1 Introduction

This paper is a continuation of a series of papers where there were proven analogous versions to WPT for finite-dimensional Jordan superalgebras [1,2,3].

One of the most classical theorems in structure theory for finite-dimensional associative algebras was given by Wedderburn [4], proving that for all finite dimensional associative algebra \(\mathcal {A}\) over an arbitrary field, with nilpotent radical \(\mathcal {N}\), there exists a subalgebra \(\mathcal {S}\subseteq \mathcal {A}\) such that \(\mathcal {S}\cong \mathcal {A}/\mathcal {N}\) and \(\mathcal {A}=\mathcal {S}\oplus \mathcal {N}\). This theorem is known in classical literature as the Wedderburn Principal Theorem (WPT).

Analogous versions of the WPT were proved for finite-dimensional Jordan algebras by Albert [5], Penico [6], and Aškinuze [7].

Superalgebras are algebras that admit a decomposition as a direct sum of vector spaces \(\mathcal {A}=\mathcal {A}_0\oplus \mathcal {A}_1\) and it satisfies the multiplicative rule \(\mathcal {A}_i\mathcal {A}_j\subseteq \mathcal {A}_{i+j\, \textrm{mod}\, 2}\). \(\mathcal {A}_0\) and \(\mathcal {A}_1\) are called respectively even and odd parts of \(\mathcal {A}\). It is well known that all associative algebra is an associative superalgebra. However, it is not a general rule. For example, if \(\mathfrak {J}\) is a Jordan superalgebra with odd part non-zero, then \(\mathfrak {J}\) is not a Jordan algebra while \(\mathfrak {J}_0\) is a Jordan algebra. It arises as a natural question to research the validity of the WPT for finite-dimensional non-associative superalgebras. Note that this is equivalent to investigating which superalgebras \(\mathcal {A}=\mathcal {A}_0\oplus \mathcal {A}_1\) with radical \(\mathcal {N}\) there exists a \(\mathcal {S}_0\)-superbimodule \(\mathcal {S}_1\subseteq \mathcal {A}_1\) such that \(\mathcal {S}_1\cong \mathcal {A}_1/\mathcal {N}_1\), \(\mathcal {A}_1 =\mathcal {S}_1\oplus \mathcal {N}_1\), \(\mathcal {A}=(\mathcal {S}_0\oplus \mathcal {S}_1)\oplus (\mathcal {N}_0\oplus \mathcal {N}_1)\), where \(\mathcal {S}_0\subseteq \mathcal {A}_0\) is a semisimple superalgebra, and \(\mathcal {A}_0/\mathcal {N}_0\cong \mathcal {S}_0\), \(\mathcal {A}_0=\mathcal {S}_0\oplus \mathcal {N}_0\).

Additionally, it is known that on the variety of finite-dimensional Jordan and alternative algebras, solubility implies nilpotence. However, in finite-dimensional Jordan and alternative superalgebras Shestakov [8] proved that there exist solvable superalgebras that are not nilpotent. In this sense as a first step, we consider solvable radical to study the validity of WPT in Jordan superalgebras. The first author [1] proved that it is possible to have an analogous version to WPT for finite-dimensional Jordan superalgebras when some conditions are imposed over the irreducible superbimodules contained in the solvable radical. First, as in the case of Jordan algebras, he proved that it is possible to reduce the problem to Jordan superalgebra \(\mathcal {A}\) with solvable radical \(\mathcal {N}\) with \(\mathcal {N}\,^2=0\) such that the quotient Jordan superalgebra \(\mathcal {A}/\mathcal {N}\) is isomorphic to a simple Jordan superalgebra \(\mathfrak {J}\). As a consequence of this result, it is possible to consider case by case of simple Jordan superalgebra. Second, he proved that if \(\mathfrak {J}\) is a simple Jordan superalgebra then it is possible to reduce the proof of WPT considering the irreducible Jordan \(\mathfrak {J}\)-superbimodules that are contained in \(\mathcal {N}\).

The classification of finite-dimensional simple Jordan superalgebras over an algebraically closed field of characteristic zero was given by Kac [9] and Kantor [10].

In this paper, Sect. 2 gives some preliminary results regarding the theory of Jordan superalgebras, including the definition of the superalgebra of the Grassmann Poisson bracket \(\mathfrak {K}\textrm{an}(2)\), and the classification of irreducible Jordan \(\mathfrak {K}\textrm{an}(2)\)– superbimodules given by Folleco and Shestakov [11]. Section 3 presents the conditions over \(\mathfrak {r}(\widetilde{x}_i\widetilde{x}_j)\in \mathcal {N}\), where \(\widetilde{x}_i\in \mathcal {A}\) is a preimage under canonical homomorphism of \(\bar{\widetilde{x}}_i\), \(\bar{\widetilde{x}}_i\in \overline{\widetilde{X}}\) and \(\overline{\widetilde{X}}\) is an additive basis of \(\mathcal {A}/\mathcal {N}\) such that \(\bar{\mathcal {S}}=\text {alg}\,\langle \overline{\widetilde{X}}_i\rangle \cong \mathfrak {K}\textrm{an}(2)\). We assume that \(\widetilde{x}_i \widetilde{x}_j=\sum _{k}\alpha _k \widetilde{x}_k +\mathfrak {r}(\widetilde{x}_i\widetilde{x}_j)_k\). Finally, Sect. 4 provides a proof of the main theorem of this paper. We shall prove that if \(\mathcal {N}\,^2=0\) and \(\mathcal {A}/\mathcal {N}\) is isomorphic to the simple Jordan superalgebra of Grassmann Poisson bracket \(\mathfrak {K}\textrm{an}(2)\), then an analogous to WPT holds.

2 Basic concepts and notation

Throughout the paper, all algebras are considered over an algebraically closed field \(\mathbb {F}\) of characteristic zero.

Recall that a superalgebra \(\mathfrak {J}=\mathfrak {J}_0\oplus \mathfrak {J}_1\) is said to be a Jordan superalgebra, if for every \(a, b, c, d\in \mathfrak {J}_0{\dot{\cup }}\mathfrak {J}_1\) the superalgebra satisfies the super identities

$$\begin{aligned}{} & {} ab=(-1)^{|a||b|}ba, \end{aligned}$$
(2.1)
$$\begin{aligned}{} & {} ((ab)c)d+(-1)^{|d|(|c|+|b|)+|b||c|}((ad)c)b+(-1)^{|a|(|b|+|c|+|d|)+|d||c|}((bd)c)a \nonumber \\{} & {} \quad =(ab)(cd)+(-1)^{|d|(|c|+|b|)}(ad)(bc)+(-1)^{|c|(|a|+|b|)}(ac)(bd). \end{aligned}$$
(2.2)

We denote the parity of a by \(|a|=i\) if \(a\in \mathfrak {J}_i\).

Let \(\mathcal {N}=\mathcal {N}_{0}\oplus \mathcal {N}_{1}\) be a superbimodule over \(\mathfrak {J}\). We say that \(\mathcal {N}\) is a Jordan \(\mathfrak {J}\)-superbimodule if the split null extension \(\mathcal {E}=\mathfrak {J}\oplus \mathcal {N}\) is a Jordan superalgebra. A regular \(\mathfrak {J}\)-superbimodule, denoted as \(\mathcal {R}\textrm{eg}\,\mathfrak {J}\), is defined on the vector super-space \(\mathfrak {J}\) with an action coinciding with the multiplication in \(\mathfrak {J}\). Besides, if \(\mathcal {N}\) is a \(\mathfrak {J}\)-superbimodule, then the superbimodule \(\mathcal {N}^\textrm{op}\) is defined as a copy of \(\mathcal {N}\) where \(\mathcal {N}^\textrm{op}_0=\mathcal {N}_1\), \(\mathcal {N}^\textrm{op}_1=\mathcal {N}_0\), and the action is defined via

$$\begin{aligned} am^\textrm{op}=(-1)^{|a|}(am)^{\textrm{op}}, \quad \quad m^\textrm{op}a=(ma)^{\textrm{op}} \end{aligned}$$
(2.3)

for all \(a\in \mathfrak {J}_0{\dot{\cup }}\mathfrak {J}_1\) and \(m\in \mathcal {N}_0{\dot{\cup }}\mathcal {N}_1\). \(\mathcal {N}^\textrm{op}\) is called the opposite of the superbimodule \(\mathcal {N}\).

Now, we consider the simple Jordan superalgebra of type Poisson Grassmann bracket \(\mathfrak {K}\textrm{an}(2):=(\mathbb {F}\cdot 1+\mathbb {F}\cdot f_1+\mathbb {F}\cdot f_2+\mathbb {F}\cdot e_{12})\oplus (\mathbb {F}\cdot u+\mathbb {F}\cdot e_1+\mathbb {F}\cdot e_2+\mathbb {F}\cdot f_{12})\) where 1 is a unity, \(f_1\bullet f_1=f_2\bullet f_2=1\), and nonzero products are given by

$$\begin{aligned}{} & {} f_2\bullet e_1=-f_1\bullet e_2=e_{12}\bullet u=f_{12}, \quad f_1\bullet f_{12}=-e_2, \quad f_2\bullet f_{12}=e_1. \end{aligned}$$
(2.4)
$$\begin{aligned}{} & {} e_1\bullet e_2=e_{12}, \ \quad e_i\bullet u=f_i, \end{aligned}$$
(2.5)

for \(i=1,2\). Products in (2.4) and (2.5) are symmetric and skew-symmetric respectively. Irreducible Jordan superbimodules over the simple Jordan superalgebra Poisson Grassmann Bracket \(\mathfrak {K}\textrm{an}(n)\) were classified by O. Folleco and I. Shestakov [11]. In particular, it was proved that if \(\mathcal {V}\) is an unital irreducible Jordan superbimodule over \(\mathfrak {K}\textrm{an}(2)\), then there exists a special element v in \(\mathcal {V}\) such that \(v \cdot e_I=v\cdot f_I=0\) where \(I\in \{1, 2, 12\}\). Besides, they proved that the action of \(\mathfrak {K}\textrm{an}(2)\) on \(\mathcal {V}(v,\alpha )\) depends on the choice of a special element \(v\in \mathcal {V}(v,\alpha )\) and a parameter \(\alpha = \mathcal {R}^2_1\in \mathbb {F}\), where \(\mathcal {R}_x\) denotes the right multiplication operator. Without loss of generality, we can assume that v is an even element. Then, for an additive basis for \(\mathcal {V}\) given by \(v, w_1, w_2, v_{12}, w\), \(v_1\), \(v_2\), and \(w_{12}\) where \(v, w_1, w_2, v_{12}\) are even elements and \(w, v_1, v_2, w_{12}\) are odd elements, the nonzero right action of \(\mathfrak {K}\textrm{an}(2)\) over \(\mathcal {V}(v,\alpha )\) is given by the following relations:

$$\begin{aligned} \begin{aligned}&v\cdot u= v_1\cdot f_1=v_2\cdot f_2=-w_{12}\cdot e_{12}=-w_1\cdot e_1=-w_2\cdot e_2=-v_{12}\cdot f_{12}=w,\\&\begin{aligned} w\cdot f_1 =w_2\cdot f_{12} =v_{12}\cdot e_2 =v_1,{} & {} w\cdot f_2 =-w_1\cdot f_{12} =-v_{12}\cdot e_1 =v_2, \\ v_{12}\cdot f_2 =v_1\cdot u =-w_{12}\cdot e_2 =w_1,{} & {} -v_{12}\cdot f_1 =v_2\cdot u =w_{12}\cdot e_1 =w_2,\\ w_1\cdot f_2=-w_2\cdot f_1=v_{12},{} & {} v_{12}\cdot u=w_{12}, \end{aligned}\\&w_1\cdot u=\alpha v_1, \quad w_2\cdot u=\alpha v_2, \quad w_{12}\cdot u=\alpha v_{12}, \quad \quad w\cdot u=w_{12}\cdot f_{12}=\alpha v,\\&v_1\cdot e_1=v_2\cdot e_2 =-v_{12}\cdot e_{12}=v, \end{aligned} \end{aligned}$$
(2.6)

where \(\alpha \in \mathbb {F}\). Note that if \(\alpha =0\), then \(\mathcal {V}(v,0)=\mathcal {R}\textrm{eg}\,\mathfrak {K}\textrm{an}(2)\).

