On Generalizations of Jacobi–Jordan Algebras

Abstract

In this paper, we present some generalizations of Jacobi–Jordan algebras. More concretely, we will focus on noncommutative Jacobi–Jordan algebras, Malcev–Jordan algebras, and general Jacobi–Jordan algebras. We adapt a method, used to classify Poisson algebras, in order to classify all general Jacobi–Jordan algebras up to dimension 4, and, in particular, all noncommutative Jacobi–Jordan algebras up to dimension 4. We present the classification of Malcev–Jordan algebras up to dimension 5. As the class of Jacobi–Jordan algebras (commutative algebras that satisfy the Jacobi identity), we find that Malcev–Jordan algebras are Jordan algebras but not necessarily nilpotent. However, we show that the classification of nilpotent Malcev–Jordan algebras is sufficient to obtain the classification of the whole class.

1. Introduction

The classification, up to isomorphism, of any class of algebras is a fundamental and very difficult problem. It is one of the most interesting problems in current research which is receiving steady attention, being a major contributor of examples. Having a healthy stock of examples not only makes the theory more tangible but also helps us to better understand the difficulties of the theory. The classification may also contribute to counterexamples of some conjectures. For instance, the long-standing conjecture, originally proposed by Albert [1] in 1948, that a commutative power-associative nilalgebra is nilpotent was disproved by Suttles [2] in 1972, who gave a counterexample of dimension 5. The classification may also give answers to some open questions. For instance, in 1954, Jacobson [3] proved that if a Lie algebra admits an invertible derivation, it must be nilpotent. Jacobson also noted that the validity of the converse was an open question. In 1956, Dixmier and Lister [4] gave a strong negative answer to this question by presenting a nilpotent Lie algebra of dimension 8 admitting only nilpotent derivations.

Lie algebras are anticommutative algebras which satisfy Jacobi identity. In [5], Jacobi–Jordan algebras as commutative algebras that satisfy Jacobi identity instead of associativity were presented. It turns out that they are a special class of Jordan nilalgebras. Those algebras are also related to Bernstein–Jordan algebras. Although they seem to be similar to Lie algebras at first sight, these algebras are quite different.

The goal of this paper is to study and present the classification of more than one generalization of the Jacobi–Jordan algebra in low dimensions. This is achieved by weakening either the commutativity condition or the Jacobi identity, or both conditions; more specifically, we deal with noncommutative Jacobi–Jordan algebras, Malcev–Jordan algebras, and general Jacobi–Jordan algebras. The classes presented here are special types of well-known algebras named Jordan algebras and noncommutative Jordan algebras. The class of noncommutative Jordan algebras is extremely large. It contains all Jordan and alternative algebras, as well as all anticommutative algebras.

A noncommutative Jacobi–Jordan algebra is a flexible algebra $\mathcal{A}$ which satisfies the Jacobi identity $J(x,y,z)=0,$ for any $x,y,z\in \mathcal{A}$. The class of noncommutative Jacobi–Jordan algebras includes Lie algebras and Jacobi–Jordan algebras. On the other hand, a general Jacobi–Jordan algebra is a flexible algebra $\mathcal{J}$ which satisfies $J(x,y,y)=0,$ for any $x,y\in \mathcal{J}$. The class of general Jacobi–Jordan algebras contains the previous class of noncommutative Jacobi–Jordan algebras, as well as anticommutative algebras (Malcev algebras, Binary Lie algebras, etc.).

We classify all general Jacobi–Jordan algebras up to dimension 4 by using a classification method similar to the one for Poisson algebras (cf. [6]). In particular, we also manage to obtain the classification of all noncommutative Jacobi–Jordan algebras up to dimension 4. The idea is to use a suitable bijective correspondence between general Jacobi–Jordan algebras and Jacobi–Jordan Poisson algebras. In fact, we apply a procedure to classify the Jacobi–Jordan Poisson algebras associated with a given Jacobi–Jordan algebra.

A Malcev–Jordan algebra is a commutative algebra $\mathcal{A}$ which satisfies $J(x,y,x·z)=J(x,y,z)·x,$ for $x,y,z\in \mathcal{A}$. It is clear that the class of Malcev–Jordan algebras is significantly wider than that of Jacobi–Jordan algebras, and therefore, Malcev–Jordan algebras are not necessarily associative. The variety of Malcev–Jordan algebras is between that of commutative associative algebras and Jordan algebras. It contains all commutative associative algebras, and therefore, Malcev–Jordan algebras are not necessarily nilpotent, although Jacobi–Jordan algebras are nilpotent. We present the classification of Malcev–Jordan algebras up to dimension 5. In fact, although Malcev–Jordan algebras are not necessarily nilpotent, we only have to classify the nilpotent Malcev–Jordan algebras to obtain the classification of the whole class. Finally, the natural problem of describing those nilpotent Jordan algebras of dimension 6 which lie beyond the limits of the narrower variety of Malcev–Jordan algebras is also addressed.

This paper is organized as follows. Section 2 contains basic notions needed in the following sections. In Section 3, we present noncommutative Jacobi–Jordan algebras and give some of their properties, and in the following section, we study general Jacobi–Jordan algebras. In Section 5, we classify all general Jacobi–Jordan algebras up to dimension 4, and we give a classification of noncommutative Jacobi–Jordan algebras up to dimension 4. In Section 6, we recall basic concepts of Jordan algebras. The last section is devoted to Malcev–Jordan algebras. The structure of this class of algebras is presented and the classification up to dimension 5 is given.

Throughout this work, we assume all vector spaces and algebras are finite-dimensional and defined over an algebraically closed field $\mathbb{F}$ with characteristic dimensions of 2 and 3.

2. Basic Definitions

An algebra A is called commutative (resp., anticommutative) if $x·y=y·x$ (resp., $x·y=-y·x$), for any $x,y\in A$. Recall that the associator of $x,y,z\in A$ is given by

$$(x,y,z):=(x·y)·z-x·(y·z).$$

(1)

An algebra A is called associative if

$$(x,y,z)=0,$$

(2)

for any $x,y,z\in A$. Also, it is said to be flexible if it satisfies $(x,y,x)=0$, for any $x,y\in A$.

Commutative algebras and anticommutative algebras are flexible. Thus, the flexibility condition is a weaker condition than commutativity and anticommutativity. The Jacobian of $x,y,z$ is given by

$$J(x,y,z):=x·(y·z)+y·(z·x)+z·(x·y),$$

(3)

for any $x,y,z\in A$.

An algebra A is called power-associative if the subalgebra generated by any element x in A is associative. A power-associative algebra A is a nilalgebra (with a nilindex of n) if there exists $n\in \mathbb{N}$ such that $x^n=0$ for any $x\in A$, and there exists $y\in A$ with $y^{n-1}\ne 0$.

We introduce the following sequences of subspaces:

$$\begin{array}{ccc}{A}^{\langle 1\rangle }:=A,\hfill & & A^{\langle n+1\rangle }:=A^{\langle n\rangle }·A;\hfill \\ A^1:=A,\hfill & & A^{n+1}:={\displaystyle \sum _{i=1}^n}{A}^i·A^{n+1-i},\hfill \end{array}$$

for $n\ge 1$. An algebra A is called the following:

• Right nilpotent if there exists $n_0\in \mathbb{N}$ such that $A^{\langle n_0\rangle }=0$;
• Nilpotent if there exists $n_0\in \mathbb{N}$ such that $A^{n_0}=0$.

Note that any commutative (anticommutative) algebra is right nilpotent if and only if it is nilpotent (see (Proposition 1 in [7])).

The annihilator of an algebra $\left(A,·\right)$ is the ideal $Ann\left(A\right):=\left\{x\in A:x·A=A·x=0\right\}$.

Lemma 1.

Let $A\ne 0$ be an algebra.

i. 

If A is nilpotent, then $Ann\left(A\right)\ne 0$.

ii. 

A is nilpotent if and only if $A/Ann\left(A\right)$ is nilpotent.

Let $\left(A,{·}_A\right)$ and $\left(B,{·}_B\right)$ be two algebras. A linear map $\phi :A\to B$ is called a homomorphism of algebras if

$$\phi \left(x{·}_{A}y\right)=\phi \left(x\right){·}_B\phi \left(y\right),$$

for $x,y\in A.$ We say that a homomorphism $\varphi :A\to B$ is an isomorphism if it is bijective. In this case, we call A and B isomorphic and denote $A\cong B$. An automorphism is a bijective homomorphism from A to itself. The set of automorphisms of the algebra $(A,·)$ is denoted by $\mathrm{Aut}\left(A\right)$.

We consider the following two new products on the underlying vector space A defined by

$$x\circ y:={\displaystyle \frac{1}{2}}(x·y+y·x),[x,y]:={\displaystyle \frac{1}{2}}(x·y-y·x),$$

(4)

for $x,y\in A.$ Let us denote $A^{+}:=(A,\circ )$ and $A^{-}:=(A,[·,·])$, named plus algebra and minus algebra, respectively. In the algebra $A^{+}$, the Jacobian of $x,y,z$ is given by

$$J_{A^{+}}(x,y,z):=x\circ \left(y\circ z\right)+y\circ \left(z\circ x\right)+z\circ \left(x\circ y\right),$$

(5)

for $x,y,z\in A$.

Lemma 2.

Let A and B be two algebras. If $A\cong B$, then $A^{+}\cong B^{+}$ and $A^{-}\cong B^{-}$.

3. Noncommutative Jacobi–Jordan Algebras

The aim of this section is to study the class of noncommutative Jacobi–Jordan algebras. We recall that a commutative algebra A is called a Jacobi–Jordan algebra if $J(x,y,z)=0,$ for all $x,y,z\in A$.

Definition 1.

An algebra $(\mathcal{A},·)$ is called a noncommutative Jacobi–Jordan algebra if it satisfies the following two identities:

$$(x·y)·x-x·(y·x)=0,$$

(6)

$$J(x,y,z)=x·(y·z)+y·(z·x)+z·(x·y)=0,$$

(7)

for any $x,y,z\in \mathcal{A}$.

In other words, noncommutative Jacobi–Jordan algebras are flexible algebras which satisfy the Jacobi identity. Every Lie algebra is a noncommutative Jacobi–Jordan algebra. Also, any Jacobi–Jordan algebra is a noncommutative Jacobi–Jordan algebra.

Definition 2.

An algebra $\mathcal{A}$ is called a Jacobi–Jordan admissible algebra if ${\mathcal{A}}^{+}$ is a Jacobi–Jordan algebra.

Lemma 3.

If $\mathcal{A}$ is a noncommutative Jacobi–Jordan algebra, then ${\mathcal{A}}^{+}$ is a Jacobi–Jordan algebra (so, $\mathcal{A}$ is a Jacobi–Jordan admissible algebra).

Proof. 

