2-generated axial algebras of Monster type (2β, β)
aa r X i v : . [ m a t h . R A ] J a n -GENERATED AXIAL ALGEBRAS OF MONSTER TYPE (2 β, β ) CLARA FRANCHI, MARIO MAINARDIS, SERGEY SHPECTOROV
Abstract.
In this paper we prove that 2-generated primitive axial algebrasof Monster type (2 β, β ) over a ring R in which 2 and β are invertible can begenerated as R -module by 8 vectors. We then completely classify 2-generatedprimitive axial algebras of Monster type (2 β, β ) over any field of characteristicother than 2. Introduction
Axial algebras constitute a class of commutative non-associative algebras gen-erated by certain idempotent elements (called axes ) such that their adjoint actionis semisimple and the relative eigenvectors satisfy a prescribed fusion law. Let R be a ring, { α, β } ⊆ R \ { , } and α = β . An axial algebra over R is called ofMonster type ( α, β ) if it satisfies the fusion law M ( α, β ) given in Table 1. Thismeans that the adjoint action of every axis has spectrum { , , α, β } and, for anytwo eigenvectors v γ , v δ with relative eigenvalues γ, δ ∈ { , , α, β } , the product v γ · v δ is a sum of eigenvectors relative to eigenvalues contained in γ ⋆ δ . This classwas introduced by J. Hall, F. Rehren and S. Shpectorov [8] in order to axiomatisesome key features of many important classes of algebras, such as the weight-2 com-ponents of OZ-type vertex operator algebras, Jordan algebras and Matsuo algebras(see the introductions of [8], [15] and [5]). They are also of particular interest forfinite group theorists as most of the finite simple groups, or their automorphs, canbe faithfully and effectively represented as automorphism groups of these algebras.In [15, 14], F. Rehren proved that every 2-generated primitive axial algebras ofMonster type ( α, β ) over a ring R in which 2, α , β , α − β , α − β , and α − β areinvertible can be generated as R -module by 8 vectors and computed the structureconstants with respect to these elements. This result has been re-proved in a slightlysimpler way by the authors in [3], where a bound for the special case of 2-generatedprimitive axial algebras of Monster type (4 β, β ) over a field of odd characteristicwas also obtained under the hypothesis that β = 1 / ⋆ α β ∅ α β ∅ α βα α α , ββ β β β , , α Table 1.
Fusion law M ( α, β ) algebra. On the other hand, in [4] an example of a 2-generated symmetric primitiveaxial algebra of Monster type (2 , ) of infinite dimension is constructed.In this paper, we focus on 2-generated primitive axial algebras of Monster type(2 β, β ) and we assume that R has characteristic other than 2. Denote by R theprime subring of R and let R [ , β, β ][ x, y, z, t ] be the polynomial ring in 4 variablesover R [ , β, β ]. We prove the following result. Theorem 1.1.
There exists a subset T ⊆ R [ , β, β ][ x, y, z, t ] of size , dependingonly on R and β , such that every -generated primitive axial algebra of Monstertype (2 β, β ) over R is completely determined, up to homomorphic images, by aquadruple ( x , y , z , t ) ∈ R which is a common zero of all the elements of T . Inparticular, every -generated primitive axial algebra of Monster type (2 β, β ) over R is linearly spanned by at most vectors. In Section 4, we give explicitly some of the polynomials of the set T and showhow to compute those that are too long to be written down here. In the symmetriccase, the knowledge of the set T is enough to obtain all quadruples ( x , y , z , t )corresponding to primitive axial algebras of Monster type (2 β, β ) over any field F of characteristic either than 2. In Section 5 we give a complete classification ofthese algebras (Theorem 5.7). Note that, in this case, our results confirm those ofT. Yabe [17] but are independent on them.In Section 6, we classify the non-symmetric algebras. In [5], A. Galt, V. Joshi,A. Mamontov, S. Shpectorov and A. Staroletov introduced the concept of doubleaxis in a Matsuo algebra M η (Γ). They proved that double axes satisfy the fusionlaw M (2 η, η ) and classified the primitive subalgebras of M η (Γ) generated by twodouble axes or by an axis and a double axis: besides the algebras of Jordan type1 A , 2 B , 3 C ( η ) and 3 C (2 η ), they found three new algebras of dimensions 4, 5 and8 which we refer to by Q ( η ), V ( η ), and V ( η ), respectively. Theorem 1.2.
Let V be a -generated primitive axial algebra of Monster type (2 β, β ) over a field F of characteristic other than . Then, either V is symmetric(and it is described in Theorem 5.7), or V is isomorphic to Q ( β ) or to its -dimensional quotient when β = − . Basics
We start by recalling the definition and basic features of axial algebras. Let R be a ring with identity where 2 is invertible and let S be a finite subset of R with1 ∈ S . A fusion law on S is a map ⋆ : S × S → S . An axial algebra over R with spectrum S and fusion law ⋆ is a commutative non-associative R -algebra V generated by a set A of nonzero idempotents (called axes )such that, for each a ∈ A ,(Ax1) ad ( a ) : v av is a semisimple endomorphism of V with spectrum containedin S ;(Ax2) for every λ, µ ∈ S , the product of a λ -eigenvector and a µ -eigenvector ofad a is the sum of δ -eigenvectors, for δ ∈ λ ⋆ µ .Furthermore, V is called primitive if(Ax3) V = h a i . An axial algebra over R is said to be of Monster type ( α, β ) if it satisfies the fusionlaw M ( α, β ) given in Table 1, with α, β ∈ F \ { , } , with α = β .Let V be an axial algebra of Monster type ( α, β ) and let a ∈ A . Let S + := { , , α } and S − := { β } . The partition {S + , S − } of S induces a Z -grading on S which, on turn, induces, a Z -grading { V a + , V a − } on V where V a + := V a + V a + V aα and V a − = V aβ . It follows that, if τ a is the map from R ∪ V to R ∪ V such that τ a | V isthe nultilication by − V aβ and leaves invariant the elements of V a + and τ a | R is theidentity, then τ a is an involutory automorphism of V (see [8, Proposition 3.4]). Themap τ a is called the Miyamoto involution associated to the axis a . By definition of τ a , the element av − βv of V is τ a -invariant and, since a lies in V a + ≤ C V ( τ a ), also av − β ( a + v ) is τ a -invariant. In particular, by symmetry, Lemma 2.1.
Let a and b be axes of V . Then ab − β ( a + b ) is fixed by the 2-generatedgroup h τ a , τ b i . If V is generated by the set of axes A := { a , a } , for i ∈ { , } , let τ i be theMiyamoto involutions associated to a i . Set ρ := τ τ , and for i ∈ Z , a i := a ρ i and a i +1 := a ρ i . Since ρ is an automorphism of V , for every j ∈ Z , a j is an axis.Denote by τ j := τ a j the corresponding Miyamoto involution. Lemma 2.2.
For every n ∈ N , and i, j ∈ Z such that i ≡ j mod n we have a i a i + n − β ( a i + a i + n ) = a j a j + n − β ( a j + a j + n ) , Proof.
This follows immediately from Lemma 2.1. (cid:3)
For n ∈ N and r ∈ { , . . . , n − } set(1) s r,n := a r a r + n − β ( a r + a r + n ) . If { , , α, β } are pairwise distinguishable in R , i.e. α , β , α − β −
1, and α − β are invertible in R , by [3, Proposition 2.4], for every a ∈ A , there is a function λ a : V → R , such that every v ∈ V can be written as v = λ a ( v ) a + u with u ∈ M δ =1 V aδ . For i ∈ Z , let(2) a i = λ a ( a i ) a + u i + v i + w i be the decomposition of a i into ad a -eigenvectors, where u i is a 0-eigenvector, v i isan α -eigenvector and w i is a β -eigenvector. From now on we assume 0 , , α, β arepairwise distinguishable in R . Lemma 2.3.
With the above notation,(1) u i = α (( λ a ( a i ) − β − αλ a ( a i )) a + ( α − β )( a i + a − i ) − s ,i ) ;(2) v i = α (( β − λ a ( a i )) a + β ( a i + a − i ) + s ,i ) ;(3) w i = ( a i − a − i ) . Lemma 2.4.
Let I be an ideal of V , a an axis of V , x ∈ V and let x = x + x + x α + x β be the decomposition of x as sum of ad a -eigenvectors. If x ∈ I , then x , x , x α , x β ∈ I . Moreover, I is τ a -invariant. CLARA FRANCHI, MARIO MAINARDIS, SERGEY SHPECTOROV
Proof.
Suppose x ∈ I . Then I contains the vectors x − ax = x + (1 − α ) x α + (1 − β ) x β ,a ( x − ax ) = α (1 − α ) x α + β (1 − β ) x β ,a ( a ( x − ax )) − βa ( x − ax ) = α ( α − β )(1 − α ) x α . Since, 0 , , α, β are pairwise distinguishable in R , it follows that I contains x , x , x α , x β . Since x τ a = x + x + x α − x β ∈ I , the last assertion follows. (cid:3) The multiplication table
From now on we assume that α = 2 β , { , , β, β } is a set of pairwise distinguish-able elements in R , and 2 is invertible in R . Let V be the universal 2-generatedprimitive axial algebra over the ring R as defined in [3] and let A := { a , a } be itsgenerating set of axes. That is, R and V are defined as follows- D is the polynomial ring Z [ x i , y i , w i , t | i, j ∈ { , } , i < j ] , where x i , y i , w, z i,j , t are algebraically independent indeterminates over Z ,for i, j ∈ { , } , with i < j ;- L is the ideal of D generated by the setΣ := { x i y i − , (1 − x i ) w i − , t − , x − x | i ∈ { , }} ;- ˆ D := D/L . For d ∈ D , we denote the element L + d by ˆ d .- W is the free commutative magma generated by the elements of A subjectto the condition that every element of A is idempotent;- ˆ R := ˆ D [Λ] is the ring of polynomials with coefficients in ˆ D and indetermi-nates set Λ := { λ c,w | c ∈ A , w ∈ W, c = w } where λ c,w = λ c ′ ,w ′ if and onlyif c = c ′ and w = w ′ .- ˆ V := ˆ R [ W ] is the set of all formal linear combinations P w ∈ W γ w w of theelements of W with coefficients in ˆ R , with only finitely many coefficientsdifferent from zero. Endow ˆ V with the usual structure of a commutativenon associative ˆ R -algebra.For i ∈ Z , set λ i := λ a ( a i ) . By Corollary 3.7 in [3], the permutation that swaps a with a induces a auto-morphism f of V such that λ a ( a ) = λ f , and λ a ( a − ) = λ f . Set T := h τ , τ i and T := h τ , f i . Lemma 3.1.
The groups T and T are dihedral groups, T is a normal subgroupof T such that | T : T | ≤ . For every n ∈ N , the set { s ,n , . . . , s n − ,n } is invariantunder the action of T . In particular, if K n is the kernel of this action, we have(1) K = T ;(2) K = T , in particular s f , = s , ;(3) T /K induces the full permutation group on the set { s , , s , , s , } withpoint stabilisers generated by τ K , τ K and f K , respectively. In partic-ular s f , = s , and s τ , = s , .Proof. This follows immediately from the definitions. (cid:3)
For i, j ∈ { , , } , with the notation fixed before Lemma 2.3, set P ij := u i u j + u i v j and Q ij := u i v j − α s ,i s ,j . Lemma 3.2.
For i, j ∈ { , , } we have (3) s ,i · s ,j = α ( a P ij − αQ ij ) . Proof.
Since u i and v j are a 0-eigenvector and an α -eigenvector for ad a , respec-tively, by the fusion rule, we have a P ij = α ( u i · v j ) and the result follows. (cid:3) The following polynomial will play a crucial rˆole in the classification of the nonsymmetric algebras in Section 6: Z ( x, y ) := 2 β x + (2 β − β y − (4 β − β . Lemma 3.3. In V the following equalities hold: s , = − β a − + a ) + βZ ( λ , λ f )( a + a − ) − h Z ( λ , λ f )( λ − β ) − ( λ − β ) i a + 2 Z ( λ , λ f ) s , . and s , = − β a − + a ) + βZ ( λ f , λ )( a + a ) − h Z ( λ f , λ )( λ f − β ) − ( λ f − β ) i a + 2 Z ( λ f , λ ) s , . Proof.
Since α = 2 β , from the first formula in [3, Lemma 4.7] we deduce theexpression for s , . The expression for s , follows by applying f . (cid:3) Lemma 3.4. In V we have a = a − − Z ( λ , λ f )( a − − a )+ 1 β h Z ( λ , λ f ) ( λ − β ) − (2 λ − β ) i ( a − a ) and a − = a − Z ( λ f , λ )( a − a − )+ 1 β h Z ( λ f , λ ) (cid:16) λ f − β (cid:17) − (cid:16) λ f − β (cid:17)i ( a − a − ) . Proof.
Since s , is invariant under τ , we have s , − s τ , = 0. On the other hand,in the expression s , − s τ , obtained from the first formula of Lemma 3.3, thecoefficient of a is − β/
2, which is invertible in R . Hence we deduce the expressionfor a . By applying the map f to the expression for a we get the expression for a − . (cid:3) CLARA FRANCHI, MARIO MAINARDIS, SERGEY SHPECTOROV
Lemma 3.5. In V we have s , s , =+ β Z ( λ f , λ )( a − + a )+ 12 (cid:20) − β − λ + λ f ) − (8 β − β + 1) β λ λ f + (16 β − β + 1) λ +(14 β − β + 1) λ f − β (14 β − β + 1) i ( a − + a )+ (cid:20) β − β λ + (8 β − β + 1) β λ λ f + 2(2 β − β λ λ f − (18 β − λ − β − β + 1) β λ λ f − β − λ f − (2 β − λ λ − βλ f λ + (54 β − β + 1)2 λ + (9 β − β + 1) λ f + β (5 β − λ − β λ f − β (24 β − β + 1)2 (cid:21) a + (cid:20) − β − β λ − (6 β − β + 1) β λ λ f − β − β λ f + (16 β − β + 1) β λ + (10 β − β + 1) β λ f − β λ f − (57 β + 26 β − (cid:21) s , + β s , . Proof.
By Lemma 3.3, Lemma 3.4, and Lemma 4.3 in [3], we may compute theexpression on the right hand side of the formula in Lemma 3.2, with i = j = 1, andthe result follows. (cid:3) Lemma 3.6. In V we have s , = s , + βZ ( λ f , λ ) a − − βZ ( λ , λ f ) a + 1 β h − β (2 β − λ + λ f ) − β − β + 1) λ λ f + 2 β (15 β − β + 1) λ + β (26 β − β + 2) λ f − β (24 β − β + 2) i a − + 1 β h β (2 β − λ + 4(8 β − β + 1) λ λ f + 8 β (2 β − λ λ f − β (15 β − λ − β (32 β − β + 3) λ λ f + 4 β λ f − β (2 β + 1) λ λ − β λ f λ + 2 β (40 β − λ + 2 β (2 β − β + 1) λ f + 2 β (5 β − λ − β λ f − β (5 β − i a + 1 β h − β (2 β − λ λ f − β − β + 1) λ λ f − β (2 β − λ f − β λ +2 β (32 β − β + 3) λ λ f + 4 β (16 β − λ f + 4 β λ λ f +2 β (2 β − λ f λ f − β (2 β + 5 β + 1) λ − β (40 β − λ f + 2 β λ − β (5 β − λ f + 4 β (5 β − i a + 1 β h β (2 β − λ + 2(8 β − β + 1) λ λ f + β (8 β − λ f − β (26 β − β + 2) λ − β (15 β − β + 1) λ f + β (24 β − β + 2) i a + 1 β h − λ − λ f ) + 24 β ( λ − λ f ) + 2 β ( λ − λ f ) i s , . Similarly, s , belongs to the linear span of the elements a − , a − , a , a , a , a , s , , and s , .Proof. Since, by Lemma 3.1, s , is invariant under f , we have s , s , − ( s , s , ) f =0. Comparing the expressions for s , s , and ( s , s , ) f obtained from Lemma 3.5,we deduce the expression for s , . By applying the map τ to the expression for s , we get the expression for s , and the last assertion follows from Lemma 3.4. (cid:3) As a consequence of the resurrection principle [11, Lemma 1.7], we can now provethe following result, which, by Theorem 3.6 and Corollary 3.8 in [3], implies thesecond part of Theorem 1.1. We use double angular brackets to denote algebrageneration and singular angular brackets for linear span.
Proposition 3.7. V = h a − , a − , a , a , a , a , s , , s , i .Proof. Set U := h a − , a − , a , a , a , a , s , , s , i . By Lemma 3.4, a , a − ∈ U , andby Lemma 3.6 also s , and s , belong to U . It follows that U is invariant under themaps τ , τ , and f . Hence, a i belongs to U , for every i ∈ Z . Now we show that U isclosed under the algebra product. Since it is invariant under the maps τ , τ , and f , it is enough to show that it is invariant under the action of ad a and it contains s , s , , s , s , , and s , s , . The products a a i , for i ∈ {− , − , , , , } belongto U by the definition of U and by Lemma 3.3. By [3, Lemma 4.3], U contains a s , and a s , . The product s , s , belongs to U by Lemma 3.5 and similarly,by Lemma 2.3 and Lemma 3.2, the products s , s , , and s , s , belong to U .Hence U is a subalgebra of V and, since it contains the generators a and a , weget U = V . (cid:3) Remark 3.8.
Note that the above proof gives a constructive way to compute thestructure constants of the algebra V relative to the generating set B . This has beendone with the use of GAP [6] . The explicit expressions however are far too long tobe written here. Corollary 3.9. R is generated as a ˆ D -algebra by λ , λ , λ f , and λ f .Proof. By Proposition 3.7, V is generated as R -module by the set B := { a − , a − ,a , a , a , a , s , , s , } . Since λ a ( v ) = ( λ a ( v f )) f , λ a is a linear function, and R = R f , we just need to show that λ a ( v ) ∈ ˆ D [ λ , λ f , λ , λ f ] for every v ∈ B . Bydefinition we have λ a ( a ) = 1 , λ a ( a ) = λ , λ a ( a ) = λ , and λ a ( a ) = λ . Since τ fixes a and is an R -automorphism of V , we get λ a ( a − ) = λ a (( a ) τ ) = λ ,λ a ( a − ) = λ a (( a ) τ ) = λ , and λ a ( s , ) = λ a ( a a − βa − βa ) = λ − β − βλ , CLARA FRANCHI, MARIO MAINARDIS, SERGEY SHPECTOROV and λ a ( s , ) = λ a ( aa − βa − βa ) = λ − β − βλ . We conclude the proof by showing that λ ∈ ˆ D [ λ , λ f , λ , λ f ]. Set φ := u u − v v − λ a ( u u − v v ) a and z := φ − β − λ ) u . Then, by the fusion law, φ is a 0-eigenvector for ad a and so z is a 0-eigenvectorfor ad a as well. Since s , is τ -invariant, it lies in V a + and the fusion law impliesthat the product zs , is a sum of a 0- and an α -eigenvector for ad a . In particular, λ a ( zs , ) = 0. By Remark 3.8 we can compute explicitly the product zs , : zs , = − β a + β h βλ − λ f − β ( β − i a − + (cid:20) − β λ − (2 β + β − λ λ f − (2 β − β + 1)2 β λ f + (4 β − β − β )2 λ +(2 β − β + 1) λ f + β λ − β λ f + β (4 β − (cid:21) a − + (cid:20) β − λ + (2 β − β λ λ f − (10 β − β + 1) λ + ( − β + 3 β − λ λ f − β (2 β − λ λ + β (2 β − λ + β (2 β − λ f + β ( β − λ − β (4 β − (cid:21) a + (cid:20) − β ) λ − ( β + 3)(2 β − λ λ f − (6 β − β + 1)2 β λ f + β (4 β + 3 β − λ +(8 β − β + 1) λ f + bt λ + β λ f − β (17 β − β + 2)4 (cid:21) a + (cid:20) β (3 β − λ + β (4 β − λ f − β (3 β − (cid:21) a + (cid:20) − βλ − (2 β − λ λ f + β (4 β + 1) λ + (2 β − λ f + β λ − β (9 β + 2)2 (cid:21) s , . Since λ a ( zs , ) = 0, taking the image under λ a of both sides, we get λ = 8( β − β λ − β + β − β λ λ f − β − β λ λ f − β − β + 1) β λ + 16(2 β − β λ λ f + 6 β λ λ + 2(2 β − β λ f λ + ( β − β + 4) β λ − β − β λ f − β + 1) β λ + 2(5 β − β . (cid:3) We conclude this section with some relations in V and R which will be useful inthe sequel for the classification of the algebras. Set d := s f , − s , , d := d τ , and, for i ∈ { , } , D i := d iτ − d i ; e := u τ v τ and E := a e − βe. Lemma 3.10.
