1 Introduction↩︎

The structure theory of nonassociative algebras and rings has been generally based on consideration of a concrete variety \(\mathcal{M}\) of algebras and then on the description or a search of (finite-dimensional) simple algebras from \(\mathcal{M}\). We may say that the structure theory of Jordan, alternative, Malcev etc. algebras was developed more or less in such way.

Recently, a new approach to get plenty of simple nonassociative algebras appears. The notion of axial algebra was proposed by J.I. Hall, F. Rehren and S. Shpectorov in 2015 [1]. It defines a class of (non-associative) commutative algebras generated by specific idempotents, and the product in an axial algebra satisfies some restrictions depending on its type. Roughly speaking, we may say that these restrictions generalize the ones originated from Pierce decomposition fulfilled on associative, alternative, or Jordan algebras. Axial algebras of Jordan type are close to Jordan algebras, while axial algebras of Monster type generalize the Griess algebra. The latter has the Monster group exactly as its automorphism group.

In the direction of axial algebras, different classifications of algebras with a small number of generators were stated. One of them, obtained by T. Yabe in 2020 (published in 2023 [2]), provides a list of all 2-generated axial algebras of Monster type \((\xi,\eta)\) admitting a flip between generating axes. One of the algebras from the Yabe’s list was denoted as \(S(\alpha,E)\) and the properties of this algebra were studied by J. McInroy and S. Shpectorov in 2022 [3]. The authors called them as split spin factor algebras by analogy with spin factor algebra, the simple Jordan algebras of special form.

The main goal of the current work to study the identities fulfilled on the split spin factors \(S(\alpha,t,E)\), where \(E\) is any vector space of dimension at least two endowed with a symmetric nondegenerate form \(\langle\cdot,\cdot\rangle\) and \(\alpha,t\) are parameters from the ground field \(F\). We show that there are no identities of degree 3 and 4 on \(S(\alpha,t,E)\), \(\alpha,t\notin\{0,1\}\), which do not follow from commutativity. Further, we prove that all identities on \(S(\alpha,E)\), \(\alpha\notin\{-1,0,1/2,1,2\}\), of degree 5 follows from commutativity and \[((a,b,c),d,b) + ((c,b,d),a,b) + ((d,b,a),c,b) = 0,\] where \((a,b,c) = (ab)c - a(bc)\). We name it as the three associators identity.

In 1965, J.M. Osborn gave the list of all irreducible relative to commutativity identities of degree 5 [4]. The fourth of these five identities [4] with \(\delta_2 = -\delta_1\neq0\) is the three associators identity with one of the three variables \(a,c,d\) equal to \(b\), e. g., \(d = b\).

There is a construction of a Jordan algebra by any sharped cubic form \((N,\#,c)\) [5]. For the proof, you need to verify dozens of relations in terms of \(N\), its derived maps \(T,S\), \(\#\), the triple product \(\{\cdot,\cdot,\}\) and the \(U\)-operator, see [6]. This construction, in particular, allows to build simple Jordan algebras of Albert type. In the work, we consider generalized sharped cubic form and prove the analogues of the known relations, which hold for sharped cubic forms. With the help of these relations (Lemmas 1–7), we more or less reduce checking the three associators identity on the algebra \(S(\alpha,t,E)\) to the properties of the Psi-map \(\Psi(r,s,q)\), which is defined via an associator, see §5. Further, in Lemma 9, we show that \(\Psi(r,s,q)\) may be expressed in terms of the projections of \(r,s,q\) on \(E\) and its bilinear form \(\langle \cdot,\cdot\rangle\). Hence, \(\Psi(r,s,q)\) defines a Lie triple product on \(S(\alpha,t,E)\) and it is connected with the simple pre-Lie algebra from [7]. Finally, we prove in Theorem 3 that the three associators identity holds on \(S(\alpha,t,E)\).

Let us provide a short outline of the work. In §2, we give the definition of the split spin factor algebra \(S(\alpha,E)\) and its natural generalization \(S(\alpha,t,E)\). In §3, we recall the results about sharped cubic form and induced Jordan algebra. In §4, we introduce a generalized sharped cubic form and prove the main relations devoted to it. In §5, we define the Psi-map and derive the equalities on it. The goal of §6 is to prove that the three associators identity holds on \(S(\alpha,t,E)\). In §7, with the help of computer algebra, we prove that all identities of degree not greater than 5 fulfilled on \(S(\alpha,E)\) follow from commutativity and the three associators identity. In the general case of \(S(\alpha,t,E)\), there are identities of degree 5, which do not follow from commutativity and the three associators identity. In §8, we formulate open problems concerning generalized sharped cubic form and induced algebras. In particular, the following question remains to be open: what identity holds on all algebras induced by a generalized sharped cubic form?

We assume that the ground field \(F\) is of characteristic not 2 and not 3.

2 Split spin factor algebra↩︎

Let \(F\) be quadratically closed, i. e. roots of any quadratic equation over \(F\) lie in it.

Below, we put the multiplication table for \(S(\alpha,E) = Fz_1+Fz_2+E\), where \(\dim E = 2\) and there exists a nondegenerate bilinear form \(\langle\cdot,\cdot\rangle\) on \(E\): \[\label{Split-Spin-Prod-Def} \begin{gather} z_1^2 = z_1, \quad z_2^2 = z_2, \quad z_1z_2 = 0, \quad ez_1 = \alpha e, \quad ez_2 = (1-\alpha)e, \\ ef = -\langle e,f\rangle(\alpha(\alpha-2)z_1+(\alpha^2-1)z_2),\;e,f\in E. \end{gather}\tag{1}\]

For \(\alpha\neq-1,2\), the algebra \(S(\alpha,E)\) admits the nondegenerate invariant bilinear form given by \[\label{Split-Spin-Inv-Form} \begin{gather} (z_1,z_1) = \alpha+1, \quad (z_2,z_2) = 2-\alpha, \quad (z_1,z_2) = 0, \\ (e,f) = (\alpha+1)(2-\alpha)\langle e,f\rangle, \quad (e,z_i)= 0, \;e,f\in E. \end{gather}\tag{2}\] Invariancy means that \((ab,c) = (a,bc)\) for all \(a,b,c\in S(\alpha,E)\).

Note that in 1 and 2 we may consider any vector space \(E\), not necessarily of dimension 2. More generally, we may study the algebra \(S(\alpha,t,E)\) depending on two parameters \(\alpha,t\), where \(E\) is a vector space of any dimension endowed with a nondegenerate bilinear form. Then the product of elements \(e,f\in E\) is defined by the formula \[ef = \langle e,f\rangle(z_1+tz_2).\] The split spin factor algebra \(S(\alpha,E)\) is an algebra \(S(\alpha,t,E)\) with \(t = (\alpha^2-1)/\alpha(\alpha-2)\). Surely, we require that \(\alpha\neq0,2\).

Proposition 1. Let \(E\) has a finite dimension \(n\ge 1\). Then \(S(\alpha,t,E)\) is a simple algebra if and only if \(\alpha\notin \{0,1\}\), \(t\neq 0\).

Proof. Let \(\alpha\notin \{0,1\}\), \(t\neq 0\). There is a basis \(\{e_1,\dots,e_n\}\) in \(E\) so that \(\langle e_i,e_j\rangle = \delta_{i,j}\). Let us show that \(S(\alpha,t,E)\) is simple. If \(I\) is a nonzero ideal in \(S(\alpha,t,E)\), then it contains a nonzero element \(x=\beta z_1 + \gamma z_2 + \sum\limits_{i=1}^n\alpha_i e_i\).

If \(\alpha_k\neq 0\) for some \(k>0\), then \(I\) contains an element \[xe_k = \alpha_k (z_1 + tz_2) + (\alpha\beta + (1-\alpha)\gamma ) e_k.\] Hence, \(y=z_1+tz_2+\delta e_k\in I\) for \(\delta = (\alpha\beta-\alpha\gamma+\gamma)/\alpha_k\). Then \(I\) contains \[(1-\alpha)yz_1-\alpha yz_2=(1-\alpha)z_1-\alpha tz_2.\] It means that \((1-\alpha)z_1\in I\) and \(\alpha tz_2\in I\). Thus, \(z_1,z_2\in I\). We also have \[I\ni z_1e_i = \alpha e_i\] for all \(1\le i\le n\). Therefore, \(I = S(\alpha,t,E)\).

If \(\alpha_i=0\) for all \(1\le i\le n\), then \(0\neq \beta z_1 + \gamma z_2\in I\). It means that \(z_i\in I\) for some \(i\in\{1,2\}\). Then \(I\ni z_ie_k\) and we have \(e_k\in I\) for \(1\le k\le n\) by assumptions. But it means that \(e_k^2 = z_1+tz_2\in I\) and \(z_1,z_2\in I\) as above.

So, \(I=S(\alpha,t,E)\) and \(S(\alpha,t,E)\) is a simple algebra.

If \(\alpha = 0\), then \(Fz_1\) is a proper ideal in \(S(\alpha,t,E)\). If \(\alpha = 1\), then \(Fz_2\) is a proper ideal in \(S(\alpha,t,E)\). If \(t=0\), then \(Fz_1+\sum\limits_{i=1}^n Fe_i\) is a proper ideal in \(S(\alpha,t,E)\). \(\square\)

We will use a notation \(O(E)\) for a subgroup of \(\mathrm{Aut}(S(\alpha,t,E))\) obtained by an extension of the orthogonal group of \(E\), which elements fix \(z_1\) and \(z_2\).

Proposition 2. Let \(E\) has a finite dimension \(n\ge 2\) and \(\alpha\notin \{0,1\}\), \(t\neq 0\).

• If \(\alpha\neq 1/2\) or \(t\neq \pm 1\), then \(\mathrm{Aut}(S(\alpha,t,E)) \cong O(E)\).

• If \(\alpha = 1/2\) and \(t=\pm 1\), then \(\mathrm{Aut}(S(\alpha,t,E)) \cong \mathbb{Z}_2\times O(E)\).

Proof. Let \(A=S(\alpha,t,E)\) and \(\varphi\in\mathrm{Aut}(A)\). There is a basis \(\{e_1,\dots,e_n\}\) in \(E\) so that \(\langle e_i,e_j\rangle = \delta_{i,j}\). We want to describe all \(x\in A\) with \(\dim\mathrm{Ann}(x)=n\). Let \(x\in A\) so that \(\dim\mathrm{Ann}(x)=n\) and \(x=\beta z_1 + \gamma z_2 + \alpha_1 e_1 +\ldots +\alpha_n e_n\). Then \[xe_i = \alpha_i(z_1+tz_2) + (\beta\alpha + \gamma (1-\alpha)) e_i.\] If \(\beta\alpha+\gamma (1-\alpha) \neq 0\), then \(xe_1,\dots, xe_n\) are linearly independent. Moreover, we have \[\begin{gather} xz_1 = \beta z_1 + \alpha (\alpha_1e_1+\dots + \alpha_n e_n), \\ xz_2 = \gamma z_2 + (1-\alpha) (\alpha_1e_1+\dots+\alpha_n e_n). \end{gather}\] An assumption \(\beta\alpha + \gamma(1-\alpha)\neq 0\) means that \(\beta\neq 0\) or \(\gamma\neq 0\). If \(\beta\neq 0\), then \(xz_1,xe_1,\dots,xe_n\) are linearly independent and \(\dim\mathrm{Ann}(x)\le 1\), a contradiction. If \(\gamma\neq 0\), then \(xz_2,xe_1,\dots,xe_n\) are linearly independent and \(\dim\mathrm{Ann}(x)\le 1\), a contradiction.

So, \(\beta\alpha + \gamma (1-\alpha) = 0\) and \(\gamma = \alpha\beta/(\alpha - 1)\).

Suppose that \(\beta\neq 0\). If \(\alpha_i\neq 0\) for some \(i\), then \(xe_i\), \(xz_1\) and \(xz_2\) are linearly independent and \(\dim\mathrm{Ann}(x) < n\), a contradiction. Thus, \(x=\beta (z_1 - \frac{\alpha}{1-\alpha}z_2)\). If \(\beta = 0\), then \(\gamma = 0\) and \(x\in E\).

Let us denote \(U=F\cdot (z_1-\frac{\alpha}{1-\alpha}z_2)\). So, \(x\in A\) and \(\dim\mathrm{Ann}(x)=n\) if and only if \(x\in E\cup U\) and \(x\neq 0\). It means that \(\varphi(x)\in E\cup U\) for any \(x\in E\cup U\).

Suppose that \(\varphi(e)\in U\) for some \(0\neq e\in E\). Since \(\dim E \ge 2\), there exists \(0\neq f\in E\) such that \(\varphi (f)\in E\). It means that \(\varphi(e+f)\notin E\cup U\), a contradiction. Hence, \(\varphi (E)=E\), \(\varphi(U)=U\). Therefore, \(\varphi(z_1-\frac{\alpha}{1-\alpha}z_2)=\delta(z_1-\frac{\alpha}{1-\alpha}z_2)\) for some nonzero \(\delta\).

We have \(\varphi(z_1+z_2)=z_1+z_2\), so \[\varphi(z_1)=(\alpha + \delta(1-\alpha))z_1+(\alpha-\delta\alpha)z_2.\] Since \(z_1\) is an idempotent, hence \(\varphi(z_1)\) is an idempotent. It means that \[\alpha+\delta(1-\alpha),\alpha-\delta\alpha\in\{0,1\}.\] If \(\alpha-\delta\alpha = 1\), then \(\delta = (\alpha-1)/\alpha\). We have two cases:

1. \(\alpha + \delta(1-\alpha) = 0\). It means that \(\delta = \alpha/(\alpha - 1)\) and \(\alpha = 1/2\) by above. Hence, \(\delta = -1\) and \(\varphi(z_1-z_2)=-z_1+z_2\). It is easy to see that \(\varphi(z_1)=z_2\), \(\varphi(z_2)=z_1\). We have \[tz_1+z_2 = \varphi (z_1+tz_2) = \varphi(e_1^2) = \varphi(e_1)^2 = \gamma (z_1+tz_2)\] for some \(\gamma\in F\). Hence, \(\gamma = t\) and \(t^2=1\), \(t=\pm 1\).