For $x,y,z\in \mathcal{A}$, we have

$$\begin{array}{ccc}\hfill J_{{\mathcal{A}}^{+}}(x,y,z)& =& x\circ (y\circ z)+y\circ (z\circ x)+z\circ (x\circ y)\hfill \\ & =& {\displaystyle \frac{1}{4}}(x·(y·z+z·y)+(y·z+z·y)·x+y·(z·x+x·z)\hfill \\ & & +(z·x+x·z)·y+z·(x·y+y·x)+(x·y+y·x)·z)\hfill \\ & =& {\displaystyle \frac{1}{4}}(J(x,y,z)+J(x,z,y)+(y·z+z·y)·x+(z·x+x·z)·y+(x·y+y·x)·z),\hfill \end{array}$$

and since $\mathcal{A}$ is flexible, we have

$$\left(y·z+z·y\right)·x+\left(z·x+x·z\right)·y+\left(x·y+y·x\right)·z=x·\left(y·z+z·y\right)+y·\left(z·x+x·z\right)+z·\left(x·y+y·x\right).$$

Hence,

$$J_{{\mathcal{A}}^{+}}(x,y,z)={\displaystyle \frac{1}{2}}(J(x,y,z)+J(x,z,y)),$$

(8)

and by (7), we conclude that ${\mathcal{A}}^{+}$ is a Jacobi–Jordan algebra. □

We recall that any flexible algebra (possibly commutative) $\mathcal{A}$ satisfying the Jordan identity (that is, $(x^2,y,x)=0$ for $x,y\in \mathcal{A}$) is a noncommutative Jordan algebra.

Lemma 4.

Any noncommutative Jacobi–Jordan algebra is a noncommutative Jordan algebra.

Proof. 

Suppose $\mathcal{A}$ is a noncommutative Jacobi–Jordan algebra; then, by Lemma 3, ${\mathcal{A}}^{+}$ is a Jacobi–Jordan algebra. Since any Jacobi–Jordan algebra is a Jordan algebra, then ${\mathcal{A}}^{+}$ is a Jordan algebra. Further, it is known that any flexible algebra is a noncommutative Jordan algebra if and only if the corresponding plus algebra is a Jordan algebra (see [8]). Therefore, $\mathcal{A}$ is flexible, so it is a noncommutative Jacobi–Jordan algebra, and therefore, $\mathcal{A}$ is a noncommutative Jordan algebra. □

Corollary 1.

Any noncommutative Jacobi–Jordan algebra is power-associative (that is, the subalgebra generated by any element is associative).

Lemma 5.

Any noncommutative Jacobi–Jordan algebra is a nilalgebra with a nilindex of at most 3.

Remark 1.

Let $\mathcal{A}$ be a noncommutative Jacobi–Jordan algebra. By Lemma 5, $\mathcal{A}$ is a nilalgebra with a nilindex of at most 3, and therefore, ${\mathcal{A}}^{+}$ is a commutative nilalgebra with a nilindex of at most 3. Since the class of Jacobi–Jordan algebras and commutative nilalgebras with a nilindex of at most 3 coincide, we conclude that ${\mathcal{A}}^{+}$ is a Jacobi–Jordan algebra. This is another proof of Lemma 3.

We reiterate that an algebra A is simple if $A^2\ne 0$ and its unique ideals are 0 and A.

Lemma 6.

Any simple noncommutative Jacobi–Jordan algebra is a Lie algebra.

Proof. 

Let $\mathcal{A}$ be a simple noncommutative Jacobi–Jordan algebra. It is known that any finite-dimensional simple noncommutative Jordan nilalgebra with a characteristic difference of 2 is anticommutative (see [9]). So, since $\mathcal{A}$ is simple, then it is an anticommutative algebra satisfying the Jacobi identity. □

Remark 2.

In general, the converse of Lemmas 3 and 5 is not true. For instance, the 7-dimensional simple non-Lie Malcev algebra is a simple noncommutative Jordan nilalgebra but not a noncommutative Jacobi–Jordan algebra because noncommutative Jacobi–Jordan algebras are Lie algebras if they are anticommutative.

Proposition 1.

Any noncommutative Jacobi–Jordan algebra $\mathcal{A}$ is simple if and only if ${\mathcal{A}}^{-}$ is simple.

Proof. 

By Lemma 6, if $\mathcal{A}$ is simple, then ${\mathcal{A}}^{-}$ is also simple. To prove the converse, we consider a non-zero ideal I of $\mathcal{A}$. For $x\in \mathcal{A}$ and $y\in I$, we have $[x,y]={\displaystyle \frac{1}{2}}(x·y-y·x)\in I$, so I is also an ideal of ${\mathcal{A}}^{-}$. Thus, if ${\mathcal{A}}^{-}$ is simple, then so is $\mathcal{A}$. □

The next result gives the classification of a noncommutative Jacobi–Jordan algebra $\mathcal{A}$ such that ${\mathcal{A}}^{-}$ is simple.

Corollary 2.

Let $\mathcal{A}$ be a noncommutative Jacobi–Jordan algebra. If ${\mathcal{A}}^{-}$ is simple, then $\mathcal{A}$ is anticommutative (i.e., $[x,y]=x·y$ and $x\circ y=0$, for $x,y\in \mathcal{A}$).

4. General Jacobi–Jordan Algebras

In this section, we introduce the general Jacobi–Jordan algebras and study some of their properties.

Definition 3.

An algebra $(\mathcal{J},·)$ is said to be a general Jacobi–Jordan algebra if it satisfies the following two identities:

$$(x·y)·x-x·(y·x)=0,$$

(9)

$$J(x,y,y)=x·y^2+y·(y·x)+y·(x·y)=0,$$

(10)

for any $x,y\in \mathcal{J}$.

Equivalently, general Jacobi–Jordan algebras are flexible algebras which satisfy $J(x,y,y)=0,$ for any $x,y\in \mathcal{J}$.

Example 1.

Every noncommutative Jacobi–Jordan algebra is a general Jacobi–Jordan algebra. Therefore, Lie algebras and Jacobi–Jordan algebras are general Jacobi–Jordan algebras. Any anticommutative algebra is a general Jacobi–Jordan algebra and thus Malcev algebras and Binary Lie algebras are general Jacobi–Jordan algebras.

General Jacobi–Jordan algebras can be characterized as follows:

Proposition 2.

A flexible algebra $\mathcal{J}$ is a general Jacobi–Jordan algebra if and only if ${\mathcal{J}}^{+}$ is a Jacobi–Jordan algebra.

Proof. 

As a consequence of (8), we have $J_{{\mathcal{J}}^{+}}(x,y,z)=0$ if and only if $J(x,y,z)+J(x,z,y)=0$ for any $x,y,z\in \mathcal{J}$. Since the last identity is a linearization of (10), the proof is finished. □

The previous result can be rephrased as follows:

Corollary 3.

A flexible algebra is a general Jacobi–Jordan algebra if and only if it is a Jacobi–Jordan admissible.

Lemma 7.

Any general Jacobi–Jordan algebra is a noncommutative Jordan algebra.

Proof. 

Since any general Jacobi–Jordan algebra is Jacobi–Jordan admissible, then it is also a Jordan admissible and therefore a noncommutative Jordan algebra. □

Corollary 4.

Any general Jacobi–Jordan algebra is power-associative.

Lemma 8.

Any general Jacobi–Jordan algebra is a nilalgebra with a nilindex of at most 3.

Lemma 9.

Any simple general Jacobi–Jordan algebra is anticommutative.

Proof. 

Let $\mathcal{J}$ be a simple general Jacobi–Jordan algebra. By Lemmas 7 and 8, $\mathcal{J}$ is a simple noncommutative Jordan nilalgebra. Since any simple noncommutative Jordan nilalgebra is anticommutative (see [9]), $\mathcal{J}$ is anticommutative. □

Proposition 3.

Let $\left(\mathcal{J},·\right)$ be a general Jacobi–Jordan algebra. Then, $\mathcal{J}$ is simple if and only if ${\mathcal{J}}^{-}$ is simple.

The next result also characterizes general Jacobi–Jordan algebras:

Proposition 4.

The class of general Jacobi–Jordan algebras and flexible nilalgebras with a nilindex of at most 3 coincide.

Proof. 

By Lemma 8, if $\mathcal{J}$ is a general Jacobi–Jordan algebra, then $\mathcal{J}$ is a flexible nilalgebra with a nilindex of at most 3. Conversely, let $\mathcal{J}$ be a flexible nilalgebra with a nilindex of at most 3. Then, for any $x,y$ in $\mathcal{J}$, we have

$$\begin{array}{ccc}\hfill J\left(x,y,y\right)+J\left(y,x,x\right)& =& x·y^2+y·\left(y·x\right)+y·\left(x·y\right)+y·x^2+x·\left(x·y\right)+x·\left(y·x\right)\hfill \\ & =& \left(x+y\right)·\left(x^2+x·y+y·x+y^2\right)\hfill \\ & =& \left(x+y\right)·{\left(x+y\right)}^2\hfill \\ & =& {\left(x+y\right)}^3\hfill \\ & =& 0,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill J\left(x,y,y\right)-J\left(y,x,x\right)& =& x·y^2+y·\left(y·x\right)+y·\left(x·y\right)-y·x^2-x·\left(x·y\right)-x·\left(y·x\right)\hfill \\ & =& \left(x-y\right)·\left(x^2-x·y-y·x+y^2\right)\hfill \\ & =& \left(x-y\right)·{\left(x-y\right)}^2\hfill \\ & =& {\left(x-y\right)}^3\hfill \\ & =& 0.\hfill \end{array}$$

Hence, $2J\left(x,y,y\right)=0$ for any $x,y\in \mathcal{J}$, i.e., $\mathcal{J}$ is a general Jacobi–Jordan algebra as required. □

We can infer several consequences from Proposition 4.

Corollary 5.

The class of general Jacobi–Jordan algebras and noncommutative Jordan nilalgebras with a nilindex of at most 3 coincide.

Corollary 6.

Any flexible nilalgebra with a nilindex of at most 3 is a noncommutative Jordan algebra.

Corollary 7.

The class of commutative general Jacobi–Jordan algebras and Jacobi–Jordan algebras coincide.

5. The Classification of General Jacobi–Jordan Algebras

In this section, we classify all general Jacobi–Jordan algebras up to dimension 4 by using a classification method similar to the one for Poisson algebras (see [6]). The idea is to use a suitable bijective correspondence between Jacobi–Jordan algebras and Jacobi–Jordan Poisson algebras. Indeed, we apply a procedure to classify the Jacobi–Jordan Poisson algebras associated with a given Jacobi–Jordan algebra.

Definition 4.

A Jacobi–Jordan Poisson algebra is a vector space $\mathcal{A}$ over an arbitrary field $\mathbb{F}$ equipped with two bilinear operations:

• A Jacobi–Jordan multiplication denoted by $\circ :\mathcal{A}\times \mathcal{A}\to \mathcal{A}$;
• An anticommutative multiplication denoted by $[·,·]:\mathcal{A}\times \mathcal{A}\to \mathcal{A}$.

