\ne 0$, a contradiction, since $n+m>n$. So $s_m(e)=0$. Now select $e'\in L$ such that $s_m(e')\ne 0$. Again, by (${{\mathcal {H}}}$2), $r_{n+m}(ee')=\frac{3}{8}r_n(e)s_m(e')\ne 0$. The contradiction shows that R and S cannot both be nonempty.

If R is empty and S is not, choose $e\in L$ with $s_m(e)\ne 0$, where again $m=\max (S)$. By (${{\mathcal {H}}}$3), $s_{2m}(e^2)=-\frac{3}{8}s_m(e)^2\ne 0$; which is a contradiction. Finally, if S is empty and R is not, let $n:=\max (R)$ and $k:=\min (R)$. If $n=k$, then L is 1-dimensional generated by $a_n$. So suppose that $k<n$. Again, select $e\in L$ such that $r_n(e)\ne 0$. If also $r_k(e)\ne 0$ then $s_{n-k}(e^2)=2r_n(e)r_k(e)\ne 0$ by (${{\mathcal {H}}}$1). This is a contradiction since $S=\emptyset$. Finally, if $r_k(e)=0$, choose $e'\in L$ with $r_k(e')\ne 0$. Now $s_{n-k}(ee')=r_n(e)r_k(e')\ne 0$ by (${{\mathcal {H}}}$1), and this is a contradiction in this final case. $\square$

Corollary 2.5

If $\mathrm {char}({{\mathbb {F}}})\ne 3$, the elements $\{a_i\,|\,i\in {{\mathbb {Z}}}\}$ are the only nontrivial idempotents in ${{\mathcal {H}}}$.

The next observation to make is that ${{\mathcal {H}}}$ is quite symmetric. Let $D:=\langle \tau , \pi \rangle$ be the infinite dihedral group acting naturally on ${{\mathbb {Z}}}$, where, for all $i\in {{\mathbb {Z}}}$, $\tau :i \mapsto -i$ and $\pi :i\mapsto -i+1$. For $\rho \in D$, let $\phi _\rho$ be the linear map that fixes all $\sigma _j$ and sends $a_i$ to $a_{i^\rho }$. Then $\phi _\rho$ is an automorphism of ${{\mathcal {H}}}$ and, since the map $\rho \mapsto \phi _\rho$ defines a faithful representation of D as an automorphism group of ${{\mathcal {H}}}$, with an abuse of notation, from now on we identify $\phi _\rho$ with $\rho$. In particular,

Proposition 2.6

If $\mathrm {char}({{\mathbb {F}}})\ne 3$, then $Aut({{\mathcal {H}}})\cong D$.

Proof

Let $\varphi \in \mathrm{Aut}({{\mathcal {H}}})$. By Corollary 2.5, $\varphi$ induces a permutation on the set $\{a_i\mid i\in {{\mathbb {Z}}}\}$. Let $j\in {{\mathbb {Z}}}_+$ and set $a_h:=a_0^\varphi$ and $a_k:=a_j^\varphi$. Then, by $({{\mathcal {H}}}1)$,

$$\begin{aligned} \sigma _j^\varphi =\left( a_0a_j-\frac{1}{2}(a_0+a_j)\right) ^\varphi =a_0^\varphi a_j^\varphi -\frac{1}{2}(a_0^\varphi +a_j^\varphi )= a_ha_k-\frac{1}{2}(a_h+a_k)=\sigma _{|h-k|}, \end{aligned}$$

whence $\varphi$ induces a permutation also on the set $\{\sigma _j\mid j\in {{\mathbb {Z}}}_+\}$. Now observe that the action on the latter set has to be trivial. Indeed, for $j\in {{\mathbb {Z}}}_+$, define the graph $\Gamma _j$ with vertices $\{a_i,\mid i\in {{\mathbb {Z}}}\}$, where $a_i$ is adjacent to $a_l$ if and only if $\sigma _j=a_ia_l-\frac{1}{2}(a_i+a_l)$. Thus, for all $h,k\in {{\mathbb {Z}}}_+$ such that $(\sigma _h)^\varphi =\sigma _k$, $\varphi$ induces a graph isomorphism from $\Gamma _h$ to $\Gamma _k$. On the other hand, by $({{\mathcal {H}}}1)$, $a_i$ is adjacent to $a_l$ in $\Gamma _j$ if and only if $|i-l|=j$. Hence, $\Gamma _j$ has exactly j distinct connected components, corresponding to the congruence classes of ${{\mathbb {Z}}}$ modulo j. In particular, the graph $\Gamma _h$ is isomorphic to $\Gamma _k$ if and only if $h=k$.

In particular, the entire group $\mathrm{Aut}({{\mathcal {H}}})$ fixes $\sigma _1$, and so it acts on the infinite string graph $\Gamma _{1}$. Since this action is faithful, we conclude that $\mathrm{Aut}({{\mathcal {H}}})\cong D$. $\square$

We are aiming to show that the $a_i$’s are axes satisfying the fusion law ${\mathcal {M}}(2,\frac{1}{2})$. Since D is transitive on the $a_i$’s, it suffices to check this for just one of them, say $a=a_0$. We start with the eigenvalues and eigenspaces of $\mathrm{ad}_a$.