Now, let \(\mathcal {A}\) be a finite-dimensional Jordan superalgebra with radical \(\mathcal {N}\) such that \(\mathcal {N}\,^2=0\), and \(\mathcal {A}/\mathcal {N}\cong \mathfrak {K}\textrm{an}(2)\). By [1] it follows that an analog to WPT is valid for \(\mathcal {A}\) if it is valid for each irreducible \(\mathfrak {K}\textrm{an}(2)\)-superbimodule. Due to [11], we just need to consider two cases, \(\mathcal {N}\cong \mathcal {V}(v,\alpha )\), and \(\mathcal {N}\cong \mathcal {V}(v,\alpha )^{\textrm{op}}\).

As a consequence of WPT for Jordan algebras it follows that there exist \(\widetilde{1}\), \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12} \in \mathcal {A}_0\) such that \(\mathcal {S}_0=\textrm{alg} \langle \widetilde{1}, \widetilde{f}_1, \widetilde{f}_2, \widetilde{e}_{12} \rangle \cong (\mathfrak {K}\textrm{an}(2))_0\), and \(\mathcal {A}_0=\mathcal {S}_0\oplus \mathcal {N}_0\). Since \((\mathcal {A}/\mathcal {N})_1\cong \mathcal {A}_1/\mathcal {N}_1\cong (\mathfrak {K}\textrm{an}(2))_1\), then there exist elements \(\widetilde{u}\), \(\widetilde{e}_1\), \(\widetilde{e}_2\) and \(\widetilde{f}_{12}\in \mathcal {A}_1\) such that \(\textrm{vec}\langle \overline{\widetilde{u}}, \overline{\widetilde{e}_1}, \overline{\widetilde{e}_2}, \overline{\widetilde{f}_{12}} \rangle \cong (\mathfrak {K}\textrm{an}(2))_1\), where \(\overline{\widetilde{x}}\) denotes the image of \(\widetilde{x}\) under canonical homomorphism. In the following, we identify elements \(\widetilde{x}\in \mathcal {A}\) with elements \(x\in \mathfrak {K}\textrm{an}(2)\). In the sequel, we denote the products in \(\mathcal {A}\) by juxtaposition, i.e., for \(\widetilde{x}\) and \(\widetilde{y}\in \mathcal {A}\) we write \(\widetilde{x}\widetilde{y}\). Note that products \(\widetilde{x}\widetilde{y}\in \mathcal {A}\) can be written as \(\widetilde{x}\widetilde{y}=\widetilde{x\bullet y}+\mathfrak {r}(\widetilde{x}\widetilde{y})\) where \(\mathfrak {r}(\widetilde{x}\widetilde{y})\in \mathcal {N}\). \(\mathfrak {r}(\widetilde{x}\widetilde{y})\) is called the radical component of the product \(\widetilde{x}\widetilde{y}\). So, given \(\widetilde{n}\in \mathcal {N}\) and \(\widetilde{x}\in \mathcal {A}\), we denote \(\widetilde{n}\widetilde{x}=\widetilde{n\cdot x}\), where \(\widetilde{n}\) is identified with \(n\in \mathcal {V}(v,\alpha )\) (\(n\in \mathcal {V}(v,\alpha )^{\textrm{op}}\)). Without loss of generality, we can assume the following products in \(\mathcal {A}\): skew-symmetric products are given by

$$\begin{aligned} \begin{aligned} \widetilde{u}\widetilde{u}=&\mathfrak {r}(\widetilde{u}\widetilde{u}),{} & {} \widetilde{e}_i\widetilde{e}_i=\mathfrak {r}(\widetilde{e}_i\widetilde{e}_i),{} & {} \widetilde{f}_{12}\widetilde{f}_{12}=\mathfrak {r}({\widetilde{f}_{12}\widetilde{f}_{12}}), \\ \widetilde{e}_1\widetilde{f}_{12}&=\mathfrak {r}({\widetilde{e}_1\widetilde{f}_{12}}),{} & {} \widetilde{e}_2\widetilde{f}_{12}=\mathfrak {r}({\widetilde{e}_2\widetilde{f}_{12}}),{} & {} \widetilde{f}_{12}\widetilde{u}=\mathfrak {r}({\widetilde{f}_{12}\widetilde{u}}), \\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+\mathfrak {r}({\widetilde{e}_1\widetilde{u}}),{} & {} \widetilde{e}_2\widetilde{u}=\widetilde{f}_2+\mathfrak {r}({\widetilde{e}_2\widetilde{u}}),{} & {} \widetilde{e}_1\widetilde{e}_2=\widetilde{e}_{12}+\mathfrak {r}({\widetilde{e}_1\widetilde{e}_2}), \end{aligned} \end{aligned}$$
(2.7)

while symmetric products are given by

$$\begin{aligned} \begin{aligned} \widetilde{f}_i\widetilde{f}_i=&\widetilde{1}\widetilde{1}=\widetilde{1},{} & {} \widetilde{1}\widetilde{f}_i=\widetilde{f}_i,{} & {} \widetilde{1}\widetilde{e}_{12}=\widetilde{e}_{12},\\ \widetilde{1}\widetilde{u}=&\widetilde{u}+\mathfrak {r}(\widetilde{1}\widetilde{u}),{} & {} \widetilde{1}\widetilde{e}_i=\widetilde{e}_i+\mathfrak {r}(\widetilde{1}\widetilde{e}_i),{} & {} \widetilde{1}\widetilde{f}_{12}=\widetilde{f}_{12}+ \mathfrak {r}(\widetilde{1}\widetilde{f}_{12}), \\ \widetilde{f}_i\widetilde{u}=&\mathfrak {r}{(\widetilde{f}_i\widetilde{u})},{} & {} \widetilde{f}_1\widetilde{e}_1=\mathfrak {r}({\widetilde{f}_1\widetilde{e}_1}),{} & {} \widetilde{f}_2\widetilde{e}_2=\mathfrak {r}({\widetilde{f}_2\widetilde{e}_2}), \\ \widetilde{e}_{12}\widetilde{f}_{12}=&\mathfrak {r}{(\widetilde{e}_{12}\widetilde{f}_{12})},{} & {} \widetilde{f}_1\widetilde{f}_{12}=-\widetilde{e}_{2}+\mathfrak {r}({\widetilde{f}_1\widetilde{f}_{12}}),{} & {} \widetilde{f}_2\widetilde{f}_{12}=\widetilde{e}_{1}+\mathfrak {r}({\widetilde{f}_2\widetilde{f}_{12}}), \\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+\mathfrak {r}({\widetilde{f}_1\widetilde{e}_2}),{} & {} \widetilde{f}_2\widetilde{e}_1=\widetilde{f}_{12}+\mathfrak {r}({\widetilde{f}_2\widetilde{e}_1}),{} & {} \widetilde{e}_{12}\widetilde{u}=\widetilde{f}_{12}+\mathfrak {r}({\widetilde{e}_{12}\widetilde{u}}), \end{aligned} \end{aligned}$$
(2.8)

for \(i=1,2\), where \(\mathfrak {r}(\widetilde{x}\widetilde{y})\in \mathcal {N}_0{\dot{\cup }}\mathcal {N}_1\), and \(\mathfrak {r}(\widetilde{x}\widetilde{y})=(-1)^{|\widetilde{x}||\widetilde{y}|}\mathfrak {r}(\widetilde{y}\widetilde{x})\).

Now, using the identity (2.2) and the fact that \(\mathcal {N}\) is a unitary irreducible bimodule over \(\mathcal {A}\), it follows that for all \(\widetilde{a}\in \mathcal {A}_1\) holds \(2((\widetilde{1}a)\widetilde{1})\widetilde{1}+\widetilde{1}\widetilde{a}=3(\widetilde{1}\widetilde{a})\widetilde{1}\). So, using this equation and assuming \(\widetilde{1}\widetilde{a}=\widetilde{a}+\widetilde{r}\) where \(\widetilde{r}\in \mathcal {N}\), as a consequence, we conclude that \(\widetilde{r}=0\). Then, we write \(\widetilde{1}\widetilde{u}=\widetilde{u}\), \(\widetilde{1}\widetilde{e}_i=\widetilde{e}_i\), and \(\widetilde{1}\widetilde{f}_{12}=\widetilde{f}_{12}\) for \(i=1,2\). Therefore, we obtain that \(\widetilde{1}\) is the unity in \(\mathcal {A}\).

3 Conditions on the radical component of products in \(\mathcal {A}\)

In this section, we investigate the conditions that are satisfied by the radical components of products \(\widetilde{x}\widetilde{y}\in \mathcal {A}\). We need to consider two cases \(\mathcal {N}\cong \mathcal {V}(v,\alpha )\), and \(\mathcal {N}\cong \mathcal {V}(v,\alpha )^{\textrm{op}}\). From now on, let \(\widetilde{1}\), \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\), \(\widetilde{u}\), \(\widetilde{e}_1\), \(\widetilde{e}_2\) and \(\widetilde{f}_{12}\in \mathcal {A}\) such that (2.7) and (2.8) holds.

3.1 Case 1

Assume that \(\mathcal {N}_0=\textrm{span}\langle v, w_1,w_2,v_{12}\rangle\), and \(\mathcal {N}_1=\textrm{span}\langle w, v_1,v_2,w_{12}\rangle\) such that \(\mathcal {N}\cong \mathcal {V}(v,\alpha )\) with nonzero actions given by (2.6).

Initially, we present some lemmas used mainly to make short the proof of WPT in this case.