These two operations are compatible in the sense that they satisfy the following Leibniz identity

$$[x\circ y,z]=[x,z]\circ y+x\circ [y,z],$$

for any $x,y,z\in \mathcal{A}$.

Proposition 5.

Consider an algebra $(\mathcal{A},·)$ whose underlying vector space $\mathcal{A}$ is endowed with the two new products ∘ and $[·,·]$ defined as in (4). Then, $(\mathcal{A},·)$ is a general Jacobi–Jordan algebra if and only if $(\mathcal{A},\circ ,[·,·\left]\right)$ is a Jacobi–Jordan Poisson algebra.

Proof. 

Since the flexible law is equivalent to $[x\circ y,z]=x\circ [y,z]+y\circ [x,z]$ for $x,y,z\in \mathcal{A}$, the proof follows immediately from Proposition 2. □

Definition 5.

Let $({\mathcal{A}}_1,{\circ }_1,{[·,·]}_1)$ and $({\mathcal{A}}_2,{\circ }_2,{[·,·]}_2)$ be two Jacobi–Jordan Poisson algebras. A Jacobi–Jordan Poisson algebra homomorphism (or just homomorphism, when the context is clear), is a linear map $\varphi :{\mathcal{A}}_1\to {\mathcal{A}}_2$ preserving the products, that is,

$$\varphi \left(x{\circ }_{1}y\right)=\varphi \left(x\right){\circ }_2\varphi \left(y\right),\varphi \left({[x,y]}_1\right)={[\varphi \left(x\right),\varphi \left(y\right)]}_2,$$

for all $x,y\in {\mathcal{A}}_1$.

Let $({\mathcal{A}}_1,{·}_1)$ and $({\mathcal{A}}_2,{·}_2)$ be two general Jacobi–Jordan algebras and let $({\mathcal{A}}_1,{\circ }_1,{[·,·]}_1)$ and $({\mathcal{A}}_2,{\circ }_2,{[·,·]}_2)$ be their associated Jacobi–Jordan Poisson algebras, respectively (see Proposition 5). If $({\mathcal{A}}_1,{·}_1)$ and $({\mathcal{A}}_2,{·}_2)$ are isomorphic, then the Jacobi–Jordan Poisson algebras $({\mathcal{A}}_1,{\circ }_1,{[·,·]}_1)$ and $({\mathcal{A}}_2,{\circ }_2,{[·,·]}_2)$ are isomorphic. Conversely, we can show that if the Jacobi–Jordan Poisson algebras $({\mathcal{A}}_1,{\circ }_1,{[·,·]}_1)$ and $({\mathcal{A}}_2,{\circ }_2,{[·,·]}_2)$ are isomorphic, then the general Jacobi–Jordan algebras $({\mathcal{A}}_1,{·}_1)$ and $({\mathcal{A}}_2,{·}_2)$ are isomorphic. So, we have the following result:

Proposition 6.

Every Jacobi–Jordan Poisson algebra $(\mathcal{A},\circ ,[·,·\left]\right)$ is associated with precisely one general Jacobi–Jordan algebra $(\mathcal{A},·)$ defined in Proposition 5. That is, we have a bijective correspondence between general Jacobi–Jordan algebras and Jacobi–Jordan Poisson algebras.

5.1. The Classification Method

(The present method, in the case of Poisson algebras, is described in more detail in [6]). Given an arbitrary Jacobi–Jordan algebra, we may consider all the Jacobi–Jordan Poisson structures defined over this algebra. This notion is captured in the following definition.

Definition 6.

Let $(\mathcal{A},\circ )$ be a Jacobi–Jordan algebra. Let $Z^2(\mathcal{A},\mathcal{A})$ be the set of all skew-symmetric bilinear maps $\theta :\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ such that

$$\theta (x\circ y,z)-\theta (x,z)\circ y-x\circ \theta (y,z)=0,$$

for $x,y,z$ in $\mathcal{A}$.

We observe that $Z^2(\mathcal{A},\mathcal{A})\ne \varnothing$ since $\theta =0\in Z^2(\mathcal{A},\mathcal{A})$. For $\theta \in Z^2(\mathcal{A},\mathcal{A})$, we define a product ${[·,·]}_{\theta }:\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ on $\mathcal{A}$ by

$${[x,y]}_{\theta }:=\theta (x,y),$$

(11)

for any $x,y$ in $\mathcal{A}$.

Lemma 10.

Let $(\mathcal{A},\circ )$ be a Jacobi–Jordan algebra and $\theta \in Z^2(\mathcal{A},\mathcal{A})$. Then, $(\mathcal{A},\circ ,{[·,·]}_{\theta })$ is a Jacobi–Jordan Poisson algebra endowed with the product defined in (11).

Proof. 

Since $(\mathcal{A},\circ )$ is Jacobi–Jordan algebra, we find that $(\mathcal{A},{[·,·]}_{\theta })$ is an anticommutative algebra. Moreover, since $\theta \in Z^2(\mathcal{A},\mathcal{A})$, we have

$${[x\circ y,z]}_{\theta }-{[x,z]}_{\theta }\circ y-x\circ {[y,z]}_{\theta }=\theta (x\circ y,z)-\theta (x,z)\circ y-x\circ \theta (y,z)=0,$$

for $x,y,z$ in $\mathcal{A}$, as required. □

Let $(\mathcal{A},\circ )$ be a Jacobi–Jordan algebra and $\theta \in Z^2(\mathcal{A},\mathcal{A})$. Consider in the underlying vector space $\mathcal{A}$ the product given as

$$x{·}_{\theta }y:=x\circ y+{[x,y]}_{\theta },$$

(12)

for any $x,y$ in $\mathcal{A}$. Thus, $(\mathcal{A},{·}_{\theta })$ is a general Jacobi–Jordan algebra.

In the reverse direction, if $(\mathcal{A},\circ ,[·,·\left]\right)$ is a Jacobi–Jordan Poisson algebra, then there exists $\theta \in Z^2(\mathcal{A},\mathcal{A})$ such that $(\mathcal{A},\circ ,{[·,·]}_{\theta })$ and $(\mathcal{A},\circ ,[·,·\left]\right)$ are isomorphic. Indeed, consider the skew-symmetric bilinear map $\theta :\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ defined by $\theta (x,y):=[x,y]$ for all $x,y$ in $\mathcal{A}$. Then, $\theta \in Z^2(\mathcal{A},\mathcal{A})$ and $(\mathcal{A},\circ ,{[·,·]}_{\theta })=(\mathcal{A},\circ ,[·,·])$.

Now, let $(\mathcal{A},\circ )$ be a Jacobi–Jordan algebra and $\mathrm{Aut}\left(\mathcal{A}\right)$ be the automorphism group of $\mathcal{A}$ with respect to product ∘. Then, $\mathrm{Aut}\left(\mathcal{A}\right)$ acts on $Z^2(\mathcal{A},\mathcal{A})$ by

$$(\theta \ast \varphi )(x,y):={\varphi }^{-1}\left(\theta \left(\varphi \left(x\right),\varphi \left(y\right)\right)\right),$$

where $\varphi \in \mathrm{Aut}\left(\mathcal{A}\right)$ and $\theta \in Z^2(\mathcal{A},\mathcal{A})$, for any $x,y$ in $\mathcal{A}$.

Lemma 11.

Let $(\mathcal{A},\circ )$ be a Jacobi–Jordan algebra and θ, $\vartheta \in Z^2(\mathcal{A},\mathcal{A})$. Then, the Jacobi–Jordan Poisson algebras $(\mathcal{A},\circ ,{[·,·]}_{\theta })$ and $(\mathcal{A},\circ ,{[·,·]}_{\vartheta })$ are isomorphic if and only if there exists $\varphi \in \mathrm{Aut}\left(\mathcal{A}\right)$ satisfying $\theta \ast \varphi =\vartheta$.

Proof. 

Suppose $\varphi :(\mathcal{A},\circ ,{[·,·]}_{\vartheta })\to (\mathcal{A},\circ ,{[·,·]}_{\theta })$ is an isomorphism of Jacobi–Jordan Poisson algebras. Then, $\varphi \in \mathrm{Aut}\left(\mathcal{A}\right)$, and by Definition 5, we have $\varphi \left({[x,y]}_{\vartheta }\right)={[\varphi \left(x\right),\varphi \left(y\right)]}_{\theta }$ for $x,y\in \mathcal{A}$. That is, $\varphi \left(\vartheta \right(x,y\left)\right)=\theta \left(\varphi \right(x),\varphi (y\left)\right)$ for $x,y\in \mathcal{A}$. Hence, $\vartheta (x,y)={\varphi }^{-1}\left(\theta \left(\varphi \left(x\right),\varphi \left(y\right)\right)\right)=(\theta \ast \varphi )(x,y)$ for $x,y\in \mathcal{A}$.

To prove the converse, if $\theta \ast \varphi =\vartheta$, then $\varphi :(\mathcal{A},\circ ,{[·,·]}_{\vartheta })\to (\mathcal{A},\circ ,{[·,·]}_{\theta })$ is an isomorphism since $\varphi \left(\vartheta \right(x,y\left)\right)=\theta \left(\varphi \right(x),\varphi (y\left)\right)$ for any $x,y\in \mathcal{A}$. □

Hence, we have a procedure to classify the Jacobi–Jordan Poisson algebras (and therefore the general Jacobi–Jordan algebras) associated with a given Jacobi–Jordan algebra $(\mathcal{A},\circ )$. It consists of three steps:

Step 1.

Compute $Z^2(\mathcal{A},\mathcal{A})$.

Step 2.

Find the orbits of $\mathrm{Aut}\left(\mathcal{A}\right)$ on $Z^2(\mathcal{A},\mathcal{A})$.

Step 3.

Choose a representative $\theta$ from each orbit and then construct the Jacobi–Jordan Poisson algebra $(\mathcal{A},\circ ,{[·,·]}_{\theta })$ (the general Jacobi–Jordan algebra $(\mathcal{A},{·}_{\theta })$).

