Automorphisms, derivations and gradings of the split quartic Cayley algebra

arXiv:2302.02411v1 [math.RA] 5 Feb 2023 AUTOMORPHISMS, DERIVATIONS AND GRADINGS OF THE SPLIT QUARTIC CAYLEY ALGEBRA. VICTOR BLASCO AND ALBERTO DAZA-GARCIA Abstract. The split quartic Cayley algebra is a structurable algebra which has been used to give constructions of Lie algebras of type D4. Here, we calculate it’s group of automorphisms, it’s algebra of derivations and it’s gradings. 1. Introduction Structurable algebras are a class of algebras with involution introduced by Allison in 1978 [Ali78] as a generalization of Jordan algebras. They are a generalization in the sense that they also have a Tits-Kantor-Koecher (TKK) construction of a Lie algebra. One of these algebras is the split quartic Cayley algebra which is used for example in [Ali91] to give constructions of Lie algebras of type D4. Here, we calculate its group of automorphism, the algebra of derivations and it’s gradings up to isomorphism. The structure is as follows: in section 2 we define the split quartic Cayley algebra and give a multiplication table, in section 3 we calculate it’s group of automorphisms and it’s algebra of derivations and in section 4 we calculate it’s automorphisms. We are going to work over an algebraically closed field Fof characteristic different from 2,3 and 5. Groups are going to be considered abelian and it’s neutral element will be denoted by e, unless we work with specific groups with their own notation. 2. The split quartic Cayley algebra This section is devoted to introduce the split Cayley algebra. In order to do so, we recall a modified Cayley Dickson process introduced in [AF84] starting with the algebra B=F⊕F⊕F⊕F. Take µ∈F×. We denote by tthe trace of B. Define bθ=−b+1 2t(b)1 for all b∈B. Let A=B⊕sB={b1+sb2|b1, b2∈B}. We define a product and an involution in Aby: (b1+sb2)(b3+sb4) = (b1b3+µ(b2bθ 4)θ) + s(bθ 1b4+ (bθ 2bθ 3)θ) b1+sb2=b1−sbθ 2 We call this algebra CD(B, µ). Notice, that since we are in an algebraically closed field, the morphism b1+sb27→ b1+√µsb2is an isomorphism from CD(B, µ) to CD(B,1). Hence, from now on, we are going to work with the algebra CD(B,1). We call this algebra the split quartic Cayley algebra check with the isomorphism in [AF84, Proposition 6.5] and the definition in [Ali90]. Call x1= (1,1,−1,−1), x2= (1,−1,1,−1), x3= (1,−1,−1,1). Call K=F1⊕Fs which is a subalgebra of Aisomorphic to F×Fvia the automorphism given by 2020 Mathematics Subject Classification. Primary 17A01; Secondary 17A30; 17A36. Key words and phrases. Structurable algebras; automorphisms; gradings. The second author acknowledges support by the F.P.I. grant PRE2018-087018. He also aknowledges support by grant MTM2017-83506-C2-1-P (AEI/FEDER, UE) and by grant PID2021123461NB-C21, funded by MCIN/AEI/10.13039/501100011033 and by ”ERDF A way of making Europe”. 