Select $j\in {{\mathbb {Z}}}_+$ and set $U=\langle a,a_{-j},a_j,\sigma _j\rangle$. It is immediate to see that U is invariant under $\mathrm{ad}_a$ and, with respect to the basis $(a,a_{-j},a_j,\sigma _j)$ the restriction of $\mathrm{ad}_a$ to U is represented by the following matrix:

$$\begin{aligned} \left( \begin{array}{rrrr} 1&{}0&{}0&{}0\\ \frac{1}{2}&{}\quad \frac{1}{2}&{}\quad 0&{}\quad 1\\ \frac{1}{2}&{}\quad 0&{}\quad \frac{1}{2}&{}\quad 1\\ -\frac{3}{4}&{}\quad \frac{3}{8}&{}\quad \frac{3}{8}&{}\quad \frac{3}{2} \end{array} \right) \end{aligned}$$

This has characteristic polynomial $x^4-\frac{7}{2}x^3+\frac{7}{2}x^2-x= (x-1)x(x-2)(x-\frac{1}{2})$ and eigenspaces $U_1=\langle a\rangle$, $U_0=\langle u_j\rangle$, $U_2=\langle v_j\rangle$, $U_{\frac{1}{2}}=\langle w_j\rangle$, where

$$\begin{aligned}&u_j:=6a-3(a_{-j}+a_j)+4\sigma _j \nonumber \\&\quad v_j:= 2a-(a_{-j}+a_j)-4\sigma _j \nonumber \\&\quad w_j:=a_{-j}-a_j. \end{aligned}$$

(1)

The only exception to this statement arises when $\mathrm {char}({{\mathbb {F}}})=3$. In this case, $2=\frac{1}{2}$ and so $U_2$ and $U_{\frac{1}{2}}$ merge into a single 2-dimensional eigenspace $U_2=U_{\frac{1}{2}}=\langle v_j,w_j\rangle$.

In all cases, we can write that $U=\langle a,u_j,v_j,w_j\rangle$, that is, $a_{-j}$, $a_j$ and $\sigma _j$ can be expressed via these vectors. From this, we deduce the following.

Lemma 2.7

The adjoint map $\mathrm{ad}_a$ is semisimple on ${{\mathcal {H}}}$ and

• (a) if $\mathrm {char}({{\mathbb {F}}})\ne 3$ then the spectrum of $\mathrm{ad}_a$ is $\{1,0,2,\frac{1}{2}\}$ and the eigenspaces are ${{\mathcal {H}}}_1=\langle a\rangle$, ${{\mathcal {H}}}_0=\langle u_j\mid j\in {{\mathbb {Z}}}_+\rangle$, ${{\mathcal {H}}}_2=\langle v_j\mid j\in {{\mathbb {Z}}}_+\rangle$, and ${{\mathcal {H}}}_{\frac{1}{2}}=\langle w_j\mid j\in {{\mathbb {Z}}}_+\rangle$;

• (b) if $\mathrm {char}({{\mathbb {F}}})=3$ then the spectrum is $\{1,0,\frac{1}{2}\}$ and the eigenspaces are ${{\mathcal {H}}}_1=\langle a\rangle$, ${{\mathcal {H}}}_0=\langle u_j\mid j\in {{\mathbb {Z}}}_+\rangle$, and ${{\mathcal {H}}}_{\frac{1}{2}}=\langle v_j,w_j\mid j\in {{\mathbb {Z}}}_+\rangle$.

In order to avoid the complication arising in characteristic 3, we will use the notation ${{\mathcal {H}}}_u:=\langle u_j\mid j\in {{\mathbb {Z}}}_+\rangle$, ${{\mathcal {H}}}_v:=\langle v_j\mid j\in {{\mathbb {Z}}}_+\rangle$, and ${{\mathcal {H}}}_w=\langle w_j\mid j\in {{\mathbb {Z}}}_+\rangle$ calling these subspaces the u-, v-, and w-parts of ${{\mathcal {H}}}$, respectively. A similar terminology will be used for sums of these subspaces. Thus, in all characteristics, we have the decomposition

$$\begin{aligned} {{\mathcal {H}}}=\langle a\rangle \oplus {{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v\oplus {{\mathcal {H}}}_w. \end{aligned}$$

Let us relate this decomposition to the ideal J.

Lemma 2.8

$J={{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v\oplus {{\mathcal {H}}}_w$.