2. \(\alpha + \delta(1-\alpha) = 1\). It means that \((\alpha - 1) = (\alpha - 1)\delta\) and \(\delta = 1\).

If \(\delta = 1\), then \(\varphi(z_1)=z_1\) and \(\varphi(z_2)=z_2\). If \(\delta = -1\), \(\alpha = 1/2\), and \(t=\pm 1\), then \(\varphi(z_1)=z_2\) and \(\varphi(z_2)=z_1\). We have proved that \(\varphi(e)\in E\) for any \(e\in E\). In cases \(\delta=1\) and \(\delta=-1\), \(\alpha=1/2\), \(t=1\) the basis \(e_1,\dots,e_n\) is mapped to an orthogonal basis of \(E\), since \(\varphi(z_1+tz_2)=z_1+tz_2\). It remains to prove that \(\mathrm{Aut}(A) \cong \mathbb{Z}_2\times O(E)\), when \(\delta=t=-1\) and \(\alpha=1/2\). In the case, we have \(\mathrm{Aut}(A) = \{ (\sigma,\psi) \mid \sigma\in S_2,\, \langle \psi(e),\psi(f)\rangle = \mathrm{sgn}(\sigma) \langle e,f\rangle,\,e,f\in E \}\), where \(\mathrm{sgn}(\sigma)\) denotes the sign of a permutation \(\sigma\in S_2\). Thus, \(\pi\colon \mathrm{Aut}(A) \to \mathbb{Z}_2\times O(E)\) defined as follows, \(\pi( (\sigma,\psi) ) = (\sigma,\sqrt{-1}^{\mathrm{sgn}(\sigma)}\psi )\) is an isomorphism. \(\square\)

3 Sharped cubic form and associated Jordan algebra↩︎

In this paragraph, we recall the results concerned sharped cubic forms and induced algebras, which occur to be Jordan. We follow the monograph [6].

A map \(N\colon V\to F\) on a space \(V\) is called cubic form if for any \(\lambda\in F\) and \(x,y\in V\) \[N(x + \lambda y) = N(x) + \lambda N(x,y) + \lambda^2 N(y,x) + \lambda^3 N(x,y,z),\] where \(N(x,y)\) is quadratic in \(x\) and linear in \(y\) and \(N(x,y,z)\) is trilinear and symmetric.

Given a cubic form \(N\) on a space \(V\), one can linearize it completely as follows, \[N(v,u,w) = N(v+u+w)-N(v+u)-N(v+w)-N(u+w)+N(u)+N(v)+N(w).\] Hence, \(N(r,r,r) = 6N(r)\).

Definition 1. Let \(V\) be a vector space endowed with a cubic form \(N\) and let \(c\in V\) be such that \(N(c) = 1\) (we call \(c\) as basepoint). Denote by \(N(x,y,z)\) the complete linearization of \(N\). We introduce quadratic spur function, linear trace form, two bilinear forms and give another definition of the form \(N(x,y)\): \[\begin{gather} S(r) = N(r,r,c)/2, \quad T(r) = N(r,c,c)/2, \tag{3} \\ S(r,q) = N(r,q,c), \quad N(r,q) = N(r,r,q)/2, \tag{4} \\ (r,q) = T(r)T(q) - S(r,q). \tag{5} \end{gather}\] From the definition we derive that \[\label{T44SOnUnit} S(c) = T(c) = 3, \quad (r,c) = T(r).\tag{6}\]

Another way to define the function \(N(r,q)\) from the given norm function \(N\) is to present \[\label{Norm-62Norm2} N(r+tq) = N(r) + tN(r,q) + t^2N(q,r) + t^3N(q).\tag{7}\] Thus, \(N(r,q,s)\) is a linearization of \(N(x,y)\).

Definition 2. Let \(V\) be a vector space endowed with a cubic form \(N\) and basepoint \(c\). A sharp map \(\#\) on \(V\) for \((N,c)\) is a quadratic operator on \(V\) satisfying the following relations: \[\begin{gather} (r^{\#},q) = N(r,q), \\ (r^{\#})^{\#} = N(r)r, \\ c_{\#}r = T(r)c - r, \end{gather}\] where the sharp-product is given by the formula \[\label{sharpproduct} r_\# q = (r+q)^\# - r^\# - q^\#.\tag{8}\] Under the conditions, we call \((N,\#,c)\) as a sharped cubic form.

Due to the definitions, \(c^{\#} = c\).

Theorem 1. Let \(V\) be a vector space endowed with a sharped cubic form \((N,\#,c)\). Then

a) \(V\) under the product \[\label{product} rq = \frac{1}{2}(r_\# q + T(r)q + T(q)r - S(r,q)c)\tag{9}\] is an algebra with a unit \(c\). Moreover, every \(r\in V\) satisfies the cubic identity \[\label{cubic-identity} r^3 - T(r)r^2 + S(r)r - N(r)c = 0,\tag{10}\] and \(r^{\#} = r^2 - T(r)r + S(r)c\) (so, \(r^{\#}r = N(r)c\)).

b) \(V\) is Jordan and the operators \[\begin{gather} U_r(s) = (r,s)r - {r^\#}_\# s, \quad U_{r,q}(s) := U_{r+q}(s) - U_r(s) - U_q(s), \tag{11} \\ \{r,s,q\} := (r,s)q + (q,s)r - (r_\# q)_\# s \tag{12} \end{gather}\] coincide with the classical \(U\)-operators and Jordan triple product respectively.

In [6], Theorem 1b) is proved via the following formulas.

Proposition 3. Let \(V\) be a vector space endowed with a sharped cubic form \((N,\#,c)\). Then the following identities hold on \(V\): \[\begin{gather} S(r) = T(r^{\#}), \quad S(r,q) = T(r_{\#}q), \\ (rq,s) = (r,qs), \quad (r_{\#}q,s) = (r,q_{\#}s) = N(r,q,s), \quad (U_r(s),q) = (s,U_r(q)), \allowdisplaybreaks \\ r^{\#}{}_{\#}(q_{\#}r) = N(r)q + (r^{\#},q)r, \quad (r^{\#}{}_{\#}q)_{\#}r = N(r)q + (r,q)r^{\#}, \\ (r_{\#}q)^{\#} + r^{\#}{}_{\#}q^{\#} = (r^{\#},q)q + (r,q^{\#})r, \\ r^{\#}{}_{\#}r = -T(r)r^{\#} - T(r^{\#})r + (S(r)T(r)-N(r))c, \\ S(r^{\#},r) = S(r)T(r) - 3N(r), \quad (r^{\#},r) = 3N(r), \\ U_r(c) = r^2, \quad U_{r,q}(c) = 2rq, \quad U_r(r^{\#}) = N(r)r, \quad (U_r(q))^{\#} = U_{r^{\#}}(q^{\#}), \\ U_r U_{r^{\#}} = N(r)^2 \mathrm{id}, \quad \{r,r^{\#},q\} = 2N(r)q, \\ N(U_r(q)) = N(r)^2N(q), \quad N(r^{\#}) = N(r)^2. \end{gather}\]

4 Generalized sharped cubic form and its algebra↩︎

Now, we suggest a construction, which generalizes sharped cubic form.

Definition 3. Let \(V\) be a vector space endowed with a cubic form \(N\) and let \(c\in V\) be such that \(N(c) = 1\). We also assume that a symmetric bilinear form \(\Delta\) is defined on \(V\) in such manner that \[\label{delta40r44c41610} \Delta(r,c) = 0\tag{13}\] for all \(r\in V\). Denote by \(N(x,y,z)\) the complete linearization of \(N\). As in Definition 1, we introduce quadratic spur function \(S(r)\), linear trace form \(T(r)\), bilinear form \(S(r,q)\) by 3 , quadratic in \(r\) and bilinear in \(q\) form \(N(r,q)\) by 4 and new bilinear form \[\label{GNorm} (r,q) = T(r)T(q) - S(r,q) - \Delta(r,q).\tag{14}\] Let us call a pair \((N,\Delta)\) as a generalized cubic form. When \(\Delta = 0\), we have an ordinary cubic form.

The equalities 6 follow from the definition immediately.

Definition 4. Let \(V\) be a vector space endowed with a generalized cubic form \((N,\Delta)\) and basepoint \(c\). A sharp map \(\#\) on \(V\) for \((N,\Delta,c)\) is a quadratic operator on \(V\) satisfying the following relations: \[\begin{gather} (r_\#q,r) + (r^\#,q) = 3N(r,q), \tag{15} \\ (r^\#)^\# = (N(r) + \Delta(r^\#,r))r, \tag{16} \\ c_{\#}r = T(r)c - r, \tag{17} \end{gather}\] where the sharp-product is given by 8 . Under the conditions, we call \((N,\Delta,\#,c)\) as a generalized sharped cubic form.

As above, we conclude that \(c^{\#} = c\).

Given a generalized sharped cubic form \((N,\Delta,\#,c)\), let us define a product on \(V\) by 9 .

Proposition 4. Fix a scalar \(\lambda\in F\setminus\{-1\}\). Given a cubic form \(N\) defined on a vector space \(V\) with a basepoint \(c\), we get a generalized cubic form by the formulas \[(r,q) = \frac{1}{\lambda+1}\left((1+\lambda/3)T(r)T(q)-S(r,q)\right), \quad \Delta(r,q) = \lambda \left((r,q) - \frac{T(r)T(q)}{3}\right).\]

Proof. The identity 14 holds by the definition. Also, we get \((r,c) = \frac{T(r)+\lambda T(r)}{1+\lambda} = T(r)\), hence, \(\Delta(r,c) = 0\). \(\square\)

Let us call a generalized cubic form defined in Proposition 4 as an inner one.

Example 1. Let \(A\) be an associative commutative algebra over a field \(F\) generated by the unit of \(F\) and an element \(\lambda\) such that \(\lambda^2 = 0\). Consider the space \(V = A\otimes_F F^3\cong A^{\otimes3}\). We endow \(V\) with the form \(N((x,y,z)) = xyz\) and take \(c = (1,1,1)\). Formally, \(N\) is not a cubic form, since it maps \(A^{\otimes3}\) to \(A\) instead of \(F\). However, we get an interesting example of an algebra. Denote \(r = (x,y,z)\), \(q = (x',y',z')\). Then \[\begin{gather} T(r) = x+y+z, \quad S(r) = xy + xz + yz, \\ S(r,q) = x(y'+z') + y(x'+z') + z(x'+y'), \; N(r,q) = xyz' + xy'z + x'yz. \end{gather}\] We define \[r^\# = (yz,xz,xy) - \lambda(y^2+z^2+2x(y+z),x^2+z^2+2y(x+z),x^2+y^2+2z(x+y)).\] and get by 9 , \[rq = (1+\lambda)(xx',yy',zz') + \lambda(yz'+y'z,xz'+x'z,xy'+x'y) - \lambda T(r)T(q)c.\] By the definition, \[c^\# = (1-6\lambda)c, \quad rc = r - 2\lambda T(r)c,\quad c_{\#}r = (1-4\lambda)T(r)c - r.\]

Now, we consider two pairs of bilinear maps \((\cdot,\cdot)\) and \(\Delta\) such that 14 holds: \[(r,q) = xx'+yy'+zz' + 3\lambda S(r,q), \quad \Delta(r,q) = - 3\lambda S(r,q).\] When \(\Delta = \lambda = 0\), we get an ordinary sharped cubic form.

It is easy to check that a cubic form \((N,\Delta,\#,c)\) satisfies 15 . Instead of 16 , the following relation holds: \[(r^\#)^\# = (N(r) + \Delta(r^\#,r) + 2\lambda T(r)S(r))r + 2\lambda S(r)r^\# - 2\lambda (T(r)N(r)+S(r)^2)c.\]

There is some routine (see the code in GAP [8]) to derive the relations fulfilled for such maps (see the same or close equalities derived for all generalized sharped cubic forms below): \[\begin{gather} \Delta(r,c) = -6\lambda T(r), \quad (r,c) = (1+6\lambda)T(r), \quad (r^\#,r) = 3N(r), \\ T(r^\#) = (1-6\lambda)(S(r) - \Delta(r,r)) - 2\lambda T(r)^2, \allowdisplaybreaks \\ (r_\#q,s) - (r,q_\#s) = \Delta(r,q_\#s) - \Delta(r_\#q,s) = T(r)\Delta(q,s) - T(s)\Delta(r,q), \\ (r_\#q,s) + (q_\#s,r) + (s_\#r,q) = 3N(r,q,s), \\ N(r^\#) = N(r)(N(r) + \Delta(r^\#,r)). \end{gather}\]

Example 2. The general case of the split spin factor \(S(\alpha,t,E)\) with \[\begin{gather} N(az_1 {+} bz_2 {+}v) = ab(\alpha a+\bar{\alpha}b) - \langle v,v\rangle(\bar{\alpha}ta+\alpha b), \\ \Delta(az_1 + bz_2 + v,kz_1 + lz_2 + u) = \alpha(\alpha-1)(a-b)(k-l)-\langle u,v\rangle(\bar{\alpha}+\alpha t),\\ (az_1 + bz_2 + v)^\# = (\alpha a+\bar{\alpha}b)(bz_1+az_2) + (t-1)\langle v,v\rangle(-\bar{\alpha}z_1+\alpha z_2) - (\bar{\alpha}a {+} \alpha b)v, \label{SplitSpinSharp} \end{gather}\tag{18}\] where \(\bar{\alpha} = 1-\alpha\) and \(c = z_1 + z_2\) is a generalized sharped cubic form. Here \(a,b,k,l\in F\) and \(u,v\in E\).

