Majorana representation of A6 involving 3C-algebras

We study a possible Majorana representation $${{\mathcal R}}$$ of the alternating group A6 of degree 6 such that for some involutions s and t in A6, generating a D6-subgroup, the corresponding Majorana axes as and at generate a subalgebra of type 3C. We show that there exists at most one such representation $${{\mathcal R}}$$ and that its dimension is at most 70. The representation $${{\mathcal R}}$$ does not correspond to a subalgebra in the Monster algebra generated by a subset of the Majorana axes canonically indexed by the involutions of an A6-subgroup in the Monster.

is a homomorphism, whose image preserves ( , ) and · ; and ψ is an rule which injectively assigns to every t ∈ T a vector a t = ψ(t) which is a Majorana axis in (V, ( , ), · ) (cf. the definition in the next paragraph) and such that the Majorana involution associated with a t coincides with ϕ(t). It is further assumed that V is generated by the elements a t taken for all t ∈ T , and dim (V ) is said to be the dimension of R. It is assumed that ϕ and ψ commute in the sense that a g −1 tg = (a t ) ϕ (g) for every g ∈ G. For a subset X of vectors in V we denote the linear span of X in V and the algebra closure of X in (V, · ) by X and X , respectively. Thus X is the smallest subspace Y in V containing X such that y 1 · y 2 ∈ Y whenever y 1 , y 2 ∈ Y .
By the definition in [1] a Majorana axis a in (V, ( , ), · ) is an idempotent of length 1, whose adjoint operator is semi-simple with spectrum {1, 0, 1 4 , 1 32 } (when talking about eigenvectors of a one really means eigenvectors of ad a ). The following conditions concerning the eigenspaces are imposed. The 1-eigenvectors of a are precisely the scalar multiples of a. The Majorana involution τ (a) associated with a is the linear transformation of V which negates every 1 32 -eigenvector and centralizes the other eigenvectors. By the above definition the Majorana involution is an automorphism of (V, ( , ), · ). This condition is equivalent to the fusion rules involving 1 32 -eigenvectors: if v and u are 1 32eigenvectors of a, and if x and y are λand μ-eigenvectors for λ, μ ∈ {1, 0, 1 4 }, then v · x is a 1 32 -eigenvector, and both v · u and x · y project to zero in the 1 32 -eigenspace. The remaining fusion rules are as follows: if α 1 and α 2 are 0-eigenvectors of a, and β 1 and β 2 are 1 4 -eigenvectors of a, then where λ, μ ∈ Sp and Sp(μ, λ) is the (λ, μ)-entry in Table 1.  Table 2 (more specifically to an algebra of type N X , where N is as above and X ∈ {A, B, C} ). For an integer i and ε ∈ {0, 1} the vector a 2i+ε is the image of a ε under the i-th power of ρ (so that τ (a 2i+ε ) = ρ −i τ ε ρ i ), and the remaining vectors in the basis of Y , given in the second column in Table 2, are centralized by D. The kernel of the action of D on Y coincides with the centre of D.
The shape of a Majorana representation R is a rule which specifies the type of the Norton-Sakuma subalgebra ψ(t 0 ), ψ(t 1 ) for every pair t 0 , t 1 of involutions in T . The rule must be stable under conjugation by the elements of G and must respect the embeddings of the algebras: The Monster algebra possesses important properties which we are included sometimes into the hypothesis: (c) both a 0 , a 1 and a 2 , a 3 have type 3A; then the 3A-axial vectors u t 0 t 0 and u t 2 t 3 in the subalgebras in (c) are equal.

The alternating group A 6
Let G ∼ = A 6 . Then G contains a single class of involutions, two classes of subgroups isomorphic to the alternating group A 5 of degree 5 (say K 1 and K 2 ), two classes of subgroups of order 3 (say H (1) and H (2) ), one class of subgroups of order 5, and two classes of elements of order 5. Let K ∈ K 1 be an A 5 -subgroup in G and let = {1, 2, 3, 4, 5, 6} be the set of cosets of K in G. We assume that the subgroups in H (1) are generated by 3-cycles on and those in H (2) by products of two commuting 3-cycles. Thus a Majorana representation of A 6 satisfying conditions (2A) and (3A) has shape (2 A, 3X, 3Y, 4B, 5A), where a s , a t ∼ = 3X or 3Y if s, t ∈ H (1) or H (2) , respectively. Since the classes H (1) and H (2) are fused in the automorphism group of A 6 , a representation of shape (2 A, 3X, 3Y, 4B, 5A) twisted by a suitable automorphism of A 6 has shape (2 A, 3Y, 3X, 4B, 5A).