Proof

By inspection, the weight function is zero on ${{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v\oplus {{\mathcal {H}}}_w$, so this sum is contained in J. Now, the equality is forced because both J and ${{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v\oplus {{\mathcal {H}}}_w$ have codimension 1 in ${{\mathcal {H}}}$. $\square$

The stabilizer $D_a$ of a in D has order 2, and it is generated by the involution $\tau$ sending every $a_i$ to $a_{-i}$ (and fixing every $\sigma _j$). From this and Eq. (1), it follows that

Lemma 2.9

The involution $\tau$ acts as identity on $\langle a\rangle \oplus {{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v$ and as minus identity on ${{\mathcal {H}}}_w$. In particular, ${{\mathcal {H}}}_\tau :=\langle a\rangle \oplus {{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v$ is the fixed subalgebra of $\tau$.

Clearly, $\tau$ will turn out to be the Miyamoto involution associated to the axis a.

We find a further subalgebra by intersecting ${{\mathcal {H}}}_\tau$ with the ideal J. Namely, we set $V:={{\mathcal {H}}}_\tau \cap J={{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v$, the uv-part of ${{\mathcal {H}}}$. Clearly, V is an ideal of ${{\mathcal {H}}}_\tau$.

We want now to determine the fusion law in the subalgebra V with respect to its basis consisting of the vectors $u_j$ and $v_j$. To this aim, it is convenient to use a second basis, given by the elements $\sigma _j$ and

$$\begin{aligned} c_j:=2a-(a_{-j}+a_j), \end{aligned}$$

(2)

for all $j\in {{\mathbb {Z}}}_+$. From the definitions, we immediately get that, for all $j\in {{\mathbb {Z}}}_+$,

$$\begin{aligned} u_j=3c_j+4\sigma _j \quad \text{ and }\quad v_j=c_j-4\sigma _j \end{aligned}$$

(3)

and

$$\begin{aligned} c_j=\frac{1}{4}(u_j+v_j) \quad \text{ and }\quad \sigma _j=\frac{1}{16}u_j-\frac{3}{16}v_j. \end{aligned}$$

(4)

In particular, this means that the set of all vectors $c_j$ and $\sigma _j$ is a basis of V. In the following lemma, we compute the products for this basis. Note that we only need to compute the products $c_ic_j$ and $c_i\sigma _j$, because $\sigma _i\sigma _j$ are given in (${{\mathcal {H}}}$3). It will be convenient to use the following notation: we set $c_0:=0$, and, for $i,j\in {{\mathbb {Z}}}_+$, we let $c_{i,j}:=-2c_i-2c_j+c_{|i-j|}+c_{i+j}$ and $\sigma _{i,j}:= -2\sigma _i-2\sigma _j+\sigma _{|i-j|}+\sigma _{i+j}$. For example, (${{\mathcal {H}}}$3) can now be restated as follows: $\sigma _i\sigma _j=-\frac{3}{8}\sigma _{i,j}$.

Lemma 2.10

For $i,j\in {{\mathbb {Z}}}_+$, we have $c_ic_j=2\sigma _{i,j}$ and $c_i\sigma _j=\frac{3}{8}c_{i,j}$.

Proof

For $i\in {{\mathbb {Z}}}_+$, set $c_i^{\pm }:=a-a_{\pm i}$. Then, by Eq. (2), $c_i=c_i^++c_i^-$ and, for $\delta , \varepsilon \in \{+,-\}$, using $({\mathcal {H}}_1)$, resp. $({\mathcal {H}}_2)$, we get

$$\begin{aligned} c_i^\varepsilon c_j^\delta =-\sigma _i-\sigma _j+\sigma _{|\varepsilon i-\delta j|},\,\, \text{ resp. } \,\, c_i^\varepsilon \sigma _j=-\frac{3}{4}c_i^\varepsilon +\frac{3}{8}(-c_j+c_{i+j}^\varepsilon +c_{|i-j|}^\varepsilon ). \end{aligned}$$

In particular, for all $i,j\in {{\mathbb {Z}}}_+$, $c_i^+c_j^+= c_i^-c_j^-$ and $c_i^+c_j^-=c_j^+c_i^-$. Thus

$$\begin{aligned} c_ic_j=\,&(c_i^++c_i^-)(c_j^++c_j^-)=2(c_i^+c_j^++c_i^+c_j^-)\\ =\,&2(-\sigma _i-\sigma _j+\sigma _{| i- j|}-\sigma _i-\sigma _j+\sigma _{ i+ j})=2\sigma _{i,j} . \end{aligned}$$

Similarly

$$\begin{aligned} c_i\sigma _j= \,&(c_i^++c_i^-)\sigma _j\\ = \,&-\frac{3}{4}c_i^++\frac{3}{8}(-c_j+c_{i+j}^++c_{|i-j|}^+)-\frac{3}{4}c_i^-+\frac{3}{8}(-c_j+c_{i+j}^-+c_{|i-j|}^-)\\ =&-\frac{3}{4}c_i -\frac{3}{4}c_j+\frac{3}{8}(c_{i+j}+c_{|i-j|})=\frac{3}{8}c_{i,j}. \end{aligned}$$

$\square$

Manifestly, all these products have a very uniform structure. Note that both $c_{i,j}$ and $\sigma _{i,j}$ are symmetric in i and j. In particular, it follows that $c_i\sigma _j=\frac{3}{8}c_{i,j}= \frac{3}{8}c_{j,i}=c_j\sigma _i$.