Lemma 3.1

Let \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_1\), \(\widetilde{e}_2\), \(\widetilde{e}_{12}\), \(\widetilde{f}_{12}\), and \(\widetilde{u}\) be as equations (2.7) and (2.8) given in 2. Then there exist scalars \(\xi _0\), \(\eta _0\), \(\alpha _0\), \(\delta _0\) \(\tau _0\), \(\sigma _0\), \(\epsilon _0\) \(\xi _1\), \(\alpha _1\), \(\delta _2\), such that \(\widetilde{f}_1 \widetilde{u}=\xi _0 w+\xi _1v_1\), \(\widetilde{f}_2 \widetilde{u}=\eta _0w+\xi _1v_2\), \(\widetilde{f}_1\widetilde{e}_1=\alpha _0w+\alpha _1 v_1\), \(\widetilde{f}_2\widetilde{e}_2=\delta _0w+\delta _2v_2\), \(\widetilde{e}_{12}\widetilde{e}_1=\tau _0w\), \(\widetilde{e}_{12}\widetilde{e}_2=\sigma _0w\), and \(\widetilde{e}_{12}\widetilde{f}_{12}=\epsilon _0w\).

Proof

Let us consider \(a=c=d=\widetilde{f}_i\), and \(b=\widetilde{x}\in \mathcal {A}_1\) in the equation (2.2), we conclude that \(((\widetilde{f}_i\widetilde{x})\widetilde{f}_i)\widetilde{f}_i=\widetilde{f}_i\widetilde{x}\). Since \(\widetilde{f}_i\widetilde{x}\in \mathcal {N}_1\), we write \(\widetilde{f}_i\widetilde{x}=Aw+B_1v_1+B_2v_2+Cw_{12}\) where \(A, B_i,\) and \(C\in \mathbb {F}\) are arbitrary scalars for \(i=1,2\). By linearly independence of elements w, and \(v_i\) we obtain \(\widetilde{f}_i\widetilde{x}=Aw+B_iv_i\) for \(i=1,2\). Thus, we have that there exist scalars \(\xi _i\), \(\alpha _i\), \(\eta _j\), \(\delta _j\in \mathbb {F}\) such that \(\widetilde{f}_1 \widetilde{u}=\xi _0 w+\xi _1v_1\), \(\widetilde{f}_1\widetilde{e}_1=\alpha _0w+\alpha _1 v_1\), \(\widetilde{f}_2 \widetilde{u}=\eta _0w+\eta _2v_2\), and \(\widetilde{f}_2\widetilde{e}_2=\delta _0w+\delta _2v_2\).

Now, taking \(a=\widetilde{f}_1\), \(b=\widetilde{u}\), and \(c=d=\widetilde{f}_2\) in the equation (2.2), we get that \((\mathfrak {r}(\widetilde{f}_1\widetilde{u})\widetilde{f}_2)\widetilde{f}_2+(\mathfrak {r}(\widetilde{f}_2\widetilde{u})\widetilde{f}_2)\widetilde{f}_1-\mathfrak {r}(\widetilde{f}_1\widetilde{u})=0\). Consequently, we conclude that \(\eta _2=\xi _1\). It is easy to see that if \({r}\in \mathcal {N}_1\), then \(({r}\widetilde{f}_i)\widetilde{e}_{12}=0\). So, substituting \(a=c=\widetilde{f}_1\), \(b=\widetilde{e}_i\), and \(d=\widetilde{e}_{12}\) in the equation (2.2), we have that \(((\widetilde{f}_i\widetilde{e}_i)\widetilde{f}_i)\widetilde{e}_{12}+((\widetilde{e}_{12}\widetilde{e}_i)\widetilde{f}_i)\widetilde{f}_i=\widetilde{e}_{12}\widetilde{e}_i\). Noting that \(\widetilde{f}_i\widetilde{e}_i\in \mathcal {N}_1\), it is clear that equality \(\mathfrak {r}((\widetilde{e}_{12}\widetilde{x})\widetilde{f}_i)\widetilde{f}_i-\mathfrak {r}(\widetilde{e}_{12}\widetilde{x})=0\) holds, and therefore we have scalars \(\tau _0\), \(\tau _1\), \(\sigma _0\), and \(\sigma _2\) such that \(\widetilde{e}_{12}\widetilde{e}_1=\tau _0w+\tau _1v_1\), and \(\widetilde{e}_{12}\widetilde{e}_2=\sigma _0w+\sigma _2v_2\).

Considering \(a=d=\widetilde{e}_{12}\), \(b=\widetilde{u}\), and \(c=\widetilde{f}_1\) in the equation (2.2), we obtain \(((\widetilde{e}_{12}\widetilde{u})\widetilde{f}_1)\widetilde{e}_{12}=0\). Expanding this equation, we have that \((\mathfrak {r}(\widetilde{e}_{12}\widetilde{u})\widetilde{f}_1)\widetilde{e}_{12}+\mathfrak {r}(\widetilde{f}_1\widetilde{f}_{12})\widetilde{e}_{12}-\mathfrak {r}(\widetilde{e}_{12}\widetilde{e}_2)=0\). From this, and using the fact that \((\mathfrak {r}(\widetilde{e}_{12}\widetilde{u})\widetilde{f}_1)\widetilde{e}_{12}=0\), we conclude that \(\mathfrak {r}(\widetilde{f}_1\widetilde{f}_{12})\widetilde{e}_{12}-\mathfrak {r}(\widetilde{e}_{12}\widetilde{e}_2)=0\). Now, let \(\widetilde{f}_j\widetilde{f}_{12}=A_jw+B_{j1}v_1+B_{j2}v_2+C_jw_{12}\) with \(A_j, B_{ji}, C_j\in \mathbb {F}\), for \(i,j=1,2\); substituting this in the last equality with \(j=1\), we determine that \(-C_1w-(\sigma _0w+\sigma _2v_2)=0\). Then, by linear independence of elements, we conclude that \(\sigma _2=0\), \(C_1=-\sigma _0\). Similarly, with \(j=2\), we deduce that \(\tau _1=0\) and \(C_2=\tau _0.\) Thus, we can write

$$\begin{aligned} \widetilde{e}_{12}\widetilde{e}_1=\tau _0 w, \quad \text { and } \quad \widetilde{e}_{12}\widetilde{e}_2=\sigma _0w. \end{aligned}$$
(3.1)

Putting \(a=\widetilde{f}_i\), \(b=\widetilde{u}\), and \(c=d=\widetilde{e}_{ij}\) in the equation (2.2), we have that \(((\widetilde{f}_i\widetilde{u})\widetilde{e}_{ij})\widetilde{e}_{ij}+((\widetilde{e}_{ij}\widetilde{u})\widetilde{e}_{ij})\widetilde{f}_i=0\), and it follows that

$$\begin{aligned} \mathfrak {r}((\widetilde{e}_{ij}\widetilde{u})\widetilde{e}_{ij})\widetilde{f}_i+\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{f}_i=0. \end{aligned}$$
(3.2)

Let \(\mathfrak {r}(\widetilde{e}_{12}\widetilde{u})=\varphi _0w+\varphi _1v_1+\varphi _2v_2+\varphi _{12}w_{12}\), and \(\mathfrak {r}(\widetilde{e}_{12}\widetilde{f}_{12})=\epsilon _0w+\epsilon _1v_1+\epsilon _2v_2+\epsilon _{12}w_{12}\), with \(\varphi _i\), \(\epsilon _i\in \mathbb {F}\) for \(i\in \{0,1,2,12\}\). Replacing these equalities in the equation (3.2) and using the fact that w, \(v_1\), \(v_2\), and \(w_{12}\) are linearly independent, we get that \(\epsilon _{1}=\epsilon _2=0\), and \(\epsilon _0=\varphi _{12}\). Finally, substituting \(a=c=d=\widetilde{e}_{12}\), and \(b=\widetilde{u}\) in the equation (2.2), we obtain that \(((\widetilde{e}_{12}\widetilde{u})\widetilde{e}_{12})\widetilde{e}_{12}=0\). Thus, by simplifying the last equation, we get that \(\mathfrak {r}(\widetilde{e}_{12}\widetilde{f}_{12})\widetilde{e}_{12}=0\), hence \(\epsilon _{12}=0\). Therefore, we can write

$$\begin{aligned} \widetilde{e}_{12}\widetilde{f}_{12}=\epsilon _0w, \end{aligned}$$
(3.3)

which completes the proof. \(\square\)

Lemma 3.2

If \(\widetilde{x}\in \{\widetilde{e}_1,\widetilde{e}_2,\widetilde{f}_{12}\}\), then \(\widetilde{x}\widetilde{x}=0\). If \(\alpha \ne 0\) then \(\widetilde{u}\widetilde{u}=0\); otherwise, there exists \(\Lambda _u\in \mathbb {F}\) such that \(\widetilde{u}\widetilde{u}=\Lambda _u v_{12}\).

Proof

Assume that \(\widetilde{x}\widetilde{x}=A_xv+B_x w_1+C_xw_2+D_xv_{12}\), where \(A_x\), \(B_x\), \(C_x\), and \(D_x\in \mathbb {F}\). Putting \(a=b=\widetilde{x}\in \mathcal {A}_1\), \(c=d=\widetilde{f}_i\) in the equation (2.2) using linear independence of v, \(w_i\), and \(v_{ij}\), we get \(((\widetilde{x}\widetilde{x})\widetilde{f}_i)\widetilde{f}_i=\widetilde{x}\widetilde{x}\). Then, we write \(\mathfrak {r}(\widetilde{x}\widetilde{x})=D_xv_{12}\). Now, taking \(a=d=\widetilde{e}_{ij}\), \(b=\widetilde{u}\), and \(c=\widetilde{f}_{ij}\) in the equation (2.2), we conclude that \(((\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_{ij})\widetilde{e}_{ij}-(\widetilde{e}_{ij}\widetilde{u})(\widetilde{e}_{ij}\widetilde{f}_{ij})=0\). Thus, it follows that \((\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_{ij})\widetilde{e}_{ij}+\mathfrak {r}(\widetilde{f}_{ij}\widetilde{f}_{ij})\widetilde{e}_{ij}+\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{f}_{ij}=0\). Observe that \((\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_{ij})\widetilde{e}_{ij}=\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{f}_{ij}=0\). Therefore \(\mathfrak {r}(\widetilde{f}_{ij}\widetilde{f}_{ij})\widetilde{e}_{ij}=0\) and using the linear independence of v, \(w_1\), \(w_2\), and \(v_{12}\) we obtain that \(D_{12}=0\). Besides, replacing \(a=\widetilde{f}_i\), \(b=\widetilde{f}_{ij}\), \(c=\widetilde{e}_j\), and \(d=\widetilde{e}_{ij}\) in (2.2) we get \(((\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_j)\widetilde{e}_{ij}+((\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{e}_j)\widetilde{f}_i-(\widetilde{f}_i\widetilde{f}_{ij})(\widetilde{e}_{ij}\widetilde{e}_j)-(\widetilde{e}_{ij}\widetilde{f}_{ij})(\widetilde{f}_i\widetilde{e}_j)=0.\) Solving this equation and using equations (3.1) and (3.3), we conclude that \(\mathfrak {r}(\widetilde{e}_j\widetilde{e}_j)\widetilde{e}_{ij}=0\). Then \(D_j=0\) for \(j=1,2\).