Let us introduce the following notations. Let $\{e_1,\dots ,e_n\}$, with $n\ge 1$, be a fixed basis of a Jacobi–Jordan algebra $(\mathcal{A},\circ )$. Define ${\Lambda }^2(\mathcal{A},\mathbb{F})$ as the space of all skew-symmetric bilinear forms on $\mathcal{A}$, that is, ${\Lambda }^2(\mathcal{A},\mathbb{F}):={\mathrm{span}}_{\mathbb{F}}\{{\Delta }_{i,j}:1\le i<j\le n\}$, where ${\Delta }_{i,j}$ is the skew-symmetric bilinear form ${\Delta }_{i,j}:\mathcal{A}\times \mathcal{A}\to \mathbb{F}$ defined by

$${\Delta }_{i,j}(e_l,e_m):=\left\{\begin{array}{cc}1& \mathrm{if} (i,j)=(l,m),\hfill \\ -1& \mathrm{if} (i,j)=(m,l),\hfill \\ 0& \mathrm{otherwise}.\hfill \end{array}\right.$$

Now, if $\theta \in Z^2(\mathcal{A},\mathcal{A})$, then $\theta$ can be uniquely written as $\theta (x,y)={\sum }_{i=1}^n{B}_i(x,y)e_i$, where $B_1,\dots ,B_n$ are skew-symmetric bilinear forms on $\mathcal{A}$. Also, we may write $\theta =(B_1,\dots ,B_n)$. Let ${\varphi }^{-1}\in \mathrm{Aut}\left(\mathcal{A}\right)$ be given by the matrix $\left(b_{ij}\right)$. If $(\theta \ast \varphi )(x,y)={\sum }_{i=1}^n{B}_i^{\prime }(x,y)e_i$, then $B_i^{\prime }={\sum }_{j=1}^n{b}_{ij}{\varphi }^t{B}_j\varphi$, whenever $i\in \{1,\dots ,n\}$.

Remark 3.

By Lemma 2, given two non-isomorphic Jacobi–Jordan algebras $\mathcal{A}$ and ${\mathcal{A}}^{\prime }$, we have that any flexible structure on $\mathcal{A}$ is not isomorphic to any flexible structure on ${\mathcal{A}}^{\prime }$.

Remark 4.

Let $X=\left(\begin{array}{cc}\alpha & \beta \end{array}\right)\in {\mathcal{M}}_{1\times 2}\left(\mathbb{F}\right)$ and $X\ne 0$. Then, there exists an invertible matrix $A\in {\mathcal{M}}_{2\times 2}\left(\mathbb{F}\right)$ such that $XA=\left(\begin{array}{cc}1& 0\end{array}\right)$. To see this, suppose first that $\alpha \ne 0$. Then, $\left(\begin{array}{cc}\alpha & \beta \end{array}\right)\left(\begin{array}{cc}{\alpha }^{-1}& -\beta \\ 0& \alpha \end{array}\right)=\left(\begin{array}{cc}1& 0\end{array}\right)$. Assume now that $\alpha =0$. Then, $\left(\begin{array}{cc}0& \beta \end{array}\right)\left(\begin{array}{cc}0& 1\\ {\beta }^{-1}& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\end{array}\right)$.

5.2. Classification in Dimensions 1 and 2

Theorem 1.

Any 1-dimensional general Jacobi–Jordan algebra is trivial.

Proof. 

Let $\mathcal{A}$ be a 1-dimensional general Jacobi–Jordan algebra. Then, $\mathcal{A}$ is commutative. By Corollary 7, $\mathcal{A}$ is a Jacobi–Jordan algebra and therefore $\mathcal{A}$ is trivial since any 1-dimensional Jacobi–Jordan algebra is trivial.

In another way, let $\mathcal{A}=〈e_1〉$. Suppose that $e_1^2=\alpha e_1$ for some $\alpha \in \mathbb{F}$. Then, $e_1^3=\alpha e_1·e_1={\alpha }^2{e}_1$. Since $\mathcal{A}$ is a nilalgebra with a nilindex of at most 3, $e_1^3=0$ and therefore $\alpha =0$. Thus, $\mathcal{A}$ is trivial. □

Proposition 7

([5,10]). Any Jacobi–Jordan algebra of dimension 2 over an algebraically closed field of characteristic differences of 2 and 3 is isomorphic to exactly one of the following algebras:

• $A_{2,1}:$ trivial algebra.
• $A_{2,2}:e_1\circ e_1=e_2.$

Theorem 2.

Let $\mathcal{J}$ be a complex general Jacobi–Jordan algebra of dimension 2. Then, $\mathcal{J}$ is isomorphic to one of the following algebras:

• ${\mathcal{J}}_{2,1}$ trivial algebra.
• ${\mathcal{J}}_{2,2}:e_1·e_2=e_2,e_2·e_1=-e_2.$
• ${\mathcal{J}}_{2,3}:e_1·e_1=e_2.$

Proof. 

By Proposition 7, we may assume ${\mathcal{J}}^{+}\in \{A_{2,1},A_{2,2}\}$. If ${\mathcal{J}}^{+}=A_{2,1}$, then $\mathcal{J}$ is an anticommutative algebra. Since any 2-dimensional anticommutative algebra is a Lie algebra, we obtain the algebras ${\mathcal{J}}_{2,1}$ and ${\mathcal{J}}_{2,2}$. Assume now that ${\mathcal{J}}^{+}=A_{2,2}$. Then $Z^2({\mathcal{J}}^{+},{\mathcal{J}}^{+})=0$. So, we obtain the algebra ${\mathcal{J}}_{2,3}$. □

Theorem 3.

Any 2-dimensional general Jacobi–Jordan algebra is a noncommutative Jacobi–Jordan algebra.

5.3. Classification in Dimension 3

Proposition 8

([5,10]). Any Jacobi–Jordan algebra of dimension 3 over an algebraically closed field of characteristic differences of 2 and 3 is isomorphic to exactly one of the following algebras:

• $A_{3,1}:$ trivial algebra.
• $A_{3,2}:e_1\circ e_1=e_2.$
• $A_{3,3}:e_1\circ e_2=e_3.$

Theorem 4.

Let $\mathcal{J}$ be a complex general Jacobi–Jordan algebra of dimension 3. Then, $\mathcal{J}$ is isomorphic to one of the following algebras (with $\alpha ,\beta \in \mathbb{C}$):

• ${\mathcal{J}}_{3,1}:$ trivial algebra.
• ${\mathcal{J}}_{3,2}:e_2·e_3=e_1,e_3·e_2=-e_1.$
• ${\mathcal{J}}_{3,3}:e_1·e_3=e_1,e_3·e_1=-e_1,e_2·e_3=e_2,e_3·e_2=-e_2.$
• ${\mathcal{J}}_{3,4}^{\alpha }:e_1·e_3=e_1+e_2,e_3·e_1=-e_1-e_2,e_2·e_3=\alpha e_2,e_3·e_2=-\alpha e_2.$
• ${\mathcal{J}}_{3,5}:e_1·e_2=e_3,e_2·e_1=-e_3,e_1·e_3=-e_2,e_3·e_1=e_2,e_2·e_3=e_1,e_3·e_2=-e_1.$
• ${\mathcal{J}}_{3,6}^{\alpha }:e_1·e_2=e_3,e_2·e_1=-e_3,e_1·e_3=e_1+e_3,e_3·e_1=-e_1-e_3,e_2·e_3=\alpha e_2,e_3·e_2=-\alpha e_2.$
• ${\mathcal{J}}_{3,7}:e_1·e_2=e_1,e_2·e_1=-e_1,e_2·e_3=e_2,e_3·e_2=-e_2.$
• ${\mathcal{J}}_{3,8}:e_1·e_2=e_3,e_2·e_1=-e_3,e_1·e_3=e_1,e_3·e_1=-e_1,e_2·e_3=e_2,e_3·e_2=-e_2.$
• ${\mathcal{J}}_{3,9}:e_1·e_1=e_2.$
• ${\mathcal{J}}_{3,10}:e_1·e_1=e_2,e_1·e_3=e_3,e_3·e_1=-e_3.$
• ${\mathcal{J}}_{3,11}:e_1·e_1=e_2,e_1·e_3=e_2,e_3·e_1=-e_2.$
• ${\mathcal{J}}_{3,12}^{\alpha }:e_1·e_2=\left(\alpha +1\right)e_3,e_2·e_1=\left(1-\alpha \right)e_3.$

Between these algebras, there are precisely the following isomorphisms:

• ${\mathcal{J}}_{3,4}^{\alpha }\cong {\mathcal{J}}_{3,4}^{\beta }$ if and only if $(\alpha -\beta )(\alpha \beta -1)=0$.
• ${\mathcal{J}}_{3,6}^{\alpha }\cong {\mathcal{J}}_{3,6}^{\beta }$ if and only if $(\alpha -\beta )(\alpha \beta -1)=0$.
• ${\mathcal{J}}_{3,12}^{\alpha }\cong {\mathcal{J}}_{3,12}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.

Proof. 

By Proposition 8, we may assume ${\mathcal{J}}^{+}\in \{A_{3,1},A_{3,2},A_{3,3}\}$. If ${\mathcal{J}}^{+}=A_{3,1}$, then $\mathcal{J}$ is an anticommutative algebra. The classification of 3-dimensional anticommutative complex algebras is given in [11]. From here, $\mathcal{J}$ is isomorphic to one of the algebras ${\mathcal{J}}_{3,1},\dots ,{\mathcal{J}}_{3,8}$. Assume now that ${\mathcal{J}}^{+}\in \{A_{3,2},A_{3,3}\}$. Then, we have the following cases:

${\mathcal{J}}^{+}=A_{3,2}$. Let $\theta =(B_1,B_2,B_3)$ be an arbitrary element of $Z^2({\mathcal{J}}^{+},{\mathcal{J}}^{+})$. Then, $\theta =(0,\alpha {\Delta }_{1,3},\beta {\Delta }_{1,3})$ for some $\alpha ,\beta \in \mathbb{C}$. The automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{ccc}{a}_{11}& 0& 0\\ a_{21}& a_{11}^2& a_{23}\\ a_{31}& 0& a_{33}\end{array}\right).$$

Let $\varphi \in \mathrm{Aut}\left({\mathcal{J}}^{+}\right)$. Then, $\theta \ast \varphi =(0,{\alpha }^{\prime }{\Delta }_{1,3},{\beta }^{\prime }{\Delta }_{1,3})$ where

$$\begin{array}{ccc}\hfill {\alpha }^{\prime }& =& {\displaystyle \frac{1}{a_{11}}}(\alpha a_{33}-\beta a_{23}),\hfill \\ \hfill {\beta }^{\prime }& =& \beta a_{11}.\hfill \end{array}$$

Let us distinguish two cases:

• $(\alpha ,\beta )=(0,0)$. In this case, $\theta =0$ and so, we obtain the algebra ${\mathcal{J}}_{3,9}$.
• $(\alpha ,\beta )\ne (0,0)$. If $\beta \ne 0$, we define $\varphi$ to be the following automorphism:

  $$\varphi =\left(\begin{array}{ccc}{\displaystyle \frac{1}{\beta }}& 0& 0\\ 0& {\displaystyle \frac{1}{{\beta }^2}}& {\displaystyle \frac{\alpha }{\beta }}\\ 0& 0& 1\end{array}\right).$$
  Then, $\theta \ast \varphi =(0,0,{\Delta }_{1,3})$. So, we obtain the representative $(0,0,{\Delta }_{1,3})$ and therefore, we obtain the algebra ${\mathcal{J}}_{3,10}$. On the other hand, if $\beta =0$, we define $\varphi$ to be the following automorphism:

  $$\varphi =\left(\begin{array}{ccc}\alpha & 0& 0\\ 0& {\alpha }^2& 0\\ 0& 0& 1\end{array}\right).$$
  Then, $\theta \ast \varphi =(0,{\Delta }_{1,3},0)$. Hence, we obtain the representative $(0,{\Delta }_{1,3},0)$ and thus, we obtain the algebra ${\mathcal{J}}_{3,11}$.