Indeed, \(N(c) = 1\), \(\Delta(r,c) = 0\). Further, for \(r = az_1 + bz_2 + v\), we have \[\begin{gather} 2T(r) = N(r+2c) - 2N(r+c) - N(2c) + 2N(c) + N(r) \\ = (a+2)(b+2)(\alpha(a+2)+\bar{\alpha}(b+2)) - 2(a+1)(b+1)(\alpha(a+1)+\bar{\alpha}(b+1)) \\ + ab(\alpha a+\bar{\alpha}b) - 6 = 2( (1+\alpha)a + (2-\alpha)b ), \end{gather}\] and 17 holds, since \[\begin{gather} r_\# c = (r+c)^\# - r^\# - c^\# \\ = (\alpha(a+1)+\bar{\alpha}(b+1))((b+1)z_1+(a+1)z_2) - (\alpha a+\bar{\alpha}b)(bz_1+az_2) - v - c \\ = ( (1+\alpha)a + (2-\alpha)b )(z_1+z_2) - az_1 - bz_2 - v = T(r)c - r. \end{gather}\]

Now, we compute due to the definition, \[\Delta(r,r^\#) = \alpha(\alpha-1)(a-b)( (\alpha a+\bar{\alpha}b)(b-a) - (t-1)\langle v,v\rangle ) + (\bar{\alpha}a +\alpha b)\langle v,v\rangle(\bar{\alpha}+\alpha t).\]

It is not difficult to show that \[N(r) + \Delta(r,r^\#) = (\bar{\alpha}a+\alpha b)( (\alpha a+\bar{\alpha}b)^2 + (2\alpha-1)(t-1)\langle v,v\rangle).\] On the other hand, the projection of \((r^\#)^\#\) on \(E\) equals by 18 \[(\bar{\alpha}a + \alpha b)\big( (\alpha a+\bar{\alpha}b)(\bar{\alpha}b + \alpha a) + (2\alpha-1)(t-1)\langle v,v\rangle \big)v = (N(r) + \Delta(r,r^\#))v.\] We leave the check that the coordinates of \((r^\#)^\#\) at \(z_1\) and \(z_2\) also equal to the ones of \((N(r) + \Delta(r,r^\#))r\). Thus, 16 follows.

Finally, we have to derive 15 . For this, we write down the required forms for \(r = az_1 + bz_2 + v\) and \(s = kz_1 + lz_2 + u\): \[\begin{gather} N(r,s) = (N(2r+s) - 2N(r+s) - N(2r) + 2N(r) + N(s))/2 \\ = \alpha a^2 l + 2\alpha ab k + 2\bar{\alpha}abl + \bar{\alpha}b^2k - \langle v,v\rangle(\bar{\alpha}t k+\alpha l) - 2\langle v,u\rangle(\bar{\alpha}t a+\alpha b), \end{gather}\]

\[\begin{gather} S(r,s) = N(r+s+c) - N(r+s) - N(r+c) - N(s+c) + N(r) + N(s) + N(c) \\ = 2(\alpha ak+al+bk+\bar{\alpha}bl) -2(\bar{\alpha}t+\alpha)\langle v,u\rangle, \end{gather}\]

\[\label{inner-product-derived} (r,s) = T(r)T(s) - S(r,s) - \Delta(r,s) = (1+\alpha)ak + (2-\alpha)bl + (1+\alpha+(2-\alpha)t)\langle v,u\rangle,\tag{19}\]

\[\begin{gather} r_\# s = (r+s)^\# - r^\# - s^\# = (\alpha a + \bar{\alpha}b)(lz_1+kz_2) + (\alpha k + \bar{\alpha}l)(bz_1+az_2) \\ + 2(t-1)(-\bar{\alpha}z_1+\alpha z_2)\langle v,u\rangle - (\bar{\alpha}a+\alpha b)u - (\bar{\alpha}k+\alpha l)v. \end{gather}\] Applying these formulas to compute \((r_\#s,r)\) and \((r^\#,s)\), we prove 15 .

We write down the product on \(S(\alpha,t,E)\) by 9 : \[\begin{gather} \label{product-derived} rs = \frac{1}{2}(r_\#s + T(r)s + T(s)r - S(r,s)c) = akz_1+blz_2+\langle v,u\rangle(z_1 + tz_2) \\ + (\alpha k +\bar{\alpha}l)v + (\alpha a+\bar{\alpha}b)u, \end{gather}\tag{20}\] which, up to rescalling the bilinear form on \(E\), coincides with the initial product on \(S(\alpha,t,E)\).

Remark 1. Denote \(\lambda = \dfrac{3\alpha(1-\alpha)}{(1+\alpha)(\alpha-2)}\). Then the generalized sharped cubic form on the split spin factor \(S(\alpha,E)\) is inner with \(\lambda\), since for \(r = az_1+bz_2+v\) and \(s = kz_1 + lz_2 + u\) we have \[\begin{gather} (r,s) - T(r)T(s)/3 \\ = (az_1 + bz_2 + v,kz_1 + lz_2 + u) - \frac{( (1+\alpha)a + (2-\alpha)b )( (1+\alpha)k + (2-\alpha)l )}{3} \\ = \frac{1}{3} \bigg( ak(1+\alpha)(3-(1+\alpha)) + bl(2-\alpha)(3-(2-\alpha)) - (1+\alpha)(2-\alpha)(al+bk) + \frac{3(1+\alpha)}{\alpha}\langle v,u\rangle \bigg) \\ = \frac{(1+\alpha)(2-\alpha)}{3}\left( (a-b)(k-l) + \frac{3\langle v,u\rangle}{\alpha(2-\alpha)} \right) = \frac{1}{\lambda} \Delta(r,s). \end{gather}\]

Let us return to generalized sharped cubic forms and relations concerned with them.

Lemma 1. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Then the following identities hold: \[\begin{gather} T(r^\#) = S(r) - \Delta(r,r), \tag{21} \\ (r^\#,r) = 3N(r), \tag{22} \\ S(r^\#,r) = T(r)(S(r)-\Delta(r,r)) - 3N(r) - \Delta(r^\#,r), \tag{23} \\ {r^\#}_\#(r_\# q) = (N(r)+\Delta(r^\#,r))q + (N(r,q)+\Delta(r^\#,q)+\Delta(r,r_\#q))r, \tag{24} \end{gather}\]

\[\begin{gather} \label{Adjoint3939} (r_{\#}q)^\#+{r^\#}_{\#}q^\# = (N(q,r)+\Delta(q,r_\#q)+\Delta(r,q^\#))r \\ + (N(r,q)+\Delta(r^\#,q)+\Delta(r,r_\#q))q, \end{gather}\tag{25}\] \[\label{sharpproduct40sharp40r4144r41} {r^\#}_\# r = -T(r)r^\# - T(r^\#)r + (T(r)(S(r)-\Delta(r,r))-N(r) - \Delta(r^\#,r))c.\tag{26}\]

Proof. Involving 14 , 15 , 17 , and 6 , we get \[T(r^\#) = (r^\#,c) = 3N(r,c) - (r_\#c,r) = 3S(r) - T(r)^2 + (r,r) = S(r) - \Delta(r,r).\]

By 15 , we rewrite \[9N(r) = 3N(r,r) = (r_\#r,r) + (r^\#,r) = 3(r^\#,r),\] since \(r_{\#}r = (2r)^{\#} - r^\# - r^\# = 2r^\#\). Thus, we derive 22 .

Applying 14 and already proved formulas, we deduce \[S(r^\#,r) = T(r)T(r^\#) - (r^\#,r) - \Delta(r^\#,r) = T(r)(S(r)-\Delta(r,r)) - 3N(r) - \Delta(r^\#,r).\]

Let us put \(r+tq\) instead of \(r\) into 16 , where \(t\in F\). Joint with 7 , we get \[\begin{gather} ((r+tq)^{\#})^{\#} = (r^{\#}+t^2q^{\#}+tr_{\#}q)^{\#} = (r^{\#}+tr_{\#}q)^{\#} + t^4(q^{\#})^{\#} + t^2(r^{\#}+tr_{\#}q)_{\#}q^{\#} \\ = (r^{\#})^{\#} + t^2(r_\#q)^{\#}+t{r^{\#}}_{\#}(r_{\#}q)+t^4(q^{\#})^{\#} +t^2{r^\#}_{\#}q^\#+t^3(r_{\#}q)_\#q^\#; \end{gather}\]

\[\begin{gather} (N(r+tq)+\Delta(r+tq,(r+tq)^\#))(r+tq) \\ = (N(r) + tN(r,q) + t^2N(q,r) + t^3N(q))(r+tq) + \Delta(r+tq,r^\#+t^2q^\#+tr_\#q)(r+tq). \end{gather}\] Comparing coefficients at \(t\) and at \(t^2\), we derive 24 and 25 respectively.

Finally, we apply 17 twice and then 21 , 24 with \(q=c\): \[\begin{gather} {r^\#}_\# r = {r^\#}_\# (T(r)c - r_\# c) = T(r)(T(r^\#)c-r^\#) - {r^\#}_\# (r_\# c) \\ = T(r)(S(r)-\Delta(r,r))c - T(r)r^\# - (N(r)+\Delta(r^\#,r))c - (N(r,c)+\Delta(r,T(r)c-r))r \\ = (T(r)(S(r)-\Delta(r,r)) - N(r) - \Delta(r^\#,r) )c - T(r)r^\# - T(r^\#)r. \qquad \square \end{gather}\]

Theorem 2. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Define a product \(\cdot\) on \(V\) by 9 . Then the conclusion of Theorem 1a) holds for \((V,\cdot)\).

Proof. Let us check that \(c\) is a unit by 3 , 4 , 6 , 17 : \[2rc = r_{\#}c + T(r)c + T(c)r - S(r,c)c = 2T(r)c - r + 3r - 2T(r)c = 2r.\]

Due to the definitions, \[r^2 = \frac{1}{2}(r_{\#}r + 2T(r)r - S(r,r)c) = r^\# + T(r)r - S(r)c.\]

It remains to show 10 . For this, we apply 23 , 26 : \[\begin{gather} 2rr^\# = r_\# r^\# + T(r)r^\# + T(r^\#)r - S(r,r^\#)c \\ = (T(r)(S(r)-\Delta(r,r)) - N(r) - \Delta(r^\#,r) )c - (T(r)(S(r)-\Delta(r,r)) - 3N(r) - \Delta(r^\#,r))c \\ = 2N(r)c, \end{gather}\] i. e. \(rr^\# = N(r)c\), which is equivalent to 10 . \(\square\)

Let us state further relations fulfilled on an algebra endowed with a generalized sharped cubic.

Lemma 2. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Then the following identities hold: \[\begin{gather} T(r_\#q) = S(r,q) - 2\Delta(r,q), \tag{27} \\ N(r^\#) = N(r)(N(r) + \Delta(r^\#,r)), \tag{28} \\ (r_\#q,s) + (q_\#s,r) + (s_\#r,q) = 3N(r,q,s), \tag{29} \\ {r^\#}_\# r^\# = 2(N(r)+\Delta(r^\#,r)))r, \tag{30}\\ U_c(r) = r, \quad U_r(c) = r^2 + \Delta(r,r)c, \quad \frac{1}{2}U_{r,q}(c) = rq + \Delta(r,q)c, \tag{31} \\ U_r(r) = r^3 - 2\Delta(r,r)r + (\Delta(r^\#,r) + T(r)\Delta(r,r))c, \tag{32} \\ U_r(r^\#) = (N(r)-2\Delta(r^\#,r))r. \tag{33} \end{gather}\]

Proof. By 21 and the definition of the sharp product, we get 27 : \[\begin{gather} T(r_\#q) = T((r+q)^\#) - T(r^\#) - T(q^\#) \\ = S(r+q) - \Delta(r+q,r+q) - S(r) + \Delta(r,r) - S(q) + \Delta(q,q) = S(r,q) - 2\Delta(r,q). \end{gather}\]

We use 22 and the axiom 16 to compute \[3N(r^\#) = ((r^\#)^\#,r^\#) = ((N(r) + \Delta(r^\#,r))r,r^\#) = 3(N(r) + \Delta(r^\#,r))N(r),\] hence, 28 is proved.

Linearization of 15 implies 29 . Since \(q_\# q = 2q^\#\), the equality 30 follows from 16 .

By the definition of the \(U\)-operator, 6 , 17 , and 21 , we have \[\begin{gather} U_c(r) = (c,r)c - {c^\#}_\# r = T(r)c - c_\# r = r, \\ U_r(c) = (c,r)r - {r^\#}_\# c = T(r)r - T(r^\#)c + r^\# = r^2 - T(r^\#)c + S(r)c = r^2 + \Delta(r,r)c, \end{gather}\] and the third equality from 31 follows immediately.

Applying 10 , 14 , 21 , 26 , we derive 32 : \[\begin{gather} U_r(r) - r^3 + 2\Delta(r,r)r - (\Delta(r^\#,r) + T(r)\Delta(r,r))c \\ = (r,r)r - {r^\#}_\# r - T(r)r^2 + S(r)r - N(r)c + 2\Delta(r,r)r - (\Delta(r^\#,r) + T(r)\Delta(r,r))c \\ = (r,r)r + T(r)r^\# + T(r^\#)r - (T(r)(S(r)-\Delta(r,r))-N(r) - \Delta(r^\#,r))c \\ - T(r)r^2 + S(r)r - N(r)c + 2\Delta(r,r)r - (\Delta(r^\#,r) + T(r)\Delta(r,r))c \\ = ((r,r) + S(r) - T(r)^2 + \Delta(r,r))r = 0. \end{gather}\]

Finally, the formula 33 holds by 22 and 30 . \(\square\)

Corollary 1. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Then \((r,s) = T(rs)\) for all \(r,s\in V\).

Proof. By the definition 9 , we write down \[2T(rs) = T(r_\#s) + 2T(r)T(s) - S(r,s)T(c) \mathop{=}\limits^{\eqref{T40sharpproduct40r44q4141}} 2(T(r)T(s) - S(r,s) - \Delta(r,s)) \mathop{=}\limits^{\eqref{GNorm}} 2(r,s).\] Corollary is proved. \(\square\)

Lemma 3. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Then the following identities are equivalent: \[\begin{gather} (r_\#q,s) = N(r,q,s) + \frac{1}{3}(T(r)\Delta(q,s)+T(q)\Delta(r,s) - 2T(s)\Delta(r,q)), \tag{34} \\ (r_\#q,s) = (r,q_\#s) + T(r)\Delta(q,s) - T(s)\Delta(r,q), \tag{35} \\ (rq,s) = (r,qs). \tag{36} \end{gather}\]

Proof. Note that 34 immediately implies 35 by 29 . Suppose that 35 holds. Then by 29 , we have \[\begin{gather} 3N(r,q,s) = (r_\#q,s) + (q_\#s,r) + (s_\#r,q) \\ = 3(r_\#q,s) - T(r)\Delta(q,s) - T(q)\Delta(r,s) + 2T(s)\Delta(r,q), \end{gather}\] and 34 is fulfilled.