The following identities hold in V , for i ∈ { , } :(1) d i = 0 , , D i = 0 , E = 0 ;(2) there exists an element t ( λ , λ f , λ , λ f ) ∈ R such that t ( λ , λ f , λ , λ f ) a + 2 β ( λ − λ f ) h βλ + ( β − λ f − β ) i ( a − + a + 2 β s , ) = 0 . Proof.
Identities involving the d i ’s and D i ’s follow from Lemma 3.1. By the fusionlaw, the product u u is a 0-eigenvector for ad a and the product u τ v τ is a 2 β -eigenvector for ad a . The last claim follows by an explicit computation of theproduct a ( u u ), which gives the left hand side of the equation. (cid:3) Lemma 3.11.
In the ring R the following holds:(1) λ a ( a a − a ) = 0 ,(2) λ a ( d ) = 0 ,(3) λ a ( d ) = 0 ,(4) λ a ( d ) = 0 .Proof. The first equation follows from the fact that a is an idempotent. Theremaining follow from Lemma 3.10. (cid:3) Strategy for the classification
By Remark 3.8, the four expressions on the left hand side of the identities inLemma 3.11 can be computed explicitly and produce respectively four polynomials p i ( x, y, z, t ) for i ∈ { , . . . , } in ˆ D [ x, y, z, t ] (with x, y, z, t indeterminates on ˆ D ),that simultaneously annihilate on the quadruple ( λ , λ f , λ , λ f ). We define also,for i ∈ { , , } , q i ( x, z ) := p i ( x, x, z, z ). The polynomials p i ’s are too long to bedisplayed here but can be computed using [1] or [6], while the polynomials q i arethe following q ( x, z ) =128 β ( − β + 608 β − β + 114 β − β + 1) x + 64 β (4352 β − β + 2992 β − β − β + 23 β − x + 64 β (64 β − β + 52 β − β + 1) x z + 16 β ( − β + 42912 β − β − β + 3477 β − β + 44) x + 16 β ( − β + 4832 β − β + 747 β − β + 4) x z + 32 β (8 β − β + 1) x z + 8 β (84832 β − β − β + 55573 β − β + 3262 β − x
40 CLARA FRANCHI, MARIO MAINARDIS, SERGEY SHPECTOROV + 8 β (19792 β − β + 17700 β − β + 647 β − x z + 16 β ( − β + 62 β − β + 1) x z + 8 β ( − β − β + 119184 β − β + 27054 β − β + 240) x + 4 β ( − β + 81156 β − β + 13527 β − β + 96) x z + 32 β (48 β − β + 12 β − xz + 4 β (2 β − z + 4 β (19648 β + 114384 β − β + 128262 β − β + 5598 β − x + 8 β (16288 β − β + 14904 β − β + 593 β − xz + 2 β ( − β + 301 β − β + 8) z + 8 β ( − β + 40040 β − β + 6959 β − β + 56) x + 2 β ( − β + 23658 β − β + 4110 β − β + 32) z + 4 β (7632 β − β + 6932 β − β + 286 β − ,q ( x, z ) == − β − β + 1) β x + (160 β − β − β + 8) β x + (8 β − β x z − (96 β + 96 β − β + 20) β x − (44 β − β + 4) β xz + (140 β − β + 16) β x + (36 β − β + 4) β z − (36 β − β + 4) β ,q ( x, z ) =( − β + 160 β − β + 8) β x + (64 β − β + 8) β x z + (288 β − β + 20 β + 40 β − β x + ( − β + 48 β + 12 β − β x z − (8 β − β xz + ( − β + 8 β + 228 β − β + 20) β x + (12 β + 70 β − β + 8) β xz + (8 β − β z + (148 β − β + 118 β − β x + (36 β − β + 34 β − β z + ( − β + 62 β − β + 4) β . We can now prove the following result that implies the first part of Theorem 1.1.
Theorem 4.1.
Let V be a -generated primitive axial algebra of Monster type (2 β, β ) over a ring R in which is invertible and the elements , , β and β arepairwise distinguishable. Then, V is completely determined, up to homomorphicimages, by a quadruple ( x , y , z , t ) ∈ R , which is a solution of the system p ( x, y, z, t ) = 0 p ( x, y, z, t ) = 0 p ( x, y, z, t ) = 0 p ( x, y, z, t ) = 0 . (4) Proof.
Let V be a primitive axial algebra of Monster type (2 β, β ) over a ring R asin the statement, generated by the two axes ¯ a and ¯ a . Then, by [3, Corollary 3.8], V is a homomorphic image of V ⊗ ˆ D R and R is a homomorphic image of R ⊗ ˆ D R .We identify the elements of ˆ D with their images in R so that the polynomials p i and q i are considered as polynomials in R [ x, y, z, t ] and R [ x, z ], respectively. For each i ∈ Z , let ¯ a i be the image of the axis a i . By Proposition 3.7 and Corollary 3.9, thealgebra V is completely determined, up to homomorphic images, by the quadruple( λ ¯ a (¯ a ) , λ ¯ a (¯ a ) , λ ¯ a (¯ a ) , λ ¯ a (¯ a − )) . This quadruple is the homomorphic image in R of the quadruple ( λ , λ f , λ , λ f )defined in Section 3 and so it is a solution of the system (4) by the definition of thepolynomials p i ’s as claimed. (cid:3) If V satisfies the hypothesis of Theorem 4.1 and, in addition, it is symmetric,then λ ¯ a (¯ a ) = λ ¯ a (¯ a ) and λ ¯ a (¯ a ) , = λ ¯ a (¯ a − )and the pair ( λ ¯ a (¯ a ) , λ ¯ a (¯ a )) is a solution of the system q ( x, z ) = 0 q ( x, z ) = 0 q ( x, z ) = 0 . (5) Lemma 4.2.
For any field F , the resultant of the polynomials q ( x, z ) and q ( x, z ) with respect to z is γx ( x − x − β ) [(16 β − x + ( − β + β + 2)][(8 β − x + ( − β + 2 β )] , where γ := − β − (4 β − β . Proof.
The resultant can be computed in the ring Z [ β, β − ][ x ] using [1]. (cid:3) We set S := (cid:26)(cid:18) β , β (cid:19) , ( β, , (cid:18) β, β (cid:19)(cid:27) , S := S ∪ { (1 , , (0 , , ( β, } , S := S ∪ (cid:26)(cid:18) (18 β − β − β − , (48 β − β + 7 β − β − β (8 β − (cid:19)(cid:27) if β = 38 , S := S ∪ (cid:26)(cid:18) (9 β − β )2(4 β − , (9 β − β )2(4 β − (cid:19)(cid:27) if β = 14 . Lemma 4.3.