Finally, assume that \(\widetilde{u}\widetilde{u}=\Lambda _u v_{12}\). Let us consider \(a=b=c=d=\widetilde{u}\) in the equation (2.2), we obtain \(((\widetilde{u}\widetilde{u})\widetilde{u})\widetilde{u}=0\). Then, it follows from this that \((\mathfrak {r}(\widetilde{u}\widetilde{u})\widetilde{u})\widetilde{u}=0\). Therefore \(\alpha \Lambda _u=0\). Note that if \(\alpha \ne 0\), then \(\Lambda _u=0\), which proves the Lemma. \(\square\)

Lemma 3.3

Let \(\widetilde{e}_1\), \(\widetilde{e}_2\), \(\widetilde{f}_{12}\), and \(\widetilde{e}_{12}\) as equations (2.7) and (2.8) in Sect. 2. Let \(\epsilon _0\) be a scalar as Lemma 3.1. Then there exist \(\nabla _0\) and \(\Omega _0\in \mathbb {F}\) such that \(\widetilde{e}_1\widetilde{f}_{12}=\Omega _0v+\epsilon _0w_2\), and \(\widetilde{e}_2\widetilde{f}_{12}=\nabla _0v-\epsilon _0w_1\).

Proof

By Lemma 3.1, we have that \(\widetilde{e}_{12}\widetilde{f}_{12}=\epsilon _0w\). Let \(\widetilde{e}_i\widetilde{f}_{12}=A_iv+B_{i1}w_1+B_{i2}w_2+C_i v_{12}\) with \(A_i\), \(B_{i1}\), \(B_{i2}\), \(C_i\in \mathbb {F}\) for \(i=1,2\). Now, putting \(a=d=\widetilde{f}_i\), and \(b=c=e_j\) in the equation (2.2), we get that \(((\widetilde{f}_i\widetilde{e}_j)\widetilde{e}_j)\widetilde{f}_i-(\widetilde{f}_i\widetilde{e}_j)(\widetilde{f}_i\widetilde{e}_j)=0\) for \(i=1,2\). From this, it follows that \((\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{e}_j)\widetilde{f}_i+\mathfrak {r}(\widetilde{e}_j\widetilde{f}_{ij})\widetilde{f}_i=0\) and it is clear that \((\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{e}_j)\widetilde{f}_i=0\). Consequently, we conclude that \(\mathfrak {r}(\widetilde{e}_j\widetilde{f}_{ij})\widetilde{f}_i=0\), i.e., \((A_jv+B_{j1}w_1+B_{j2}w_2+C_iv_{12})\widetilde{f}_i=0\). Now, applying action on the last equation, we obtain that \(\pm B_{ii}v_{12}\pm C_iw_j=0\), and using the linear independence of \(w_j\) and \(v_{12}\), we get that \(B_{ii}=C_i=0\) for \(i=1,2\). So, we write \(\widetilde{e}_i\widetilde{f}_{12}=A_iv+B_{ij}w_j\). Further, substituting \(a=\widetilde{f}_i\), \(b=\widetilde{e}_i\), and \(c=d=\widetilde{f}_{ij}\) in identities (2.2) we obtain that \(((\widetilde{f}_i\widetilde{e}_i)\widetilde{f}_{ij})\widetilde{f}_{ij}-((\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_{ij})\widetilde{e}_i-((\widetilde{e}_i\widetilde{f}_{ij})\widetilde{f}_{ij})\widetilde{f}_i=0\). From this, it is clear to see that \(\mathfrak {r}(\widetilde{e}_j\widetilde{f}_{ij})\widetilde{e}_i-(\mathfrak {r}(\widetilde{e}_i\widetilde{f}_{ij})\widetilde{f}_{ij})\widetilde{f}_i=0\). Therefore, we obtain that \(B_{12}=-B_{21}\).

Finally, substituting \(a=\widetilde{e}_{12}\), \(b=\widetilde{e}_1\), \(c=\widetilde{f}_1\), and \(d=\widetilde{f}_{12}\) in the equation (2.2) and making the calculations, we obtain that \((\mathfrak {r}(\widetilde{e}_{12}\widetilde{f}_{12})\widetilde{f}_1)\widetilde{e}_1+(\mathfrak {r}(\widetilde{e}_1\widetilde{f}_{12})\widetilde{f}_1)\widetilde{e}_{12}=0\). Therefore, we conclude that \(\epsilon _0=B_{12}\), and the proof is complete. \(\square\)

Note that, the following identities (3.4)–(3.15) hold as a consequence of Lemma’s  3.1, 3.2, and 3.3.

$$\begin{aligned}&\mathfrak {r}(\widetilde{e}_i\widetilde{e}_j)\widetilde{e}_{ij}=0. \end{aligned}$$
(3.4)
$$\begin{aligned}&( \mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{f}_i)\widetilde{e}_i-\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_i=0\end{aligned}$$
(3.5)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{f}_i+\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_i-\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})=0.\end{aligned}$$
(3.6)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{e}_{ij})\widetilde{u}-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{u}-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_{ij}=0\end{aligned}$$
(3.7)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{u})\widetilde{f}_i+\mathfrak {r}(\widetilde{f}_{ij}\widetilde{u})\widetilde{f}_i=0 \end{aligned}$$
(3.8)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{f}_j+\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_j-\mathfrak {r}(\widetilde{f}_j\widetilde{e}_j)=0. \end{aligned}$$
(3.9)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{e}_i+\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_i+\mathfrak {r}(\widetilde{e}_i\widetilde{e}_j) -(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{e}_i)\widetilde{f}_i)\widetilde{u}-(\mathfrak {r}(\widetilde{e}_i\widetilde{u})\widetilde{f}_i)\widetilde{e}_{ij}=0. \end{aligned}$$
(3.10)
$$\begin{aligned}&\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_j-(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{e}_j)\widetilde{f}_i)\widetilde{u}-(\mathfrak {r}(\widetilde{e}_j\widetilde{u})\widetilde{f}_i)\widetilde{e}_{ij} +(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})+ \mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{f}_{ij}=0. \end{aligned}$$
(3.11)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{e}_i)\widetilde{f}_{ij}-2\mathfrak {r}(\widetilde{e}_{i}\widetilde{f}_{ij})\widetilde{f}_{ij}+\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{f}_i=0.\end{aligned}$$
(3.12)
$$\begin{aligned}&2\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_{ij}-2\mathfrak {r}(\widetilde{e}_{j}\widetilde{f}_{ij})-(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{f}_i)\widetilde{u}-(\mathfrak {r}(\widetilde{f}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{e}_{ij}+\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{e}_j=0.\end{aligned}$$
(3.13)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{u})\widetilde{u}-\mathfrak {r}(\widetilde{f}_{ij}\widetilde{u})\widetilde{u}-(\mathfrak {r}(\widetilde{e}_{j}\widetilde{u})\widetilde{u}) \widetilde{f}_i-\mathfrak {r}(\widetilde{f}_j\widetilde{u})\widetilde{f}_i+\mathfrak {r}(\widetilde{u}\widetilde{u})\widetilde{f}_{ij}=0.\end{aligned}$$
(3.14)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{u})\widetilde{u}-\mathfrak {r}(\widetilde{e}_j\widetilde{u})\widetilde{u}-\mathfrak {r}(\widetilde{f}_{j}\widetilde{u})-(\mathfrak {r}(\widetilde{f}_{i}&\widetilde{u})\widetilde{u})\widetilde{f}_{ij} \nonumber \\&-(\mathfrak {r}(\widetilde{f}_{ij}\widetilde{u})\widetilde{u})\widetilde{f}_i+\mathfrak {r}(\widetilde{u}\widetilde{u})\widetilde{e}_j=0. \end{aligned}$$
(3.15)

The proof of these identities is not completely presented. Identities (3.4)–(3.15) are obtained via appropriated replacements in the equation (2.2). The proofs of identities (3.4) and (3.6) are only sketched. First, putting \(a=\widetilde{f}_i\), \(b=\widetilde{f}_{ij}\), \(c=\widetilde{e}_i\), and \(d=\widetilde{e}_{ij}\) in the equation (2.2), we get \(((\widetilde{f}_i\widetilde{f}_{ij}\widetilde{e}_i)\widetilde{e}_{ij})+((\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{e}_i)\widetilde{f}_i-(\widetilde{f}_i\widetilde{f}_{ij})(\widetilde{e}_{ij}\widetilde{e}_i)-(\widetilde{e}_{ij}\widetilde{f}_{ij})(\widetilde{f}_i\widetilde{e}_i)=0\). An easy computation shows that \(\mathfrak {r}(\widetilde{e}_i\widetilde{e}_j)\widetilde{e}_{ij}=0\), which completes the proof of the identity (3.4). Now, taking \(a=\widetilde{e}_{ij}\), \(b=\widetilde{u}\), and \(c=d=\widetilde{f}_i\) in the equation (2.2) we obtain that \(((\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{f}_i+((\widetilde{f}_i\widetilde{u})\widetilde{f}_i)\widetilde{u}_{ij}-\widetilde{e}_{ij}\widetilde{u}=0\). By Lemma 3.1, we have that \((\mathfrak {r}(\widetilde{f}_i\widetilde{u})\widetilde{f}_i)\widetilde{e}_{ij}=0\), then \((\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{f}_i+\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_i-\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})=0\); i.e., identity (3.6) is proved. Similarly, identities (3.5), (3.7)–(3.15) are proved.

Lemma 3.4

Let \(\widetilde{1}\), \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\), \(\widetilde{u}, \widetilde{e}_1, \widetilde{e}_2, \widetilde{f}_{12}\) be as equations (2.7) and (2.8) in Sect. 2. Let \(\xi _0\), \(\alpha _0\), \(\delta _0\), \(\eta _0\), \(\tau _0\), \(\sigma _0\), \(\epsilon _0\), \(\xi _1\), \(\alpha _1\), \(\delta _2\), \(\Omega _0\), \(\nabla _0\), \(\Lambda _u\in \mathbb {F}\) as Lemmas 3.1, 3.2, and 3.3. Then, there exist scalars \(\beta _0\), \(\theta _0\), \(\varphi _0\), \(\lambda _0\), \(\gamma _0\), \(\chi _0\) such that the following equalities hold: symmetry products

$$\begin{aligned} \begin{aligned} \widetilde{f}_1\widetilde{f}_{12}=&-\widetilde{e}_2+\gamma _0w+\beta _0v_1+\delta _0v_2-\sigma _0w_{12},\\ \widetilde{f}_2\widetilde{f}_{12}=&\widetilde{e}_1+\lambda _0v-\alpha _0w_1-\theta _0w_2+\tau _0w_{12}, \\ \widetilde{e}_{12}\widetilde{u}=&\widetilde{f}_{12}+\varphi _0w+(\Omega _0+\alpha \tau _0)v_1+(\nabla _0+\alpha \sigma _0)v_2+\epsilon _0w_{12}, \\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+\beta _0w+\gamma _0v_1-(\nabla _0+\alpha \sigma _0)v_2-\epsilon _0w_{12},\\ \widetilde{f}_2\widetilde{e}_1=&\widetilde{f}_{12}+ \theta _0w+(\Omega _0+\alpha \tau _0)v_1-\lambda _0v_2+\epsilon _0w_{12}, \end{aligned} \end{aligned}$$

and skew-symmetric products

$$\begin{aligned} \begin{aligned} \widetilde{e}_1\widetilde{e}_2=&\widetilde{e}_{12}+\chi _0v+\tau _0w_1+\sigma _0w_2, \\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+ (\alpha (-\lambda _0-\nabla _0-\alpha \sigma _0)-\xi _0)v +\alpha _0w_1 +(\chi _0+\varphi _0+\beta _0)w_2-\alpha \tau _0v_{12}, \\ \widetilde{e}_2\widetilde{u}=&\widetilde{f}_2+(\alpha (\gamma _0+\Omega _0+\alpha \tau _0)-\eta _0)v-(\chi _0+\varphi _0 -\theta _0)w_1+\delta _0w_2-\alpha \sigma _0v_{12}, \\ \widetilde{f}_{12}\widetilde{u}=&\alpha (\beta _0-\theta _0+\varphi _0+\chi _0)v-(\Omega _0+\alpha \tau _0)w_1 +(\nabla _0+\alpha \sigma _0)w_2, \end{aligned} \end{aligned}$$

and \(\Lambda _u=0\), \(\delta _2=\gamma _0+\Omega _0+\alpha \tau _0\), \(\alpha _1=-\lambda _0-\nabla _0-\alpha \sigma _0\), \(\alpha \epsilon _0=0\).