Now, we prove the last equivalency: \[\begin{gather} 2(rq,s) - 2(r,qs) = (r_\#q,s) + T(r)(q,s) + T(q)(r,s) - S(r,q)(c,s) \\ - (r,q_\#s) - T(q)(r,s) - T(s)(r,q) + S(q,s)(r,c) \\ = (r_\#q,s) - (r,q_\#s) - T(r)\Delta(s,q) + T(s)\Delta(r,q) \\ + T(r)((q,s)+\Delta(q,s)+S(q,s)) - T(s)((r,q)+\Delta(r,q)+S(r,q)) \\ \mathop{=}\limits^{\eqref{GNorm}} (r_\#q,s) - (r,q_\#s) - T(r)\Delta(s,q) + T(s)\Delta(r,q) + T(r)T(q)T(s)-T(s)T(r)T(q). \end{gather}\] Hence, 35 and 36 are equivalent. \(\square\)

Remark 2. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized cubic form satisfying 16 and 17 . Then \(V\) is a generalized sharped cubic form with invariant form \((\cdot,\cdot)\) if and only if \(V\) satisfies \[(r^\#,q) = N(r,q) + (T(r)\Delta(r,q) - T(q)\Delta(r,r))/3, \label{def:GSharp1-old}\tag{37}\] for all \(r,q\in V\). Indeed, suppose that \(V\) satisfies 37 . Then a linearization of 37 implies 34 . Taking 34 with \(s = r\), we get \[(r_\#q,r) = 2N(r,q) + \frac{1}{3}(T(q)\Delta(r,r)-T(r)\Delta(r,q)),\] which sum with 37 gives 15 . Conversely, the identity 37 holds on every generalized sharped cubic form with invariant form \((\cdot,\cdot)\), since 37 follows from 34 with \(q=r\).

Now we state identities concerned the triple product \(\{r,s,q\}\) defined by 12 .

Lemma 4. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Then the following identities hold: \[\label{triple40r44r44q41} \{r,r,q\} {=} (2r^2{-}\Delta(r,r))q {-} 3\Delta(r,q)r {+} \big(2T(r)\Delta(r,q) {-} (r^\#,q) {+} \Delta(r_\# q,r) {+} N(r,q)\big)c,\tag{38}\]

\[\begin{gather} \label{triple40r44s44q4143triple40s44r44q41} \{r,s,q\} + \{s,r,q\} = (4(rs) - 2\Delta(r,s))q -3\Delta(r,q)s - 3\Delta(s,q)r \\ + \big(2T(r)\Delta(s,q) + 2T(s)\Delta(r,q) - (r_\# s,q) + \Delta(s_\# q, r) + \Delta(r_\# q, s) + N(r,s,q)\big)c, \end{gather}\tag{39}\]

\[\begin{gather} \label{40r44s44q41} (r,s,q) = \frac{1}{4}\big( \{s,r,q\} - \{s,q,r\} + \Delta(q,s)r - \Delta(r,s)q \\ -(\Delta(q_\#s,r) - \Delta(r_\#s,q) + 2T(r)\Delta(s,q) - 2T(q)\Delta(r,s) - (r_\# s,q) + (r,s_\# q))c\big). \end{gather}\tag{40}\]

Proof. Denote \[\eta(r,q) = (2T(r)\Delta(r,q) + T(q)\Delta(r,r) + \Delta(r^\#,q) + \Delta(r_\# q,r) + N(r,q) - T(r)S(r,q))c.\] Then linearization of 26 gives \[\begin{gather} 0 = (r_\# q)_\# r + {r^\#}_\# q + T(q)r^\#+T(r)(r_\# q) + T(r^\#)q+T(r_\# q)r - T(q)S(r)c + \eta(r,q). \end{gather}\] By 9 , 21 , and 27 , we have \[\begin{gather} 0 = (r_\# q)_\# r + r^2_\# q + S(r)(c_\# q) + T(q)(r^\# - S(r)c) + S(r)q - \Delta(r,r)q + S(r,q)r \\ - 2\Delta(r,q)r + \eta(r,q). \end{gather}\] The identity 17 implies \[\begin{gather} 0 = (r_\# q)_\# r + r^2_\# q + S(r)T(q)c + T(q)r^2 -T(r)T(q)r + S(r,q)r - 2\Delta(r,q)r \\ - \Delta(r,r)q + \eta(r,q) = (r_\# q)_\# r + r^2_\# q + S(r)T(q)c + T(q)r^2 - T(r)T(q)r \\ + S(r,q)r - 2\Delta(r,q)r - \Delta(r,r)q + \eta(r,q). \end{gather}\]

The identity \(T(r^2)=(r,r)\) and 9 imply \[\begin{gather} 0 = (r_\# q)_\# r + 2r^2q - T(r^2)q - T(q)r^2 + S(r^2,q)c + S(r)T(q)c \\ + T(q)r^2 - T(r)T(q)r + S(r,q)r - 2\Delta(r,q)r - \Delta(r,r)q + \eta(r,q) \\ = (r_\# q)_\# r + 2r^2q + S(r^\#,q)c - (r,r)q + T(r)S(r,q)c - S(r)S(c,q)c \\ + S(r)T(q)c - T(r)T(q)r + S(r,q)r - 2\Delta(r,q)r - \Delta(r,r)q + \eta(r,q). \end{gather}\] By 12 and 14 , we have \[\begin{gather} 0 = -\{r,r,q\} + (r,q)r + 2r^2q - T(r)T(q)r +S(r,q)r - S(r)T(q)c + S(r^\#,q)c + T(r)S(r,q)c \\ - 2\Delta(r,q)r - \Delta(r,r)q + \eta(r,q) = -\{r,r,q\} + 2r^2q - 3\Delta(r,q)r - \Delta(r,r)q \\ + \big(2T(r)\Delta(r,q) + T(q)\Delta(r,r) + \Delta(r^\#,q) + \Delta(r_\# q,r) + N(r,q) - S(r)T(q) + S(r^\#,q)\big)c. \end{gather}\] By 14 , we have \[\begin{gather} 0 = -\{r,r,q\} + (2r^2-\Delta(r,r))q - 3\Delta(r,q)r \\ + \big(2T(r)\Delta(r,q) + T(q)\Delta(r,r) + T(r^\#)T(q) - (r^\#,q) + \Delta(r_\# q,r) + N(r,q) - S(r)T(q) \big)c. \end{gather}\] The relation 21 implies \[0 = -\{r,r,q\} + (2r^2-\Delta(r,r))q - 3\Delta(r,q)r + \big(2T(r)\Delta(r,q) - (r^\#,q) + \Delta(r_\# q,r) + N(r,q)\big)c.\]

So, we have proved the identity 38 .

The identity 39 is a linearization of 38 , while 40 is a consequence of 39 . \(\square\)

5 \(\Psi\)-map↩︎

Define \(\Psi(r,s,q)\) as follows, \[\begin{gather} \label{PsiDef} \Psi(r,s,q) = (r,s,q) - \Delta(q,s)r + \Delta(r,s)q \\ + 1/4(\Delta(q_\#s,r) - \Delta(r_\#s,q) + 2T(r)\Delta(s,q) - 2T(q)\Delta(r,s) + (q_\# s,r) - (r_\# s, q))c. \!\!\! \end{gather}\tag{41}\] By the definition, \(\Psi(r,s,q) + \Psi(q,s,r) = 0\) for all \(r,s,q\) and \(\Psi(r,s,q) = 0\) if either of \(r,s,q\) equals to \(c\).

The equalities 40 and 41 joint imply \[4\Psi(r,s,q) = (r_\# s)_\# q - (s_\# q)_\# r + ((r,s) + 3\Delta(r,s))q - ((q,s) + 3\Delta(q,s))r.\]

We introduce the following notations: \[\widetilde{(r,q)} = (r,q) + 3\Delta(r,q), \quad (r,s,q)_\# = (r_\# s)_\# q - r_\#(s_\# q).\] Then the last relation obtained has the form \[\label{PsiViaSharpAsso} 4\Psi(r,s,q) = (r,s,q)_\# + \widetilde{(r,s)}q - \widetilde{(s,q)}r.\tag{42}\] Similarly, we derive that \[4\Psi(r,s,q) = U_{q,s}(r) - U_{r,s}(q) + 3(\Delta(r,s)q - \Delta(q,s)r).\]

We may rewrite 27 as follows, \[T(r_\# q) = T(r)T(q) - \widetilde{(r,q)}. \label{T40sharp41TildeInner}\tag{43}\]

Lemma 5. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Then the following identities are equivalent: \[\begin{gather} \widetilde{(r_\# s,q)} = \widetilde{(r,s_\# q)}, \tag{44} \\ \{r,r^\#,q\} = (2N(r)-\Delta(r,r^\#))q - 3\Delta(r^\#,q)r, \tag{45} \\ T(\Psi(r,s,q)) = 0. \tag{46} \end{gather}\]

Proof. We deduce 45 : \[\begin{gather} \{r,r^\#,q\} \mathop{=}^{\eqref{tripleproduct}} (r,r^\#)q + (r^\#,q)r - {r^\#}_\#(r_\# q) \\ \mathop{=}^{\eqref{Adjoint39}} (r,r^\#)q + (r^\#,q)r - (N(r)+\Delta(r^\#,r))q - (N(r,q)+\Delta(r^\#,q)+\Delta(r,r_\#q))r \\ \mathop{=}^{\eqref{def:GSharp1},\,\eqref{40sharp40r4144r41}} (2N(r)-\Delta(r,r^\#))q - (-2/3(r^\#,q)+1/3(r_\#q,r) +\Delta(r^\#,q)+\Delta(r,r_\#q) )r, \end{gather}\] which is equal to the right-hand side of 45 if and only 44 holds for \(s = r\). To show that 44 and 45 are equivalent, it remains to derive 44 from itself fulfilled for \(s = r\). A linearization of \[\label{InvFormUnderTildeInnerPartial} (r_\#q,r) + 3\Delta(r_\#q,r) = (r_\#r,q) + 3\Delta(r_\#r,q)\tag{47}\] gives \[(r_\#q,s) + (s_\#q,r) + 3\Delta(r_\#q,s) + 3\Delta(s_\#q,r) = 2(r_\#s,q) + 6\Delta(r_\#s,q).\] We may rewrite the last expression with the help of 29 : \[N(r,s,q) + \Delta(r_\#q,s) + \Delta(s_\#q,r) + \Delta(r_\#s,q) = (r_\#s,q) + 3\Delta(r_\#s,q).\] Because of the symmetry, 44 follows.

Due to 41 and to Corollary 1, we have \[\begin{gather} 4T(\Psi(r,s,q)) = 4(rs,q) - 4(r,sq) - 4T(r)\Delta(q,s) + 4T(q)\Delta(r,s) \\ + 3(\Delta(q_\#s,r) - \Delta(r_\#s,q) + 2T(r)\Delta(q,s) - 2T(q)\Delta(r,s) + (r,s_\#q) - (r_\# s, q)) \\ \mathop{=}\limits^{\eqref{product}} 2(r_\#s,q) + 2T(r)(s,q) - 2S(r,s)T(q) - 2(r,s_\# q) - 2T(q)(r,s) + 2S(q,s)T(r) \\ + 2T(r)\Delta(q,s) - 2T(q)\Delta(r,s) + 3(\Delta(q_\#s,r) - \Delta(r_\#s,q) + (r,s_\#q) - (r_\# s, q)) \\ \mathop{=}\limits^{\eqref{GNorm}} (r,s_\# q) - (r_\#s,q) + 2T(r)T(s)T(q) - 2T(s)T(r)T(q) + 3(\Delta(q_\#s,r) - \Delta(r_\#s,q)) \\ = \widetilde{(r,s_\# q)} - \widetilde{(r_\# s,q).} \end{gather}\] Hence, 44 and 46 are equivalent. \(\square\)

Remark 3. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\) such that the form \((\cdot,\cdot)\) is invariant, i. e. 36 holds. Then the form \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant if and only if the equality \[\Delta(r_\#q,s) - \Delta(r,q_\#s) = \frac{T(s)\Delta(r,q) - T(r)\Delta(q,s)}{3} \label{InvFormUnderDelta}\tag{48}\] if fulfilled for all \(r,s,q\in V\). When \(N\) is inner, then invariancy of \((\cdot,\cdot)\) implies 48 and so implies \(\#\)-invariancy of \(\widetilde{(\cdot,\cdot)}\).

Lemma 6. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\) such that the form \((\cdot,\cdot)\) is invariant. Then the following identities are fulfilled: \[\begin{gather} (U_r(q),s) = (q,U_r(s)) + T(s)\Delta(r^\#,q)-T(q)\Delta(r^\#,s), \label{UopInvariancy} \end{gather}\tag{49}\]

\[\begin{gather} \label{Ur40Usharp40r4141} U_r(U_{r^\#}(q)) = [-3(r^\#,q)\Delta(r,r^\#) + (N(r)+\Delta(r,r^\#))(\Delta(r^\#,q)+\Delta(r,r_\# q) \\ +2/3(T(r)\Delta(r,q)-T(q)\Delta(r,r)))]r + (N(r)+\Delta(r,r^\#))^2q, \end{gather}\tag{50}\]

Proof. The formula 49 holds by 35 .