1 2 V. BLASCO AND A. DAZA-GARCIA 17→ (1,1), s7→ (1,−1). Then, the action ◦:K×A→Agiven by g◦x=xg for all g∈K, x ∈A, endows Awith a structure of left K-module, which is a free K-module spanned by 1, x1, x2, x3. If we identify Kwith F×Fand call ex the involution given by ex(x, y) = (y, x),the multiplication and the involution follows from the following rules: (f1)(g1) = (fg)1,(gxi)(f1) = (fg)xi= (f1)(fxi) (fxi)(gxi) = (fg)1,(fxi)(gxj) = (fgxk)(2.1) for all f, g ∈Kand {i, j, k}={1,2,3}and f01 + f1x1+f2x2+f3x3= ex(f0)1 + f1x1+f2x2+f3x3(2.2) for all f0, f1, f2, f3∈K. Remark 2.1.Notice that if we define the subspaces S={x∈A|x=−x},H= {x∈A|x=x},M={x∈H|sx +xs = 0}, we get that: S=Fs, H=F1⊕ 3 M i=1 Kxi!,M= 3 M i=1 Kxi,AlgF(S) = K Remark 2.2.There is a Z2 2grading of Agiven by A(¯ 0,¯ 0) =K,A(¯ 0,¯ 1) =Kx1, A(¯ 1,¯ 0) =Kx2and A(¯ 1,¯ 1) =Kx3. We call this grading the standard quartic grading and denote it by ΓSQ. 3. Automorphisms and derivations In this section we calculate the groups of automorphisms and the algebra of derivations of (A,−) (i.e. those automorphisms and derivations which commute with the involution). We begin with some easy properties: Lemma 3.1. Let ϕ∈Aut(A,−)and d∈Der(A,−) (1) ϕ(S) = S,ϕ(H) = H,d(S)⊆Sand d(H)⊆H (2) ϕ(K) = K,ϕ(M) = M,d(K) = 0 and d(M)⊆M. Proof. Each conteinment ’⊆’ in (1) is due to the fact that the involution commutes with ϕand d. The equalities follow from the fact that ϕis invertible. Since ϕ(S) = S, There is λ∈F×such that ϕ(s)e=λs. Since ϕ(1) = 1, we get ϕ(K) = K. If m∈Hand sm +ms = 0, applying ϕwe get that λ(sϕ(m) + ϕ(m)s) = 0. Hence ϕ(M)⊆M. We get the equality since ϕis invertible. Since dis a derivation d(1) = 0. Using (1), there is βsuch that d(s) = βs. Since 0 = d(1) = d(s2) = 2β1, we get that d(K) = 0. Finally, if m∈M0 = d(sm +ms) = sd(m) + d(m)s. Using (1) it follows d(M)⊆M.  Now, we will start calculating the automorphisms. In order to do so, we let S3 be the symmetric on 3 elements, and we will need the following lemma. Lemma 3.2. Let ϕ∈Aut(A,−). There is a permutation in S3which we denote σϕsuch that ϕ(Kxi) = Kxσϕ(i)for all i∈ {1,2,3}. Proof. Due to lemma 3.1 there are r1, r2, r3∈Ksuch that ϕ(xi) = r1x1+r2x2+ r3x3. Let ibe such that ri6= 0. Then since 1 = ϕ(xi)2=r1r1+r3r3+r3r3+r2r3x1+ r1r3x2+r1r2x3. That, due to remark 2.2 means that r1r2=r2r3=r3r1= 0 since up to scalar, the only zero divisors in F×Fare (1,0) and (0,1), this implies that rj, rk= 0 for {i, j, k}={1,2,3}. Hence ϕ(x1) = rixiand riri= 1. Since due to lemma 3.1 ϕ(Kxi) = ϕ(K)ϕ(xi) = Krixi, then we have proved that there is a map AUTOMORPHISMS, DERIVATIONS AND GRADINGS OF THE SPLIT QUARTIC CAYLEY ALGEBRA.3 σϕ:{1,2,3} → {1,2,3}such that ϕ(Kxi) = Kxσ(i)for all i∈ {1,2,3}. Since ϕis invertible this map is a permutation.  Remark 3.3.Let σbe a permutation in S3. We denote by fσ:A→Athe map defined as fσ(r01 + r1x1+r2x2+r3x3) = r01 + r1xσ(1) +r2xσ(2) +r3xσ(3). Using (2.1) is not hard to check that this is an automorphism of (A,−). Moreover the map θ:S3→Aut(A,−) defined by σ7→ fσis a monomorphism of groups and we denote it’s image by H. If we have an algebra with involution (B,−), a group Gand a grading Γ: B= Lg∈GBg, we denote Aut(B,Γ,−) := {ϕ∈Aut(B,−)|ϕ(Bg) = Bg∀g∈G}, Lemma 3.4. Aut(A,−)∼ =Aut(A,ΓSQ,−)⋊H Proof. Let ϕ∈Aut(A,−). We are going to show that ϕ◦f−1 σϕ∈Aut(A,ΓSQ,−). Since θas defined in Remark 3.3 is an automorphism f−1 σϕ=fσ−1 ϕ. By definition ϕ(Kxi) = Kxσϕ(i)for all i. Hence, ϕ◦f−1 σϕ∈Aut(A,ΓSQ,−). Therefore, Aut(A,−) = Aut(A,ΓSQ,−)H. Finally, fσ∈Aut(A,ΓSQ,−) if and only if σ= id. therefore Aut(A,ΓSQ,−)∩H={id}finally it is not hard to show that Aut(A,ΓSQ,−) is a normal subgroup so the result follows.  