${\mathcal{J}}^{+}=A_{3,3}$. If $\theta =(B_1,B_2,B_3)$ is an arbitrary element of $Z^2({\mathcal{J}}^{+},{\mathcal{J}}^{+})$, then $\theta =(0,0,\alpha {\Delta }_{1,2})$ for some $\alpha \in \mathbb{C}$. Moreover, the automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{ccc}{a}_{11}& a_{12}& 0\\ a_{21}& a_{22}& 0\\ a_{31}& a_{32}& a_{11}{a}_{22}+a_{12}{a}_{21}\end{array}\right):a_{12}=a_{21}=0\mathrm{or}{a}_{11}=a_{22}=0.$$

Assume that $\theta \ast \varphi =(0,0,{\alpha }^{\prime }{\Delta }_{1,2})$. Then,

$${\alpha }^{\prime }={\displaystyle \frac{1}{a_{11}{a}_{22}+a_{12}{a}_{21}}}(\alpha a_{11}{a}_{22}-\alpha a_{12}{a}_{21}).$$

Since $a_{12}=a_{21}=0\mathrm{or}{a}_{11}=a_{22}=0$, we have ${\alpha }^{\prime 2}={\alpha }^2$. So, we obtain the representatives ${\theta }^{\alpha }=\left(0,0,\alpha {\Delta }_{1,2}\right)$. Moreover, we have ${\theta }^{\alpha }$ and ${\theta }^{{\alpha }^{\prime }}$ in the same orbit if and only if ${\alpha }^{\prime 2}={\alpha }^2$. We denote the algebras corresponding to the representatives ${\theta }^{\alpha }$ by ${\mathcal{J}}_{3,12}^{\alpha }$. □

Theorem 5.

Let $\mathcal{A}$ be a complex noncommutative Jacobi–Jordan algebra of dimension 3. Then, $\mathcal{A}$ is isomorphic to one of the following algebras (with $\alpha ,\beta \in \mathbb{C}$):

• ${\mathcal{A}}_{3,1}:$ trivial algebra.
• ${\mathcal{A}}_{3,2}:e_1·e_2=e_3,e_2·e_1=-e_3.$
• ${\mathcal{A}}_{3,3}:e_1·e_2=e_2,e_2·e_1=-e_2,e_1·e_3=e_2+e_3,e_3·e_1=-e_2-e_3.$
• ${\mathcal{A}}_{3,4}^{\alpha }:e_1·e_2=e_2,e_2·e_1=-e_2,e_1·e_3=\alpha e_3,e_3·e_1=-\alpha e_3.$
• ${\mathcal{A}}_{3,5}:e_1·e_2=e_3,e_2·e_1=-e_3,e_1·e_3=-2e_1,e_2·e_3=2e_2,e_3·e_2=-2e_2.$
• ${\mathcal{A}}_{3,6}:e_1·e_1=e_2.$
• ${\mathcal{A}}_{3,7}:e_1·e_1=e_2,e_1·e_3=e_3,e_3·e_1=-e_3.$
• ${\mathcal{A}}_{3,8}:e_1·e_1=e_2,e_1·e_3=e_2,e_3·e_1=-e_2.$
• ${\mathcal{A}}_{3,9}^{\alpha }:e_1·e_2=\left(\alpha +1\right)e_3,e_2·e_1=\left(1-\alpha \right)e_3.$

Between these algebras, there are precisely the following isomorphisms:

• ${\mathcal{A}}_{3,4}^{\alpha }\cong {\mathcal{A}}_{3,4}^{\beta }$ if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$.
• ${\mathcal{A}}_{3,9}^{\alpha }\cong {\mathcal{A}}_{3,9}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.

5.4. Classification in Dimension 4

Proposition 9

([5,10]). Any Jacobi–Jordan algebra of dimension 4 over an algebraically closed field of characteristic differences from 2 and 3 is isomorphic to exactly one of the following algebras:

• $A_{4,1}$ trivial algebra.
• $A_{4,2}:e_1\circ e_1=e_2.$
• $A_{4,3}:e_1\circ e_2=e_3.$
• $A_{4,4}:e_1\circ e_1=e_3,e_2\circ e_2=e_4.$
• $A_{4,5}:e_1\circ e_1=e_3,e_1\circ e_2=e_4.$
• $A_{4,6}:e_1\circ e_2=e_4,e_3\circ e_3=e_4.$

Theorem 6.

Let $\mathcal{J}$ be a complex general Jacobi–Jordan algebra of dimension 4. Then, $\mathcal{J}$ is anticommutative or it is isomorphic to one of the following algebras (with $\alpha ,\beta \in \mathbb{C}$):

• ${\mathcal{J}}_{4,1}:e_1·e_1=e_2.$
• ${\mathcal{J}}_{4,2}:e_1·e_1=e_2,e_1·e_3=e_2,e_3·e_1=-e_2.$
• ${\mathcal{J}}_{4,3}:e_1·e_1=e_2,e_3·e_4=e_2,e_4·e_3=-e_2.$
• ${\mathcal{J}}_{4,4}^{\alpha }:e_1·e_1=e_2,e_3·e_4=e_2,e_4·e_3=-e_2,e_1·e_3=e_3,e_3·e_1=-e_3,e_1·e_4=\alpha e_4,e_4·e_1=-\alpha e_4.$
• ${\mathcal{J}}_{4,5}^{\alpha }:e_1·e_1=e_2,e_1·e_3=e_3,e_3·e_1=-e_3,e_1·e_4=\alpha e_4,e_4·e_1=-\alpha e_4.$
• ${\mathcal{J}}_{4,6}:e_1·e_1=e_2,e_1·e_4=e_2,e_4·e_1=-e_2,e_1·e_3=e_3,e_3·e_1=-e_3.$
• ${\mathcal{J}}_{4,7}:e_1·e_1=e_2,e_1·e_3=e_3,e_3·e_1=-e_3,e_1·e_4=e_3+e_4,e_4·e_1=-e_3-e_4.$
• ${\mathcal{J}}_{4,8}:e_1·e_1=e_2,e_3·e_4=e_2,e_4·e_3=-e_2,e_1·e_3=e_3,e_3·e_1=-e_3,e_1·e_4=e_3+e_4,e_4·e_1=-e_3-e_4.$
• ${\mathcal{J}}_{4,9}:e_1·e_1=e_2,e_3·e_4=e_2,e_4·e_3=-e_2,e_1·e_4=e_3,e_4·e_1=-e_3.$
• ${\mathcal{J}}_{4,10}:e_1·e_1=e_2,e_1·e_4=e_3,e_4·e_1=-e_3.$
• ${\mathcal{J}}_{4,11}:e_1·e_1=e_2,e_1·e_3=e_2,e_3·e_1=-e_2,e_1·e_4=e_3,e_4·e_1=-e_3.$
• ${\mathcal{J}}_{4,12}:e_1·e_1=e_2,e_3·e_4=e_3,e_4·e_3=-e_3,e_1·e_3=e_4,e_3·e_1=-e_4.$
• ${\mathcal{J}}_{4,13}:e_1·e_1=e_2,e_1·e_4=e_2,e_4·e_1=-e_2,e_3·e_4=e_3,e_4·e_3=-e_3,e_1·e_3=e_4,e_3·e_1=-e_4.$
• ${\mathcal{J}}_{4,14}:e_1·e_1=e_2,e_3·e_4=e_3,e_4·e_3=-e_3,e_1·e_4=e_4,e_4·e_1=-e_4.$
• ${\mathcal{J}}_{4,15}:e_1·e_1=e_2,e_1·e_3=e_2,e_3·e_1=-e_2,e_3·e_4=e_3,e_4·e_3=-e_3,e_1·e_4=e_4,e_4·e_1=-e_4.$
• ${\mathcal{J}}_{4,16}:e_1·e_1=e_2,e_3·e_4=e_3,e_4·e_3=-e_3.$
• ${\mathcal{J}}_{4,17}:e_1·e_1=e_2,e_1·e_4=e_2,e_4·e_1=-e_2,e_3·e_4=e_3,e_4·e_3=-e_3.$
• ${\mathcal{J}}_{4,18}:e_1·e_1=e_2,e_1·e_3=e_2,e_3·e_1=-e_2,e_3·e_4=e_3,e_4·e_3=-e_3.$
• ${\mathcal{J}}_{4,19}^{\alpha }:e_1·e_2=\left(1+\alpha \right)e_3,e_2·e_1=\left(1-\alpha \right)e_3,e_1·e_4=e_4,e_4·e_1=-e_4,e_2·e_4=e_4,e_4·e_2=-e_4.$
• ${\mathcal{J}}_{4,20}:e_1·e_2=e_3,e_2·e_1=e_3,e_1·e_4=e_3+e_4,e_4·e_1=-e_3-e_4,e_2·e_4=e_4,e_4·e_2=-e_4.$
• ${\mathcal{J}}_{4,21}^{\alpha }:e_1·e_2=\left(1+\alpha \right)e_3,e_2·e_1=\left(1-\alpha \right)e_3,e_1·e_4=e_4,e_4·e_1=-e_4.$
• ${\mathcal{J}}_{4,22}:e_1·e_2=e_3,e_2·e_1=e_3,e_2·e_4=e_3,e_4·e_2=-e_3,e_1·e_4=e_4,e_4·e_1=-e_4.$
• ${\mathcal{J}}_{4,23}:e_1·e_2=e_3,e_2·e_1=e_3,e_1·e_4=e_3,e_4·e_1=-e_3,e_2·e_4=e_3,e_4·e_2=-e_3.$
• ${\mathcal{J}}_{4,24}:e_1·e_2=e_3+e_4,e_2·e_1=e_3-e_4,e_1·e_4=e_3,e_4·e_1=-e_3,e_2·e_4=e_3,e_4·e_2=-e_3.$
• ${\mathcal{J}}_{4,25}:e_1·e_2=e_3,e_2·e_1=e_3,e_1·e_4=e_3,e_4·e_1=-e_3.$
• ${\mathcal{J}}_{4,26}:e_1·e_2=e_3+e_4,e_2·e_1=e_3-e_4,e_1·e_4=e_3,e_4·e_1=-e_3.$
• ${\mathcal{J}}_{4,27}:e_1·e_2=e_3+e_4,e_2·e_1=e_3-e_4.$
• ${\mathcal{J}}_{4,28}^{\alpha }:e_1·e_2=\left(1+\alpha \right)e_3,e_2·e_1=\left(1-\alpha \right)e_3.$
• ${\mathcal{J}}_{4,29}:e_1·e_1=e_3,e_2·e_2=e_4.$
• ${\mathcal{J}}_{4,30}^{\alpha }:e_1·e_1=e_3,e_2·e_2=e_4,e_1·e_2=e_3+\alpha e_4,e_2·e_1=-e_3-\alpha e_4.$
• ${\mathcal{J}}_{4,31}:e_1·e_1=e_3,e_1·e_2=e_2+e_4,e_2·e_1=e_4-e_2,e_1·e_4=e_4,e_4·e_1=-e_4,e_2·e_3=-2e_4,e_3·e_2=2e_4.$
• ${\mathcal{J}}_{4,32}:e_1·e_1=e_3,e_1·e_2=e_4+e_3,e_2·e_1=e_4-e_3.$
• ${\mathcal{J}}_{4,33}^{\alpha }:e_1·e_1=e_3,e_1·e_2=\left(1+\alpha \right)e_4,e_2·e_1=\left(1-\alpha \right)e_4.$
• ${\mathcal{J}}_{4,34}:e_1·e_2=e_3+e_4,e_2·e_1=e_4-e_3,e_3·e_3=e_4,e_1·e_3=-e_1,e_3·e_1=e_1,e_2·e_3=e_2,e_3·e_2=-e_2.$
• ${\mathcal{J}}_{4,35}:e_1·e_2=e_4,e_2·e_1=e_4,e_3·e_3=e_4,e_1·e_3=-e_4,e_3·e_1=e_4.$
• ${\mathcal{J}}_{4,36}^{\alpha }:e_1·e_2=\left(1+\alpha \right)e_4,e_2·e_1=\left(1-\alpha \right)e_4,e_3·e_3=e_4.$