Let F be a field of characteristic other than , β ∈ F . If Then, theset of the solutions of the system of equations (5) is(1) S ∪ { ( µ, | µ ∈ F } , if β = ;(2) S ∪ S , if either β ∈ { , , } or β (cid:8) , (cid:9) and (6) (16 β − β − β + 46 β − β − β − β + β − β + 2 β −
1) = 0; (3) S in all the remaining cases.Proof. Using [1], it is straightforward to check that the possible values x for asolution ( x , z ) of the system (5) are given by Lemma 4.2 when β = and can becomputed directly when β = . Since q ( x, z ) is linear in z , for every value of x there is at most a solution of the system. The elements of S are indeed solutions.When x = (18 β − β − β − , we solve q ( x , z ) = 0 obtaining the given correspondingvalue for z . On the other hand, the value of q ( x, z ) computed in Z [ β ] on this pairis a non-zero polynomial in β which vanishes exactly when either β ∈ { , , } , or β = and Equation (6) is satisfied. (cid:3) In order to classify primitive axial algebras of Monster type (2 β, β ) over F gener-ated by two axes ¯ a and ¯ a we can proceed, similarly as we did in [3], in the followingway. We first solve the system (5) and classify all symmetric algebras. Then weobserve that, the even subalgebra hh ¯ a , ¯ a ii and the odd subalgebra hh ¯ a − , ¯ a ii aresymmetric, since the automorphisms τ and τ respectively, swap the generatingaxes. Hence, from the classification of the symmetric case, we know all possibleconfigurations for the subalgebras hh ¯ a , ¯ a ii and hh ¯ a − , ¯ a ii and from the relationsfound in Section 3, we derive the structure of the entire algebra.5. The symmetric case
In this and in the following section we let V be a primitive axial algebra ofMonster type (2 β, β ) over a field F of characteristic other than 2, generated by thetwo axes ¯ a and ¯ a . By [3, Corollary 3.8], V is a homomorphic image of V ⊗ ˆ D F .For every element v ∈ V , we denote by ¯ v its image in V . In particular ¯ a and¯ a are the images of a and a and all the formulas obtained from the ones inLemmas 3.2, 3.3, 3.4, 3.10 with ¯ a i and ¯ s r,j in the place of a i and s r,j , respectively,hold in V . With an abuse of notation we identify the elements of R with theirimages in F , so that in particular λ = λ ¯ a (¯ a ), λ f = λ ¯ a (¯ a ), λ = λ a (¯ a ) , and λ f = λ ¯ a (¯ a ).We begin with a quick overview of the known 2-generated primitive symmetricalgebras of Monster type (2 β, β ). Among these, there are(1) the algebras of Jordan type 1 A , 2 B , 3 C ( β ) and 3 C (2 β ) which we denoteby 2 A ( β ) (see [9]).(2) the algebra 3 A (2 β, β ) defined in [15].(3) the algebras V ( β ) and V ( β ) defined in [5].(4) the 3-dimensional algebra V ( β ) with basis (¯ a , ¯ a , ¯ a ) and the multipli-cation defined as in Table 2. Note that it coincides with the algebra III ( ξ, − ξ ξ − , × defined by Yabe in [17], with ξ = 2 β .(5) the 5-dimensional algebra Y ( β ) with basis (¯ a , ¯ a , ¯ a , ¯ a , ¯ s ) and multipli-cation table Table 3. Note that it coincides with the algebra IV ( ξ, β, µ )defined by Yabe [17], when β = − ξ and ξ = 2 β . ¯ a − ¯ a ¯ a ¯ a − ¯ a − β (¯ a + ¯ a − ) + β ¯ a β (¯ a − + ¯ a ) + β ¯ a ¯ a β (¯ a + ¯ a − ) + β ¯ a ¯ a β (¯ a + ¯ a ) + β ¯ a − ¯ a β (¯ a − + ¯ a ) + β ¯ a β (¯ a + ¯ a ) + β ¯ a − ¯ a Table 2.
Multiplication table for the algebra V ( β )It is immediate that the values of ( λ , λ ) corresponding to the trivial algebra 1 A and to the algebra 2 B are (1 ,
1) and (0 , V ( β ). Lemma 5.1.
Let F be a field of characteristic other than and β ∈ F such that β − β − . The algebra V ( β ) is a -generated symmetric Frobenius axialalgebra satisfying the fusion law M (2 β, β ) and such that for every i ∈ {− , , } , ad ¯ a i has eigenvalues , β , and β . In particular it not an axial algebra of Jordantype. Moreover, λ = λ = β +14 . Furthermore(1) if ch F = 3 , the algebra V ( β ) is primitive and simple;(2) if ch F = 3 , then β = 2 and V ( β ) is neither primitive nor simple. Ithas a -dimensional quotient over the ideal F (¯ a + ¯ a − + ¯ a ) isomorphic to C ( − × and a quotient isomorphic to A (over the ideal h ¯ a − ¯ a , ¯ a − ¯ a i ).Proof. If ch F = 3, then ¯ a − ¯ a − and − β +18 ¯ a + β (¯ a + ¯ a − ) are respectively a β -and 2 β -eigenvector for ad ¯ a and¯ a = 3 β + 18 β ¯ a + 1 β ( − β + 18 ¯ a + β a + ¯ a − ) + β a − ¯ a − ) , whence λ = β +18 β = β +14 . The Frobenius form is defined by (¯ a i , ¯ a i ) = 1 and(¯ a i , ¯ a j ) = λ , for i, j ∈ {− , , } and i = j . The projection graph (see [10] for thedefinition) has ¯ a and ¯ a as vertices and an edge between them since (¯ a , ¯ a ) = 0.Thus it is connected and so by [10, Corollary 4.15 and Corollary 4.11] every properideal of V is contained in the radical of the form. Since the determinant of theGram matrix of the Frobenius form with respect to the basis (¯ a − , ¯ a , ¯ a ) is alwaysnon-zero, the algebra is simple.If ch F = 3, then condition 18 β − β − β = 2. Hence 2 β = 1and ¯ a and ¯ a are both 1-eigenvectors for ad ¯ a . All the other properties are easilyverified. (cid:3) Lemma 5.2.
Let F be a field of characteristic other than and β ∈ F \{ , , } . Thealgebra A (2 β, β ) is a -generated symmetric Frobenius axial algebra of Monstertype (2 β, β ) with λ = λ = β (9 β − β − . It is simple except when (18 β − β − β − β + 2)(5 β −
1) = 0 , in which case one of the following holds(1) β = , ch F = 3 , and there is a unique quotient of maximal dimensionwhich is isomorphic to C ( β ) ;(2) β − β − , ch F = 3 , and there is a unique non trivial quotient whichis isomorphic to V ( β ) ;(3) β − β + 2 = 0 , ch F = 3 , and there is a unique non trivial quotientwhich is isomorphic to A ; (4) ch F = 3 , β = − and there are four non trivial quotients isomorphicrespectively to C ( − , C ( − × , V ( − , A (see [9, (3.4)] for the definitionof C ( − × ).Proof. Let V be the algebra 3 A (2 β, β ). Then V has a Frobenius form and theprojection graph (see [10] for the definition) has ¯ a and ¯ a as vertices and an edgebetween them since (¯ a , ¯ a ) = 0. Thus it is connected and so by [10, Corollary 4.15and Corollary 4.11] every proper ideal of V is contained in the radical of the form.The Gram matrix of the Frobenius form with respect to the basis (¯ a , ¯ a , ¯ a , ¯ s , )can be computed easily and has determinant − β (9 β − β + 2) (18 β − β − β − β − . Suppose first ch F = 3. When β = we see that the radical is generated by thevector β (¯ a + ¯ a + ¯ a )+ ¯ s , and hence the quotient over the radical is isomorphic tothe algebra 3 C ( β ). If (18 β − β −
1) = 0, then the radical is generated by the vector − β (¯ a +¯ a +¯ a )+ ¯ s , and it follows that the quotient over the radical is isomorphicto the algebra V ( β ), which by Lemma 5.1 is simple. Finally, if (9 β − β + 2) = 0,then the radical is three dimensional, with generators¯ a − ¯ a , ¯ a − ¯ a , (2 β − a + ¯ s , . It is immediate to see that the quotient over the radical is the trivial algebra 1 A .Using Lemma 2.4, it is straightforward to prove that the radical is a minimal ideal.Now assume ch F = 3. Then the radical of the form is three dimensional, withgenerators ¯ a − ¯ a , ¯ a − ¯ a , ¯ s , and it is straightforward to see that it contains properly the non-zero ideals F (¯ a +¯ a + ¯ a + ¯ s , ), F (¯ a + ¯ a + ¯ a − ¯ s , ), and F (¯ a + ¯ a + ¯ a ). Claim (4) follows. (cid:3) Lemma 5.3.
Let F be a field of characteristic other than and β ∈ F \ { , , } .(1) The algebra A has ( λ , λ ) = ( β, . It is simple except when β = − , inwhich case it has a non trivial quotient of dimension , denoted by C ( − ) × (see [9, (3.4)] ).(2) The algebra C ( β ) has λ = λ = β . It is simple except when β = − , inwhich case it has a non trivial quotient of dimension , denoted by C ( − × (see [9, (3.4)] ).(3) The algebra V ( β ) has ( λ , λ ) = ( β, . It is simple except when β = − ,in which case it has a unique non trivial quotient over the ideal generatedby ¯ a − + ¯ a + ¯ a + ¯ a − β ¯ s , , which is a simple algebra of dimension .(4) The algebra V ( β ) has ( λ , λ ) = (cid:16) β, β (cid:17) . It is simple provided β
6∈ { , − } .If β = − , the algebra has a unique non trivial quotient over theideal F (¯ a − + ¯ a − + ¯ a + ¯ a + ¯ a + ¯ a − β ¯ s , − β ¯ s , ) which is asimple algebra of dimension .If β = 2 , then it has a unique non trivial quotient which is iso-morphic to A ( β ) .Proof. (1) and (2) are proved in [9, (3.4)]. Let V ∈ { V ( β ) , V ( β ) } . Then, V is asubalgebra of a Matsuo algebra and so it is endowed of a Frobenius form. As in theproof of Lemma 5.2, every proper ideal of V is contained in the radical of the form. ¯ a ¯ a ¯ a ¯ a ¯ s ¯ a ¯ a ¯ s + β (¯ a + ¯ a ) 4 β ¯ s − β − (¯ a + ¯ a ) ¯ s + β (¯ a + ¯ a ) β ¯ s + β (¯ a + ¯ a )¯ a ¯ a ¯ s + β (¯ a + ¯ a ) 4 β ¯ s − β − (¯ a + ¯ a ) β ¯ s + β (¯ a + ¯ a )¯ a ¯ a ¯ s + β (¯ a + ¯ a ) β ¯ s + β (¯ a + ¯ a )¯ a ¯ a β ¯ s + β (¯ a + ¯ a )¯ s β ¯ s + β (¯ a + ¯ a ) β − (4¯ s − ¯ a − ¯ a − ¯ a − ¯ a ) Table 3.