We rewrite with the help of 16 and 33 : \[\begin{gather} U_r(U_{r^\#}(q)) = U_r( (r^\#,q)r^\# - {(r^\#)^\#}_\# q ) \\ = (r^\#,q)(N(r)-2\Delta(r^\#,r))r - (N(r) + \Delta(r^\#,r))U_r(r_\# q). \end{gather}\] Further, we apply 24 and 35 , \[\begin{gather} U_r(r_\# q) = (r,r_\# q)r - {r^\#}_\#(r_\# q) = ((q,r_\# r) + T(q)\Delta(r,r) - T(r)\Delta(r,q))r \\ - (N(r)+\Delta(r^\#,r))q - (N(r,q)+\Delta(r^\#,q)+\Delta(r,r_\#q))r. \end{gather}\] Thus, \[U_r(U_{r^\#}(q)) = Ar + (N(r) + \Delta(r^\#,r))^2 q,\] where again by 35 we reduce \[\begin{gather} A = (r^\#,q)(N(r)-2\Delta(r^\#,r)) - (N(r) + \Delta(r^\#,r))( 2(r^\#,q) + T(q)\Delta(r,r) - T(r)\Delta(r,q) \allowdisplaybreaks \\ - N(r,q) - \Delta(r^\#,q) - \Delta(r,r_\#q) ) = -3(r^\#,q)\Delta(r^\#,r) \\ + (N(r) + \Delta(r^\#,r))(N(r,q) - (r^\#,q) + \Delta(r^\#,q)+\Delta(r,r_\# q) + T(r)\Delta(r,q) - T(q)\Delta(r,r) ). \end{gather}\] It remains to use 37 to prove 50 . \(\square\)

Let us prove some further properties of \(\Psi\).

Lemma 7. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\) such that \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant. Then the following identities for \(\Psi\) are fulfilled: \[\begin{gather} \Psi(r,s,q) + \Psi(s,q,r) + \Psi(q,r,s) = 0, \tag{51} \\ \widetilde{(\Psi(r,s,q),x)} + \widetilde{(\Psi(q,s,x),r)} + \widetilde{(\Psi(x,s,r),q)} = 0. \tag{52} \end{gather}\] If, additionally, \((\cdot,\cdot)\) is invariant, then \[\Delta(\Psi(r,s,q),x) + \Delta(\Psi(q,s,x),r) + \Delta(\Psi(x,s,r),q) = 0. \label{Delta-Psi}\tag{53}\]

Proof. The equality 51 follows by 42 .

Based on 42 , we rewrite and get \[\begin{gather} 4(\widetilde{(\Psi(r,s,q),x)} + \widetilde{(\Psi(q,s,x),r)} + \widetilde{(\Psi(x,s,r),q)}) \\ = \widetilde{((r,s,q)_\#,x)} + \widetilde{((q,s,x)_\#,r)} + \widetilde{((x,s,r)_\#,q)} \\ + \widetilde{(r,s)}\widetilde{(q,x)} - \widetilde{(s,q)}\widetilde{(r,x)} + \widetilde{(q,s)}\widetilde{(x,r)} - \widetilde{(s,x)}\widetilde{(q,r)} + \widetilde{(x,s)}\widetilde{(r,q)} - \widetilde{(s,r)}\widetilde{(x,q)} \\ \mathop{=}\limits^{\eqref{InvFormUnderTildeInner}} \widetilde{(r_\# s,x_\# q)} - \widetilde{(s_\# q,x_\# r)} + \widetilde{(q_\# s,r_\# x)} - \widetilde{(s_\# x,r_\# q)} + \widetilde{(x_\# s,q_\# r)} - \widetilde{(s_\# r,q_\# x)} = 0, \end{gather}\] as required.

Let us prove 53 , for this, we write down \[\begin{gather} 4\Delta(\Psi(r,s,q),x) \mathop{=}\limits^{\eqref{PsiViaSharpAsso}} \Delta((r_\#s)_\#q - r_\#(s_\# q) + \widetilde{(r,s)}q - \widetilde{(q,s)}r,x) \\ \mathop{=}\limits^{\eqref{InvFormUnderDelta}} \Delta(r_\# s,q_\#x) - \Delta(s_\# q,r_\#x) + \widetilde{(r,s)}\Delta(q,x) - \widetilde{(q,s)}\Delta(r,x) \allowdisplaybreaks \\ + \frac{T(x)\Delta(r_\# s,q)-T(r_\#s)\Delta(q,x)-T(x)\Delta(r,s_\# q)+T(s_\#q)\Delta(r,x)}{3} \\ \mathop{=}\limits^{\eqref{InvFormUnderDelta}} \Delta(r_\# s,q_\#x) - \Delta(s_\#q,r_\#x) + \widetilde{(r,s)}\Delta(q,x) - \widetilde{(q,s)}\Delta(r,x) \\ + \frac{T(x)(T(q)\Delta(r,s)-T(r)\Delta(s,q))}{9} + \frac{T(s_\#q)\Delta(r,x)-T(r_\#s)\Delta(q,x)}{3}. \end{gather}\] Analogously, we have \[\begin{gather} 4\Delta(\Psi(q,s,x),r) = \Delta(q_\#s,r_\#x) - \Delta(s_\#x,r_\#q) + \widetilde{(s,q)}\Delta(r,x) - \widetilde{(s,x)}\Delta(r,q) \\ + \frac{T(r)(T(x)\Delta(s,q)-T(q)\Delta(s,x))}{9} + \frac{T(s_\#x)\Delta(r,q)-T(q_\#s)\Delta(r,x)}{3}, \end{gather}\]

\[\begin{gather} 4\Delta(\Psi(x,s,r),q) = \Delta(x_\#s,r_\#q) - \Delta(s_\#r,q_\#x) + \widetilde{(x,s)}\Delta(r,q) - \widetilde{(r,s)}\Delta(q,x) \\ + \frac{T(q)(T(r)\Delta(s,x)-T(x)\Delta(r,s))}{9} + \frac{T(s_\#r)\Delta(q,x)-T(x_\#s)\Delta(r,q)}{3}. \end{gather}\] The sum of the three expressions equals 0. \(\square\)

Lemma 8. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be a generalized sharped cubic form on \(V\). Suppose that \((\cdot,\cdot)\) is invariant and nondegenerate, \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant, and \(\dim V\geq2\). Then \((N,\Delta,\#,c)\) is inner if and only if \[\label{Delta-Psi-Sharp} \Delta(s,\Psi(r,s,q)_\# x + \Psi(q,s,x)_\#r + \Psi(x,s,r)_\#q) = 0\tag{54}\] holds for all \(r,s,q,x\in V\).

Proof. Let us rewrite 54 in more convenient form. With the help of 42 and 48 , we get \[\begin{gather} \label{Delta-Psi-Sharp-help} 4\Delta(s,\Psi(r,s,q)_\# x + \Psi(q,s,x)_\#r + \Psi(x,s,r)_\#q) = \Delta(s,(r,s,q)_\# x + \widetilde{(r,s)}q_\# x - \widetilde{(q,s)}r_\# x \\ + (q,s,x)_\#r + \widetilde{(q,s)}x_\# r - \widetilde{(x,s)}q_\# r + (x,s,r)_\#q +\widetilde{(x,s)}r_\# q - \widetilde{(r,s)}q_\#x ) \\ \allowdisplaybreaks = \Delta((r,s,q)_\#, x_\#s) + \frac{T(s)}{3}\Delta((r,s,q)_\#,x) - \frac{T((r,s,q)_\#)}{3}\Delta(x,s) \\ + \Delta((q,s,x)_\#, r_\#s) + \frac{T(s)}{3}\Delta((q,s,x)_\#,r) - \frac{T((q,s,x)_\#)}{3}\Delta(r,s) \\ + \Delta((x,s,r)_\#, q_\#s) + \frac{T(s)}{3}\Delta((x,s,r)_\#,q) - \frac{T((x,s,r)_\#)}{3}\Delta(q,s). \end{gather}\tag{55}\] The sum of the three summands at \(T(s)/3\) is zero due to 53 . Further, \[\begin{gather} \Delta((r_\#s)_\#q,x_\#s) - \Delta((q_\#(s_\#x),r_\#s) \mathop{=}\limits^{\eqref{InvFormUnderDelta}} \frac{1}{3}( T(x_\#s)\Delta(r_\#s,q) - T(r_\#s)\Delta(x_\#s,q) ) \\ \mathop{=}\limits^{\eqref{T40sharp41TildeInner}} \frac{1}{3}( (T(x)T(s) - \widetilde{(x,s)})\Delta(r_\#s,q) - (T(r)T(s) - \widetilde{(r,s)})\Delta(x_\#s,q) ). \end{gather}\] Hence, \[\begin{gather} \Delta((r,s,q)_\#, x_\#s) + \Delta((q,s,x)_\#, r_\#s) + \Delta((x,s,r)_\#, q_\#s) \\ = \frac{1}{3}( (T(x)T(s) - \widetilde{(x,s)})(\Delta(r_\#s,q)-\Delta(r,s_\#q)) - (T(r)T(s) - \widetilde{(r,s)})(\Delta(x_\#s,q)-\Delta(x,s_\#q) \\ + (T(q)T(s) - \widetilde{(q,s)})(\Delta(x_\#s,r)-\Delta(x,s_\#r)) ) \allowdisplaybreaks \\ \mathop{=}\limits^{\eqref{InvFormUnderDelta}} \frac{1}{9}( (T(x)T(s) - \widetilde{(x,s)})(T(q)\Delta(r,s)-T(r)\Delta(q,s)) \\ - (T(r)T(s) - \widetilde{(r,s)})(T(q)\Delta(x,s)-T(x)\Delta(s,q)) \\ + (T(q)T(s) - \widetilde{(q,s)})(T(r)\Delta(x,s)-T(x)\Delta(r,s)) ), \end{gather}\] where the last expression equals the following one \[\begin{gather} \label{Delta-Psi-Sharp-Equi} T(q)\Delta(s,x)(r,s)-T(r)\Delta(s,x)(s,q)+T(x)\Delta(r,s)(s,q)-T(q)\Delta(r,s)(s,x)\\ + T(r)\Delta(s,q)(s,x)-T(x)\Delta(s,q)(r,s) = 0 \end{gather}\tag{56}\] with coefficient \(1/9\).

The rest summands of 55 give by 42 and 46 : \[\begin{gather} \frac{1}{3}( (T(q)\widetilde{(r,s)} - T(r)\widetilde{(q,s)})\Delta(x,s) + (T(x)\widetilde{(q,s)} - T(q)\widetilde{(x,s)})\Delta(r,s) \\ + (T(r)\widetilde{(x,s)} - T(x)\widetilde{(r,s)})\Delta(q,s) ), \end{gather}\] which is equal to 56 with coefficient \(1/3\).

Therefore, we have showed that 54 is equivalent to 56 . It is easy to check that if \((N,\Delta,\#,c)\) is inner, then 56 holds.

Now, we want to show that if 56 is true, then the cubic form is inner. Putting \(q = c\) in 56 , we derive \[\label{CubicFormInnerEqui} \Delta(s,x)\bigg((r,s)-\frac{T(r)T(s)}{3}\bigg) = \Delta(s,r)\bigg((x,s)-\frac{T(x)T(s)}{3}\bigg).\tag{57}\] Analogously, we write down \[\label{CubicFormInnerEqui39} \Delta(s,r)\bigg((r,t)-\frac{T(r)T(t)}{3}\bigg) = \Delta(r,t)\bigg((r,s)-\frac{T(r)T(s)}{3}\bigg).\tag{58}\]

Multiplying 57 by \(\Delta(r,t)\) and adding 58 multiplied by \(\Delta(s,x)\), we get \[\Delta(r,s)\bigg( \Delta(s,x)\bigg((r,t)-\frac{T(r)T(t)}{3}\bigg) - \Delta(r,t)\bigg((x,s)-\frac{T(x)T(s)}{3}\bigg) \bigg) = 0.\]

If \(\Delta\equiv0\), then \((N,\Delta,\#,c)\) is inner.

Otherwise, take \(r,s\) such that \(\Delta(r,s)\neq0\). We may assume that \(T(r) = 0\), since \(\Delta(c,s) = 0\). Now, we find \(t\in V\) with the property \((r,t)\neq0\). Denote \(\lambda = \Delta(r,t)/(r,t)\). Hence, \(\Delta(s,x) = \lambda \big((x,s)-\frac{T(x)T(s)}{3}\big)\) for all \(x\) and all \(s\) satisfying \(\Delta(r,s)\neq0\) with fixed \(r\). In particular, \(\Delta(s,r) = \lambda \big((r,s)-\frac{T(r)T(s)}{3}\big)\). Consider \(s\) such that \(\Delta(r,s) = 0\). Then by 58 , \(\Delta(r,p)(r,s) = 0\) for all \(p\). Hence, \((r,s) = 0\) and again \[\label{criterion-inner} \Delta(r,s) = \lambda ((r,s)-T(r)T(s)/3)\tag{59}\] holds.

If for every \(r\neq0\) such that \(T(r) = 0\), one may find a corresponding \(s\) with the property \(\Delta(r,s)\neq0\), then we have 59 . Hence, every \(a \in V\) may be written as \(\mu c + r\), where \(\mu\in F\) and \(T(r) = 0\). Then \(\Delta(a,b) = \lambda \big((a,b)-\frac{T(a)T(b)}{3}\big)\), where \(\lambda\neq0\) depends on \(a\) and some \(t\). From 57 , we conclude that \(\lambda\) is a constant.

If we may find \(r\neq0\) such that \(T(r) = 0\) and \(\Delta(r,s) = 0\) for all \(s\in V\), then by 57 , \(\Delta(s,x) = 0\) for all \(x\) and \(s\not\perp r\) with respect to the form \((\cdot,\cdot)\). Let us take any \(a\) orthogonal to \(r\) and fixed \(s\not\perp r\). Then \(\Delta(a,x) = \Delta(a+s,x) - \Delta(s,x) = 0\), so \(\Delta \equiv 0\), a contradiction. Thus, \((N,\Delta,\#,c)\) is inner. \(\square\)

Recall that the generalized sharped cubic form on the split spin factor \(S(\alpha,E)\) is inner with \(\lambda = \dfrac{3\alpha(1-\alpha)}{(1+\alpha)(\alpha-2)}\). Therefore, the relations 53 and 54 are fulfilled on \(S(\alpha,E)\). In §6, we will prove that these identities hold on \(S(\alpha,t,E)\).