Between these algebras, there are precisely the following isomorphisms:

• ${\mathcal{J}}_{4,4}^{\alpha }\cong {\mathcal{J}}_{4,4}^{\beta }$ if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$.
• ${\mathcal{J}}_{4,5}^{\alpha }\cong {\mathcal{J}}_{4,5}^{\beta }$ if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$.
• ${\mathcal{J}}_{4,19}^{\alpha }\cong {\mathcal{J}}_{4,19}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{J}}_{4,21}^{\alpha }\cong {\mathcal{J}}_{4,21}^{\beta }$ if and only if $\alpha =\beta$.
• ${\mathcal{J}}_{4,28}^{\alpha }\cong {\mathcal{J}}_{4,28}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{J}}_{4,30}^{\alpha }\cong {\mathcal{J}}_{4,30}^{\beta }$ if and only if $\alpha =\beta$.
• ${\mathcal{J}}_{4,33}^{\alpha }\cong {\mathcal{J}}_{4,33}^{\beta }$ if and only if $\alpha =\beta$.
• ${\mathcal{J}}_{4,36}^{\alpha }\cong {\mathcal{J}}_{4,36}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.

Proof. 

By Proposition 9, we may assume ${\mathcal{J}}^{+}\in \left\{A_{4,1},\dots ,A_{4,6}\right\}$. If ${\mathcal{J}}^{+}=A_{4,1}$, then $\mathcal{J}$ is an anticommutative algebra. Assume now that ${\mathcal{J}}^{+}\in \left\{A_{4,2},\dots ,A_{4,6}\right\}$. Then, we have the following five cases:

${\mathcal{J}}^{+}=A_{4,2}$. The automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ a_{21}& a_{11}^2& a_{23}& a_{24}\\ a_{31}& 0& a_{33}& a_{34}\\ a_{41}& 0& a_{43}& a_{44}\end{array}\right).$$

Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left({\mathcal{J}}^{+},{\mathcal{J}}^{+}\right)$. Then,

$$\begin{array}{ccc}\hfill B_1& =& 0,\hfill \\ \hfill B_2& =& {\alpha }_1{\Delta }_{1,3}+{\alpha }_4{\Delta }_{1,4}+{\alpha }_7{\Delta }_{3,4},\hfill \\ \hfill B_3& =& {\alpha }_2{\Delta }_{1,3}+{\alpha }_5{\Delta }_{1,4}+{\alpha }_8{\Delta }_{3,4},\hfill \\ \hfill B_4& =& {\alpha }_3{\Delta }_{1,3}+{\alpha }_6{\Delta }_{1,4}+{\alpha }_9{\Delta }_{3,4},\hfill \end{array}$$

for some ${\alpha }_1,\dots ,{\alpha }_9\in \mathbb{C}$. Let $\varphi =\left(a_{ij}\right)\in \mathrm{Aut}\left({\mathcal{J}}^{+}\right)$ and write

$$\theta \ast \varphi =(0,{\alpha }_1^{\prime }{\Delta }_{1,3}+{\alpha }_4^{\prime }{\Delta }_{1,4}+{\alpha }_7^{\prime }{\Delta }_{3,4},{\alpha }_2^{\prime }{\Delta }_{1,3}+{\alpha }_5^{\prime }{\Delta }_{1,4}+{\alpha }_8^{\prime }{\Delta }_{3,4},{\alpha }_3^{\prime }{\Delta }_{1,3}+{\alpha }_6^{\prime }{\Delta }_{1,4}+{\alpha }_9^{\prime }{\Delta }_{3,4}).$$

Then,

$$\begin{array}{ccc}\hfill {\alpha }_8^{\prime }& =& {\alpha }_8{a}_{44}-{\alpha }_9{a}_{34},\hfill \\ \hfill {\alpha }_9^{\prime }& =& {\alpha }_9{a}_{33}-{\alpha }_8{a}_{43}.\hfill \end{array}$$

By Remark 4, we may assume $\left({\alpha }_8,{\alpha }_9\right)\in \left\{\left(0,0\right),\left(1,0\right)\right\}$. Assume first that ${\alpha }_8={\alpha }_9=0$. Then, we have

$$\begin{array}{ccc}\hfill {\alpha }_2^{\prime }& =& {\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_2{a}_{33}{a}_{44}-{\alpha }_3{a}_{33}{a}_{34}+{\alpha }_5{a}_{43}{a}_{44}-{\alpha }_6{a}_{34}{a}_{43}\right),\hfill \\ \hfill {\alpha }_3^{\prime }& =& {\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_3{a}_{33}^2-{\alpha }_5{a}_{43}^2-{\alpha }_2{a}_{33}{a}_{43}+{\alpha }_6{a}_{33}{a}_{43}\right),\hfill \\ \hfill {\alpha }_5^{\prime }& =& {\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_3{a}_{34}^2-{\alpha }_5{a}_{44}^2-{\alpha }_2{a}_{34}{a}_{44}+{\alpha }_6{a}_{34}{a}_{44}\right),\hfill \\ \hfill {\alpha }_6^{\prime }& =& {\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_2{a}_{34}{a}_{43}-{\alpha }_3{a}_{33}{a}_{34}+{\alpha }_5{a}_{43}{a}_{44}-{\alpha }_6{a}_{33}{a}_{44}\right).\hfill \end{array}$$

This then implies that $\left(\begin{array}{cc}{\alpha }_2^{\prime }& {\alpha }_5^{\prime }\\ {\alpha }_3^{\prime }& {\alpha }_6^{\prime }\end{array}\right)=a_{11}{\left(\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right)}^{-1}\left(\begin{array}{cc}{\alpha }_2& {\alpha }_5\\ {\alpha }_3& {\alpha }_6\end{array}\right)\left(\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right)$. Thus, we may assume

$$\left(\begin{array}{cc}{\alpha }_2& {\alpha }_5\\ {\alpha }_3& {\alpha }_6\end{array}\right)\in \left\{\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 0& \alpha \end{array}\right),\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\right\}.$$

So, the following cases arise:

• $\left(\begin{array}{cc}{\alpha }_2& {\alpha }_5\\ {\alpha }_3& {\alpha }_6\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$.

  –

  ${\alpha }_7=0$.

  *

  ${\alpha }_1={\alpha }_4=0$. Then, $\theta =0$ and we obtain the algebra ${\mathcal{J}}_{4,1}$.

  –

  $\left({\alpha }_1,{\alpha }_4\right)=\left(0,0\right)$. Let $\varphi$ be the first of the following matrices if ${\alpha }_4=0$ or the second if ${\alpha }_4\ne 0$:

  $$\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& {\alpha }_1^2& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_4}}& -{\displaystyle \frac{{\alpha }_1}{{\alpha }_4}}\end{array}\right)$$

  Then, $\theta \ast \varphi =\left(0,{\Delta }_{1,3},0,0\right)$ and therefore, we have the algebra ${\mathcal{J}}_{4,2}$.

  –

  ${\alpha }_7\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -{\displaystyle \frac{{\alpha }_4}{{\alpha }_7}}& 0& 1& 0\\ {\displaystyle \frac{{\alpha }_1}{{\alpha }_7}}& 0& 0& {\displaystyle \frac{1}{{\alpha }_7}}\end{array}\right)$$

  Then, $\theta \ast \varphi =\left(0,{\Delta }_{3,4},0,0\right)$. So, we obtain the algebra ${\mathcal{J}}_{4,3}$.

• $\left(\begin{array}{cc}{\alpha }_2& {\alpha }_5\\ {\alpha }_3& {\alpha }_6\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& \alpha \end{array}\right)$.

  –

  ${\alpha }_7\ne 0$. We define $\varphi$ to be the above automorphism. Then, $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{1,3},\alpha {\Delta }_{1,4}\right)$. So, we obtain the algebras ${\mathcal{J}}_{4,4}^{\alpha }$. Moreover, the algebras ${\mathcal{J}}_{4,4}^{\alpha }$ and ${\mathcal{J}}_{4,4}^{\beta }$ are isomorphic if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$.

  –

  ${\alpha }_7=0$.

  *

  $\alpha \ne 0$. Choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& {\displaystyle \frac{{\alpha }_4}{\alpha }}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,{\Delta }_{1,3},\alpha {\Delta }_{1,4}\right)$. Hence, we obtain the algebras ${\mathcal{J}}_{4,5}^{\alpha \ne 0}$. Furthermore, the algebras ${\mathcal{J}}_{4,5}^{\alpha \ne 0}$ and ${\mathcal{J}}_{4,5}^{\beta \ne 0}$ are isomorphic if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$.

  *

  $\alpha =0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_4=0$ or the second if ${\alpha }_4\ne 0$:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_4}}\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,{\Delta }_{1,3},0\right)$ if ${\alpha }_4=0$ or $\theta \ast \varphi =\left(0,{\Delta }_{1,4},{\Delta }_{1,3},0\right)$ if ${\alpha }_4\ne 0$. So, we obtain the algebra ${\mathcal{J}}_{4,5}^{\alpha =0}$ if ${\alpha }_4=0$ or the algebra ${\mathcal{J}}_{4,6}$ if ${\alpha }_4\ne 0$.