Multiplication table for the algebra Y ( β )When V = V ( β ), the Gram matrix, with respect to the basis ¯ a − , ¯ a , ¯ a , ¯ a , − β ¯ s , ,is 2 β β ββ β β β β ββ β β β β β β . The determinant of this matrix is 2(2 β − (4 β + 1) and so, if β = − we get thethesis. If β = − , the radical of the form is the 1-dimensional ideal F (¯ a − + ¯ a +¯ a + ¯ a − β ¯ s , ). When V = V ( β ), the Gram matrix, with respect to the basis¯ a , ¯ a , ¯ a − , ¯ a , ¯ a − , ¯ a , − β ¯ s , , − β ¯ s , given in [5, Table 8], is β β β β β β ββ β β β β β ββ β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β . The determinant of this matrix is − β − ( β − (7 β +1) and so, if β
6∈ { , − } ,the algebra is simple. If β = − , then the radical of the form is the 1-dimensionalideal F (¯ a − + ¯ a − + ¯ a + ¯ a + ¯ a + ¯ a − β ¯ s , − β ¯ s , ) and the result follows. Finallysuppose β = 2. Then the radical of the form is 5-dimensional with basis¯ a − ¯ a , ¯ a − ¯ a − , ¯ a − ¯ a − , ¯ a − ¯ a , ¯ s , − ¯ s , and the quotient over the radical is an algebra of type 2 A ( β ). Using Lemma 2.4, itis straightforward to prove that the radical is a minimal ideal. (cid:3) Lemma 5.4.
Let F be a field of characteristic other than and β ∈ F such that β + 2 β − . The algebra Y ( β ) is a -generated primitive symmetric Frobeniusaxial algebra of Monster type (2 β, β ) , with λ = β + and λ = β . It is simple,except when ch F = 11 and β = 4 , in which case it has a unique non-trivial quotientover the ideal F (¯ a + ¯ a + ¯ a + ¯ a + 3¯ s ) .Proof. All the properties are easily verified. Note that the Frobenius form is definedby (¯ a i , ¯ a i ) = 1, (¯ a i , ¯ a j ) = λ , for i, j ∈ { , , , } such that i − j ≡
1, (¯ a , ¯ a ) = (¯ a , ¯ a ) = λ , and (¯ a i , ¯ s ) = β for i ∈ { , , , } . Then, the Frobenius form hasGram matrix, with respect to the basis (¯ a , ¯ a , ¯ a , ¯ a , ¯ s ), β + β β +
14 14 ββ + β + β ββ β + β +
14 14 ββ + β β + β β β β β β with determinant β ( β − (2 β − β +3). For β = − , condition 4 β +2 β − ch F = 11 and we get that the radical of the form is one-dimensionalgenerated by ¯ a + ¯ a + ¯ a + ¯ a + 3¯ s . The result follows with the argument alreadyused to prove Lemma 5.3. (cid:3) Lemma 5.5.
Let V be a symmetric primitive axial algebra of Monster type ( α, β ) over a field F , generated by two axes ¯ a and ¯ a . Suppose there exists A ∈ F suchthat ¯ a = ¯ a − + A (¯ a − ¯ a ) . Then, one of the following holds(1) A = 0 and ¯ a = ¯ a − ;(2) A = 1 , ¯ a = ¯ a − and V is spanned by ¯ a , ¯ a , ¯ s , .Proof. If A = 0, the claim is trivial. Suppose A = 0. By the symmetries of thealgebra, we get ¯ a − = ¯ a + A (¯ a − ¯ a − ) and ¯ a = ¯ a + A (¯ a − ¯ a ) . By substituting the expression for ¯ a in the definition of ¯ s , we get0 = ¯ s , − ¯ s τ , = A (1 − β )(¯ a − ¯ a ) . Then, since β = 1 /
2, we have ¯ a = ¯ a and, by the symmetry, ¯ a − = ¯ a . ByLemma 4.3 in [3], (2) holds. (cid:3) Proposition 5.6.
Let V be a symmetric primitive axial algebra of Monster type (2 β, β ) over a field F of characteristic other than , generated by two axes ¯ a and ¯ a . If V has dimension at most , then either V is an algebra of Jordan type β or β , or β − β − in F and V is isomorphic to the algebra V ( β ) .Proof. Since V is symmetric, ad ¯ a and ad ¯ a have the same eigenvalues. Since 1 isan eigenvalue for ad ¯ a , it follows from the fusion law that if 0 is an eigenvalue forad ¯ a , or V has dimension at most 2, then V is of Jordan type β or 2 β . Let usassume that 0 is not an eigenvalue for ad ¯ a . Then ¯ u = 0 (recall the definition of u in Section 2) and we get¯ s , = [ λ (1 − β ) − β ]¯ a + β a + ¯ a − ) . Since we have also ¯ u f = 0 we deduce¯ a = ¯ a − + (cid:20) β ( λ (1 − β ) − β ) − (cid:21) (¯ a − ¯ a ) . Thus we can apply Lemma 5.5. If claim (2) or (3) holds, then V has dimension atmost 2 and we are done. Suppose claim (1) holds, that is β ( λ (1 − β ) − β ) − s , = β a + ¯ a + ¯ a − ) and ¯ a ¯ a = 32 β (¯ a + ¯ a ) + β a − , whence we get that V satisfies the multiplication given in Table 2 and so it isisomorphic to a quotient of V ( β ). Since by hypothesis β
6∈ { , } , by Lemma 5.1, V ( β ) is simple and V ∼ = V ( β ). The vector v := 3 β ¯ a + (2 β − a − + ¯ a ) is a2 β -eigenvector for ad ¯ a and, in order to satisfy the fusion law (in particular v · v must be a 1-eigenvector for ad ¯ a ), β must be such that 18 β − β − (cid:3) Theorem 5.7.
Let V be a primitive symmetric axial algebra of Monster type (2 β, β ) over a field F of characteristic other than , generated by two axes ¯ a and ¯ a . Then,one of the following holds:(1) V is an algebra of Jordan type β or β ;(2) β − β − in F and V is an algebra of type V ( β ) ;(3) V is an algebra of type A (2 β, β ) ;(4) V is an algebra of type V ( β ) ;(5) V is an algebra of type V ( β ) ;(6) β + 2 β − in F and V is an algebra of type Y ( β ) ;(7) β = − and V is isomorphic to the quotient of V ( β ) over the one-dimensionalideal generated by ¯ a − + ¯ a + ¯ a + ¯ a − β ¯ s , ;(8) β = − and V is isomorphic to the quotient of V ( β ) over the one-dimensionalideal generated by ¯ a − + ¯ a − + ¯ a + ¯ a + ¯ a + ¯ a − β ¯ s , − β ¯ s , ;(9) ch F = 11 , β = 4 and V is isomorphic to the quotient of Y ( β ) over theone-dimensional ideal generated by ¯ a + ¯ a + ¯ a + ¯ a + β ¯ s , .Proof. By the remark after Theorem 4.1, V is determined, up to homomorphicimages, by the pair ( λ , λ ), which must be a solution of (5). By [3, Corollary 3.8 ]and Proposition 3.7, V is spanned on F by the set ¯ a − , ¯ a − , ¯ a , ¯ a ,¯ a ,¯ a , ¯ s , , and¯ s , .Assume first λ = β . Then, by Lemma 3.6, we get ¯ s , = ¯ s , = ¯ s , . If( λ , λ ) = ( β, β ), we see that the algebra satisfies the multiplication table of thealgebra V ( β ). Hence V is isomorphic to a quotient of V ( β ) and by Lemma 5.3 weget that either (5) or (8) holds. Assume λ = β . We compute¯ E = (2 β − λ − β )4 [¯ s , + β (¯ s , − ¯ a − + ¯ a )] , hence, since (2 β − λ − β ) = 0, we get ¯ s , = β (¯ a − − ¯ a − ¯ s , ). Then, fromthe identity ¯ s , − ¯ s τ , = 0 we get ¯ a = ¯ a − , and so ¯ s , = ¯ s , and ¯ a − = ¯ a .Hence the dimension is at most 5. If ( λ , λ ) = ( β,
0) we see that V satisfies themultiplication table of V ( β ) and either (4) or (7) holds. Finally, if ( λ , λ ) = ( β, Z ( β, β ) = 0 and so by Lemma 3.3 and Equation (1) we get ¯ a ¯ a = ¯ a , that is¯ a is a 1-eigenvector for ad ¯ a . By primitivity, this implies ¯ a = ¯ a . Consequently,we have ¯ a − = ¯ a f = ¯ a f = ¯ a and from the multiplication table we see that V is aquotient of the algebra 2 A ( β ). Thus V an axial algebra of Jordan type 2 β . Now assume λ = β . We have¯ D = − β − λ − β ) β h ( β −
1) (¯ a − − ¯ a )+ (cid:18) λ (4 β − λ − β ) β + 10 β − − λ (cid:19) (¯ a − − ¯ a ) (cid:21) . By Lemma 3.10, ¯ D = 0. Since λ = β and β
6∈ { , } , the coefficient of ¯ a − in ¯ D is non zero and we get(7) ¯ a − = ¯ a + 1( β − (cid:20) λ (4 β − λ − β ) β + 10 β − − λ (cid:21) (¯ a − ¯ a − ) . Since V is symmetric, the map f swapping ¯ a and ¯ a is an algebra automorphismand so(8) ¯ a = ¯ a − + 1( β − (cid:20) λ (4 β − λ − β ) β + 10 β − − λ (cid:21) (¯ a − ¯ a ) . It follows also ¯ s , ∈ h ¯ a − , ¯ a , ¯ a , ¯ a , ¯ s , i and hence V = h ¯ a − , ¯ a , ¯ a , ¯ a , ¯ s , i .Moreover, equation ¯ d = 0 of Lemma 3.10 becomes(9) B (¯ a − − ¯ a ) + C (¯ a − ¯ a ) = 0with B and C in F .Assume β = and ( λ , λ ) = (cid:16) (9 β − β )2(4 β − , (9 β − β )2(4 β − (cid:17) (note that λ = β since β = 0). Then the identities ¯ a − = ¯ a − and ¯ a = ¯ a give the equations2 β (18 β − β −
1) (¯ a − ¯ a − ) = 0 and 2(18 β − β + 1)(4 β −
1) (¯ a − ¯ a − ) = 0 . Since, in any field F , the two polynomials (18 β −