We denote by S1the subgroup of K×whose underlying set is {r∈K×|rr= 1} and we denote by C2the ciclic group of order 2 generated by σ. We can define an action on Kby σ(s) = −s. Like this we identify C2with Aut(K). Lemma 3.5. Aut(A,ΓSQ,−)∼ =(S1×S1)⋊C2with product given by (r1, r2, g)⋆ (s1, s2, h) = (r1g(s1), r2g(s2), gh). Proof. Consider the morphism θ: (S1×S1)⋊Aut(K)→Aut(A,ΓSQ,−) given by θ(r1, r2, ψ)(s0+s1x1+s2x2+s3x3) = ψ(s0) + ψ(s1)(r1x1) + ψ(s2)(r2x2) + ψ(s3)(r3x3). Where r3=r1r2. Using (2.1) and (2.2) it is clear that θ(r1, r2, ψ) is an automorphism. Since ri(rixi) = xiwe can check that θ(r1, r2, ψ)−1= θ(ψ−1(r1), ψ−1(r2), ψ−1). Clearly θis injective. Moreover, if ϕis an element of Aut(A,ΓSQ,−), then, let ψ=ϕ|K1,ϕ(x1) = r1x1,ϕ(x2) = r2x2and ϕ(x3) = r3x3. Since x2 i= 1 for i= 1,2, we get that riri= 1. Since x1x2=x3we get that r3=r1r2. Hence, it’s easy to show that θ(r1, r2, ψ) = ϕ. Since Aut(K) consist on the identity and the involution sending sto −s, it’s easy to check that it is isomorphic to C2. We can finish calculating the automorphisms with the following proposition: Theorem 3.6. Aut(A,−)∼ =((S1×S1)⋊C2)⋊S3 Proof. This is a consequence of Lemma 3.4 and Lemma 3.5  Finally, we calculate the derivations. In order to do so, for two given numbers λ, β ∈Fwe define the map d(λ,β):A→Aby d(r0+r1x1+r2x2+r3x3) = λr1(sx1) + βr2(sx2)−(λ+β)r3(sx3). Theorem 3.7. Der(A,−) = {dλ,β |λ, β ∈F} Proof. From Lemma 3.1 we get that for any d∈Der(A,−), d(M)⊆M. Therefore, for i, j, k ={1,2,3}we get that d(xi) = rixi+rjxj+rkxkfor some ri, rj, rk∈K. Since 0 = d(1) = d(x2 i) = xid(xi) + d(xi)xiwe get that ri+ri+ 2(rjxk+rkxj) = 0. Therefore, there is some λi∈Fsuch that d(xi) = λi(sxi). Moreover, since λ3(sx3) = d(x3) = d(x1x2) = x1d(x2) + d(x1)x2, it follows that λ3=−λ1−λ2. Finally, since d(K) = 0 and using the properties of the derivations, it follows that d(r0+r1x1+r2x2+r3x3) = λ1r1(sx1)+λ2r2(sx2)−(λ1+λ2)r3(sx3). Now, checking 4 V. BLASCO AND A. DAZA-GARCIA that d1,0and d0,1are derivations is easy and since they span {dλ,β |λ, β ∈F}we get the equality.  4. Gradings Given an algebra with involution (A,−) and a group G, a G-grading Γ on A is a vector space decomposition: Γ: A=M g∈G Ag satisfying that AgAh⊆Agh and Ag⊆Agfor all g, h ∈G. If the grading is fixed we refer to Aas a G-graded algebra with involution. We say that an element x is homogeneous if there is some g∈Gsuch that x∈Ag. In this case we say that xhas degree gand we denote it as deg(x) = g. We say that a subspace Vof Ais a graded subspace if V=Lg∈G(V∩Ag) in this case we will denote Vg=V∩Ag. Remark 4.1.For a Ggrading Γ, Sand Hare graded subspaces (see [AC20, lemma 3.8]). Moreover, since S=sFand s2= 1, we have that deg(s)2=ewhere eis the neutral element of G. Given two G-graded algebras with involution (A,−) and (B,−) we say that they are isomorphic if there exist an isomorphism of algebras with involution ϕ:A→Bsatisfying that ϕ(Ag) = Bg. Given a G-grading Γ and a H-grading Γ′of (A,−) we say that Γ′is a coarsening of Γ (or that Γ is a refinement of Γ′) if for every h∈Hthere is a g∈G such that Ag⊆Ah. The basic facts about gradings can be found in [EK13]. Example 4.2. Given the split quartic Cayley algebra (A,−) and {i, j, k}= {1,2,3}we can define the Z2-grading Γi S:A=A¯ 0⊕A¯ 1with A¯ 0=K⊕Kxi and A¯ 1=Kxj⊕Kxk. These gradings are a coarsening of the standard quartic grading ΓSQ. Moreover, given i6=jand a permutation σwith σ(i) = jwe get that Γi sis isomorphic to Γj via the automorphism fσwith the notation of 3.3 In this section we are going find up to isomorphism the gradings on (A,−). We start with a lemma: Lemma 4.3. For a G-grading Γ: A=Lg∈GAgon (A,−), the subspaces Kand Mare graded subspaces. Proof. In any algebra with involution Sand F1 are graded subspaces. Hence, K=F1⊕Sis a graded subspace. Let m∈Mand let ag∈Agbe such that m=Pg∈Gag. Let g0be the degree of sand for every g∈Gdenote by πgthe projection on Agwith respect to the decomposition given by the grading. Since 0 = sm +ms and 0 = πg(sm +ms) = sag+ags, we get that for every g∈G,ag∈M. Therefore, Mis a graded subspace.  Since s2= 1, it’s easy to deduce using (2.1) that for any m∈Ms(sm) = m. Therefore, we can define two subspaces of M: Mσ={m∈M|sm =σm}for σ=± And M=M+⊕M−. AUTOMORPHISMS, DERIVATIONS AND GRADINGS OF THE SPLIT QUARTIC CAYLEY ALGEBRA.5 Lemma 4.4. For a Ggrading Γ,deg(s) = eif and only if M+and M−are graded subspaces. Proof. Let deg(s) = e. In this case, if m∈Mg, for some g∈Gthen, there are m+∈M+and m−∈M−such that m=m++m−. Since sm =m+−m−∈Mgwe get that mσ1 2(m+(σsm)) ∈Mgfor σ=±. Hence, Mg= (M+∩Mg)⊕(M−∩Mg). From that is easy to check that M+and M−are graded. If M+and M−are graded, let g= deg(s). Then, let m∈(M+) for some h∈Git should happen that h= deg(m) = deg(sm) = gh and so g=e. Remark 4.5.Notice that M+=1 2(1 + s)Mand M−=1 2(1 −s)Mso we are going to call e+=1 2(1 + s) and e−=1 2(1 −s). We denote as b:M×M→Fthe bilinear form which satisfies xy =b(x, y)1 + λs +mfor λ∈Fand m∈M. Lemma 4.6. For any Ggrading on A,bis a non-degenerate homogeneous bilinear form (i.e. b(Mg,Mh) = 0 if and only if gh =e). Proof. In order to show that bis non degenerate, we take m∈M. then, there are r1, r2, r3∈Ksuch that m=r1x1+r2x2+r3x3. Without loss of generality, we suppose that r16= 0. Then, either r1=βeσfor σ=±or r1r1=βx1in both cases with β6= 0. In the first case b(x, e−σx1) = βand in the second case b(x, r1x1) = β. In order to show you that it is homogeneous, we take x∈Mgand y∈Mh. Then, xy ∈Agh. Suppose that gh 6=e. If deg(s) = gh, then, Agh =Fs⊕M∩Agh and in other case Agh =M∩Agh  Example 4.7. Let Gbe an abelian group, i∈Fsuch that i2=−1 and ζ∈Fa primitive cubic root of unit. (1) For g1, g2∈Gdenote by ΓSQ(G, g1, g2) the grading on (A,−) given by deg(s) = e, deg(e+x1) = g1, deg(e+x2) = g2, deg(e+x3) = (g1g2)−1, deg(e−x1) = g−1 1, deg(e−x2) = g−1 2and deg(e−x3) = g1g2. (2) For g, g1, g2∈Gwith gan element of order 2, denote by ΓSQ(G, g, g1, g2) the grading given by deg(s) = g,deg(x1) = g1,deg(x2) = g2,deg(x3) = g1g2,deg(sx1) = gg1,deg(sx2) = gg2and deg(sx3) = gg1g2 (3) For λ∈F×and h, g, f ∈Gsuch that g2=f2=h−1and g6=f, we denote by ΓS(G, λ, h, g, f) the grading in which deg(s) = e, deg(e+x1) = h= deg(e−x1)−1, deg(e+(x2+λx3)) = g, deg(e+(−λ−1x2+x3)) = f (4) For h, g ∈Gwith hof order 2, we denote by Γ1 S(G, h, g) the grading induced by deg(s) = h, deg(e+x2+e−x3) = gand deg(e−x2+e+x3) = g−1. (5) For h, g ∈Gsuch that hhas order 2 and ghas order 4 we denote by Γ2 S(G, h, g) the grading for which deg(s) = h, deg(x1) = g2and deg(x2+ ix3) = g. (6) For h, g, f ∈Gwith h, g and fof order 2, we denote by Γ3 S(G, h, g, f) the grading for which deg(s) = hand deg(x2+x3) = gand deg(x2−x3) = f. (6) For g1, g2∈Gof order 3 and g16=g26= (g1g2)−1, we denote by Γ(G, g1, g2) the grading given by deg(s) = e, deg(e+(x1+ζx2+ζ2x3)) = g1, deg(e+(x1+ ζ2x2+ζx3)) = g2, deg(e+(x1+x2+x3)) = (g1g2)−1 (7) For h, g1∈Gsuch that hhas order 2 and ghas order 3, we denote by Γ(G, h, g) the grading given by deg(s) = hand deg(x1+ζx2+ζ2x3) = g. Given a G-grading Γ: A=Lg∈GAgand a H-grading Γ′:A=Lh∈HAh, we say that the gradings are compatible if A=L(g,h)∈G×HAg∩Ah. 6 V. BLASCO AND A. DAZA-GARCIA Proposition 4.8. If Γis a grading is compatible with ΓSQ, then it is isomorphic to either ΓSQ(G, g1, g2)for some g1, g2∈Gas in example 4.7or to ΓSQ(G, g, g1, g2) for some g, g1, g2as in example 4.7. Proof. If Γ: A=Lg∈GAgis such a grading with deg(s) = e. Since it is compatible with ΓSQ, for every i= 1,2,3 there should be a gisuch that Agi∩Kxi=Kxior giand gi′such that (Agi∩Kxi)⊕(Ag′ i∩Kxi). In the first case, deg(e+xi) = deg(e−xi) = gi. In the second case, since s(Ah∩Kxi) = (Ah∩Kxi) for h=gior h=g′ i, we can assume that Agi∩Kxi=e+xiand that Ag′ i∩Kxi=e−xi. Since (e+xi)(e−xi) = e+, we get that in both cases deg(e+xi) deg(e−xi) = e. Finally, since (e−x1)(e−x2) = e+x3, we get that g3=g−1 1g−1 2. Hence Γ = ΓSQ(G, g1, g2). If Γ: A=Lg∈GAgis such a grading with deg(s) = gfor gan order 2 element. Since each Kxiare graded, if for σ=±,eσxiis homogeneous, deg(eσxi) = deg(s(eσxi)) = gdeg(eσxi). Hence, for every i= 1,2,3 there is a group element gi there is an invertible ri∈Ksuch that deg(rixi) = gisince the field is algebraically closed, we can assume that riri= 1. Since (rixi)2= 1, g2 i=e. Moreover, since (r1x1)(r2x2) = (r1r2)x3, we can assupe that r3=r1r2and that g3=g1g2. Hence, Γ is isomorphic to ΓSQ(G, g, g1, g2) via the morphism θ(r1, r2, Id) with the notation of 3.5.  Proposition 4.9. If Γ: A=Lg∈GAgis a grading compatible with Γi Sfor some i= 1,2,3but not with ΓSQ then either it is isomorphic to ΓS(G, λ, h, g, f)with elements as in example 4.7 or it is isomorphic to Γi S(G, h, g)for i= 1,2with the notation as in example 4.7. Proof. Up to isomorphism we can suppose that it is compatible with Γ1 S. Due to lemma 4.3, Kx1is a graded subspace. If deg(s) = e, since sKx1=Kx1, then e+x1and e−x2are homogeneous. Let deg(e+x1) = h. Since (e+x1)(e−x1), we get that deg(e−x1) = h−1. We are going to prove that e+x2cannot be homogeneous. We prove it by contradiction. If it is homogeneous of degree g, (e+x2)(e+x1) = e−x3is homogeneous of degree gh. Necesarily, there should be a λ∈Fsuch that e+(λx2+x3) is homogeneous of degree f. Necessarily f6=gotherwise e+x3is homogeneous and multiplying by e+x1, we get that e−x1is homogeneous and that means that the grading is compatible with ΓSQ. Now, multiplying by e+x1we get that