Proof

Let \(\xi _i\), \(\alpha _i\) for \(i=0,1\); \(\delta _j\), \(\sigma _j\), \(\eta _j\) for \(j=0,2\); \(\varphi _k\) for \(k=0,1,2\); \(\epsilon _0\), \(\tau _0\), \(\sigma _0\), \(\Lambda _u\), \(\nabla _0\), and \(\Omega _0\) be as in Lemmas 3.1, 3.2, and 3.3.

Assume that there exist some scalars \(\beta _i\), \(\theta _i\), \(\gamma _i\), \(\lambda _i\), \(\Delta _i\), \(\kappa _i\), \(\rho _i\), \(\varphi _i\), and \(\chi _i\in \mathbb {F}\), \(i=0,1,2,12\) such that

$$\begin{aligned} \begin{aligned} \widetilde{f}_{12} \widetilde{u}=&\Delta _0v+\Delta _1w_1+\Delta _2w_2+\Delta _{12}v_{12},{} & {} \widetilde{e}_{12}\widetilde{u}=\widetilde{f}_{12}+\varphi _0w+\varphi _1v_1+\varphi _2v_2+\epsilon _0w_{12}\\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+\beta _0w+\beta _1v_1+\beta _2v_2+\beta _{12}w_{12},{} & {} \widetilde{f}_2\widetilde{e}_1=\widetilde{f}_{12}+\theta _0w+\theta _1v_1+\theta _2v_2+\theta _{12}w_{12}, \\ \widetilde{f}_1\widetilde{f}_{12}=&-\widetilde{e}_2+\gamma _0w+\gamma _1v_1+\gamma _2v_2-\sigma _0w_{12},{} & {} \widetilde{f}_2\widetilde{f}_{12}= \widetilde{e}_1+\lambda _0w+\lambda _1v_1+\lambda _2v_2+\tau _0w_{12},\\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+\rho _0v+\rho _1w_1+\rho _2w_2+\rho _{12}v_{12},{} & {} \widetilde{e}_2\widetilde{u}=\widetilde{f}_2+\kappa _0v+\kappa _1w_1+\kappa _2w_2+\kappa _{12}v_{12},\\ \widetilde{e}_1\widetilde{e}_2=&\widetilde{e}_{12}+\chi _0v+\chi _1w_1+\chi _2w_2+\chi _{12}v_{12}. \end{aligned} \end{aligned}$$

Now, we use identities (3.4)–(3.15) to complete our proof. By the identity (3.4), it is clear that \(\chi _{12}=0\). Now, considering the identity (3.5) we have that \(\chi _1=\tau _0\), \(\chi _2=\sigma _0\), \(\lambda _2=-\theta _0\), and \(\gamma _1=\beta _0\). Taking the identity (3.6) we have that \(\gamma _0=\beta _1\), \(\varphi _2=-\beta _2\), \(\lambda _0=-\theta _2\), \(\varphi _1=\theta _1\), and \(\epsilon _0=\theta _{12}=-\beta _{12}\). Note that the identity (3.7) implies that \(\alpha \epsilon _0=0\). Observe that if \(\alpha \ne 0\), then \(\epsilon _0=0\). Later, using the identity (3.8) we obtain that \(\Delta _i+\varphi _i=0,\) for \(i=1,2\), and \(\Delta _{12}=-\alpha \epsilon _0=0\). Combining these last two equalities, it is clear that \(\varphi _1=-\Delta _1=\theta _1\), and \(\varphi _2=-\Delta _2=-\beta _2\). Further, considering the identity (3.9), we get \(\delta _2=\gamma _0+\varphi _1\), \(\alpha _1=-\lambda _0-\varphi _2\), \(\gamma _2=\delta _0\), and \(\lambda _1=-\alpha _0\). Besides, taking the identity (3.10), we have that \(\rho _2=\chi _0+\varphi _0+\beta _0\), and \(\kappa _1=-(\chi _0+\varphi _0-\theta _0)\). Later, from the identity (3.11), it follows that \(\kappa _2=\gamma _2=\delta _0\), and \(\rho _1=-\lambda _1=\alpha _0\). Further, by the equation (3.13) we get that \(\varphi _1=\Omega _0+\alpha \tau _0\), and \(\varphi _2=\nabla _0+\alpha \sigma _0\). Thus, we conclude that \(\delta _2=\gamma _0+\Omega _0+\alpha \tau _0\), and \(\alpha _1=-\lambda _0-\nabla _0-\alpha \sigma _0\). Besides, using the identity (3.14), we obtain that \(\kappa _0=\alpha (\beta _1-\Delta _1)-\eta _0=\alpha (\gamma _0+\Omega _0+\alpha \tau _0)-\eta _0\), \(\rho _0=\alpha (\theta _2+\Delta _2)-\xi _0=\alpha (-\lambda _0-\nabla _0-\alpha \sigma _0)\), and \(\Delta _0=\alpha \beta _0-\alpha \kappa _1-\Lambda _u=\alpha \rho _2-\alpha \theta _0-\Lambda _u\). Thus, \(\Delta _0=\alpha (\chi _0+\varphi _0+\beta _0-\theta _0)-\Lambda _u\). Finally, since the identity (3.15) we have that \(\Delta _0=\alpha \gamma _1-\alpha \kappa _1+\Lambda _u\), \(\kappa _{12}=-\alpha \sigma _0\), and \(\rho _{12}=-\alpha \tau _0\). Note that \(\gamma _1=\beta _0\) and comparing this with the identity (3.14), we conclude \(\Lambda _u=0\), and the proof is complete. \(\square\)

Note that by Lemmas 3.1 and 3.4 we have proved that:

Theorem 3.5

Let \(\mathcal {A}\) be a finite-dimensional Jordan superalgebra with solvable radical \(\mathcal {N}\) such that \(\mathcal {N}\,^2=0\), \(\mathcal {A}/\mathcal {N}\cong \mathfrak {K}\textrm{an}(2)\), and \(\mathcal {N}\) is isomorphic to irreducible \(\mathfrak {K}\textrm{an}(2)\)-superbimodule \(\mathcal {V}(v,\alpha )\). Then there exists \(\widetilde{1}\in \mathcal {A}_0\) such that \(\widetilde{1}\) is a unity of \(\mathcal {A}\) and there exist elements \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\in \mathcal {A}_0\), and \(\widetilde{u}\), \(\widetilde{e}_1\), \(\widetilde{e}_2\), \(\widetilde{f}_{12}\in \mathcal {A}_1\) such that \(\textrm{alg}\langle \widetilde{1},\widetilde{f}_1, \widetilde{f}_2, \widetilde{e}_{12}\rangle \cong (\mathfrak {K}\textrm{an}(2))_0\) and \(\textrm{span} \langle \overline{\widetilde{u}}, \overline{\widetilde{e}_1}, \overline{\widetilde{e}_2}, \overline{\widetilde{f}}_{12} \rangle \cong (\mathfrak {K}\textrm{an}(2))_1\) where \(\overline{\widetilde{x}}\) is the image of \(\widetilde{x}\) under canonical homomorphism. Moreover, there exist scalars \(\xi _0\), \(\xi _1\), \(\alpha _0\), \(\delta _0\), \(\sigma _0\), \(\eta _0\), \(\tau _0\), \(\sigma _0\), \(\epsilon _0\) \(\beta _0\), \(\theta _0\), \(\varphi _0\), \(\lambda _0\), \(\gamma _0\), \(\chi _0\) such that \(\alpha \epsilon _0=0\) and the non-zero products are given by: symmetric products

$$\begin{aligned} \widetilde{f}_i\widetilde{f}_i=&1, \quad i=1,2; \\ \widetilde{f}_1\widetilde{u}=&\xi _0w+\xi _1v_1, \quad \widetilde{f}_2\widetilde{u}=\xi _0w+\xi _1v_2,\\ \widetilde{f}_1\widetilde{e}_1=&\alpha _0w+(-\lambda _0-\nabla _0-\alpha \sigma _0)v_1, \\ \widetilde{f}_2\widetilde{e}_2=&\delta _0w+(\gamma _0+\Omega _0+\alpha \tau _0)v_2,\\ \widetilde{e}_{12}\widetilde{e}_1=&\tau _0w,\quad \widetilde{e}_{12}\widetilde{e}_2=\sigma _0w, \quad \widetilde{e}_{12}\widetilde{f}_{12}=\epsilon _0w\\ \widetilde{f}_1\widetilde{f}_{12}=&-\widetilde{e}_2+\gamma _0w+\beta _0v_1+\delta _0v_2-\sigma _0w_{12},\\ \widetilde{f}_2\widetilde{f}_{12}=&\widetilde{e}_1+\lambda _0v-\alpha _0w_1-\theta _0w_2+\tau _0w_{12}, \\ \widetilde{e}_{12}\widetilde{u}=&\widetilde{f}_{12}+\varphi _0w+(\Omega _0+\alpha \tau _0)v_1+(\nabla _0+\alpha \sigma _0)v_2+\epsilon _0w_{12}, \\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+\beta _0w+\gamma _0v_1-(\nabla _0+\alpha \sigma _0)v_2-\epsilon _0w_{12},\\ \widetilde{f}_2\widetilde{e}_1=&\widetilde{f}_{12}+ \theta _0w+(\Omega _0+\alpha \tau _0)v_1-\lambda _0v_2+\epsilon _0w_{12}, \end{aligned}$$

and skew-symmetric products,

$$\begin{aligned} \widetilde{f}_{12}\widetilde{e}_1=&\Omega _0v+\epsilon _0w_2, \quad \widetilde{f}_{12}\widetilde{e}_2=\nabla _0v-\epsilon _0w_1\\ \widetilde{e}_1\widetilde{e}_2=&\widetilde{e}_{12}+\chi _0v+\tau _0w_1+\sigma _0w_2, \\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+ (\alpha (-\lambda _0-\nabla _0-\alpha \sigma _0)-\xi _0)v +\alpha _0w_1 +(\chi _0+\varphi _0+\beta _0)w_2-\alpha \tau _0v_{12}, \\ \widetilde{e}_2\widetilde{u}=&\widetilde{f}_2+(\alpha (\gamma _0+\Omega _0+\alpha \tau _0)-\eta _0)v-(\chi _0+\varphi _0 -\theta _0)w_1+\delta _0w_2-\alpha \sigma _0v_{12}, \\ \widetilde{f}_{12}\widetilde{u}=&\alpha (\beta _0-\theta _0+\varphi _0+\chi _0)v-(\Omega _0+\alpha \tau _0)w_1 +(\nabla _0+\alpha \sigma _0)w_2. \end{aligned}$$

3.2 Case 2

Assume that \(\mathcal {N}_0=\textrm{span}\langle w, v_1,v_2,w_{12}\rangle\) and \(\mathcal {N}_1=\textrm{span}\langle v, w_1,w_2,v_{12}\rangle\) such that \(\mathcal {N}\cong \mathcal {V}(v,\alpha )^{\textrm{op}}\). We present a series of lemmas to simplify the proof in this case.