5) and (18 β − β + 1) have nocommon roots, we have ¯ a = ¯ a − , whence ¯ a − = ¯ a , and it is straightforward tosee that V satisfies the multiplication table of the algebra 3 A (2 β, β ). Hence theresult follows from Lemma 5.2.Now assume ( λ , λ ) = (cid:16) β , β (cid:17) . Then we have B = ( β − β − β and C = 0 . Moreover, the identity ¯ a − = ¯ a − give the equation( β + 2 β − β (¯ a − ¯ a − ) = 0Suppose B = 0, or ( β + 2 β − = 0: we get ¯ a − = ¯ a and consequently ¯ s , = ¯ s , .From the identity ¯ s , − ¯ s , = 0 we get(5 β − β (cid:20) ¯ s , + β a + ¯ a + ¯ a − ) (cid:21) = 0 . If β = , it follows that the dimension is at most 3 and in fact we get a quotientof the algebra 3 C ( β ) which is an algebra of Jordan type β . If β = 1 /
5, we have β = (9 β − β )2(4 β − and V is a quotient of the algebra 3 A (2 β, β ). Finally, if B = 0 =( β + 2 β − ch F = 7 and β = 2. In this case, by Equation (7), wehave ¯ a − = ¯ a + ¯ a − ¯ a − . Computing the vectors ¯ u and ¯ v with Lemma 2.3, we get ¯ v = 0 and hence 0 = ¯ a − λ ¯ a − u − w = 2(¯ a − ¯ a − ). Thus, ¯ a − = ¯ a and¯ a = ¯ a , V has dimension at most 3 and the result follows from Proposition 5.6.Assume ( λ , λ ) = (0 , B = C = − (9 β − β + 2 β − β ( β − , the identity ¯ a = ¯ a gives the equation(10) 5(2 β − β ( β − (cid:2) − B (¯ a − ¯ a − ) + (18 β + 27 β − β + 4)(¯ a − ¯ a ) (cid:3) = 0 , and the identity ¯ D = 0 of Lemma 3.10 gives(4 β − B (¯ a − − ¯ a ) = 0Thus, if (4 β − B = 0, we get ¯ a − = ¯ a . Then, ¯ a = ¯ a f − = ¯ a f = ¯ a andhence ¯ s , = (1 − β )¯ a . Further, from the first equation of Lemma 3.3, we alsoget ¯ s , = − β (¯ a + ¯ a ), whence ¯ a ¯ a = 0. Thus V is isomorphic to the algebra2 B . Suppose B = 0. If (18 β + 27 β − β + 4) = 0, from Equation (10), weget ¯ a = ¯ a and, by the symmetry, also ¯ a = ¯ a − . Thus the dimension is atmost 3 and we conclude by Proposition 5.6. If (18 β + 27 β − β + 4) = 0, thenit follows ch F = 31 and β = 9. In this case ¯ E − ¯ E τ = 0: a contradiction toLemma 3.10. Now assume β = and B = 0 (thus, in particular, ch F = 3).From Equation (7) we get ¯ a − = ¯ a + (¯ a − ¯ a − ). Further, ¯ v = 0 and we get0 = ¯ a − λ ¯ a − u − w = (¯ a − ¯ a − ). Thus, if ch F = 5, we get ¯ a − = ¯ a and¯ a = ¯ a ; V has dimension at most 3 and the result follows from Proposition 5.6. If ch F = 5, then B = −
1, Equation (9) gives ¯ a = ¯ a − − (¯ a − ¯ a ) and so Lemma 5.5implies that V has dimension at most 3 and we conclude as above.Suppose ( λ , λ ) = (1 , D = 0 is equal to2(4 β − β + 11 β − β − β + 2)( β − β − β (¯ a − − ¯ a ) = 0Hence, if (4 β − β + 11 β − β − β + 2)( β − = 0, we get ¯ a − = ¯ a ,by the symmetry, ¯ a = ¯ a and V has dimension at most 3 and we conclude byProposition 5.6. So let us assume(11) (4 β − β + 11 β − β − β + 2)( β −
2) = 0 . Further, we have B = − β (cid:0) β − β + 127 β − β + 6 (cid:1) and C = − β (4 β + 2 β − β − β + 2)( β − . If (4 β − β + 11 β −
2) = 0, then B and C are not zero and by Lemma 5.5, V has dimension at most 3 and we conclude by Proposition 5.6. Moreover, if β = 2then ( λ , λ ) = ( β , β ) and we reduce to the previous case. Hence we assume(9 β − β + 2) = 0 . Then C = 0. If B = 0, from Equation (9) we get ¯ a = ¯ a − and by symmetry,¯ a = ¯ a − . Then, Equation (7) becomes¯ a − ¯ a − = 1( β − (cid:20) β − − β ) β + 10 β − (cid:21) (¯ a − − ¯ a ) , whence, either ¯ a = ¯ a − and again V has dimension at most 3 and we conclude byProposition 5.6, or 1( β − (cid:20) β − − β ) β + 10 β − (cid:21) = − β − β + 22 β − β ( β −
1) = 0 . It is now straightforward to check that the two polynomials (9 β − β + 2) and(11 β − β + 22 β −
4) have no common roots in any field of characteristic otherthan 2 and we get a contradiction. We are now left to consider the case when B = 0. Then the two polynomials (9 β − β + 2) and B have a common rootif and only if ch F ∈ { , } and the common root if β = −
1. In both cases weget ( λ , λ ) = (1 ,
1) = (9 β − β )2(4 β − and so V is a quotient of the algebra 3 A (2 β, β ) asalready proved.Suppose now β = and ( λ , λ ) = ( µ, µ ∈ F \ { , , β } . In particular,note that ch F = 3, since β = 1. Then Equations (7) and (8) become¯ a − = ¯ a + 103 (¯ a − − ¯ a ) and ¯ a = ¯ a − + 103 (¯ a − ¯ a )and the identity ¯ D = 0 gives the relation283 (4 µ − a − ¯ a − ) = 0 . Since we are assuming λ = β , (4 µ − = 0. So, if ch F = 7, we get ¯ a = ¯ a − .Hence V ha dimension at most 3 and we conclude by Proposition 5.6. If ch F = 7,then β = 2 and we conclude, with the same argument used above for the case when( λ , λ ) = ( β , β ), ch F = 7, β = 2.Finally assume that β = , β ∈ { , , } or β satisfies Equation (11), and( λ , λ ) = (cid:16) (18 β − β − β − , (48 β − β +7 β − β − β (8 β − (cid:17) . First of all, note that the onlycommon solution of Equation (11) and of the equation 2 β + 5 β − β = 1when ch F = 5. Since we are assuming β = 1, under our hypotheses 2 β + 5 β − F : in particular this implies λ = β . The identity ¯ D = 0becomes(2 β + 5 β − β + 2 β − β + β − β − β − β (8 β − ( β −
1) (¯ a − ¯ a − ) = 0 . If (4 β +2 β − β + β − β − = 0 in F , we deduce ¯ a = ¯ a − and, by symmetry,¯ a = ¯ a . Thus V has dimension at most 3 and we conclude by Proposition 5.6. If(5 β + β −
1) = 0, then λ = λ = β and we are in a case considered above. If(4 β + 2 β −
1) = 0, we get λ = β + and λ = β . From the identity ¯ E = 0 we get¯ a = ¯ a − , consequently ¯ a − = ¯ a , and it follows that V satisfies the multiplicationtable of the algebra Y ( β ). Hence, by Lemma 5.4, either (6) or (9) hold. Finallyassume ch F = 7 and β = . Then λ = λ = and, since β = 1, we have also ch F = 5. From identity ¯ d = 0, we get ¯ a − = ¯ a and consequently ¯ a = ¯ a − . If further ch F = 3, identity ¯ E = 0 implies ¯ a − = ¯ a . Then ¯ a = ¯ a , but λ = = 1,a contradiction. Hence ch F = 3, ( λ , λ ) = ( β , β ) and we conclude as in the caseconsidered above. (cid:3) The non symmetric case
This section is devoted to the proof of Theorem 1.2. Let V be generated by thetwo axes ¯ a and ¯ a . We set V e := hh ¯ a , ¯ a ii and V o := hh ¯ a − , ¯ a ii . As noticed atthe end of Section 4, V e and V o are symmetric, since the automorphisms τ and τ respectively, swap their generating axes. Hence, from Theorem 5.7 we get thepossible values for the pair ( λ , λ f ) and the structure of those subalgebras. Notethat V is symmetric if and only if λ = λ f and λ = λ f in F . Lemma 6.1. If V has dimension , then V ∼ = V ( β ) .Proof. Suppose V has dimension 8. Then the generators ¯ a − , ¯ a − , ¯ a , ¯ a , ¯ a , ¯ a , ¯ s , ,and ¯ s , are linearly independent. We express ¯ d defined before Lemma 3.10 as alinear combination of the basis vectors and since ¯ d = 0 in V , every coefficientmust be zero. In particular, considering the coefficients of ¯ a − , ¯ a , and ¯ s , we get,respectively, the equations(6 β − β λ + (2 β − β λ f − (8 β −
3) = 0(2 β − β λ + (6 β − β λ f − (8 β −
3) = 08 β ( λ − λ f ) − β ( λ − λ f ) − β ( λ − λ f ) = 0whose common solutions have λ = λ f and λ = λ f . Hence V is symmetric andthe result follows from Theorem 5.7. (cid:3) Lemma 6.1 and Theorem 5.7 imply that if V is non-symmetric, then the dimen-sions of the even subalgebra and of the odd subalgebra are both at most 5. As aconsequence, from Lemma 3.4 we derive some relations between the odd and evensubalgebras. Lemma 6.2. If ¯ a − = ¯ a , then Z ( λ f , λ )(¯ a − ¯ a + ¯ a − − ¯ a ) =(12) = 1 β h Z ( λ f , λ ) (cid:16) λ f − β (cid:17) − (cid:16) λ f − β (cid:17)i (¯ a − − ¯ a ) . If ¯ a − = ¯ a , then Z ( λ , λ f )(¯ a − − ¯ a + ¯ a − − ¯ a ) =(13) = 1 β h Z ( λ , λ f ) ( λ − β ) − ( λ − β ) i (¯ a − − ¯ a ) . If ¯ a − = ¯ a , then Z ( λ f , λ )(¯ a − ¯ a − ) =(14) = 1 β h Z ( λ f , λ ) (cid:16) λ f − β (cid:17) − (cid:16) λ f − β (cid:17)i (¯ a − ¯ a − ) . If ¯ a − = ¯ a , then Z ( λ , λ f )(¯ a − − ¯ a ) =(15) = 1 β h Z ( λ , λ f ) ( λ − β ) − ( λ − β ) i (¯ a − ¯ a ) . If ¯ a − = ¯ a and ¯ a = ¯ a − , then h βZ ( λ , λ f ) − λ f − β ) i (¯ a − ¯ a − ) =(16) = 1 β h β − βZ ( λ , λ f )( λ f − β ) − β ( λ − λ f ) (cid:16) β − λ − λ f (cid:17) +2(2 β − ( λ f − β ) − β ( λ − λ f ) i (¯ a − ¯ a − ) . Proof.