Corollary 2. Given a vector space \(V\), let \((N,\Delta,\#,c)\) be an inner generalized sharped cubic form on \(V\). Suppose that \((\cdot,\cdot)\) is invariant and nondegenerate, \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant, and \(\dim V\geq2\). Then

a) \(\Delta(s,x)(r,s) = \Delta(r,s)(s,x)\), when either \(T(s) = 0\) or \(T(r) = T(x) = 0\),

b) \(\Delta(s,\Psi(r,s,q)) = 0\) for all \(r,s,q\in V\).

Proof. a) The relation 57 is equivalent by 6 to \(\Delta(s,x)(r,s) = \Delta(r,s)(s,x)\), when either \(T(s) = 0\) or \(T(r) = T(x) = 0\).

b) We consider 54 with \(x = c\). Since \(\Psi(q,s,c) = \Psi(c,s,r) = 0\) and by 46 , we get \(\Delta(s,\Psi(r,s,q)) = 0\). \(\square\)

6 Sum of the three associators identity↩︎

Now, we are ready to prove that \(S(\alpha,t,E)\) satisfies the identity \[\label{Wb-identity} W_b(a,c,d) := ((a,b,c),d,b) + ((c,b,d),a,b) + ((d,b,a),c,b) = 0.\tag{60}\]

First, we show that \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant on \(S(\alpha,t,E)\). By the proof of Lemma 5, it is enough to check 47 . We compute for \(r = az_1 + bz_2 + v\) and \(q = gz_1 + hz_2 + w\): \[\begin{gather} (r_\#r,q) + 3\Delta(r_\#r,q) = 2(1+\alpha)g( (\alpha a +\bar{\alpha}b )b + (\alpha-1)(t-1)\langle v,v\rangle) \\ + 2(2-\alpha)h( (\alpha a +\bar{\alpha}b )a + \alpha(t-1)\langle v,v\rangle) - 2(1+\alpha+(2-\alpha)t)(\bar{\alpha}a+\alpha b)\langle v,w\rangle \\ + 6\alpha(\alpha-1)(g-h)( (\alpha a + \bar{\alpha}b)(b-a)-(t-1)\langle v,v\rangle ) + 6(\bar{\alpha}+\alpha t)(\bar{\alpha}a+\alpha b)\langle v,w\rangle; \end{gather}\]

\[\begin{gather} (r_\#q,r) + 3\Delta(r_\#q,r) = (1+\alpha)a(h(\alpha a+\bar{\alpha}b) + b(\alpha g+\bar{\alpha}h) + 2(\alpha-1)(t-1)\langle v,w\rangle) \\ + (2-\alpha)b(g(\alpha a+\bar{\alpha}b) + a(\alpha g+\bar{\alpha}h) + 2\alpha(t-1)\langle v,w\rangle) \\ - (1+\alpha +(2-\alpha)t)( (\bar{\alpha}a+\alpha b)\langle v,w\rangle + (\bar{\alpha}g+\alpha h)\langle v,v\rangle ) \allowdisplaybreaks \\ + 3\alpha(\alpha-1)(a-b)( (\alpha a + \bar{\alpha}b)(h-g) + (\alpha g + \bar{\alpha}h)(b-a) - 2(t-1)\langle v,w\rangle) \\ + 3(\bar{\alpha}+\alpha t)( (\bar{\alpha}a+\alpha b)\langle v,w\rangle + (\bar{\alpha}g+\alpha h)\langle v,v\rangle ). \end{gather}\] In both \(\widetilde{(r_\#r,q)}\) and \(\widetilde{(r_\#q,r)}\), the coefficients at \(\langle v,v\rangle\) equal \(2(2\alpha-1)(t-1)(\bar{\alpha}g+\alpha h)\), at \(\langle v,w\rangle\) equal \(4(2\alpha-1)(t-1)(\bar{\alpha}a+\alpha b)\). The rest summands equal to the same expression \[2(\alpha a+\bar{\alpha}b)( (1+\alpha)bg + (2-\alpha)ah + 3\alpha(\alpha-1)(a-b)(h-g)).\]

Denote the coefficient at \(c\) in 41 as \(\Phi(r,s,q)\). If \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant, then \[\label{PhiSimple} \Phi(r,s,q) = 1/2(T(r)\Delta(s,q) - T(q)\Delta(r,s) + \Delta(r_\#s,q) - \Delta(q_\#s,r) ).\tag{61}\] Below, we apply that \(\widetilde{(\cdot,\cdot)}\) is \(\#\)-invariant on \(S(\alpha,t,E)\).

By 41 , \[\begin{gather} ((r,s,q),x,s) + ((q,s,x),r,s) + ((x,s,r),q,s) \\ = \Delta(s,q)(r,x,s) - \Delta(r,s)(q,x,s) + (\Psi(r,s,q),x,s) + \Delta(s,x)(q,r,s) - \Delta(q,s)(x,r,s) \allowdisplaybreaks \\ + (\Psi(q,s,x),r,s) + \Delta(r,s)(x,q,s) - \Delta(s,x)(r,q,s) + (\Psi(x,s,r),q,s) \allowdisplaybreaks \\ = \Delta(s,q)((r,s,x)+\Psi(x,s,r)) + \Delta(r,s)((x,s,q)+\Psi(q,s,x)) + \Delta(s,x)((q,s,r)+\Psi(r,s,q)) \\ - ( \Delta(\Psi(r,s,q),x) + \Delta(\Psi(q,s,x),r) + \Delta(\Psi(x,s,r),q) )s \\ - ( \Phi(\Psi(r,s,q),x,s) + \Phi(\Psi(q,s,x),r,s) + \Phi(\Psi(x,s,r),q,s) )c + \Psi_0. \end{gather}\] where \[\label{Psi0} \Psi_0 = \Psi(\Psi(r,s,q),x,s) + \Psi(\Psi(q,s,x),r,s) + \Psi(\Psi(x,s,r),q,s).\tag{62}\]

With the help of 41 and Lemma 5, we rewrite the last expression as follows, \[\begin{gather} ((r,s,q),x,s) + ((q,s,x),r,s) + ((x,s,r),q,s) \\ = \Delta(s,q)\left(-\Delta(r,s)x+\Delta(x,s)r + \frac{T(x)\Delta(r,s) - T(r)\Delta(x,s) + \Delta(x_\#s,r) - \Delta(r_\#s,x)}{2}c\right) \\ + \Delta(r,s)\left(-\Delta(x,s)q+\Delta(q,s)x + \frac{T(q)\Delta(s,x) - T(x)\Delta(s,q) + \Delta(q_\#s,x) - \Delta(x_\#s,q)}{2}c\right) \\ + \Delta(s,x)\left(-\Delta(s,q)r+\Delta(r,s)q + \frac{T(r)\Delta(s,q) - T(q)\Delta(r,s) + \Delta(r_\#s,q) - \Delta(q_\#s,r)}{2}c\right) \\ - ( \Delta(\Psi(r,s,q),x) + \Delta(\Psi(q,s,x),r) + \Delta(\Psi(x,s,r),q) )s \\ - ( \Phi(\Psi(r,s,q),x,s) + \Phi(\Psi(q,s,x),r,s) + \Phi(\Psi(x,s,r),q,s) )c + \Psi_0 \\ = \frac{1}{2}( \Delta(s,q)(\Delta(x_\#s,r) - \Delta(r_\#s,x)) + \Delta(r,s)( \Delta(q_\#s,x) - \Delta(x_\#s,q) ) \\ + \Delta(s,x)( \Delta(r_\#s,q) - \Delta(q_\#s,r) ) - \frac{1}{2}\Delta(s,\Psi(r,s,q)_\# x + \Psi(q,s,x)_\#r + \Psi(x,s,r)_\#q) \\ \allowdisplaybreaks + \frac{1}{2}( \Delta(\Psi(r,s,q),x_\#s) + \Delta(\Psi(q,s,x),r_\#s) + \Delta(\Psi(x,s,r),q_\#s) ) \\ + ( \Delta(\Psi(r,s,q),x) + \Delta(\Psi(q,s,x),r) + \Delta(\Psi(x,s,r),q) )((1/2)T(s)c-s) + \Psi_0. \end{gather}\] Further, we will show that \(\Psi_0 = 0\), the identities 53 and 54 are fulfilled on \(S(\alpha,t,E)\) as well as \[\label{Wb:additional} \Delta(\Psi(r,s,q),x_\#s) + \Delta(\Psi(q,s,x),r_\#s) + \Delta(\Psi(x,s,r),q_\#s) = 0,\tag{63}\]

\[\begin{gather} \label{Long-Delta-OnS} \Delta(s,q)(\Delta(x_\#s,r) - \Delta(r_\#s,x)) + \Delta(r,s)( \Delta(q_\#s,x) - \Delta(x_\#s,q) ) \\ + \Delta(s,x)( \Delta(r_\#s,q) - \Delta(q_\#s,r) ) = 0. \end{gather}\tag{64}\]

Now, let us explain that the identity 60 does not hold in general even for inner cubic forms. Consider the trivial case \(\Delta \equiv 0\), which may be interpreted as an inner case with \(\lambda = 0\). Then the product coming from a sharped cubic form \((N,\#,c)\) is known to be Jordan [6]. To check if 60 is fulfilled for the Jordan algebra, it is enough to study the case of a special Jordan algebra, since the identity has the degree five.

Let \(J\) be a special Jordan algebra, i. e. \(J\) is a subalgebra of \(A^{(+)}\), where \(A\) is an associative algebra and the product \(\circ\) in \(A^{(+)}\) is defined as follows, \(a\circ b = ab + ba\). Then \((a,b,c)_\circ = bac - bca + cab - acb\). Further, \[\begin{gather} ((a,b,c)_\circ,d,b)_\circ = b(a,b,c)_\circ d + d(a,b,c)_\circ b - db(a,b,c)_\circ - (a,b,c)_\circ bd \\ \allowdisplaybreaks = b^2(acd-cad) + b(ca - ac)bd + db(ac - ca)b + d(ca-ac)b^2 \\ - db^2(ac-ca) + db(ca-ac)b - b(ac-ca)bd - (ca-ac)b^2 d \\ = b^2(acd-cad) + (dca-dac)b^2 - db^2(ac-ca) - (ca-ac)b^2 d. \end{gather}\] Thus, \[((a,b,c)_\circ,d,b)_\circ {+} ((c,b,d)_\circ,a,b)_\circ {+} ((c,b,d)_\circ,a,b)_\circ {=} b^2(acd-cad + cda-dca + dac-adc ) {+} {\ldots},\] where the first two letters of all rest summands differ from \(b^2\). Hence, this expression is nonzero in the case of any associative algebra \(A\), which does not satisfy any identity of degree less than 6. For example, the matrix algebra \(M_3(F)\) is such an algebra [9]. The space \(M_3(F)\) is equipped with the identity matrix as a basepoint, the determinant as a norm, and a sharp map sends a matrix to its adjoint. Then the associated algebra is isomorphic to \(M_3(F)^{(+)}\). Slightly different sharped cubic form on \(H_3(C)\), the Hermitian matrices over a Cayley—Dickson algebra \(C\), defines the simple Jordan algebra of Albert type [6].

To derive the identity 60 , we need the following result.

Lemma 9. In \(S(\alpha,t,E)\), we have \[\label{PsiFormula} \Psi(r,s,q) = (2\alpha-1)(t-1)(\langle u,w\rangle v - \langle u,v\rangle w),\tag{65}\] where \(r = r_0 + v\), \(s = s_0 + u\), \(q = q_0 + w\) f or \(r_0,s_0,q_0\in Fz_1+Fz_2\) and \(v,u,w\in E\).

Proof. Let us express the associator of the elements \(r = az_1 + bz_2 + v\), \(s = kz_1 + lz_2 + u\) and \(q = gz_1 + hz_2 + w\). We compute \((rs)q\) applying 20 : \[\begin{gather} (rs)q = (ak + \langle v,u\rangle)gz_1 + (bl +t\langle v,u\rangle)hz_2 + ((\alpha k + \bar{\alpha}l)\langle v,w\rangle \\ + (\alpha a + \bar{\alpha}b)\langle u,w\rangle)(z_1+tz_2) + (\alpha g + \bar{\alpha}h)(\alpha k + \bar{\alpha}l)v \\ + (\alpha g + \bar{\alpha}h)(\alpha a + \bar{\alpha}b)u + (\alpha(ak + \langle v,u\rangle) + \bar{\alpha}(bl +t\langle v,u\rangle))w. \end{gather}\] Analogously, we have \[\begin{gather} (qs)r = (gk + \langle w,u\rangle)az_1 + (hl +t\langle w,u\rangle)bz_2 + ((\alpha k + \bar{\alpha}l)\langle v,w\rangle \\ + (\alpha g + \bar{\alpha}h)\langle u,v\rangle)(z_1+tz_2) + (\alpha a + \bar{\alpha}b)(\alpha k+\bar{\alpha}l)w \\ + (\alpha g + \bar{\alpha}h)(\alpha a+\bar{\alpha}b)u + (\alpha(kg + \langle w,u\rangle) + \bar{\alpha}(lh +t\langle w,u\rangle))v. \end{gather}\] Therefore, \[\begin{gather} (r,s,q) = (g\langle v,u\rangle - a\langle w,u\rangle)z_1 + t(h\langle v,u\rangle - b\langle w,u\rangle)z_2 \\ + ( (\alpha a + \bar{\alpha}b)\langle u,w\rangle - (\alpha g + \bar{\alpha}h)\langle u,v\rangle))(z_1+tz_2) \\ - ( \alpha(\alpha-1)(a-b)(k-l) - (\alpha+\bar{\alpha}t)\langle v,u\rangle )w + ( \alpha(\alpha-1)(k-l)(g-h) - (\alpha+\bar{\alpha}t)\langle w,u\rangle )v. \end{gather}\]