Lemma 3.6

Let \(\widetilde{f}_i\), \(\widetilde{e}_i\), \(\widetilde{e}_{12}\), \(\widetilde{f}_{12}\), and \(\widetilde{u}\) be as Eqs. (2.7) and (2.8) in Sect. 2. Then there exist scalars \(\xi _2\), \(\xi _{12}\), \(\eta _{12}\), \(\alpha _2\), \(\alpha _{12}\), \(\delta _1\), \(\delta _{12}\), and \(\epsilon _0\) such that \(\widetilde{f}_1 \widetilde{u}=\xi _2 w_2+\xi _{12}v_{12}\), \(\widetilde{f}_2 \widetilde{u}=-\xi _2 w_1+\eta _{12}v_{12}\), \(\widetilde{f}_1\widetilde{e}_1=\alpha _2w_2+\alpha _{12} v_{12}\), \(\widetilde{f}_2\widetilde{e}_2=\delta _1w_1+\delta _{12}v_{12}\), \(\widetilde{e}_{12}\widetilde{e}_1=\alpha _2v\), \(\widetilde{e}_{12}\widetilde{e}_2=-\delta _1v\), and \(\widetilde{e}_{12}\widetilde{f}_{12}=\epsilon _0v\).

Proof

We can now proceed analogously to the proof of Lemma 3.1 using the same replacements. \(\square\)

Lemma 3.7

For all \(\widetilde{x}\in \mathcal {A}_1\), \(\widetilde{x}\widetilde{x}=0\).

Proof

Let \(\widetilde{x}\in \mathcal {A}_1\) be. Using the Eq. (2.2), it is easy to see that \(((\widetilde{x}\widetilde{x})\widetilde{f}_i)\widetilde{f}_i=\widetilde{x}\widetilde{x}\). Then, we get that \(((\widetilde{x}\widetilde{x})\widetilde{f}_i)\widetilde{e}_i=(\widetilde{x}\widetilde{x})(\widetilde{f}_i\widetilde{e}_i)\), and, we conclude that \(\widetilde{x}\widetilde{x}=0\). \(\square\)

By Lemmas 3.6 and 3.7, we are now in a position to show valid equations in \(\mathcal {A}\).

$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_j)\widetilde{f}_{i}+\mathfrak {r}(\widetilde{f}_j\widetilde{f}_{ij})\widetilde{f}_i+\mathfrak {r}(\widetilde{f}_i\widetilde{e}_i)=0. \end{aligned}$$
(3.16)
$$\begin{aligned}&( \mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{f}_i)\widetilde{e}_i-\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_i-(\mathfrak {r}(\widetilde{f}_i\widetilde{e}_i)\widetilde{f}_i)\widetilde{e}_j -(\mathfrak {r}(\widetilde{e}_i\widetilde{e}_j)\widetilde{f}_i)\widetilde{f}_i-2\mathfrak {r}(\widetilde{f}_i\widetilde{e}_i)\widetilde{f}_{ij}=0. \end{aligned}$$
(3.17)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{u})\widetilde{f}_i)\widetilde{e}_{ij}+(\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{f}_i +\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_i-\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})=0. \end{aligned}$$
(3.18)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{u})\widetilde{f}_i)\widetilde{e}_i-(\mathfrak {r}(\widetilde{e}_i\widetilde{u})\widetilde{f}_i)\widetilde{f}_i -(\mathfrak {r}(\widetilde{f}_i\widetilde{e}_i)\widetilde{f}_i)\widetilde{u}+\mathfrak {r}(\widetilde{e}_i\widetilde{u})=0. \end{aligned}$$
(3.19)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{e}_{j}\widetilde{u})\widetilde{f}_j)\widetilde{f}_{i}+(\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{f}_j)\widetilde{u}-\mathfrak {r}(\widetilde{f}_j\widetilde{f}_{ij})\widetilde{u}-\mathfrak {r}(\widetilde{e}_i\widetilde{u})-\mathfrak {r}(\widetilde{f}_j\widetilde{u})\widetilde{f}_{ij}=0. \end{aligned}$$
(3.20)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{u})\widetilde{f}_i)\widetilde{f}_{ij}-(\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_i)\widetilde{u}+\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{u}-(\mathfrak {r}(\widetilde{f}_{ij}\widetilde{u})\widetilde{f}_i)\widetilde{f}_i+2\mathfrak {r}(\widetilde{f}_i\widetilde{u})\widetilde{e}_j=0. \end{aligned}$$
(3.21)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{e}_{ij})\widetilde{u}-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{f}_{ij})\widetilde{u}+(\mathfrak {r}(\widetilde{e}_j\widetilde{u})\widetilde{e}_{ij})\widetilde{f}_i -\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)\widetilde{f}_{ij}-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{f}_{ij}=0 \end{aligned}$$
(3.22)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_i)\widetilde{u}+\mathfrak {r}(\widetilde{e}_i\widetilde{e}_j)\widetilde{u}+\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})-(\widetilde{f}_{ij}\widetilde{u})\widetilde{e}_i)\widetilde{f}_i-\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_i\nonumber \\&-\mathfrak {r}(\widetilde{e}_i\widetilde{u})\widetilde{e}_j+\mathfrak {r}(\widetilde{f}_i\widetilde{e}_j)=0 \end{aligned}$$
(3.23)
$$\begin{aligned}&(\mathfrak {r}(\widetilde{f}_{ij}\widetilde{u})\widetilde{e}_{ij})\widetilde{f}_i+(\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{e}_{ij})\widetilde{u}-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{e}_j)\widetilde{u}-\mathfrak {r}(\widetilde{f}_i\widetilde{f}_{ij})\widetilde{f}_{ij}\nonumber \\&-\mathfrak {r}(\widetilde{e}_{ij}\widetilde{u})\widetilde{e}_j-\mathfrak {r}(\widetilde{e}_j\widetilde{f}_{ij})=0 \end{aligned}$$
(3.24)

In the same manner as equalities (3.4)–(3.15) with suitable replacement in the identity (2.2), we can see those equations (3.16)–(3.24) hold in \(\mathcal {A}\).

Lemma 3.8

Let \(\widetilde{1}\), \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\), \(\widetilde{u}, \widetilde{e}_1, \widetilde{e}_2\), and \(\widetilde{f}_{12}\) be as relations eqrefpro: A1A1 and (2.8) in Sect. 2. Let \(\xi _2\), \(\xi _{12}\), \(\eta _{12}\), \(\alpha _2\), \(\alpha _{12}\), \(\delta _1\), \(\delta _{12}\), \(\epsilon _0\in \mathbb {F}\) be as Lemma 3.6. Then, there exist scalars \(\gamma _0\), \(\lambda _0\), \(\beta _{12}\), \(\theta _{12}\), \(\varphi _0\), \(\varphi _1\), \(\varphi _2\) such that the following equalities hold: with symmetric products

$$\begin{aligned} \begin{aligned} \widetilde{e}_{12}\widetilde{u}=&\widetilde{f}_{12}+\varphi _0v+\varphi _1w_1+\varphi _2w_2 + \epsilon _0v_{12},\\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+(\xi _2-\varphi _0)v-\varphi _1w_1-(\delta _1+\varphi _2)w_2+\beta _{12}v_{12},\\ \widetilde{f}_2\widetilde{e}_1=&\widetilde{f}_{12}+(-\xi _0+\varphi _0)v+(-\alpha _2+\varphi _1)w_1+\varphi _2w_2+\theta _{12}v_{12},\\ \widetilde{f}_1\widetilde{f}_{12}=&-\widetilde{e}_2+\gamma _0v+\delta _{12}w_1-\beta _{12}w_2+(\delta _1+\varphi _2)v_{12},\\ \widetilde{f}_2\widetilde{f}_{12}=&\widetilde{e}_1+\lambda _0v-\theta _{12}w_1+\alpha _{12}w_2+(\alpha _2-\varphi _1)v_{12}, \\ \end{aligned} \end{aligned}$$

and skew-symmetric products

$$\begin{aligned} \begin{aligned} \widetilde{e}_1\widetilde{f}_{12}=&(\alpha _2-\varphi _1)w-\alpha _{12}v_1-\theta _{12}v_2,\\ \widetilde{e}_2\widetilde{f}_{12}=&-(\delta _1+\varphi _1)w+\beta _{12}v_1-\delta _{12}v_2,\\ \widetilde{f}_{12}\widetilde{u}=&-\varphi _0w+(\xi _{12}-\alpha \varphi _1)v_1+(\eta _{12}-\alpha \varphi _2)v_2-\epsilon _0w_{12},\\ \widetilde{e}_1\widetilde{e}_2=&\widetilde{e}_{12}+(\alpha _{12}+\delta _{12})w+\alpha _2v_1+\delta _1v_2 \\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+(-\lambda _0+\eta _{12})w-\alpha (\epsilon _0-\theta _{12})v_1-(\epsilon _2+\alpha \alpha _{12})v_2-\alpha _2w_{12}, \\ \widetilde{e}_2\widetilde{u}=&\widetilde{f}_2+(-\gamma _0+\xi _{12})w+(\epsilon _2+\alpha \delta _{12})v_1-\alpha (\epsilon _0+\beta _{12})v_2+\delta _1w_{12}.\\ \end{aligned} \end{aligned}$$