Theorem 1
The following assertions hold:

Inner products
In this section by R AC or R CC we denote a representation satisfying the hypothesis in Theorem 1 (i) or (ii), so that k ) = 0 one gets 1 70-dimensional representation R CC was constructed by Ákos Seress (private communication of February 9, 2011).
and the result follows.
Thus the Norton inequality condition fails under the hypothesis of Theorem 1 (i) and from now on we only deal with the representation R CC (where all 3-elements are of type 3C). The Majorana representations of A 5 were classified in [5]. There only one such representation which satisfying the (2 A) condition has shape (2 A, 3C, 5A). Thus the restriction of R CC to an A 5 -subgroup is known.
Recall that Out (A 6 ) is elementary abelian of order 4, and if A 6 < X < Aut (A 6 ), then X ∼ = S 6 , PG L 2 (9) or M 10 . In S 6 all the elements of order 5 are conjugate, but there are still two classes of elements/subgroups of order 3; in PG L 2 (9) there are two classes of elements of order 5, but all elements of order 3 are conjugate, while M 10 enjoys both the conjugation properties: a single class of 5-elements and a single class of 3-elements.
The following lemma is a direct consequence of the Majorana axioms and the shape of R CC .  6 and let f be an element of order 5 in A 6 . Then the possible values of (a r , w f ) are described by Table 3, where f = (1, 2, 3, 4, 5).

Lemma 3.3 Let r be an involution in A
Proof Since the restriction of R CC to an A 5 -subgroup is known, it is sufficient to justify the zero entries in the last column of the last two rows. For the sixth row consider t = (2, 5) (3,4). Then t inverts f and generates with r a D 8 -subgroup. Therefore (which is a 1 4 -eigenvector of a t in a t , w f ∼ = 5A) and (which is a 0-eigenvector of a t in a t , a r ∼ = 4B) are perpendicular. Since t stabilizes w f and swaps a r with a trt , we have Now, making use of Lemma 3.1 and substituting the known entries from the last columns of the second and third rows in Table 3, we deduce the equality (w f , a r ) = 0. For the seventh row the calculations are similar.
Next we determine the inner products between w f 's. Before doing that we describe the A 6 -orbits on the pairs of its subgroups of order 5. The set of these subgroups forms a 6 × 6 grid with rows corresponding to the A 5 -subgroups in K 1 and columns corresponding to the A 5 -subgroups in K 2 . Every subgroup of F = f of order 5 in A 6 is contained in a unique subgroup K 1 ( f ) ∈ K 1 and in a unique subgroup K 2 ( f ) ∈ K 2 , so that The following (partially filled up) Table 4 contains generators of the subgroups of order 5 in two members of K 1 (the first and second rows) and in two members of K 2 (the first and the second columns). The generators are chosen to be A 6 -conjugate, and the subgroup generated by the element in the i-th row and j-th column will be denoted by F i j , so that F 11 = (1, 2, 3, 4, 5) . 6 . Then the following assertions hold:  Table 4; the subgroups in (ii) normalized by t are F 11 , F 12 , F 21 and F 22 . Since N A 6 (t) ∼ = D 8 and for t ∈ A 5 we have N A 5 (t) ∼ = 2 2 , (iii) follows. The pair {F 11 , F 22 } provides an example of two subgroups of order 5 normalized by a common involution, but not contained in a common A 5 -subgroup. In (iv) if F = F 11 and K corresponds to the second row (so that K is the stabilizer in A 6 of 1 ∈ ) then t = (2, 5)(3, 4) is the unique involution with the required properties.

Lemma 3.4 Let t be an involution in A
Lemma 3.5 Let f and e be elements of order 5 in A 6 contained in the same conjugacy class, such that F = f and E = e are distinct. Then the following assertions hold: 6 , then (w f , w e ) = 0 whenever f and e are inverted by a common involution, and (w f , w e ) = 5 2 ·7 2 21 , otherwise.
In order to prove the above lemma we will make use of the following restatement of Lemma 5.4 from [5].

Lemma 3.6 Let e be an element of order 5 in K ∼ = A 5 and t be an involution in K which does not normalize e . Then
where f 1 and f 2 are K -conjugates of e inverted by t and generating distinct subgroups, and for d = f i for i = 1 or 2 is a 1 4 -eigenvector of a t in a t , w f ∼ = 5A.
An immediate consequence of Lemma 3.6 and the Majorana axioms is the following.

Lemma 3.7
With e and t being as in Lemma 3.6 the projections of w e to 1-, 1 32 -, 1 4 and 0-eigenspaces of a t are equal to f 2 ), and respectively.
Now suppose that f ∈ K . Then by Lemma 3.4 (iv) we can assume without loss of generality that f is inverted by t, which provides us with an algorithm of calculating (w f , w e ) from the orthogonality relation In what follows we will demonstrate an easier method how to get (w f , w e ). At this stage we just summarize an important consequence of the orthogonality relation.