Similarly to the above, for $i,j\in {{\mathbb {Z}}}_+$, we introduce $u_{i,j}:=-2u_i-2u_j+u_{|i-j|}+u_{i+j}$ and $v_{i,j}:= -2v_i-2v_j+v_{|i-j|}+v_{i+j}$, where also $u_0:=0, v_0:=0$. By Eq. (3) we have

$$\begin{aligned} u_{i,j}=3c_{i,j}+4\sigma _{i,j} \, \text{ and } \,v_{i,j}=c_{i,j}-4\sigma _{i,j}. \end{aligned}$$

(5)

Now we compute the products of the vectors $u_j$ and $v_j$.

Lemma 2.11

For all $i,j\in {{\mathbb {Z}}}_+$, we have that $u_iu_j=3u_{i,j}$, $u_iv_j=-3v_{i,j}$, and $v_iv_j=-u_{i,j}$.

Proof

By Eq. (3),

$$\begin{aligned} u_iu_j=(3c_i+4\sigma _i)(3c_j+4\sigma _j)=9c_ic_i+12c_i\sigma _j+12\sigma _ic_j+16\sigma _i\sigma _j. \end{aligned}$$

By Lemma 2.10, this becomes

$$\begin{aligned} 9(2\sigma _{i,j})+12\left( \frac{3}{8}c_{i,j}\right) +12\left( \frac{3}{8}c_{i,j}\right) +16\left( -\frac{3}{8}\sigma _{i,j}\right) =9c_{i,j}+12\sigma _{i,j}=3u_{i,j}. \end{aligned}$$

Similarly,

$$\begin{aligned} u_iv_j=\,&(3c_i+4\sigma _i)(c_j-4\sigma _j)=3c_ic_j+4\sigma _ic_j-12c_i\sigma _j-16\sigma _i\sigma _j\\ =\,&3(2\sigma _{i,j})+4\left( \frac{3}{8}c_{i,j}\right) -12\left( \frac{3}{8}c_{i,j}\right) -16\left( -\frac{3}{8}\sigma _{i,j}\right) =-3c_{i,j}+12\sigma _{i,j}=-3v_{i,j} \end{aligned}$$

and

$$\begin{aligned} v_iv_j=\,&(c_i-4\sigma _i)(c_j-4\sigma _j)=c_ic_j-4\sigma _ic_j-4c_i\sigma _j+ 16\sigma _i\sigma _j\\ =\,&2\sigma _{i,j}-4\left( \frac{3}{8}c_{i,j}\right) -4\left( \frac{3}{8}c_{i,j}\right) +16\left( -\frac{3}{8}\sigma _{i,j}\right) =-3c_{i,j}-4\sigma _{i,j}=-u_{i,j}. \end{aligned}$$

$\square$

We are now ready to establish the fusion law for the axis a.

Proof of Theorem2.2 Recall that, when $\mathrm {char}({{\mathbb {F}}})\ne 3$, we have $\langle a\rangle ={{\mathcal {H}}}_1$, ${{\mathcal {H}}}_u={{\mathcal {H}}}_0$, ${{\mathcal {H}}}_v={{\mathcal {H}}}_2$, and ${{\mathcal {H}}}_w={{\mathcal {H}}}_{\frac{1}{2}}$ are the eigenspaces of $\mathrm{ad}_a$. So we need to see how these subspaces behave under the product in ${{\mathcal {H}}}$.

First of all, clearly, $\langle a\rangle {{\mathcal {H}}}_u=0$, $\langle a\rangle {{\mathcal {H}}}_v={{\mathcal {H}}}_v$ and $\langle a\rangle {{\mathcal {H}}}_w={{\mathcal {H}}}_w$. Now focus on the products of the u-, v-, and w-parts. These parts are all contained in J, and so their products are also contained in J, avoiding $\langle a\rangle$. Furthermore, the involution $\tau$ induces a grading on J, under which $V={{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v$ is the even part and ${\mathcal {H}}_w$ is the odd part. Hence, ${{\mathcal {H}}}_u{{\mathcal {H}}}_w$ and ${{\mathcal {H}}}_v{{\mathcal {H}}}_w$ are contained in ${{\mathcal {H}}}_w$, while ${{\mathcal {H}}}_w{{\mathcal {H}}}_w\subseteq V={{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v$. It remains to check the products within V. As can be seen from Lemma 2.11, ${{\mathcal {H}}}_u{{\mathcal {H}}}_u\subseteq {{\mathcal {H}}}_u$, ${{\mathcal {H}}}_u{{\mathcal {H}}}_v\subseteq {{\mathcal {H}}}_v$, and ${{\mathcal {H}}}_v{{\mathcal {H}}}_v\subseteq {{\mathcal {H}}}_u$. Hence, when $\mathrm {char}({{\mathbb {F}}})\ne 3$, a satisfies the fusion law of Table 2, which is slightly stricter than the fusion law ${\mathcal {M}}(2,\frac{1}{2})$.