By applying the maps τ and τ to the formulas of Lemma 3.4 we findsimilar formulas for a − and a . Equations (12), (13), (14), and (15) follow. Toprove Equation 16, note that if ¯ a − = ¯ a and ¯ a = ¯ a − , then ¯ s , = ¯ s , . Thus¯ s , − ¯ s , = 0 and the claim follows from Lemma 3.6. (cid:3) In view of the above relations, it is important to investigate some subalgebras ofthe symmetric algebras.
Lemma 6.3. (1) If V is one of the algebras V ( β ) , Y ( β ) , or their four dimensional quotients,then, V = hh ¯ a − − ¯ a , ¯ a − ¯ a ii . (2) If V is one of the algebras C ( β ) , C ( − × , A (2 β, β ) , and V ( β ) , then, V = hh ¯ a − − ¯ a , ¯ a − ¯ a , ii , unless V = 3 C (2) and ch F = 5 , or ch F = 3 and V = 3 C ( − × or V = V ( − .Proof. To prove (1), set W := hh ¯ a − − ¯ a , ¯ a − ¯ a ii . Let V be equal to V ( β )or its four dimensional quotient when β = − . Then ¯ a − ¯ a = 0 = ¯ a ¯ a . Thus(¯ a − − ¯ a ) = ¯ a − + ¯ a and (¯ a − ¯ a ) = ¯ a + ¯ a , whence we get that ¯ a , ¯ a belongto W and the claim follows.Let V be equal to Y ( β ). Then, W contains the vectors(¯ a − − ¯ a ) = ¯ a − + ¯ a + (2 β − a + ¯ a ) − β ¯ s , , (¯ a − ¯ a ) = ¯ a + ¯ a + (2 β − a − + ¯ a ) − β ¯ s , , ¯ a − ¯ a = 14( β − (cid:8) (¯ a − − ¯ a ) − (¯ a − ¯ a ) − β − a − − ¯ a ) + (¯ a − ¯ a )] (cid:9) , and (¯ a − ¯ a ) = ¯ a + ¯ a − β (¯ a + ¯ a ) − s , . It is straightforward to check that the five vectors ¯ a − − ¯ a , ¯ a − ¯ a , (¯ a − − ¯ a ) ,¯ a − ¯ a , and (¯ a − ¯ a ) generate the entire algebra Y ( β ). Suppose now ch F = 11and V is the quotient of Y (4) over the ideal F (¯ a + ¯ a + ¯ a + ¯ a +3¯ s ). By proceedingas above, we get that W contains the vectors¯ a − − ¯ a , ¯ a − ¯ a , ¯ a − ¯ a , and ¯ a − ¯ a , which, again, are generators of the entire algebra V .To prove (2), set W := hh ¯ a − − ¯ a , ¯ a − ¯ a , ii . Let V be equal to 3 C ( β ). Then, W contains also (¯ a − ¯ a ) = (1 − β )(¯ a + ¯ a ) + β ¯ a − and the three vectors generate V , unless β = 2 and ch F = 5. If V = 3 A (2 β, β ), then W contains also the vectors(¯ a − ¯ a ) = (1 − β )(¯ a +¯ a ) − s , and (¯ a − ¯ a )(¯ a − − ¯ a ) = (1 − β )¯ a − ¯ s , . Thus we get four vectors that generate V . If V = V ( β ), then W contains (¯ a − ¯ a ) =(1 − β )(¯ a + ¯ a ) − β ¯ a − and we get that W contains three vectors that generate V , unless β = 2 and ch F = 3. Finally, if V = 3 C ( − × , then ¯ a − ¯ a and(¯ a − ¯ a ) = 3(¯ a + ¯ a ) generate V , unless ch F = 3. (cid:3) We start now to consider the possible configurations.
Lemma 6.4. If V is non-symmetric, then V e and V o are not isomorphic to V ( β ) , Y ( β ) or to one of their four dimensional quotients.Proof. It is enough to show that the claim holds for V e . Let us assume by contra-diction that V e is isomorphic to a quotient of V ( β ) or to a quotient of Y ( β ). Then,the vectors ¯ a − , ¯ a , ¯ a , ¯ a are linearly independent and, from the first formula inLemma 3.4, we get that Z ( λ , λ f ) = 0 and ¯ a = ¯ a − . Moreover ¯ a − = ¯ a and soEquation (13) holds.Suppose 2 Z ( λ , λ f ) ( λ − β ) − ( λ − β ) = 0. Then (¯ a − − ¯ a ) ∈ hh ¯ a − , ¯ a ii . Since V o is invariant under τ , it contains also ¯ a − ¯ a . Thus by Lemma 6.3, V = V o , acontradiction.Assume now 2 Z ( λ , λ f ) ( λ − β ) − ( λ − β ) = 0. Then, by Equation (13), wehave ¯ a − ¯ a − + ¯ a − ¯ a − = 0. Since ¯ a = ¯ a − , V o must be isomorphic to one ofthe following: 3 A (2 β, β ), 3 C ( β ), V ( β ) or 3 C ( − × . Then ¯ a − = ¯ a . Thus we get¯ a − − ¯ a = 0 and ¯ a = ¯ a − , a contradiction. (cid:3) Lemma 6.5. If V is non-symmetric, then V e and V o are not isomorphic to A .Proof. Clearly, it is enough to show that the claim holds for V e . Let us assumeby contradiction that V e is isomorphic to 1 A , so that λ = 1 and, for every i ∈ Z ,¯ a = ¯ a i and ¯ s , = (1 − β )¯ a . Then, the Miyamoto involution τ is the identityon V and the β -eigenspace for ad ¯ a is trivial. In particular, ¯ a = ¯ a − , and V o isisomorphic either to 1 A , or to 2 A ( β ), or to 2 B . Then, ¯ s , = ¯ s , and V is generatedby ¯ a − , ¯ a , ¯ a , and ¯ s , . Suppose V o ∼ = 1 A . Then, V is generated by ¯ a , ¯ a , and¯ s , . It follows that τ is the identity on V and the β -eigenspace for ad ¯ a is trivial.This implies that V is an axial algebra of Jordan type, a contradiction since V isnon-symmetric.Now suppose V o ∼ = 2 A or 2 B . If Z ( λ , λ f ) = 0, from the formula for ¯ s , inLemma 3.3 we get ¯ s , = ( λ − β )¯ a − β (¯ a +¯ a − ), whence ¯ a ¯ a = λ ¯ a + β (¯ a − ¯ a − ),a contradiction since ¯ a ¯ a is τ -invariant. Hence, Z ( λ , λ f ) = 0 and so Z ( λ f , λ ) = (4 β − β ( λ f − β ). From the formula for ¯ a − in Lemma 3.4 and Equation (16) we getthat the quadruple ( λ , λ f , , λ f ) must be a solution of the system(17) Z ( λ , λ f ) = 0(4 β − λ f − β ) = β λ f − β ( λ − λ f ) (cid:16) β − λ − λ f (cid:17) + β − β ( λ f − β ) − − λ f ) = 0 p ( λ , λ f , , λ f ) = 0 p ( λ , λ f , , λ f ) = 0 . When λ f = 0, the second equation yields that either λ f = β or β = . In theformer case, we obtain λ = λ f = β from the first equation and the third gives λ f = 1: a contradiction. In the latter case β = the system (17) is equivalent to − λ + = 016 λ = 0 − λ + 53 = 0which does not have any solution in any field F . Finally, when λ f = β , again weget a contradiction since the system (17) does not have any solution in any field F and for every β . (cid:3) Lemma 6.6.