Lemma 3.8
The inner product (w e , w f ) is the same whenever e and f are conjugate and not inverted by a common involution.
An important property of the Majorana representation of A 5 of shape (2 A, 3C, 5A) is the following (cf. Lemma 5.7 (i) in [5]).

Lemma 3.9 Let K (5) be a set of conjugate representatives of the subgroups of order
Proof of Lemma 3.7. The assertions (i) and (ii) follow from Lemmas 4.1 (vii) and 4.2 (ii) in [5], respectively. Let e and f be as in (iii). If t is an involution which simultaneously inverts e and f , then the value of (w f , w e ) can be deduced from the orthogonality relation where β (t) f is the a 1 4 -eigenvector of a t in a t , w f ∼ = 5A defined in Lemma 3.6, while is a 0-eigenvector of a t in a t , w e ∼ = 5A (cf. Table 3 in [5]). Thus it only remains to handle the case when f and e are not inverted by a common involution. Let K be an A 5 -subgroup in A 6 , let K (5) be as in Lemma 3.9, and let e an element of order 5, which is not in K , but conjugate in A 6 to members of K (5) . Then by Lemma 3.4 K (5) contains a unique element contained in a common Our goal is to prove that the algebra product · is closed on X (A 6 ). Since · is closed on X (K ) for every A 5 -subgroup K in A 6 , in order to achieve the goal it is sufficient to prove the following two assertions.
Proposition 4.2 X (A 6 ) contains w f · w e whenever f, e = A 6 .

Proof of Proposition 4.1
Throughout the subsection we assume that f = (1, 2, 3, 4, 5) and r = (1, 6)(3, 5), so that the pair ( f, r ) corresponds to the sixth row of Table 3. The situation when the pair corresponds to the seventh row can be handled in a similar way. There are five involutions in A 6 inverting f . If s j denotes the involution which inverts f and stabilizes the element j ∈ , then the information on the relationship between r and s j can be read from the following table.
For every 1 ≤ j ≤ 5 the subalgebra Z j = w f , a s j is of type 5A. Now for j being 4 or 5 we apply a variation of the resurrection principle (cf. Lemma 1.8 in [3]). The vectors are 0-and 1 4 -eigenvectors of a s j in Z j , while α (s j ) r = (a r + a s j rs j ) − 1 32 a s j is a 0-eigenvector of a s j in Y j ∼ = 3C.

Proof of Proposition 4.2
The core of the argument here is similar to that in the previous subsection. Notice that X (A 6 ) contains all the products a t · a s for t, s ∈ T by definition and all the products a t · w f by Proposition 4.1 and since the product · is closed on X (K ) for every A 5 -subgroup K in A 6 . Let f and e be two conjugate elements of order 5 in A 6 , generating the whole of A 6 . Suppose first both f and e are inverted by an involution t. In this case the product w f · w e can calculated by applying the resurrection principle to the 0-eigenvector of a t in a t , w f ∼ = 5A and the 1 4 -eigenvector of a t in a t , w e ∼ = 5A. Therefore from now on we assume that f and e are not inverted by a common involution. Let K i (e) be the unique A 5 -subgroup in K i containing e and let t i ( f, e) be the unique involution in K i (e) which inverts t (cf. Lemma 3.4 (iv)). Put t 1 = t 1 ( f, e), t 2 = t 2 ( f, e), t 3 = t 1 (e, f ), t 4 = t 2 (e, f ).

Lemma 4.6
The involutions t 1 , t 2 , t 3 and t 4 are pairwise distinct, and Proof Since K 1 (e) = K 1 (e t 1 ) and K 2 (e) = K 2 (e t 2 ), the equality t 1 = t 2 would immediately imply that t 1 inverts e. In a similar way one shows that t 3 = t 4 . It is also clear that t j = t k for 1 ≤ j ≤ 2 and 3 ≤ k ≤ 4 since t j inverts f , while t k inverts e, while f and e are chosen not to be inverted by a common involution. Since t 1 , t 2 ∈ N A 6 ( f ) ∼ = D 10 and t 3 , t 4 ∈ N A 6 ( e ) ∼ = D 10 the equalities in the assertion follow.  Table 5 (with s j substituted by t i ) which are contained in a t i , w f ∼ = 5A. By the fusion rules we have  Lemma 4.7 The following assertions hold: (i) for 1 ≤ i ≤ 4 the vector w f · w e is equal to the negative of (w f · w e ) ϕ(t i ) = w t i f t i · w t i et i modulo X