Table 2 Fusion law for ${{\mathcal {H}}}$

Since D is transitive on the $a_i$, all of them are axes satisfying the law in Table 2. To complete the proof of Theorem 2.2, it remains to show that ${{\mathcal {H}}}=\langle \!\langle a_0,a_1\rangle \!\rangle$. Let $H:=\langle \!\langle a_0,a_1\rangle \!\rangle$. Note that $\sigma _1=a_0a_1-\frac{1}{2}(a_0+a_1)\in H$. Also, $\frac{3}{8}a_{-1}=a_0\sigma _1+\frac{3}{4}a_0-\frac{3}{8}a_1-\frac{3}{2}\sigma _1\in H$. Assuming that $\mathrm {char}({{\mathbb {F}}})\ne 3$, this gives us that $a_{-1}\in H$. Clearly, $\langle \!\langle a_0,a_1\rangle \!\rangle$ is invariant under the involution $\pi \in D$ switching $a_0$ and $a_1$. Now, we also see that $H=\langle \!\langle a_{-1},a_0,a_1\rangle \!\rangle$ is invariant under the involution $\tau \in D_a$. Since $D=\langle \tau ,\pi \rangle$, this makes H invariant under all of D, and so H contains all axes $a_i$. Clearly, this means that $H={{\mathcal {H}}}$. This completes the proof of Theorem 2.2.$\square$

Note that, if $\mathrm {char}({{\mathbb {F}}})\ne 3$, then, for any two distinct i and j in ${{\mathbb {Z}}}$, the subalgebra $\langle \langle a_i, a_j \rangle \rangle$ is linearly spanned by the set $\{ a_n, \sigma _n| n \text{ is } \text{ a } \text{ multiple } \text{ of } \text{ g.c.d. }(i,j)\}$. In particular $\langle \langle a_i, a_j \rangle \rangle$ is isomorphic to ${{\mathcal {H}}}$.

3 The characteristic 3 case

Let us discuss what happens when $\mathrm {char}({{\mathbb {F}}})=3$. The fusion that we established for the decomposition ${{\mathcal {H}}}=\langle a\rangle \oplus {{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v\oplus {{\mathcal {H}}}_w$ remains true, but it becomes even stricter because so many coefficients are multiples of 3. For example, ${{\mathcal {H}}}_u{{\mathcal {H}}}_u=0={{\mathcal {H}}}_u{{\mathcal {H}}}_v$. On the other hand, $2=\frac{1}{2}$ in characteristic 3, so we need to merge ${{\mathcal {H}}}_v$ and ${{\mathcal {H}}}_w$ into a single eigenspace ${{\mathcal {H}}}_{\frac{1}{2}}={{\mathcal {H}}}_v\oplus {{\mathcal {H}}}_w$. Let us see what fusion we get inside this eigenspace. We already know that ${{\mathcal {H}}}_v{{\mathcal {H}}}_v\subseteq {{\mathcal {H}}}_u$.

Lemma 3.1

If $\mathrm {char}({{\mathbb {F}}})=3$ then ${{\mathcal {H}}}_v{{\mathcal {H}}}_w=0$ and ${{\mathcal {H}}}_w{{\mathcal {H}}}_w\subseteq {{\mathcal {H}}}_u$.

Proof

Indeed, for $i,j\in {{\mathbb {Z}}}_+$, $v_iw_j=(2a-(a_{-i}+a_i)-4\sigma _i)(a_{-j}-a_j)$. Taking into account that $aw_j=\frac{1}{2}w_j$ and that $\sigma _iw_j=0$ when $\mathrm {char}({{\mathbb {F}}})=3$ [see (${{\mathcal {H}}}$2)], we obtain that

$$\begin{aligned} v_iw_j= \,&(a_{-j}-a_j)-(a_{-i}+a_i)(a_{-j}-a_j)\\ =\,&(a_{-j}-a_j)-(\frac{1}{2}(a_{-i}+a_{-j})+ \sigma _{|i-j|})-\left( \frac{1}{2}(a_i+a_{-j})+\sigma _{i+j}\right) \\&+\left( \frac{1}{2}(a_{-i}+a_j)+\sigma _{i+j}\right) + \left( \frac{1}{2}(a_i+a_j)+\sigma _{|i-j|}\right) =0. \end{aligned}$$

Also,

$$\begin{aligned} w_iw_j= \,&(a_{-i}-a_i)(a_{-j}-a_j)=\left( \frac{1}{2}(a_{-i}+a_{-j})+\sigma _{|i-j|}\right) - \left( \frac{1}{2}(a_i+a_{-j})+\sigma _{i+j}\right) \\&-\,\left( \frac{1}{2}(a_{-i}+a_j)+\sigma _{i+j}\right) + \left( \frac{1}{2}(a_i+a_j)+\sigma _{|i-j|}\right) =2\sigma _{|i-j|}-2\sigma _{i+j}\\ =\,&\frac{1}{2}u_{|i-j|}- \frac{1}{2}u_{i+j}. \end{aligned}$$

The last equality holds because $u_k=4\sigma _k$ when $\mathrm {char}({{\mathbb {F}}})=3$ [see Eq. (1)]. $\square$

So we can see now that in characteristic 3 the axes $a_i$ in ${{\mathcal {H}}}$ satisfy the fusion law in Table 3.