It remains to substitute all known summands in 61 : \[\begin{gather} \Psi(r,s,q) = (r,s,q) - \Delta(q,s)r + \Delta(r,s)q \\ + 1/2(T(r)\Delta(s,q) - T(q)\Delta(r,s) + \Delta(r_\#s,q) - \Delta(q_\#s,r) )c \\ = (g\langle v,u\rangle - a\langle w,u\rangle)z_1 + t(h\langle v,u\rangle - b\langle w,u\rangle)z_2 + ( (\alpha a + \bar{\alpha}b)\langle u,w\rangle - (\alpha g + \bar{\alpha}h)\langle u,v\rangle)(z_1+tz_2) \\ - ( \alpha(\alpha-1)(a-b)(k-l) - (\alpha+\bar{\alpha}t)\langle v,u\rangle )w + ( \alpha(\alpha-1)(k-l)(g-h) - (\alpha+\bar{\alpha}t)\langle w,u\rangle )v \allowdisplaybreaks \\ + (\alpha(\alpha-1)(k-l)(g-h)-(\bar{\alpha}+\alpha t)\langle u,w\rangle)(((1+\alpha)a + (2-\alpha)b)(z_1+z_2)/2-r) \\ - (\alpha(\alpha-1)(a-b)(k-l)-(\bar{\alpha}+\alpha t)\langle v,u\rangle) (((1+\alpha)g + (2-\alpha)h)(z_1+z_2)/2-q) \\ + \big[ \alpha(\alpha-1)(g-h)\big( (\alpha a+\bar{\alpha}b)(l-k) + (\alpha k+\bar{\alpha}l)(b-a) - 2(t-1)\langle v,u\rangle \big) \\ - \alpha(\alpha-1)(a-b)\big( (\alpha g+\bar{\alpha}h)(l-k) + (\alpha k+\bar{\alpha}l)(h-g) - 2(t-1)\langle u,w\rangle \big) \\ + (\bar{\alpha} + \alpha t)( (\bar{\alpha}a + \alpha b)\langle u,w\rangle - (\bar{\alpha}g + \alpha h)\langle v,u\rangle) \big](z_1+z_2)/2. \end{gather}\] At \(z_1/2\) we have the coefficient \[\begin{gather} 2(g\langle v,u\rangle - a\langle w,u\rangle + (\alpha a + \bar{\alpha}b)\langle u,w\rangle - (\alpha g + \bar{\alpha}h)\langle v,u\rangle) \\ + (\alpha(\alpha-1)(k-l)(g-h) -(\bar{\alpha}+\alpha t)\langle u,w\rangle)((-1+\alpha)a + (2-\alpha)b) \\ \allowdisplaybreaks - (\alpha(\alpha-1)(a-b)(k-l)-(\bar{\alpha}+\alpha t)\langle v,u\rangle) ((-1+\alpha)g + (2-\alpha)h) \\ + \alpha(\alpha-1)(g-h)\big( (\alpha a+\bar{\alpha}b)(l-k) + (\alpha k+\bar{\alpha}l)(b-a) - 2(t-1)\langle v,u\rangle \big) \\ - \alpha(\alpha-1)(a-b)\big( (\alpha g+\bar{\alpha}h)(l-k) + (\alpha k+\bar{\alpha}l)(h-g) - 2(t-1)\langle u,w\rangle \big) \\ + (\bar{\alpha} + \alpha t)( (\bar{\alpha}a + \alpha b)\langle u,w\rangle - (\bar{\alpha}g + \alpha h)\langle v,u\rangle). \end{gather}\] At \(\langle v,u\rangle\), we have \[\begin{gather} g(2-2\alpha+(-1+\alpha)(\bar{\alpha}+\alpha t) -2\alpha(\alpha-1)(t-1)-\bar{\alpha}(\bar{\alpha}+\alpha t)) \\ + h(-2\bar{\alpha}+(2-\alpha)(\bar{\alpha}+\alpha t)+2\alpha(\alpha-1)(t-1) -\alpha(\bar{\alpha}+\alpha t)) = 0. \end{gather}\] Analogously, we have zero coefficient at \(\langle w,u\rangle\). The rest summands equal \(\alpha(\alpha-1)\) multiplied by \[\begin{gather} (k-l)(g-h)((-1+\alpha)a+(2-\alpha)b) - (a-b)(k-l)((-1+\alpha)g+(2-\alpha)h) \\ + (g-h)((\alpha a+\bar{\alpha}b)(l-k) + (\alpha k+\bar{\alpha}l)(b-a)) - (a-b)((\alpha g+\bar{\alpha}h)(l-k) + (\alpha k+\bar{\alpha}l)(h-g)), \end{gather}\] which is zero.

Analogously, we have zero coordinate at \(z_2\). Finally, we have \[\Psi(r,s,q) = (-\alpha-\bar{\alpha}t+\bar{\alpha}+\alpha t)(\langle u,w\rangle v - \langle u,v\rangle w) = (2\alpha-1)(t-1)(\langle u,w\rangle v - \langle u,v\rangle w),\] as required. \(\square\)

Remark 4. It is easy to clarify, why Corollary 2b is true in \(S(\alpha,t,E)\). Indeed, by 65 , we have \(\Delta(s,\Psi(r,s,q)) = 0\) for \(r = r_0 + v\), \(s = s_0 + u\), \(q = q_0 + w\), where \(r_0,s_0,q_0\in Fz_1+Fz_2\), \(v,u,w\in E\), since \[\langle u,\langle u,w\rangle v - \langle u,v\rangle w \rangle = \langle u,v\rangle \langle u,w\rangle - \langle u,v \rangle\langle u,w\rangle = 0.\] Let \(\mu = (2\alpha-1)(t-1)\). In the case \(\dim E = 2\), take a basis \(e_1,e_2\) of \(E\) such that \(\langle e_1,e_2\rangle = 0\). Then for \(v=v_1e_1+v_2e_2\), \(u=u_1e_1+u_2e_2\), and \(w=w_1e_1+w_2e_2\), we have \[\Psi(r,s,q) = \mu(v_1w_2 - v_2w_1)(u_2e_1 - u_1e_2).\] Thus, \(\Psi(r,s,q)\) is proportional to the vector \(u^\perp = u_2 e_1 - u_1 e_2\), which is orthogonal to \(u\) with respect to \(\langle \cdot,\cdot\rangle\).

Corollary 3. In \(S(\alpha,t,E)\), the relation \(N(\Psi(r,s,q)) = 0\) holds for all \(r,s,q\). Hence, \(\Psi(r,s,q)^3 = S(\Psi(r,s,q))\Psi(r,s,q)\). An algebra \(A\), in which every element satisfies the equality \(x^3 = \varphi(x,x)x\) for some bilinear form \(\varphi\), is called pseudo-composition algebra [10].

Corollary 4. In \(S(\alpha,t,E)\), we have \(\Delta(\Psi(r,s,q),x_\#s) = \Delta(\Psi(r,s,q)_\#s,x)\). Put \(s_0 = kz_1+lz_2\). Applying the definition and Remark 4, we get \[\begin{gather} \Delta(\Psi(r,s,q)_\#s,x) = -(\bar{\alpha}k+\alpha l)\Delta(\Psi(r,s,q),x) = (\bar{\alpha}k+\alpha l)(\bar{\alpha}+\alpha t)\langle \Psi(r,s,q),y\rangle, \\ \Delta(\Psi(r,s,q),s_\#x) = -(\bar{\alpha}+\alpha t)\langle \Psi(r,s,q),s_\#x|_{E}\rangle = (\bar{\alpha}+\alpha t)(\bar{\alpha}k+\alpha l)\langle \Psi(r,s,q),y\rangle, \end{gather}\] hence, the required formula is proved.

Remark 5. Let us fix \(u\in E\), then the product \([v,w] := \Psi (v,u,w)\) is a Lie one [7], thus, \[\Psi(\Psi(v,u,w),u,x) + \Psi(\Psi(w,u,x),u,v) + \Psi(\Psi(x,u,v),u,w) = 0\] holds for all \(v,u,w,x\in E\). Further, the ternary product \([v,u,w]:= \Psi(v,w,u)\) defines a Lie triple system, i. e. the following identities for \([\cdot,\cdot,\cdot]\) hold: \[\begin{gather} [x,y,z] + [y,x,z] = 0, \quad [x,y,z] + [y,z,x] + [z,x,y] = 0, \\ [x,y,[u,v,w]] = [[x,y,u],v,w] + [u,[x,y,v],w] + [u,v,[x,y,w]], \end{gather}\] more about triple systems see [11]. We believe that such construction of a Lie triple system via an inner product is known, however, we are not able to find a suitable reference.

Theorem 3. The identity 60 holds on the algebra \(S(\alpha,t,E)\).

Proof. Denote \(\mu = (2\alpha-1)(t-1)\). The identity 53 holds on \(S(\alpha,t,E)\), since by Lemma 9, for \(r = r_0 + v\), \(s = s_0 + u\), \(q = q_0 + w\), and \(x = x_0 + y\), where \(r_0,s_0,q_0,x_0\in Fz_1+Fz_2\), \(v,u,w,y\in E\), we have \[\begin{gather} \Delta(\Psi(r,s,q),x) + \Delta(\Psi(q,s,x),r) + \Delta(\Psi(x,s,r),q) \\ = \mu( \langle u,w\rangle \Delta(v,x) - \langle u,v\rangle \Delta(w,x) + \langle u,y\rangle \Delta(w,r) - \langle u,w\rangle \Delta(y,r) \\ + \langle u,v\rangle \Delta(y,q) - \langle u,y\rangle \Delta(v,q) ) = 0. \end{gather}\]

Let us verify that 54 is fulfilled on \(S(\alpha,t,E)\). Denote \(s_0 = kz_1 + lz_2\). We apply Lemma 9: \[\begin{gather} \Psi(r,s,q)_\# x = \mu(\langle u,w\rangle v - \langle u,v\rangle w)_\# x = 2\mu(t-1)(-\bar{\alpha}z_1+\alpha z_2)( \langle u,w\rangle \langle v,y\rangle - \langle u,v\rangle \langle w,y\rangle ) \\ - \mu\chi(x)(\langle u,w\rangle v - \langle u,v\rangle w), \end{gather}\] where \(\chi(p_1 z_1+p_2 z_2 + \omega) = \bar{\alpha}p_1 + \alpha p_2\). Hence, by Remark 4, we get \[\label{delta40s44sharpproduct40Psi44x4141} \Delta(s,\Psi(r,s,q)_\# x) = -2\mu(t-1)\alpha(\alpha-1)(k-l)( \langle u,w\rangle \langle v,y\rangle - \langle u,v\rangle \langle w,y\rangle ).\tag{66}\] Define \(\pi = -2\mu(t-1)\alpha(\alpha-1)(k-l)\). Then \[\begin{gather} \Delta(s,\Psi(r,s,q)_\# x + \Psi(q,s,x)_\#r + \Psi(x,s,r)_\#q) \\ = \pi( \langle u,w\rangle \langle v,y\rangle - \langle u,v\rangle \langle w,y\rangle + \langle u,y\rangle \langle v,w\rangle - \langle u,w\rangle \langle v,y\rangle + \langle v,u\rangle \langle w,y\rangle - \langle u,y\rangle \langle v,w\rangle ) = 0. \end{gather}\]

Let us prove the relation 63 . By Lemma 9, \[\begin{gather} \Delta(\Psi(r,s,q),x_\#s) = \mu\Delta(\langle u,w\rangle v - \langle u,v\rangle w,-\chi(x)u-\chi(s)y) \\ = \mu\chi(s)(\bar{\alpha}+\alpha t) (\langle u,w\rangle \langle v,y\rangle - \langle u,v\rangle \langle w,y\rangle). \end{gather}\] As above (see 66 ), we conclude that 63 holds.

Now, we check 64 . First, we involve 44 and then 9 and 14 : \[\begin{gather} \label{DeltaDelta} - 3( \Delta(s,q)(\Delta(x_\#s,r) - \Delta(r_\#s,x)) + \Delta(r,s)( \Delta(q_\#s,x) - \Delta(x_\#s,q) ) \\ + \Delta(s,x)( \Delta(r_\#s,q) - \Delta(q_\#s,r) ) = \Delta(s,q)( (x_\#s,r) - (r_\#s,x)) \\ + \Delta(r,s)( (q_\#s,x) - (x_\#s,q) ) + \Delta(s,x)( (r_\#s,q) - (q_\#s,r) \\ = 2( \Delta(s,q)((xs,r)-(x,sr)) + \Delta(r,s)( (qs,x) - (q,sx)) + \Delta(s,x)( (rs,q) - (r,sq)) ) \\ + \Delta(s,q)( T(x)\Delta(s,r) - T(r)\Delta(s,x) ) + \Delta(s,r)( T(q)\Delta(s,x) - T(x)\Delta(s,q) ) \\ + \Delta(s,x)( T(r)\Delta(s,q) - T(q)\Delta(s,r) ) \\ = 2( \Delta(s,q)((xs,r){-}(x,sr)) + \Delta(r,s)( (qs,x) {-} (q,sx)) + \Delta(s,x)( (rs,q) {-} (r,sq)) ). \end{gather}\tag{67}\]

By 19 and 20 , we compute \[\begin{gather} \label{invariancy-deduced} (rs,q) - (r,sq) = (1+\alpha)g(ak+\langle v,u\rangle) + (2-\alpha)h(bl+t\langle v,u\rangle) \\ + (1+\alpha+(2-\alpha)t)( (\alpha a+\bar{\alpha}b)\langle u,w\rangle + (\alpha k+\bar{\alpha}l)\langle v,w\rangle ) \allowdisplaybreaks \\ - (1+\alpha)a(gk+\langle u,w\rangle) + (2-\alpha)b(hl+t\langle u,w\rangle) \\ + (1+\alpha+(2-\alpha)t)( (\alpha g+\bar{\alpha}h\langle u,v\rangle + (\alpha k+\bar{\alpha}l)\langle v,w\rangle ) \\ = (1-\alpha^2+\alpha(\alpha-2)t)( (g-h)\langle v,u\rangle - (a-b)\langle u,w\rangle ). \end{gather}\tag{68}\] Denote \(\nu = (1-\alpha^2+\alpha(\alpha-2)t)\) and let \(x_0 = mz_1 + nz_2\). Thus, \[\begin{gather} \Delta(s,x)( (rs,q) - (r,sq)) = \nu\alpha(\alpha-1)(k-l)\big( (m-n)(g-h)\langle v,u\rangle - (m-n)(a-b)\langle u,w\rangle \big) \\ + \nu(\bar{\alpha}+\alpha t)( (a-b)\langle u,w\rangle \langle u,y\rangle - (g-h)\langle u,v\rangle \langle u,y\rangle ). \end{gather}\] The analogous expressions for \(\Delta(s,q)((xs,r)-(x,sr))\) and \(\Delta(r,s)( (qs,x) - (q,sx))\) joint provide that 67 equals zero.