e(x2+λx3) is homogeneous of degree fh. Since b(e−(x2+λx3), e+x2) = 1 2and b(e−(x2+λx3), e+(λe2, e3)) = λand since ker(b(e−(x2+λx3)),·)|(Kx2⊕Kx3)∩M+has to be a graded subspace, then λ= 0. Therefore, e+x3is homogeneous and as we saw before, this leads to a contradiction with the fact that Γ is not compatible with ΓSQ. Due to the previous discussion, we can assume (because we can multiply by scalar) that there are λ, β ∈F×such that e+(x2+λx3) is homogeneous of degree gand e+(βx2+x3)) is homogeneous of degree fand both are linearly independent. Multiplying by e+x1you get that e−(λx2+x3) is homogeneous of degree gh and that e−(x2+βx3) is homogeneous of degree fh. Call ϕ=b(e−(λx2+ x3),·)|(Kx2⊕Kx3)∩M+. Since ker(ϕhas to be a graded subspace of x2⊕Kx3)∩M+ and ϕ(e+(x2+λx3)) = λ6= 0, necessarily, 0 = ϕ(e+(βx2+x3)) = 1 2(λβ + 1). In order to see that g2=f2=h−1, we see that the square of (e+(x2+λx3)) and of e+(−λ−1x2+x3) are nonzero multiples of e−x1. Therefore, β=−λ−1and therefore, Γ is isomorphic to ΓS(G, λ, h, g, f). If deg(s) = hfor h6=e, clearly h2=e. By lemma 4.4 we know that there is an invertible r1∈Ksuch that r1x1is homogeneous of degree f. Using the automorphism θ(1 √r1r1r1,1, Id) we can assume that r1= 1. We are going to prove AUTOMORPHISMS, DERIVATIONS AND GRADINGS OF THE SPLIT QUARTIC CAYLEY ALGEBRA.7 by contradiction that for no r2∈K,r2x2is homogeneous. Suppose it is. If r2is multiple of eσfor some σ=±, then hdeg(r2x2) = deg(s(r2x2)) = deg(r2x2) which would be a contradiction. If r2is invertible, since s(r2x2) = (sr2)x2,x1(r2x2) = r2x3and (sr2)x3are homogeneous, Γ would be compatible with ΓSQ. Hence, there are r2, r3∈K\0 such that r2x2+r3x3is homogeneous of degree g, then multiplying by x1we get that r3x2+r2x3is homogeneous of degree gf. If there is no r2x2+r3x3homogeneous with r2, r3invertible, then for an homogeneous element like this, (r2x2+r3x3)2= 2r2r3x1= 0 since r2r3∈Feσfor some σ=±. We can suppose then that r2=λeσand r3=βe−σ. Moreover, by scaling the element we can suppose that λ= 1. Call x=r2x2+r3x3. Since x1xhas is a linear combination of xand sx then, either x1x=xin which case β= 1 and deg(x1) = eor x1x=sx, in which case β=−1 and deg(x1) = deg(s). Using θ(s, 1, Id) if necessary, we can suppose that β= 1 and deg(x1) = e. Hence, x=e+x2+e−x3is homogeneous of degree gand since there should be an homogeneous element which doesn’t belong to span{x, sx, x1x, (sx1)x}= span{x, sx}, using the same arguments we see that y=e−x2+e+x3is homogeneous. Since xy = 1 + x1, we get that b(x, y)6= 0 and so yis homogeneous of degree g−1. Hence, Γ is isomorphic to Γ1 S(G, h, g) Finally, assume that r2x2+r3x3is homogeneous with r2, r3invertible, using the automorphism θ(1,1 √r2r2r2) and multiplying by scalar we can assume that r2= 1. Since (x2+r3x3)2= 2r3x1and it is homogeneous, we can assume that r3∈F1∪Fs. Using if necesary the automorphism θ(s, 1,id) we can suppose that r3=λ1 for some λ∈F×. If deg(x1)6=e, since b(x, x1x)6= 0 we get that b(x, x1x) = 0 and that means that λ2=−1. Since in this case −λx1x=x2−λx3, any choice of λwould be an homogeneous element. Hence the grading is isomorphic to Γ2 S(G, h, g). Finally, if deg(x1) = e,λ=±1. Since