Table 3 Fusion law for ${{\mathcal {H}}}$ in characteristic 3

In particular, ${{\mathcal {H}}}$ is an algebra of Jordan type $\frac{1}{2}$. Let us prove that, in fact, ${{\mathcal {H}}}$ is a Jordan algebra when $\mathrm {char}({{\mathbb {F}}})=3$. For this, we need to verify the Jordan identity $x(yx^2)=(xy)x^2$ for all $x,y\in {{\mathcal {H}}}$. We obtain this from the following more general result.

Proposition 3.2

Let B be a commutative baric algebra over a field of characteristic other than 2, with weight function $\omega$. Suppose that

• (a) $B=A\oplus I$, , where A is a subspace and I is an ideal,

• (b) $IB=0$, and

• (c) for all $a,b\in A$, $ab-\frac{1}{2}(\omega (b)a+\omega (a)b)\in I$.

Then, B is a Jordan algebra.

Proof

Let $\mu :B\rightarrow A$ be the projection onto A with respect to the decomposition $B=A\oplus I$. Then, every $x\in B$ can be written in a unique way as $x=\mu (x)+x_I$, where $\mu (x)\in A$ and $x_I\in I$. Let $x,y\in B$. By hypothesis (b), $xy=\mu (x)\mu (y)=\mu (x)y$ and, by hypothesis (c), $\mu (x^2)=\omega (x)\mu (x)$. Hence, using the commutativity of B, we have

$$\begin{aligned} x(yx^2)&=\mu (x)(\mu (y)\mu (x^2)) =\mu (x)(\mu (y)\omega (x)\mu (x))=\omega (x)\mu (x)(\mu (y)\mu (x)) \\&=(\mu (x)\mu (y))\omega (x)\mu (x)= (\mu (x)\mu (y))\mu (x^2)=(xy)x^2. \end{aligned}$$

Therefore B is a Jordan algebra. $\square$

Corollary 3.3

Let A and I be vector spaces over a field ${{\mathbb {F}}}$ of characteristic other than 2, let $\omega :A\rightarrow {{\mathbb {F}}}$ be a linear map and $\sigma :A\times A\rightarrow I$ a symmetric bilinear map. Let $B:=A\oplus I$ and define an algebra product on B by $BI=0$ and, for $a,b\in A$, $ab:= \frac{1}{2}(\omega (b)a+\omega (a)b)+\sigma (a,b)$. Then B is a baric Jordan algebra.

Proof

Define $\overline{\omega }:B\rightarrow {{\mathbb {F}}}$ by $\overline{\omega }(a+i)=\omega (a)$, for every $a\in A$, $i\in I$. Then, $\overline{\omega }$ is a weight function on B (note that it may not be the only choice for a weight function) and B satisfies the hypotheses of Proposition 3.2. The claim follows. $\square$

Theorem 3.4

If $\mathrm {char}({{\mathbb {F}}})=3$ then ${{\mathcal {H}}}$ is a baric Jordan algebra.

Proof

It is enough to show that ${\mathcal {H}}$ satisfies the hypotheses of Proposition 3.2. We let $A:=\langle a_i\,|\,i\in {{\mathbb {Z}}}\rangle$ and $I:=\langle \sigma _j\,|\,j\in {{\mathbb {Z}}}_+\rangle$. Then clearly ${\mathcal {H}}$ is the direct sum of A and I and, by (${{\mathcal {H}}}$2) and (${{\mathcal {H}}}$3), since $\mathrm {char}({{\mathbb {F}}})=3$, we have $I{{\mathcal {H}}}=0$ (in particular I is an ideal). Let $\mu :{\mathcal {H}} \rightarrow A$ be the projection onto A with respect to the decomposition ${\mathcal {H}}=A\oplus I$. Let $x,y\in A$ and write $x=\sum _{i\in {{\mathbb {Z}}}}r_ia_i$ and $y=\sum _{i\in {{\mathbb {Z}}}}t_ia_i$. Note that $\lambda (x)=\sum _{i\in {{\mathbb {Z}}}}r_i$ and $\lambda (y)=\sum _{i\in {{\mathbb {Z}}}}t_i$. By (${{\mathcal {H}}}$1), we have

$$\begin{aligned} \mu (xy)&=\sum _{i,j\in {{\mathbb {Z}}}}r_it_j\frac{1}{2}(a_i+a_j)\\&=\sum _{i,j\in {{\mathbb {Z}}}}r_it_j\frac{1}{2}a_i+\sum _{i,j\in {{\mathbb {Z}}}}r_it_j\frac{1}{2}a_j\\&=\frac{1}{2}\left( \sum _{j\in {{\mathbb {Z}}}}t_j\right) \left( \sum _{i\in {{\mathbb {Z}}}}r_ia_i\right) +\frac{1}{2}\left( \sum _{i\in {{\mathbb {Z}}}}r_i\right) \left( \sum _{j\in {{\mathbb {Z}}}}t_ja_j\right) \\&=\frac{\lambda (y)}{2}x+\frac{\lambda (x)}{2}y. \end{aligned}$$