• $\left(\begin{array}{cc}{\alpha }_2& {\alpha }_5\\ {\alpha }_3& {\alpha }_6\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$.
  Consider the following matrices and let $\varphi ={\varphi }_1$ if ${\alpha }_7=0$ or $\varphi ={\varphi }_2$ if ${\alpha }_7\ne 0$:

  $${\varphi }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& {\alpha }_4-{\alpha }_1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),{\varphi }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -{\displaystyle \frac{{\alpha }_4}{{\alpha }_7}}& 0& {\displaystyle \frac{1}{\sqrt{{\alpha }_7}}}& 0\\ {\displaystyle \frac{{\alpha }_1}{{\alpha }_7}}& 0& 0& {\displaystyle \frac{1}{\sqrt{{\alpha }_7}}}\end{array}\right).$$
  Then, $\theta \ast \varphi =\left(0,0,{\Delta }_{1,3}+{\Delta }_{1,4},{\Delta }_{1,4}\right)$ if ${\alpha }_7=0$ or $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{1,3}+{\Delta }_{1,4},{\Delta }_{1,4}\right)$ if ${\alpha }_7\ne 0$. Thus, we obtain the algebras ${\mathcal{J}}_{4,7}$ and ${\mathcal{J}}_{4,8}$.
• $\left(\begin{array}{cc}{\alpha }_2& {\alpha }_5\\ {\alpha }_3& {\alpha }_6\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

  –

  If ${\alpha }_7\ne 0$, then we choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}{\alpha }_7& 0& 0& 0\\ 0& {\alpha }_7^2& {\alpha }_4{\alpha }_7& 0\\ 0& 0& {\alpha }_7& 0\\ {\alpha }_1& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{1,4},0\right)$ and we obtain the algebra ${\mathcal{J}}_{4,9}$.

  –

  ${\alpha }_7=0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_1=0$ or the second if ${\alpha }_1\ne 0$:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_4& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& {\alpha }_1^2& {\alpha }_4& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_1}}\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},0\right)$ if ${\alpha }_1=0$ or $\theta \ast \varphi =\left(0,{\Delta }_{1,3},{\Delta }_{1,4},0\right)$ if ${\alpha }_1\ne 0$. Therefore, we obtain the algebras ${\mathcal{J}}_{4,10}$ and ${\mathcal{J}}_{4,11}$.

Assume now that $\left({\alpha }_8,{\alpha }_9\right)=\left(1,0\right)$. Then, we have the following cases:

• ${\alpha }_3\ne 0$. Choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\displaystyle \frac{{\alpha }_7}{{\alpha }_3}}& {\displaystyle \frac{1}{{\alpha }_3}}\left({\alpha }_1-{\alpha }_2{\alpha }_7-{\alpha }_6{\alpha }_7\right)\\ -{\displaystyle \frac{1}{{\alpha }_3}}\left({\alpha }_6^2+{\alpha }_3{\alpha }_5\right)& 0& {\displaystyle \frac{1}{{\alpha }_3}}& -{\displaystyle \frac{{\alpha }_6}{{\alpha }_3}}\\ {\alpha }_2+{\alpha }_6& 0& 0& 1\end{array}\right).$$
  Then, $\theta \ast \varphi =(0,{\alpha }_4^{\prime }{\Delta }_{1,4},{\Delta }_{3,4},{\Delta }_{1,3})$. If ${\alpha }_4^{\prime }=0$, we obtain the algebra ${\mathcal{J}}_{4,12}$. If ${\alpha }_4^{\prime }\ne 0$, we choose ${\varphi }^{\prime }$ as follows:

  $${\varphi }^{\prime }=\left(\begin{array}{cccc}{\alpha }_4^{\prime }& 0& 0& 0\\ 0& {\left({\alpha }_4^{\prime }\right)}^2& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_4^{\prime }}}& 0\\ 0& 0& 0& 1\end{array}\right)$$
  Then, $\theta \ast \varphi {\varphi }^{\prime }=(0,{\Delta }_{1,4},{\Delta }_{3,4},{\Delta }_{1,3})$. So, we obtain the algebra ${\mathcal{J}}_{4,13}$.
• ${\alpha }_3=0$.

  –

  ${\alpha }_6\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_6}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_6^2}}& {\alpha }_7& {\displaystyle \frac{{\alpha }_4-{\alpha }_5{\alpha }_7}{{\alpha }_6}}\\ -{\displaystyle \frac{{\alpha }_5}{{\alpha }_6}}& 0& 1& 0\\ {\displaystyle \frac{{\alpha }_2}{{\alpha }_6}}& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi =(0,{\alpha }_1^{\prime }{\Delta }_{1,3},{\Delta }_{3,4},{\Delta }_{1,4})$. If ${\alpha }_1^{\prime }=0$, we obtain the algebra ${\mathcal{J}}_{4,14}$. If ${\alpha }_1^{\prime }\ne 0$, we choose ${\varphi }^{\prime }$ as follows:

  $${\varphi }^{\prime }=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_1^{\prime }}}& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi {\varphi }^{\prime }=(0,{\Delta }_{1,3},{\Delta }_{3,4},{\Delta }_{1,4})$. So, we obtain the algebra ${\mathcal{J}}_{4,15}$.

  –

  ${\alpha }_6=0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_7& 0\\ -{\alpha }_5& 0& 1& 0\\ {\alpha }_2& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi =(0,{\alpha }_1^{\prime }{\Delta }_{1,3}+{\alpha }_4^{\prime }{\Delta }_{1,4},{\Delta }_{3,4},0)$. If ${\alpha }_1^{\prime }={\alpha }_4^{\prime }=0$, we obtain the algebra ${\mathcal{J}}_{4,16}$. Otherwise, let ${\varphi }^{\prime }$ be the first of the following matrices if ${\alpha }_1^{\prime }=0$ or the second if ${\alpha }_1^{\prime }\ne 0$:

  $$\left(\begin{array}{cccc}{\alpha }_4^{\prime }& 0& 0& 0\\ 0& {\left({\alpha }_4^{\prime }\right)}^2& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\alpha }_1^{\prime }& 0& 0& 0\\ 0& {\left({\alpha }_1^{\prime }\right)}^2& 0& 0\\ 0& 0& 1& -{\displaystyle \frac{{\alpha }_4^{\prime }}{{\alpha }_1^{\prime }}}\\ 0& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi {\varphi }^{\prime }=(0,{\Delta }_{1,4},{\Delta }_{3,4},0)$ if ${\alpha }_1^{\prime }=0$ and $\theta \ast \varphi {\varphi }^{\prime }=(0,{\Delta }_{1,3},{\Delta }_{3,4},0)$ if ${\alpha }_1^{\prime }\ne 0$. Hence, we obtain the algebras ${\mathcal{J}}_{4,17}$ and ${\mathcal{J}}_{4,18}$.

${\mathcal{J}}^{+}=A_{4,3}$. If $\theta =\left(B_1,B_2,B_3,B_4\right)$ is an arbitrary element of $Z^2\left({\mathcal{J}}^{+},{\mathcal{J}}^{+}\right)$, then

$$\begin{array}{ccc}\hfill B_1& =& B_2=0,\hfill \\ \hfill B_3& =& {\alpha }_1{\Delta }_{1,2}+{\alpha }_2{\Delta }_{1,4}+{\alpha }_3{\Delta }_{2,4},\hfill \\ \hfill B_4& =& {\alpha }_4{\Delta }_{1,2}+{\alpha }_5{\Delta }_{1,4}+{\alpha }_6{\Delta }_{2,4},\hfill \end{array}$$

for some ${\alpha }_1,\dots ,{\alpha }_6\in \mathbb{C}$. Moreover, the automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{cccc}{a}_{11}& a_{12}& 0& 0\\ a_{21}& a_{22}& 0& 0\\ a_{31}& a_{32}& a_{11}{a}_{22}+a_{12}{a}_{21}& a_{34}\\ a_{41}& a_{42}& 0& a_{44}\end{array}\right):a_{12}=a_{21}=0\mathrm{or}{a}_{11}=a_{22}=0.$$

Let $\varphi =\left(a_{ij}\right)\in \mathrm{Aut}\left({\mathcal{J}}^{+}\right)$ and write

$$\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{1,2}+{\beta }_2{\Delta }_{1,4}+{\beta }_3{\Delta }_{2,4},{\beta }_4{\Delta }_{1,2}+{\beta }_5{\Delta }_{1,4}+{\beta }_6{\Delta }_{2,4}\right).$$

Then,

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\displaystyle \frac{1}{a_{44}(a_{11}{a}_{22}+a_{12}{a}_{21})}}\left(\begin{array}{c}{\alpha }_1{a}_{11}{a}_{22}{a}_{44}-{\alpha }_1{a}_{12}{a}_{21}{a}_{44}-{\alpha }_4{a}_{11}{a}_{22}{a}_{34}+{\alpha }_4{a}_{12}{a}_{21}{a}_{34}+{\alpha }_2{a}_{11}{a}_{42}{a}_{44}-{\alpha }_2{a}_{12}{a}_{41}{a}_{44}\\ +{\alpha }_3{a}_{21}{a}_{42}{a}_{44}-{\alpha }_3{a}_{22}{a}_{41}{a}_{44}-{\alpha }_5{a}_{11}{a}_{42}{a}_{34}+{\alpha }_5{a}_{12}{a}_{41}{a}_{34}-{\alpha }_6{a}_{21}{a}_{42}{a}_{34}+{\alpha }_6{a}_{22}{a}_{41}{a}_{34}\end{array}\right),\hfill \\ \hfill {\beta }_2& =& {\displaystyle \frac{1}{a_{11}{a}_{22}+a_{12}{a}_{21}}}\left({\alpha }_2{a}_{11}{a}_{44}+{\alpha }_3{a}_{21}{a}_{44}-{\alpha }_5{a}_{11}{a}_{34}-{\alpha }_6{a}_{21}{a}_{34}\right).\hfill \\ \hfill {\beta }_3& =& {\displaystyle \frac{1}{a_{11}{a}_{22}+a_{12}{a}_{21}}}\left({\alpha }_2{a}_{12}{a}_{44}+{\alpha }_3{a}_{22}{a}_{44}-{\alpha }_5{a}_{12}{a}_{34}-{\alpha }_6{a}_{22}{a}_{34}\right).\hfill \\ \hfill {\beta }_4& =& {\displaystyle \frac{1}{a_{44}}}\left({\alpha }_4{a}_{11}{a}_{22}-{\alpha }_4{a}_{12}{a}_{21}+{\alpha }_5{a}_{11}{a}_{42}-{\alpha }_5{a}_{12}{a}_{41}+{\alpha }_6{a}_{21}{a}_{42}-{\alpha }_6{a}_{22}{a}_{41}\right),\hfill \\ \hfill {\beta }_5& =& {\alpha }_5{a}_{11}+{\alpha }_6{a}_{21},\hfill \\ \hfill {\beta }_6& =& {\alpha }_5{a}_{12}+{\alpha }_6{a}_{22}.\hfill \end{array}$$