Kx+K(x1x) = Fx⊕Fsx we need to complete with another homogeneous element. By the same argument it has to be y=x2−x3so the grading is Γ3 S(G, h, g, f) where deg(x2+x3) = g. Proposition 4.10. Let Γ: A=Lg∈GAgbe a grading on (A,−)which is not compatible with any Γi S. Then it is isomorphic either to Γ(G, g1, g2)for g1, g2of order 3or to Γ(G, h, g)for hof order 2and gof order 3. Proof. If deg(s) = ewe start by proving that eσxicannot be homogeneous. Since we can use the automorphisms fτand θ(1,1,ex), we can prove it for i= 1 and σ= +. In this case, ker(b(e+x1,·))∩M−=Fe−x2⊕Fe−x3is homogeneous. Then, since b is non degenerate, there should be an homogeneous element of degree g,x=e−(x1+ λ2x2+λ3x3) with λ2, λ3∈F. Since b(e+x1, x) and b(x2, x) are not 0, it follows that e+x1and x2have the same degree. Since y=1 2x2−λ2λ3e+x1=e+(λ3x2+ λ2x3), if λ2λ36= 0 then, since y2is homogeneous, then e−x1is homogeneous and then ker(b(e+x1,·)) ∩ker(b(e−x1,·)) = Kx2⊕Kx3is graded and because of that this grading is compatible with Γ1 S. If λ26= 0 but λ3= 0. Since x2(e+x1) is homogeneous, e−x2is homogeneous. Hence z=x(e−x2) = e+(λ3x1+x3) is homogeneous. Since b(x, z) = λ36= 0 it follows that zand e+x1have the same degree and so z−λ3e+x1=e+x3is homogeneous. Therefore, (e+x2)(e+x3) = e−x1 is homogeneous and it follows as before that it is not compatible with Γ1 S. If λ2=λ3= 0 we have it because of the same argument. Let x=e+(λ1x1+λ2x2+λ3x3) be an homogeneous element of degree g. It follows that λ1λ2λ36= 0. Hence, by scalar multiplication we can assume that λ1λ2λ3= 1. Take another homogeneous element y=e+(β1x1+β2x2+β3x3) of degree hwith β1β2β3= 1 such that g6=h(which should exist since the grading is not compatible with Γ1 S). We can check that (x2)2= 4xand (y2)2= 4y. That means that g3=eand h3=e. Moreover, its easy to see that b(x, x2) = 6λ1λ2λ36= 0 and 8 V. BLASCO AND A. DAZA-GARCIA b(y, y2) = 6β1β2β3. And because h2g6=ewe deduce that b(x, y2) = b(y, x2) = 0. That implies that λ1λ2β3+λ2λ3β1+λ3λ1β2=β1β2λ1+β2β3λ1+β3β1λ2= 0 (4.1) Moreover, since hg 6=g2and gh 6=h2, we deduce that xy 6=x2and xy 6=y2. Since xy =e+[(λ2β3+λ3β2)x1+ (λ1β3+λ3β1)x2+ (λ2β1+λ1β2)x3] using (4.1) we see that xy =e+(−λ2λ3β1λ−1 1x1−λ1λ3β2λ−1 2x1−λ1λ2β3λ−1 3x3). Therefore, if we call z′=−xy we can see that it’s coefficients products equals to 1. Hence, for z=1 2z′2we get that z2= 2z‘ and (z2)2= 4z. And we can check that the map sending x7→ deg(e+(x1+ζx2+ζ2x3)) = g1,y7→ deg(e+(x1+ζ2x2+ζx3)) = g2 and z7→ deg(e+(x1+x2+x3)) = (g1g2)−1is an isomorphism and so the grading is isomorphic to Γ(G, g, h). If deg(s) = h, as before, Kxicannot be a graded subspace. If all the homogeneous elements x=r1x1+r2x2+r3x3such that r1, r2and r3 are non zero, then, the projection of x2in Mis y=r2r3x1+r1r3x3+r1r2x3which is homogeneous. If r1and r2are not invertible, then this is in Mσfor σ=±and it would happen that deg(y) = deg(sy) which can’t happen unless r3= 0. If r1is not invertible but r2and r3are invertible, we use yto show a contradiction. Hence r1, r2and r3are invertible. Using the map θ(1 √r1r1r1,1 √r2r2r2) we can assume that r1and r2are scalars. If deg(x) = g, since (x2)2= (r1r2r3)x, we can assume that either