Finally, it remains to prove that \(\Psi_0 = 0\), see 62 . By Lemma 9, we express \[\Psi(\Psi(r,s,q),x,s) = \mu^2( \langle v,u\rangle \langle w,y\rangle u - \langle w,u\rangle \langle y,v\rangle u - \langle y,u\rangle \langle v,u\rangle w + \langle w,u\rangle \langle y,u\rangle v),\] Hence, \[\begin{gather} \Psi_0 = \mu^2(\langle v,u\rangle \langle w,y\rangle u - \langle w,u\rangle \langle y,v\rangle s - \langle y,u\rangle \langle v,u\rangle w + \langle w,u\rangle \langle y,u\rangle v \\ + \langle w,u\rangle \langle y,v\rangle u - \langle y,u\rangle \langle v,w\rangle u - \langle v,u\rangle \langle w,u\rangle y + \langle y,u\rangle \langle v,u\rangle w \\ + \langle y,u\rangle \langle v,w\rangle u - \langle v,u\rangle \langle w,y\rangle u - \langle w,u\rangle \langle y,u\rangle v + \langle v,u\rangle \langle w,u\rangle y ) = 0. \end{gather}\] The statement is proved. \(\square\)

Remark 6. Due to 68 , we see that the form \((\cdot,\cdot)\) is invariant on \(S(\alpha,E)\), as it was noted in [3].

Remark 7. Let us return to Example 1. We may introduce the bilinear form \(\widetilde{(r,q)} = (r,q) + \Delta(r,q)\). Then \[T(r_\# q) = (1-4\lambda)T(r)T(q) - \widetilde{(r,q)}, \quad \widetilde{(r_\# s,q)} = \widetilde{(r,s_\# q)}.\] Define the trilinear form \(\Psi\) by the formula 42 . Denote \(r = (a,b,c)\), \(s = (i,j,k)\), and \(q = (e,f,g)\). Thus, we have \[\begin{gather} \Psi(r,s,q) = \lambda( j(-ag+ce+bg-cf) + k (-af+be - bg+ cf), \\ i(ag-ce-bg+cf) + k(af-be-ag+ce), \; i(af-be+bg-cf) + j(-af+be+ag-ce)). \end{gather}\] Then the relations 51 , 53 , 54 and 62 are fulfilled. Further, the identity of the three associators holds, and the ternary product \([v,u,w]:= \Psi(v,w,u)\) defines a Lie triple system, see the code in GAP [8]. Moreover, the identity 60 is fulfilled on the space \(V = A\otimes_F F^3\cong A^{\otimes3}\), where \(A = F[\lambda]\).

7 Identities↩︎

In this section, we prove that the algebra \(S(\alpha,t,E)\) does not satisfy any polynomial identity of degrees 3 and 4, and all identities of degree 5 satisfied by \(S(\alpha,E)\) follow from commutativity and the identity 60 . In 1989, S.Yu. Vasilovsky found a basis of the \(T\)-ideal of identities fulfilled on the simple Jordan algebra of a nondegenerate form considered over a field of characteristic 0 [12]. One of them has the close form \((d,(a,b,c),b) + (a,(c,b,d),b) + (c,(d,b,a),b) = 0\).

In [4], it was proved that if a commutative (non-associative) unital algebra \(A\) over a field of characteristics not 2 or 3 satisfies an identity of degree 4 not implied by the commutative law, then \(A\) satisfies at least one of the following three identities: \[\begin{gather} (x^2x)x = x^2x^2, \tag{69} \\ 2((yx)x)x + yx^3 = 3(yx^2)x, \tag{70} \\ 2(y^2x)x + 2(x^2y)y +(yx)(yx) = 2((yx)y)x+2((yx)x)y+y^2x^2. \tag{71} \end{gather}\]

Then we have the following:

Lemma 10. Let \(E\) has a dimension \(n\ge 1\) and \(\alpha,t\notin \{0,1\}\). Then every identity of degree no more than 4 in the algebra \(S(\alpha,t,E)\) follows from commutativity.

Proof. To prove the statement, it is enough to show that the algebra \(S(\alpha,t,E)\) does not satisfy the identities 69 –71 .

First, let us show the identity 69 does not hold. Consider the left hand-side of 69 and set \(x = e\in E\) such that \(\langle e,e\rangle = 1\), then we have \[(e^2e)e=((z_1+t z_2)e)e) = (\alpha e + t (1-\alpha) e)e = (\alpha+t(1-\alpha))(z_1+t z_2).\] The right-hand side of 69 for \(x=e\) gives \[e^2e^2=(z_1+t z_2)(z_1+t z_2) = z_1+t^2z_2.\]

Since \(t\neq 0,1\), we conclude that the identity 69 does not hold.

To show that 70 does not hold, it is enough to consider \(x = e\in E\) such that \(\langle e,e\rangle = 1\) and \(y=z_1\). Then the left-hand side of 70 equals \[\label{id2subl32ez951} 2((z_1 e)e)e+z_1 e^3 = 3\alpha(\alpha + t(1-\alpha))e,\tag{72}\] while the right-hand side of 70 gives \[\label{id2subr32ez951} 3(z_1e^2)e=3\alpha e.\tag{73}\] We see that the right-hand sides of 72 and 73 are equal if and only if \(t=1\). By the conditions, \(t\neq 1\) and therefore the identity 70 does not hold.

Now we consider 71 . Define \[\phi(x,y)=2(y^2x)x + 2(x^2y)y +(yx)(yx) - 2((yx)y)x-2((yx)x)y-y^2x^2.\]

Then \(\phi(e,z_1)=(1-\alpha^2)z_1+t\alpha (2-\alpha)z_2 \neq 0\) for any \(\alpha,t\not\in\{0,1\}\). Consequently, 71 is not an identity in the algebra \(S(\alpha,t,E)\). \(\square\)

In [4], the list of all irreducible relative to commutativity identities of degree five is given. There are exactly five such identities, and the fourth of them [4] with \(\delta_2 = -\delta_1\neq0\) is nothing more than 60 with one of the three variables \(a,c,d\) equal to \(b\), e. g., \(d = b\).

The proof of the following theorem is established through computations conducted with the assistance of software programs such as Wolfram Mathematica and Albert [13].

Theorem 4. Let \(E\) has a dimension \(n\ge 2\) and \(\alpha\notin \{-1,0,1/2,1,2\}\). Every identity of degree no more than 5 in the algebra \(S(\alpha,E)\) over a field of characteristics 0 is a consequence of commutativity and 60 .

Proof. By Lemma 10, it remains to show that there are no identities in degree 5, which do not follow from commutativity and the identity 60 .

Let \(\mathcal{W}(X)\) denote a free algebra defined by identities of commutativity and 60 , which is generated by a set \(X\). Since we deal with a field of characteristics 0, then every polynomial identity is equivalent to a set of multilinear identities [14].

Let \(\mathcal{P}\) be a monomial basis of the multilinear part of degree 5 of the free commutative algebra \(\mathrm{Com}(X)\). Then \(\mathcal{P}\) consists of the 60 monomials of the type \((((**)*)*)*\), 30 monomials of the type \(((**)*)(**)\), and 15 monomials of the type \(((**)(**))*\). Define the set \[\begin{array}{ccccc} \mathcal{Z} = &\{((x_3 x_5) x_4) (x_1 x_2), & ((x_4 x_5) x_3) (x_1 x_2), &((x_2 x_5) x_4) (x_1 x_3), & ((x_4 x_5) x_2) (x_1 x_3), \\ &((x_2 x_5) x_3) (x_1 x_4), & ((x_3 x_5) x_2) (x_1 x_4), &(((x_1 x_5) x_4) x_3) x_2, & (((x_2 x_5) x_4) x_3) x_1, \\ &(((x_3 x_5) x_4) x_2) x_1, & (((x_4 x_5) x_3) x_2) x_1\}. \end{array}\]

To construct a monomial basis \(\mathcal{B}\) of the multilinear part of degree 5 of \(\mathcal{W}(X)\), where \(X=\{x_1,x_2,x_3,x_4,x_5\}\), we employ the software program Albert and obtain 95 basic monomials. We can represent the set of multilinear basic monomials as \(\mathcal{B}=\mathcal{P} \setminus \mathcal{Z}\).

If there exists a multilinear polynomial identity of degree 5 fulfilled on \(S(\alpha,t, E)\), which does not follow from commutativity and 60 , then it can be represented as a linear combination of monomials from \(\mathcal{B}\). Let us define a linear combination of elements in \(\mathcal{B}\) as \[\psi(x_1,x_2,x_3,x_4,x_5)=\sum_{b_i\in\mathcal{B}}\lambda_i b_i.\] To establish the theorem, it is necessary to demonstrate the linear independence of monomials from \(\mathcal{B}\). To achieve this, we use the Wolfram Mathematica software tool. A special code has been developed for calculating all substitutions, extracting homogeneous equations from them and solving them [8].

Since \(\dim E\geq2\), it is enough to show that all identities in degree 5 fulfilled on \(S(\alpha,t,E_0)\), where \(\dim E_0 = 2\), follow from commutativity and 60 . Let us choose a basis \(e,f\) of \(E_0\) such that \(\langle e,e\rangle = \langle f,f\rangle = 1\) and \(\langle e,f\rangle = 0\). We use the function Tuples\([\{z_1,z_2,e,f\},5]\) to generate all 1024 possible permutations of length 5 using the basic elements \(\{z_1,z_2,e,f\}\) and substitute them into \(\psi(x_1,x_2,x_3,x_4,x_5)\). Employing the function Union[], we express the obtained polynomials in terms of the coefficients \(\lambda_i\) and the basic elements \(\{z_1,z_2,e,f\}\). This yields a set of 635 polynomials.

Further, we express these polynomials by collecting coefficients at the basic elements \(\{z_1,z_2,e,f\}\) and extract the coefficients corresponding to these elements with the functions Collect[] and Coefficient[]. By employing the function Union[] once more, we reduce the number of polynomials to 498. Then we consider the system of equations formed by setting all these polynomials equal to zero. This system of equations is expressed in the coefficients \(\lambda_i\), where \(i\in \{1,\ldots, 95\}\).

The only trivial solution that emerges is \(\lambda_i=0\) for \(\alpha\notin\{-1,0,\frac{1}{2},1,2\}\), where \(i\in \{1,\ldots, 95\}\). This result demonstrates that the monomials involved in the linear combination \(\psi(x_1,x_2,x_3,x_4,x_5)\) are linearly independent. This completes the proof. \(\square\)

Remark 8. The above theorem is valid when \(t = (\alpha^2-1)/\alpha(\alpha-2)\). However, it is essential to note that for a general value of \(t\) that does not depend on \(\alpha\), the algebra \(S(\alpha,t,E)\) can have an identity of degree 5 which does not follow from commutativity and 60 .

For example, for \(t=5\) and \(\alpha=11/4\), there is an identity \[\begin{gather} ((c,a,e),b,d) + ((e,a,d),b,c) + ((d,a,c),b,e) \\ + (c,b,a)[R_d,R_e] + (d,b,a)[R_e,R_c] + (e,b,a)[R_c,R_d] = 0, \end{gather}\] which does not follow from commutativity and 60 . The validity of the identity can be checked using a program given in [15] or requiring a program from the authors.

8 Open problems↩︎

We finish the work with several open problems concerned the subject.

• Find an identity, which does not follow from commutativity and is fulfilled on every algebra associated to a generalized cubic form.

• Does the identity 48 follow from the definition of generalized sharped cubic form, or there exists a counterexample to it?

• Given a generalized sharped cubic form \((N,\Delta,\#,c)\), which satisfies 48 , is it true that \(N(\Psi(r,s,q)) = 0\) for all \(r,s,q\)?

• Find the basis of the \(T\)-ideal of identities fulfilled on \(S(\alpha,t,E)\).

9 Acknowledgments↩︎

V. Gubarev is supported by Mathematical Center in Akademgorodok under agreement No. 075-15-2022-281 with the Ministry of Science and Higher Education of the Russian Federation. A.S. Panasenko is supported by the Program of fundamental scientific researches of Russian Academy of Sciences, project FWNF-2022-0002.

The results of §2 are supported by the Program of fundamental scientific researches of Russian Academy of Sciences, project FWNF-2022-0002. The results of §4–6 are supported by Mathematical Center in Akademgorodok under agreement No. 075-15-2022-281 with the Ministry of Science and Higher Education of the Russian Federation.

Vsevolod Gubarev
Novosibirsk State University
Pirogova str. 2, 630090 Novosibirsk, Russia
Sobolev Institute of Mathematics
Acad. Koptyug ave. 4, 630090 Novosibirsk, Russia
e-mail: wsewolod89@gmail.com

Farukh Mashurov
Suleyman Demirel University
Abylai Khan Street 1/1
Kaskelen, Kazakhstan
e-mail: f.mashurov@gmail.com

Alexander Panasenko
Sobolev Institute of Mathematics
Acad. Koptyug ave. 4, 630090 Novosibirsk, Russia
Novosibirsk State University
Pirogova str. 2, 630090 Novosibirsk, Russia
e-mail: a.panasenko@g.nsu.ru

References↩︎