This proves that $xy-\frac{1}{2}(\lambda (y)x+\lambda (x)y)$ belongs to I. $\square$

4 Final remarks

Now that we have proved our main results, let us tie the loose ends. In this section, ${{\mathbb {F}}}$ is a field of characteristic other than 2.

4.1 Frobenius form

We refer to [5] for a detailed discussion of the radical of an axial algebra, projection form and projection graph.

Proposition 4.1

• (a) The form $(\cdot ,\cdot )$ is, up to a scalar factor, the only Frobenius form on ${{\mathcal {H}}}$.

• (b) Every proper ideal of ${{\mathcal {H}}}$ is contained in J, which is the radical of ${{\mathcal {H}}}$.

Proof

The form $(\cdot ,\cdot )$ that we introduced on ${{\mathcal {H}}}$ is a projection form, because it is a Frobenius form satisfying $(a_i,a_i)=1$ for all $i\in {{\mathbb {Z}}}$. The projection graph has all $a_i$ as vertices with an edge between $a_i$ and $a_j$, $i\ne j$, whenever $(a_i,a_j)\ne 0$. In fact, since $(a_i,a_j)=1$ for all $i,j\in {{\mathbb {Z}}}$, the projection graph of ${{\mathcal {H}}}$ is the complete graph. In particular, it is connected. It now follows from [5, Proposition 4.19] that ${{\mathcal {H}}}$ has only one Frobenius form up to a scalar factor. The same connectivity property implies, by [5, Corollary 4.15], that every proper ideal of ${{\mathcal {H}}}$ is contained in the radical of ${{\mathcal {H}}}$ (the largest ideal not containing axes). Finally, by [5, Corollary 4.11], the radical of ${{\mathcal {H}}}$ coincides with the radical of the projection form $(\cdot ,\cdot )$, and that is J. $\square$

4.2 Automorphisms of the subalgebra V

Recall that in Section 2, we introduced the subalgebra $V={{\mathcal {H}}}_u\oplus {{\mathcal {H}}}_v$ and its two bases $\{u_i, v_j\,|\, i,j\in {{\mathbb {Z}}}_+\}$ and $\{c_i, \sigma _j\,|\, i,j\in {{\mathbb {Z}}}_+\}$. Let $\theta$ and $\psi$ be the linear transformations $V\rightarrow V$ defined by

$$\begin{aligned} (u_j)^\theta = u_j, \,\,\, (v_j)^\theta = -v_j \end{aligned}$$

and

$$\begin{aligned} (c_j)^\psi =\frac{1}{2}v_j=\frac{1}{2}c_j-2\sigma _j,\,\,\,\, (\sigma _j)^\psi =-\frac{1}{8}u_j=-\frac{3}{8}c_j-\frac{1}{2}\sigma _j. \end{aligned}$$

Lemma 4.2

The linear transformations $\theta$ and $\psi$ defined above are automorphisms of V.

Proof

This follows from the formulae in Lemmas 2.10 and 2.11. $\square$

It is clear that $\theta$ has order 2. Note that $\psi$ is also an involution, because the matrix

$$\begin{aligned} \left( \begin{array}{rr} \frac{1}{2} &{}\quad -2\\ -\frac{3}{8} &{}\quad -\frac{1}{2} \end{array}\right) \end{aligned}$$

squares to identity, and also that $\psi \rho$ is an element of order 3. This means that $T:=\langle \theta ,\psi \rangle \cong D_6$. It is interesting that the subalgebra V has symmetries independent of the entire algebra ${{\mathcal {H}}}$. Indeed, we claim that only the identity element from T extends to an element of $\mathrm{Aut}({{\mathcal {H}}})=D$. Define the support of an element

$$\begin{aligned} e=\sum _{i\in {{\mathbb {Z}}}}r_i(e) a_i+\sum _{j\in {{\mathbb {Z}}}_+}s_j(e)\sigma _j \end{aligned}$$

as $\{i\in {{\mathbb {Z}}}\mid r_i(e)\ne 0\}$. The orbit under T of $c_1=2a-(a_{-1}+a_1)$ consists of the six elements $\pm c_1$, $\pm \frac{1}{2}v_1=\pm (\frac{1}{2}c_1-2\sigma _1)$ and $\pm (\frac{1}{2}c_1+2\sigma _1)$. All these elements have support $\{-1,0,1\}$. Hence if $t\in T$ extends to an element of $d\in D$, then d, as an automorphism of ${{\mathbb {Z}}}$, must preserve the set $\{-1,0,1\}$. Manifestly, since the automorphisms of ${{\mathbb {Z}}}$ are either translations or reflections, the identity $1_D$ and $\tau$ are the only elements of D preserving this set. Both $1_D$ and $\tau$ act trivially on V, so $t=1_T$, as claimed.