r3∈Fsor r3∈F1. Since we can scale we can suppose that r1r2r3= 1 or r1r2r3=s. In the first case and in the second case g3=e g3h=eif g3h=ewe can multiply by sand use the automorphism θ(s, s, id) and we are in the first case. Now, either there is an element like this whose degree has order 3 or there are 3 linearly independent elements whose degree is 3. In the second case, necessarily, since Khas dimension 2, there must be an element of degree 3 such that r1, r2or r3 is 0 so we don’t consider it here. Now, since the projections of x, x2, x2x, sx, s(x2x) on Mspan M, necessarily, b(x, x2) = 6r1r2r36= 0 and that implies r3∈F1. Now, up to scalar, we can suppose that r1r2r3= 1 and we can check that the map sending x→x1+ζx2+ζx3induces an isomorphism of algebras. Therefore, Γ is isomorphic to Γ(G, h, g). Finally, we will show that these are all the possibilities. Indeed, if there is an homogeneous element x=r1x1+r2x2of degree g for r1and r2different from 0, since x2= 2r1r2x3necessarily, we get that r1r2= 0. Hence, we can suppose that there is λ1, λ2∈Fsuch that x=λ1e+x1+λ2e−x2. Moreover, sx =λ1e+x1−λ2e−x2 is also homogeneous of degree gh. We can show that all homogeneous elements y=t1x1+t2x2+t3x3with t1, t2, t3∈Khave t1, t2or t3equal to 0. Otherwise, xy =e−t2+e+t3+ (e+t3)x1+ (e−t3)x2+ (e−t2+e+t3)x3so either yor xy has coefficients which are not invertible and arguing as before, this is impossible. Hence, all the homogeneous elements in Mshould be of the form λieσxi±λje−σxjfor i6=j and λi, λj∈F. We can finally show, that if x=λ1e+x1+λ2e−x2is homogeneous, there should be β2, β3∈F×such that y=β2e++β3e−is homogeneous. But since xy =λ2β2e−+λ1β2e−x3+λ2β3e+x1and that would imply that e−is homogeneous since Kand Mare homogeneous subspaces. But this would be a contradiction with the fact that deg(s)6= deg(1).  References [Ali78] B.N. Allison, A class of nonassociative algebras with involution containing the class of Jordan algebras, Math. Ann. 237 (1978), 133–156. [Ali91] B.N. Allison, Construction of 3x3 matrix Lie algebra of type D4, Journal of Algebra 143 (1991), 63–92. [Ali90] B.N. Allison, Simple structurable algebras of skew-dimension 1, Comm. Algebra 18 (1990), 1245–1279. AUTOMORPHISMS, DERIVATIONS AND GRADINGS OF THE SPLIT QUARTIC CAYLEY ALGEBRA.9 [AF84] B.N. Allison and J.R. Faulkner, A Cayley-Dickson process for a class of structurable algebras, Trans. Amer. Math. Soc. 283 (1984), 185–210. [AC20] D. Aranda-Orna and A.S. Cordova-martinez Gradings on tensor products of compositon algebras and on the Smirnov algebra , Linear algebra and it’s applications 548 (2020), 1–36. [EK13] A. Elduque and M. Kochetov, Gradings on simple Lie algebras, Mathematical Surveys and Monographs 189, American Mathematical Society, Providence, RI; Atlantic Association for Research in the Mathematical Sciences (AARMS), Halifax, NS, 2013. Departamento de Matem´ aticas e Instituto Universitario de Matem´ aticas y Aplicaciones, Universidad de Zaragoza, 50009 Zaragoza, Spain Email address:[email protected] Departamento de Matem´ aticas e Instituto Universitario de Matem´ aticas y Aplicaciones, Universidad de Zaragoza, 50009 Zaragoza, Spain Email address:[email protected]