Proof

Let \(\beta _i\), \(\theta _i\), \(\gamma _i\), \(\lambda _i\), \(\Delta _i\), \(\kappa _i\), \(\rho _i\), \(\varphi _i\), and \(\chi _i\in \mathbb {F}\), for \(i=0,1,2,12\) be such that

$$\begin{aligned} \begin{aligned} \widetilde{f}_{12} \widetilde{u}=&\Delta _0w+\Delta _1v_1+\Delta _2v_2+\Delta _{12}w_{12},{} & {} \widetilde{e}_{12}\widetilde{u}=\widetilde{f}_{12}+\varphi _0v+\varphi _1w_1+\varphi _2w_2+\epsilon _0v_{12}\\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+\beta _0v+\beta _1w_1+\beta _2w_2+\beta _{12}v_{12},{} & {} \widetilde{f}_2\widetilde{e}_1=\widetilde{f}_{12}+\theta _0v+\theta _1w_1+\theta _2w_2+\theta _{12}v_{12}, \\ \widetilde{f}_1\widetilde{f}_{12}=&-\widetilde{e}_2+\gamma _0v+\gamma _1w_1+\gamma _2w_2+\gamma _{12}v_{12},{} & {} \widetilde{f}_2\widetilde{f}_{12}= \widetilde{e}_1+\lambda _0v+\lambda _1w_1+\lambda _2w_2+\lambda _{12}v_{12},\\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+\rho _0w+\rho _1v_1+\rho _2v_2+\rho _{12}w_{12},{} & {} \widetilde{e}_2\widetilde{u}=\widetilde{f}_2+\kappa _0w+\kappa _1v_1+\kappa _2v_2+\kappa _{12}w_{12},\\ \widetilde{e}_1\widetilde{f}_{12}=&\Omega _0w+\Omega _1v_1+\Omega _2v_2+\Omega _{12}w_{12},{} & {} \widetilde{e}_2\widetilde{f}_{12}=\nabla _0w+\nabla _1v_1+\nabla _2v_2+\nabla _{12}w_{12}, \\ \widetilde{e}_1\widetilde{e}_2=&\widetilde{e}_{12}+\chi _0w+\chi _1v_1+\chi _2v_2+\chi _{12}w_{12}. \end{aligned} \end{aligned}$$

Using the identity (3.4) we get that \(\chi _{12}=0\). Later, considering the identity (3.16), we have that \(\lambda _2=\alpha _{12}\), and \(\gamma _1=\delta _{12}\). The following identities hold as a consequence of identity (3.17): \(\chi _0=\alpha _{12}+\delta _{12}\), \(\chi _1=\alpha _2\), \(\chi _2=\delta _1\), and \(\gamma _{12}=-\beta _2\), \(\lambda _{12}=-\theta _1\). Now, considering the Eq. (3.18) it follows that \(\beta _0=\xi _2-\varphi _0=-\theta _0\), \(\beta _1=-\varphi _1\), \(\gamma _2=\beta _{12}\), \(\theta _2=\varphi _2\), and \(\lambda _1=-\theta _{12}\). Later, taking the identity (3.19) we conclude that \(\rho _2=-\xi _2-\alpha \alpha _{12}\), \(\kappa _1=\xi _2+\alpha \delta _{12}\), \(\rho _{12}=-\alpha _2\), and \(\kappa _{12}=\delta _1\). On the other hand, the Eq. (3.20) implies that \(\kappa _0=-\xi _{12}+\gamma _0\), \(\rho _0=\eta _{12}-\lambda _0\), \(\rho _{12}=-\lambda _{12}+\beta _1\), and \(\kappa _{12}=-\theta _2+\gamma _{12}\) and, in consequence, \(\theta _1=-\alpha _2+\varphi _1\), and \(\beta _2=-\delta _1-\varphi _2\). Moreover, substituting \(a=\widetilde{f}_i\), \(b=c=\widetilde{u}\), and \(d=\widetilde{e}_j\) in the identity (2.2), we get that \(\rho _1=\alpha (\Delta _{12}+\theta _{12})\), and \(\kappa _2=\alpha (\Delta _{12}-\beta _{12})\). Our next claim is obtained from the Eq. (3.21) concluding that \(\Delta _0=-\varphi _0\), \(\Delta _1=\xi _{12}-\alpha \varphi _1\), and \(\Delta _2=\eta _{12}-\alpha \varphi _2\). Further, replacing \(a=\widetilde{f}_i\), \(b=\widetilde{u}\), \(c=\widetilde{e}_{ij}\), and \(d=\widetilde{e}_j\) in the Eq. (2.2) and rewriting this replacement, we obtain \(\varphi _{12}=\epsilon _0\). Besides, using the Eq. (3.23) we have that \(\Delta _{12}=-\epsilon _0\). Finally, taking the Eq. (3.24) we obtain that \(\Omega _{12}=\nabla _{12}=0\), \(\Omega _0=\alpha _2-\varphi _1\), \(\nabla _0=-(\delta _1+\varphi _2)\), \(\Omega _1=-\alpha _{12}\), \(\nabla _2=-\delta _{12}\), \(\Omega _2=-\theta _{12}\), and \(\nabla _1=\beta _{12}\), which completes the proof. \(\square\)

By Lemmas 3.6–3.8, we have proved the following theorem:

Theorem 3.9

Let \(\mathcal {A}\) be a finite-dimensional Jordan superalgebra with solvable radical \(\mathcal {N}\) such that \(\mathcal {N}\,^2=0\), \(\mathcal {A}/\mathcal {N}\cong \mathfrak {K}\textrm{an}(2)\), and \(\mathcal {N}\) is isomorphic to irreducible \(\mathfrak {K}\textrm{an}(2)\)-superbimodule \(\mathcal {V}(v,\alpha )^{\textrm{op}}\). Then there exists \(\widetilde{1}\in \mathcal {A}_0\) such that \(\widetilde{1}\) is a unity of \(\mathcal {A}\) and there exist elements \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\in \mathcal {A}_0\), and \(\widetilde{u}\), \(\widetilde{e}_1\), \(\widetilde{e}_2\), \(\widetilde{f}_{12}\in \mathcal {A}_1\) such that \(\textrm{alg}\langle \widetilde{1},\widetilde{f}_1, \widetilde{f}_2, \widetilde{e}_{12}\rangle \cong (\mathfrak {K}\textrm{an}(2))_0\) and \(\textrm{span} \langle \overline{\widetilde{u}}, \overline{\widetilde{e}_1}, \overline{\widetilde{e}_2}, \overline{\widetilde{f}}_{12} \rangle \cong (\mathfrak {K}\textrm{an}(2))_1\) where \(\overline{\widetilde{x}}\) is the image of \(\widetilde{x}\) under canonical homomorphism. Moreover, there exist scalars \(\xi _2\), \(\xi _{12}\), \(\eta _{12}\), \(\alpha _2\), \(\alpha _{12}\) \(\delta _1\), \(\delta _{12}\), \(\epsilon _0\), \(\gamma _0\), \(\lambda _0\), \(\beta _{12}\), \(\theta _{12}\), \(\varphi _0\), \(\varphi _1\), \(\varphi _2\), non-zero products are given by:symmetric products

$$\begin{aligned} \widetilde{f}_i\widetilde{f}_i=&1, \quad i=1,2; \\ \widetilde{f}_1\widetilde{u}=&\xi _2w_2+\xi _{12}v_{12}, \quad \widetilde{f}_2\widetilde{u}=-\xi _2w_1+\eta _{12}v_{12},\\ \widetilde{f}_1\widetilde{e}_1=&\alpha _2w_2+\alpha _{12}v_{12}, \quad \widetilde{f}_2\widetilde{e}_2=\delta _1w_1+\delta _{12}v_{12},\\ \widetilde{e}_{12}\widetilde{e}_1=&\alpha _2v,\quad \widetilde{e}_{12}\widetilde{e}_2=-\delta _1v, \quad \widetilde{e}_{12}\widetilde{f}_{12}=\epsilon _0v, \\ \widetilde{e}_{12}\widetilde{u}=&\widetilde{f}_{12}+\varphi _0v+\varphi _1w_1+\varphi _2w_2+\epsilon _0v_{12}, \\ \widetilde{f}_1\widetilde{e}_2=&-\widetilde{f}_{12}+(\epsilon _2-\varphi _0)v-\varphi _1w_1-(\delta _1+\varphi _2)w_2+\beta _{12}v_{12},\\ \widetilde{f}_2\widetilde{e}_1=&\widetilde{f}_{12}+(-\epsilon _0+\varphi _0)v+(-\alpha _2+\varphi _1)w_1+\varphi _2w_2+\theta _{12}v_{12},\\ \widetilde{f}_1\widetilde{f}_{12}=&-\widetilde{e}_2+\gamma _0v+\delta _{12}w_1-\beta _{12}w_2+(\delta _1+\varphi _2)v_{12}\\ \widetilde{f}_2\widetilde{f}_{12}=&\widetilde{e}_1+\lambda _0v-\theta _{12}w_1+\alpha _{12}w_2+(\alpha _2-\varphi _1)v_{12} \end{aligned}$$

and skew-symmetric products,

$$\begin{aligned} \widetilde{e}_1\widetilde{e}_2=&\widetilde{e}_{12}+(\alpha _{12}+\delta _{12})w+\alpha _2v_1+\delta _1v_2\\ \widetilde{e}_1\widetilde{u}=&\widetilde{f}_1+ (-\lambda _0+\eta _{12})w-\alpha (\epsilon _0-\theta _{12})v_1-(\epsilon _2+\alpha \alpha _{12})v_2-\alpha _2w_{12} \\ \widetilde{e}_2\widetilde{u}=&\widetilde{f}_2+(-\gamma _0+\xi _{12})w+(\epsilon _2+\alpha \delta _{12})v_1-\alpha (\epsilon _0+\beta _{12})v_2+\delta _1w_{12} \\ \widetilde{f}_{12}\widetilde{u}=&-\varphi _0w+(\xi _{12}-\alpha \varphi _1)v_1+(\eta _{12}-\alpha \varphi _2)v_2-\epsilon _0w_{12},\\ \widetilde{e}_1\widetilde{f}_{12}=&(\alpha _2-\varphi _1)w-\alpha _{12}v_1-\theta _{12}v_2, \\ \widetilde{e}_2\widetilde{f}_{12}=&-(\delta _1+\varphi _1)w+\beta _{12}v_1-\delta _{12}v_2. \end{aligned}$$

4 WPT for Jordan superalgebras of type \(\mathfrak {K}\textrm{an}(2)\)

In this section, We apply Theorems 3.5 and  3.9 proved in Sect. 3 to prove our main result.

Theorem 4.1

Let \(\mathcal {A}\) be a finite-dimensional Jordan superalgebra over an algebraically closed field, with solvable radical \(\mathcal {N}\) such that \(\mathcal {N}\,^2=0\) and \(\mathcal {A}/\mathcal {N}\cong \mathfrak {K}\textrm{an}(2)\). Then, there exists \(\mathcal {S}=\mathcal {S}_0\oplus \mathcal {S}_1\) subsuperalgebra of \(\mathcal {A}\) such that \(\mathcal {A}=\mathcal {N}\oplus \mathcal {S}\), and \(\mathcal {S}\cong \mathfrak {K}\textrm{an}(2)\), i.e., an analogous to WPT holds for \(\mathcal {A}\).

Proof

By results of the first author [1], the proof has two parts according to O. Folleco and I. Shestakov [11] results, we need to consider the cases \(\mathcal {N}\cong \mathcal {V}(v,\alpha )\), and \(\mathcal {N}\cong \mathcal {V}(v,\alpha )^{\textrm{op}}\).