Let V be non-symmetric. If ¯ a = ¯ a − , then ¯ a = ¯ a − and vice versa.Proof. Let us assume by contradiction that ¯ a = ¯ a − and ¯ a − = ¯ a . Then, Equa-tions (14) and (15) hold.If Z ( λ , λ f ) = 0 = Z ( λ f , λ ), then it follows λ = λ f and, by Equations (15)and (16), λ = λ f = β . Thus V is symmetric: a contradiction. Therefore, withoutloss of generality, we may assume Z ( λ f , λ ) = 0. Then, Equation (14) implies¯ a − ¯ a − , ¯ a − ¯ a ∈ V o , and, since β = and V = V o , Lemma 6.3 yields that ch F = 5 , β = 2 , and V e ∼ = 3 C (2) . If also Z ( λ , λ f ) = 0 or 4 Z ( λ f , λ ) (cid:16) λ f − β (cid:17) − (cid:16) λ f − β (cid:17) = 0, from Equations (15)and (14), respectively, we get that ¯ a − − ¯ a and ¯ a − − ¯ a are contained in theeven subalgebra and so, by Lemma 6.3, we get V o ∼ = 3 C (2). From the formulas inLemma 3.3 we find¯ s , = ( λ − β )¯ a − β (¯ a − + ¯ a ) and ¯ s , = ( λ f − β )¯ a − β (¯ a + ¯ a ) . Comparing the two expressions and using the invariance of ¯ s , under τ and τ ,we get λ ¯ a − β ¯ a − = λ f ¯ a − β ¯ a ( λ − β )(¯ a − ¯ a ) = β (¯ a − − ¯ a )( λ f − β )(¯ a − ¯ a − ) = β (¯ a − ¯ a − ) . From the above identities we can express ¯ a , ¯ a − , and ¯ a − as linear combinationsof ¯ a , ¯ a , ¯ a . So V has dimension at most 3 and so V = V o = V e , a contradiction.Suppose finally Z ( λ , λ f ) = 0 and 4 Z ( λ f , λ ) (cid:16) λ f − β (cid:17) − (cid:16) λ f − β (cid:17) = 0 . Then,we have λ f = 2 λ + 3, λ f = λ ( λ + 1), and p ( λ , λ + 3 , , λ ( λ + 1)) = λ ( λ − p ( λ , λ + 3 , , λ ( λ + 1)) = 2 λ − λ − . Hence, by Theorem 4.1, must be λ = 2 and we get a contradiction to our initialassumption, since Z (2 λ + 3 , λ ) = 2 − λ = 0. (cid:3) Lemma 6.7. If V is non-symmetric, then V e and V o are not isomorphic to one ofthe following: A (2 β, β ) , C ( β ) , V ( β ) , or their quotients.Proof. Let us assume by contradiction that V e is isomorphic to one of the algebras3 A (2 β, β ), 3 C ( β ), V ( β ), or their quotients. By the previous lemmas, we mayalso assume that V o is isomorphic to 2 A or 2 B . Then, ¯ a = ¯ a − , ¯ a = ¯ a − , and ¯ a = ¯ a − . From Equations (12) and (15) we get that the quadruple ( λ , λ f , λ , λ f )satisfies the following conditions(18) (cid:26) Z ( λ f , λ ) = 04 Z ( λ, λ f ) ( λ − β ) = (2 λ − β ) . Since Z ( λ f , λ ) = 0, from the second formula of Lemma 3.3, we get¯ s , = − β ¯ a − + ( λ f − β )¯ a , whence it follows that V o ∼ = 2 B and λ f = 0. In particular, (¯ a − ¯ a − ) = ¯ a + ¯ a − and so V o = hh ¯ a − ¯ a − ii . Let us now consider Equation (16). If the coefficientof ¯ a − ¯ a − in Equation (16) is not 0, then, the above remark implies V = V e , acontradiction. Thus the coefficients ¯ a − − ¯ a and ¯ a − ¯ a − in Equation (16) are 0.Then, by Theorem 4.1, ( λ , λ f , λ ,
0) is a solution of the system: (19) Z ( λ f , λ ) = 04 Z ( λ, λ f ) ( λ − β ) = (2 λ − β ) βZ ( λ , λ f ) − λ f − β ) = 0 h β − β ( λ f − β ) − β ( λ − λ f ) (cid:16) β − λ − λ f (cid:17) + β − β ( λ f − β ) − λ i = 0 p ( λ , λ f , λ ,
0) = 0 p ( λ , λ f , λ ,
0) = 0 . Now Z ( λ f , λ ) = 0 implies that Z ( λ , λ f ) = β − β ( λ − β ), whence the systemin (19) is equivalent to (20) Z ( λ f , λ ) = 0 β − β ( λ − β ) = (2 λ − β ) (4 β +2 β − β ( λ f − β ) = 0 h β − β ( λ f − β ) − β ( λ − λ f ) (cid:16) β − λ − λ f (cid:17) + β − β ( λ f − β ) − λ i = 0 p ( λ , λ f , λ ,
0) = 0 p ( λ , λ f , λ ,
0) = 0 If λ = β , then the system has the unique solution ( β, β, β , p ( β, β, β ,
0) = − β = 0. This is a contradictionto Theorem 4.1. Suppose λ = β (9 β − β − or 18 β − β − λ = β +14 , but λ = β . Using the first equation we express λ f as a polynomial in λ . Then, bythe second equation λ = β and so the third equation implies 4 β + 2 β − λ between the fourth equation andthe two last equations and between the two last equations. The three resultants arepolynomial expressions in β which vanish provided the System (19) has a solution.Comparing these expressions, we obtain that the system has no solution, in anyfield F and for any value of β . (cid:3) Proof of Theorem 1.2.
Let V be a non-symmetric 2-generated primitive axialalgebra of Monster type (2 β, β ). By the previous lemmas in this section, we mayassume that the even and the odd subalgebras are isomorphic to either 2 A or 2 B .Then, from Lemma 3.4 we get that ( λ , λ f , λ , λ f ) satisfies the following equations(21) Z ( λ , λ f )( λ − β ) = λ and(22) Z ( λ f , λ )( λ f − β ) = λ f . Suppose first that λ = λ f . Then we get Z ( λ , λ f )( λ − β ) = Z ( λ f , λ )( λ f − β ),which is equivalent to ( λ − λ f )( λ + λ f − β ) = 0 and so λ + λ f − β = 0,since λ = λ f as the algebra is non-symmetric. Then, Equations (21) and (22) areequivalent to(23) 1 β ( λ − β ) = λ β ( λ f − β ) = λ f λ = λ f = 0, we get the solution ( β, β, ,
0) which correspond to a symmetricalgebra, a contradiction. Suppose λ = λ f = β . Then it is long but straightforwardto check that there is no quadruple ( λ , λ f , β, β ) which is a common solution ofEquations (4) and (23).Finally assume that λ = β and λ f = 0. Then, by Equation (22), either Z ( λ f , λ ) = 0 or λ f = β . If Z ( λ f , λ ) = 0, we check that no quadruple ( λ , λ f , β, λ f = β .Then Equation (21) becomes ( λ − β ) = β β, β, β,
0) and ( β , β, β, λ , λ f , λ , λ f ) =( β , β, β, s , = − β (¯ a +¯ a ) and so V has dimensionat most 4. Moreover, V satisfies the same multiplication table as the algebra Q ( β ).By Theorem 8.6 in [5], for β = − the algebra Q ( β ) is simple, while it has a 3-dimensional quotient over the radical F (¯ a + ¯ a + ¯ a + ¯ a − ) when β = − . Theclaim follows. (cid:3) References [1] Decker, W.; Greuel, G.-M.; Pfister, G.; Sch¨onemann, H.:
Singular
J. Algebra [3] Franchi, C., Mainardis, M., Shpectorov, S., 2-generated axial algebras of Monster type. https://arxiv.org/abs/2101.10315. [4] Franchi, C., Mainardis, M., Shpectorov, S., An infinite dimensional 2-generated axial algebraof Monster type. https://arxiv.org/abs/2007.02430. [5] Galt, A., Joshi, V., Mamontov, A., Shpectorov, S., Staroletov, A., Double axes and subalge-bras of Monster type in Matsuo algebras. https://arxiv.org/abs/2004.11180.
Invent. Math. , , (1982), 1-102.[8] Hall, J., Rehren, F., Shpectorov, S.: Universal Axial Algebras and a Theorem of Sakuma, J. Algebra (2015), 394-424.[9] Hall, J., Rehren, F., Shpectorov, S.: Primitive axial algebras of Jordan type,
J. Algebra (2015), 79-115.[10] Khasraw, S.M.S., McInroy, J., Shpectorov, S.: On the structure of axial algebras,
Trans.Amer. Math. Soc. , (2020), 2135-2156. [11] Ivanov, A. A., Pasechnik, D. V., Seress, ´A., Shpectorov, S.: Majorana representations of thesymmetric group of degree 4, J. Algebra (2010), 2432-2463[12] Norton, S. P.: The uniqueness of the Fischer-Griess Monster. In: McKay, J. (ed.) Finitegroups-coming of age (Montreal, Que.,1982). Contemp. Math. 45, pp. 271–285. AMS, Provi-dence, RI (1985)[13] Norton, S. P.: The Monster algebra: some new formulae. In Moonshine, the Monster and re-lated topics (South Hadley, Ma., 1994), Contemp. Math. 193, pp. 297-306. AMS, Providence,RI (1996)[14] Rehren, F., Axial algebras, PhD thesis, University of Birmingham, 2015.[15] Rehren, F., Generalized dihedral subalgebras from the Monster,
Trans. Amer. Math. Soc. (2017), 6953-6986.[16] Sakuma, S.: 6-transposition property of ττ