5 Discussion

In this final section, we pose some questions related to ${{\mathcal {H}}}$. We have to warn the reader that there is a lot of work in progress on this topic and many results quoted in this section have not yet been published nor undergone a blind refereeing process.

By Yabe’s results, or by the preprint of the authors mentioned in the introduction, every symmetric 2-generated primitive algebra of Monster type $(\alpha ,\beta )$ over a field of characteristic other than 2 is at most 8-dimensional unless $(\alpha ,\beta )=(2,\frac{1}{2})$. In the non-symmetric case, such a bound has been proved only for $\alpha \ne 4\beta$ (see [16] and the paper by the authors posted in Mathematics arXiv:2101.10379).

Question 5.1

Is ${{\mathcal {H}}}$ the only infinite-dimensional primitive algebra of Monster type? Is it the only such algebra having dimension greater than 8?

In characteristic other than 5, by Yabe’s result, for the symmetric case, the answer to this question depends on the knowledge of the ideals of ${{\mathcal {H}}}$. In characteristic 5, the first two authors have shown that the algebra ${\mathcal {H}}$ admits a proper cover which is still a symmetric primitive algebra of Monster type $(2,\frac{1}{2})$ (Mathematics arXiv:2101.09506).

Question 5.2

Does ${{\mathcal {H}}}$ contain any nonzero ideal of infinite codimension? Is it possible to classify all ideals in ${{\mathcal {H}}}$?

Note that, if $\mathrm {char}( {{\mathbb {F}}})\ne 3$, J is not the only proper nonzero ideal of ${{\mathcal {H}}}$. For example, the ideal I generated by all elements $\sigma _j$ has codimension 1 in J (and hence codimension 2 in ${{\mathcal {H}}}$). Indeed, from $({\mathcal {H}}_2)$ we get that, for all $i\in {{\mathbb {Z}}}_+$,

$$\begin{aligned} a_0\sigma _i=-\frac{3}{8}c_i+\frac{3}{2}\sigma _i , \end{aligned}$$

whence $c_i\in I$, and $u_i, v_i\in I$. Further, again by $({\mathcal {H}}_2)$, we have

$$\begin{aligned} w_1\sigma _1=-\frac{3}{4}w_1+\frac{3}{8}w_2 \text{ and, } \text{ for } i>1, \,\,w_i\sigma _1=-\frac{3}{4}w_i+\frac{3}{8}(w_{i+1}-w_{i-1}) \end{aligned}$$

thus, recursively, we get that $w_i\in \langle w_1\rangle +I$ for all $i\in {{\mathbb {Z}}}_+$. Hence I has codimension at most 1 in J. On the other hand, consider the 2-dimensional commutative algebra ${{\mathbb {F}}}{\bar{a}}_0\oplus {{\mathbb {F}}}{\bar{a}}_1$ with multiplication defined by ${\bar{a}}_i{\bar{a}}_i={\bar{a}}_i$, $i=0,1$ and ${\bar{a}}_0{\bar{a}}_1=\frac{1}{2}({\bar{a}}_0+{\bar{a}}_1)$ and let $\varphi$ be the linear map sending $a_i$ to $i{\bar{a}}_1-(i-1){\bar{a}}_0$, for all $i\in {{\mathbb {Z}}}$, and $\sigma _j$ to 0, for all $j\in {{\mathbb {Z}}}_+$. Then $\varphi$ is a surjective algebra homomorphism with $I\le \ker \varphi$, whence, by comparing the codimensions, $I=\ker \varphi$ has codimension 2 in ${\mathcal {H}}$.

In addition to finding all ideals of ${{\mathcal {H}}}$, it is also interesting to find all of its subalgebras and their automorphisms.

Question 5.3

Which is the full automorphism group of ${{\mathcal {H}}}$ in characteristic 3? In any characteristic other than 2, which is the automorphism group of V? Which automorphisms of V extend to larger subalgebras of ${{\mathcal {H}}}$?

Finally, the fact that ${{\mathcal {H}}}$ is a baric algebra looks exciting. A baric axial algebra of Monster type $(\alpha ,\beta )$ satisfies the fusion law as in Table 2, but with 2 and $\frac{1}{2}$ substituted with arbitrary $\alpha$ and $\beta$. Indeed, for every $\eta \in \{0,\alpha , \beta \}$ and for every $\eta$-eigenvector v, the linearity of the weight function forces v to be in the kernel of the weight function and the claim follows since the kernel is an ideal.

Question 5.4

Are there examples of baric algebras of Monster type $(\alpha ,\beta )$ for other values of $\alpha$ and $\beta$? Is it possible to classify all such algebras?