Case 1. Let \(\mathcal {N}\) be an irreducible superbimodule over \(\mathfrak {K}\textrm{an}(2)\) such that \(\mathcal {N}\) is isomorphic to \(\mathcal {V}(v,\alpha )\). Assume that \(\widetilde{1}\), \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\), \(\widetilde{u}\), \(\widetilde{e}_1\), \(\widetilde{e}_2\), \(\widetilde{f}_{12}\in \mathcal {A}\) such that conditions of Theorem 3.5 hold. Let \(\hat{1}=\widetilde{1}\), \(\hat{f}_1=\widetilde{f}_1\), \(\hat{f}_2=\widetilde{f}_2\), and \(\hat{e}_{12}=\widetilde{e}_{12}\in \mathcal {A}_0\), and let \(\mathcal {S}_0=\textrm{alg}\langle \hat{1}, \hat{f}_1, \hat{f}_2, \hat{e}_{12}\rangle\), thus we have \(\mathcal {S}_0\cong (\mathfrak {K}\textrm{an}(2))_0\). Now, let \(\hat{u}=\widetilde{u}+\mathfrak {r}(\hat{u}),\) \(\hat{e}_1=\widetilde{e}_1+\mathfrak {r}(\hat{e}_1)\), \(\hat{e}_2\widetilde{e}_1+\mathfrak {r}(\hat{e}_2)\), and \(\hat{f}_{12}=\widetilde{f}_{12}+\mathfrak {r}(\hat{f}_{12}) \in \mathcal {A}_1\). From this, note that \(\textrm{span}\langle \overline{\hat{u}}, \overline{\hat{e}_1}, \overline{\hat{e}_2}, \overline{\hat{f}_{12}}\rangle \cong (\mathfrak {K}\textrm{an}(2))_1\).

Let A, \(B_i\), C, D, \(E_i\), F, G, \(H_i\), L, J, \(K_i\), \(M\in \mathbb {F}\) for \(i=1,2\) be such that \(\hat{u}=\widetilde{u}+Aw+B_1v_1+B_2v_2+Cw_{12}\), \(\hat{f}_{12}=\widetilde{f}_{12}+Dw+E_1v_1+E_2v_2+Fw_{12}\), \(\hat{e}_1=\widetilde{e}_1+Gw+H_1v_1+H_2v_2+L w_{12}\), and \(\hat{e}_2= \widetilde{e}_2+Jw+K_1v_1+K_2v_2+Mw_{12}\). Computing the products of elements \(\hat{x}\hat{y}\), we have the following equalities hold:

$$\begin{aligned} \begin{aligned} \hat{e}_1\hat{u}=&\widetilde{e}_1\widetilde{u}+(\alpha G-B_1)v+H_1w_1+(H_2-C)w_2+\alpha L v_{12},\\ \hat{e}_2\hat{u}=&\widetilde{e}_2\widetilde{u}+(\alpha J-B_2)v+(K_1+C)w_1+K_2w_2+\alpha M v_{12},\\ \hat{f}_{12}\hat{u}=&\widetilde{f}_{12}\widetilde{u}+\alpha (D-C)v+E_1w_1+E_2w_2+\alpha F v_{12},\\ \hat{e}_1\hat{e}_2=&\widetilde{e}_1\widetilde{e}_2+(H_2-K_1)v-Lw_1-Mw_2,\\ \hat{e}_1\hat{f}_{12}=&\widetilde{e}_1\widetilde{f}_{12}+(\alpha L-E_1)v-Fw_2,\\ \hat{e}_1\hat{f}_{12}=&\widetilde{e}_1\widetilde{f}_{12}+(\alpha M-E_2)v-Fw_1. \end{aligned} \end{aligned}$$
(4.1)

Now, by Theorem 3.5, we have that \(\hat{e}_1\hat{e}_2=\hat{e}_{12}\), \(\hat{e}_1\hat{u}=\hat{f}_1\), \(\hat{e}_2\hat{u}=\hat{f}_2\), and \(\hat{f}_{12}\hat{u}=\hat{f}_{12}\hat{e}_1=\hat{f}_{12}\hat{e}_2=0\) if and only if \(H_1=-\alpha _0\), \(K_2=-\delta _0\), \(L=\tau _0\), \(M=\sigma _0\), \(F=\epsilon _0\), \(E_1=\Omega _0+\alpha \tau _0\), \(E_2=-\nabla _0-\alpha \sigma _0\), and

$$\begin{aligned} \begin{aligned} \alpha (G-\lambda _0-\nabla _0-\alpha \sigma _0)-B_1-\xi _0=0,{} & {} \alpha (J+\gamma _0+\Omega _0+\alpha \tau _0)-\eta _0-B_2=0, \\ -(\chi _0+\varphi _0-\theta _0)+K_1+C=0,{} & {} \chi _0+\varphi _0+\beta _0+H_2-C=0,\\ \chi _0+H_2-K_1=0,{} & {} \alpha (\beta _0-\theta _0+\varphi _0+\chi _0+D-C)=0. \end{aligned} \end{aligned}$$
(4.2)

Hence, it is clear that \(2K_1=\chi _0-(\theta _0+\beta _0)\), \(2H_2=-(\theta _0+\beta _0+\chi _0)\), \(2C=\chi _0+\beta _0-\theta _0+2\varphi _0\), and \(\alpha (2D+(\chi _0+\beta _0-\theta _0))=0.\) Further, using Lemma 3.1, we have that \(\hat{f}_i\hat{u}=0\) if and only if \(A=-\xi _1\), \(B_1=\xi _0\), and \(B_2=\eta _0\); and \(\hat{f}_i\hat{e}_i=0\) if and only if \(J=-(\gamma _0+\Omega _0+\alpha \tau _0)\), and \(G=(\lambda _0+\nabla _0+\alpha \sigma _0)\). Observe that all these conditions are consistent with the first line of the equation (4.2).

It is easy to verify that \(\hat{e}_{12}\hat{e}_1=\hat{e}_{12}\hat{e}_1=\hat{e}_{12}\hat{f}_{2}=0\). To complete the proof in this case, we need to prove that \(\hat{e}_{12}\hat{u}=\hat{e}_1\hat{f}_2=-\hat{e}_2\hat{f}_1=\hat{f}_{12}\). A computation shows that

$$\begin{aligned} \begin{aligned} \hat{e}_{12}\hat{u}&=\widetilde{e}_{12}\widetilde{u}-Cw\\&=\widetilde{f}_{12}+\varphi _0w+(\Omega _0+\alpha \tau _0)v_1+(\nabla +\alpha \sigma _0)v_2+\epsilon _0w_{12} -(\frac{1}{2}(\chi _0+\beta _0-\theta _0)-\varphi _0)w\\&=\widetilde{f}_{12}+(\Omega _0+\alpha \tau _0)v_1+(\nabla +\alpha \sigma _0)v_2+\epsilon _0w_{12} -\frac{1}{2}(\chi _0+\beta _0-\theta _0)w\\&=\widetilde{f}_{12}+Dw+E_1v_1+E_2v_2+Fw_{12}\\&=\hat{f}_{12}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \hat{e}_{1}\hat{f}_2&=\widetilde{e}_{1}\widetilde{f}_2+Gv_2+H_2w\\&=\widetilde{f}_{12}+\theta _0w+(\Omega _0+\alpha \tau _0)v_1-\lambda _0v_2+\epsilon _0w_{12}+Gv_2+H_2w\\&=\hat{f}_{12}. \end{aligned} \end{aligned}$$

Similarly to the last computation, it is clear that \(\hat{e}_{1}\hat{f}_2=-\hat{f}_{12}\), \(\hat{f}_1\hat{f}_{12}=-\hat{e}_2\), and \(\hat{f}_2\hat{f}_{12}=\hat{e}_1\). Finally, taking \(\mathcal {S}_1=\textrm{alg}\langle \hat{u}, \hat{e}_1,\hat{e}_2, \hat{f}_{12}\rangle\), it follows that \(\mathcal {S}_0\oplus \mathcal {S}_1\cong \mathfrak {K}\textrm{an}(2)\), and \(\mathcal {A}=\mathcal {N}\oplus \mathcal {S}\), as a consequence, we have that an analogue to WPT holds for \(\mathcal {A}\) when \(\mathcal {N}\cong \mathcal {V}(v,\alpha )\), which completes the proof of the first case.

Case 2. Assume that \(\mathcal {N}\) is an irreducible bimodule isomorphic to \(\mathcal {N}\cong \mathcal {V}(v,\alpha )^{\textrm{op}}\), and let \(\widetilde{1}\), \(\widetilde{f}_1\), \(\widetilde{f}_2\), \(\widetilde{e}_{12}\), \(\widetilde{u}\), \(\widetilde{e}_1\), \(\widetilde{e}_2\), and \(\widetilde{f}_{12}\) be as Theorem 3.9. In the same manner, as in case 1, we can choose elements \(\hat{1}\), \(\hat{f}_1\), \(\hat{f}_2\), \(\hat{e}_{12}\in \mathcal {A}_0\) such that \(\mathcal {S}_0=\langle \hat{1}, \hat{f}_1, \hat{f}_2,he_{12}\rangle \cong (\mathfrak {K}\textrm{an}(2)_0)\). Taking elements \(\hat{u}\), \(\hat{e}_1\), \(\hat{e}_2\), \(\hat{f}_{12}\in \mathcal {A}_1\) such that

$$\begin{aligned} \hat{u}=&\widetilde{u}+Av-\eta _{12}w_1+\xi _{12}w_2+\xi _2v_{12},\\ \hat{f}_{12}=&\widetilde{f}_{12}+(\varphi _0-\xi _2)v+\varphi _1w_1+\varphi _2w_2+\epsilon _0v_{12}, \\ \hat{e}_1=&\widetilde{e}_1+\lambda _0v+(\epsilon _0-\theta _{12})w_1+\alpha _{12}w_2+\alpha _2v_{12},\\ \hat{e}_2=&\widetilde{e}_2-\gamma _0v-\delta _{12}w_1+(\epsilon _0+\beta _{12})w_2-\delta _1v_{12}. \end{aligned}$$

It is easy to see that \(\mathcal {S}_1=\textrm{vec}\langle \hat{u}, \hat{e}_1, \hat{e}_2, \hat{f}_{12}\rangle \cong (\mathfrak {K}\textrm{an}(2))_1\), and consequently, we conclude that there exists a subsuperalgebra \(\mathcal {S}\) in \(\mathcal {A}\) such that \(\mathcal {S}=\mathcal {S}_0\oplus \mathcal {S}_1\cong \mathfrak {K}\textrm{an}(2)\) and \(\mathcal {A}\cong \mathcal {S}\oplus \mathcal {N}\), which completes the proof. \(\square\)

Observe that proof of the Theorem 4.1 is independent of \(\alpha\).

Let us mention one important consequence of the Theorem 4. This result implies that the second cohomology group (SCG)

\(\mathcal {H}^2(\mathfrak {K}\textrm{an}(2),\mathcal {V}(v,\alpha ))=0\), and \(\mathcal {H}^2(\mathfrak {K}\textrm{an}(2),\mathcal {V}(v,\alpha )^{\textrm{op}})=0\).

The study of nontrivial SCG for finite-dimensional Jordan superalgebra was made by the authors in [12].