Let us consider the following cases:

• $\left({\alpha }_5,{\alpha }_6\right)\ne \left(0,0\right)$. Then, without any loss of generality, we may assume that ${\alpha }_5\ne 0$. Suppose first that ${\alpha }_6\ne 0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5=0$ or the second if ${\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5\ne 0$:

  $$\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_5}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_6}}& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_5{\alpha }_6}}& {\displaystyle \frac{{\alpha }_3{\alpha }_5}{{\alpha }_5{\alpha }_6}}\\ 0& -{\displaystyle \frac{{\alpha }_4}{{\alpha }_5{\alpha }_6}}& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_5}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_6}}& 0& 0\\ -{\displaystyle \frac{{\alpha }_3}{{\alpha }_5{\alpha }_6}}{\displaystyle \frac{{\alpha }_1{\alpha }_5-{\alpha }_2{\alpha }_4}{{\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5}}& -{\displaystyle \frac{{\alpha }_3}{{\alpha }_5{\alpha }_6}}{\displaystyle \frac{{\alpha }_1{\alpha }_5-{\alpha }_2{\alpha }_4}{{\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5}}& {\displaystyle \frac{1}{{\alpha }_5{\alpha }_6}}& {\displaystyle \frac{{\alpha }_3}{{\alpha }_6\left({\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5\right)}}\\ -{\displaystyle \frac{1}{{\alpha }_5}}{\displaystyle \frac{{\alpha }_1{\alpha }_5-{\alpha }_2{\alpha }_4}{{\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5}}& -{\displaystyle \frac{1}{{\alpha }_6}}{\displaystyle \frac{{\alpha }_1{\alpha }_6-{\alpha }_3{\alpha }_4}{{\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5}}& 0& {\displaystyle \frac{1}{{\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5}}\end{array}\right).$$
  Then, $\theta \ast \varphi =\left(0,0,\alpha {\Delta }_{1,2},{\Delta }_{1,4}+{\Delta }_{2,4}\right)$ if ${\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5=0$ or $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},{\Delta }_{1,4}+{\Delta }_{2,4}\right)$ if ${\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5\ne 0$. So, we obtain the algebras ${\mathcal{J}}_{4,19}^{\alpha }$ and ${\mathcal{J}}_{4,20}$. For any $\alpha \in \mathbb{F}$, ${\mathcal{J}}_{4,19}^{\alpha }$ is not isomorphic to ${\mathcal{J}}_{4,20}$. Furthermore, ${\mathcal{J}}_{4,19}^{\alpha }$ is isomorphic to ${\mathcal{J}}_{4,19}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
  Assume now that ${\alpha }_6=0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_3=0$ or the second if ${\alpha }_3\ne 0$:

  $$\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_5}}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_5}}& {\displaystyle \frac{{\alpha }_2}{{\alpha }_5}}\\ 0& -{\displaystyle \frac{{\alpha }_4}{{\alpha }_5}}& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_5}}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_5}}& {\displaystyle \frac{{\alpha }_2}{{\alpha }_3{\alpha }_5^2}}\\ {\displaystyle \frac{1}{{\alpha }_3{\alpha }_5^2}}\left({\alpha }_1{\alpha }_5-{\alpha }_2{\alpha }_4\right)& -{\displaystyle \frac{{\alpha }_4}{{\alpha }_5}}& 0& {\displaystyle \frac{1}{{\alpha }_3{\alpha }_5}}\end{array}\right).$$
  Then, $\theta \ast \varphi =\left(0,0,\alpha {\Delta }_{1,2},{\Delta }_{1,4}\right)$ if ${\alpha }_3=0$ or $\theta \ast \varphi =\left(0,0,{\Delta }_{2,4},{\Delta }_{1,4}\right)$ if ${\alpha }_3\ne 0$. Hence, we obtain the algebras ${\mathcal{J}}_{4,21}^{\alpha }$ and ${\mathcal{J}}_{4,22}$. For any $\alpha \in \mathbb{F}$, ${\mathcal{J}}_{4,21}^{\alpha }$ is not isomorphic to ${\mathcal{J}}_{4,22}$. Moreover, ${\mathcal{J}}_{4,21}^{\alpha }$ is isomorphic to ${\mathcal{J}}_{4,21}^{\beta }$ if and only if $\alpha =\beta$.
• ${\alpha }_5={\alpha }_6=0$.

  –

  $\left({\alpha }_2,{\alpha }_3\right)\ne \left(0,0\right)$. Then, without any loss of generality, we may assume that ${\alpha }_2\ne 0$. Suppose first that ${\alpha }_3\ne 0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_4=0$ or the second if ${\alpha }_4\ne 0$:

  $$\left(\begin{array}{cccc}{\alpha }_3& 0& 0& 0\\ 0& {\alpha }_2& 0& 0\\ 0& 0& {\alpha }_3{\alpha }_2& 0\\ {\alpha }_1& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_2{\alpha }_4}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_3{\alpha }_4}}& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_2{\alpha }_3{\alpha }_4^2}}& 0\\ {\displaystyle \frac{{\alpha }_1}{{\alpha }_2{\alpha }_3{\alpha }_4}}& 0& 0& {\displaystyle \frac{1}{{\alpha }_2{\alpha }_3{\alpha }_4}}\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4}+{\Delta }_{2,4},0\right)$ if ${\alpha }_4=0$ or $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4}+{\Delta }_{2,4},{\Delta }_{1,2}\right)$ if ${\alpha }_4\ne 0$. Therefore, we obtain the algebras ${\mathcal{J}}_{4,23}$ and ${\mathcal{J}}_{4,24}$.

  Assume now that ${\alpha }_3=0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_4=0$ or the second if ${\alpha }_4\ne 0$:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& -{\displaystyle \frac{{\alpha }_1}{{\alpha }_2}}& 0& {\displaystyle \frac{1}{{\alpha }_2}}\end{array}\right),\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_2{\alpha }_4}}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_2{\alpha }_4}}& 0\\ 0& -{\displaystyle \frac{{\alpha }_1}{{\alpha }_2}}& 0& {\displaystyle \frac{1}{{\alpha }_2}}\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},0\right)$ if ${\alpha }_4=0$ or $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},{\Delta }_{1,2}\right)$ if ${\alpha }_4\ne 0$. So, we obtain the algebras ${\mathcal{J}}_{4,25}$ and ${\mathcal{J}}_{4,26}$.

  –

  ${\alpha }_2={\alpha }_3=0$. If ${\alpha }_4\ne 0$, we choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& {\alpha }_1\\ 0& 0& 0& {\alpha }_4\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,2}\right).$ If ${\alpha }_4=0$, then we have $\theta =\left(0,0,\alpha {\Delta }_{1,2},0\right)$. Thus, we obtain the algebras ${\mathcal{J}}_{4,27}$ and ${\mathcal{J}}_{4,28}^{\alpha }$. Moreover, ${\mathcal{J}}_{4,28}^{\alpha }$ is isomorphic to ${\mathcal{J}}_{4,28}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.

${\mathcal{J}}^{+}=A_{4,4}$. If $\theta =\left(B_1,B_2,B_3,B_4\right)$ is an arbitrary element of $Z^2\left({\mathcal{J}}^{+},{\mathcal{J}}^{+}\right)$, then $\theta =\left(0,0,{\alpha }_1{\Delta }_{1,2},{\alpha }_2{\Delta }_{1,2}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. Moreover, the automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{cccc}ϵa_{11}& \left(1-ϵ\right)a_{12}& 0& 0\\ \left(1-ϵ\right)a_{21}& ϵa_{22}& 0& 0\\ a_{31}& a_{32}& ϵa_{11}^2& \left(1-ϵ\right)a_{12}^2\\ a_{41}& a_{42}& \left(1-ϵ\right)a_{21}^2& ϵa_{22}^2\end{array}\right):ϵ\in \left\{0,1\right\}.$$

Let $\varphi =\left(a_{ij}\right)\in \mathrm{Aut}\left({\mathcal{J}}^{+}\right)$ and write $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{1,2},{\beta }_2{\Delta }_{1,2}\right)$. Then, ${\beta }_1={\displaystyle \frac{a_{22}}{a_{11}}}{\alpha }_1,{\beta }_2={\alpha }_2{\displaystyle \frac{a_{11}}{a_{22}}}$ if $ϵ=1$ or ${\beta }_1=-{\alpha }_2{\displaystyle \frac{a_{12}}{a_{21}}},{\beta }_2=-{\displaystyle \frac{{\alpha }_1}{a_{12}}}{a}_{21}$ if $ϵ=0$. We have the following cases:

• ${\alpha }_1={\alpha }_2=0$. Then, we obtain the Jacobi–Jordan algebra ${\mathcal{J}}_{4,29}$.
• $\left({\alpha }_1,{\alpha }_2\right)\ne \left(0,0\right)$. Then, we may assume ${\alpha }_1\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& {\alpha }_1^2& 0\\ 0& 0& 0& 1\end{array}\right).$$
  Hence, we obtain the representatives ${\theta }^{\alpha }=\left(0,0,{\Delta }_{1,2},\alpha {\Delta }_{1,2}\right)$. Moreover, ${\theta }^{\alpha }$ and ${\theta }^{\beta }$ are in the same orbit if and only if $\alpha =\beta$. Hence, we obtain the algebras ${\mathcal{J}}_{4,30}^{\alpha }$.

${\mathcal{J}}^{+}=A_{4,5}$. If $\theta =\left(B_1,B_2,B_3,B_4\right)$ is an arbitrary element of $Z^2({\mathcal{J}}^{+},{\mathcal{J}}^{+})$, then

$$\theta =(0,{\alpha }_1{\Delta }_{1,2},{\alpha }_2{\Delta }_{1,2},{\alpha }_3{\Delta }_{1,2}+{\alpha }_1{\Delta }_{1,4}-2{\alpha }_1{\Delta }_{2,3})$$

for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. Moreover, the automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ a_{21}& a_{22}& 0& 0\\ a_{31}& a_{32}& a_{11}^2& 0\\ a_{41}& a_{42}& 2a_{11}{a}_{21}& a_{11}{a}_{22}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in \mathrm{Aut}\left({\mathcal{J}}^{+}\right)$ and write $\theta \ast \varphi =\left(0,{\beta }_1{\Delta }_{1,2},{\beta }_2{\Delta }_{1,2},{\beta }_3{\Delta }_{1,2}+{\beta }_1{\Delta }_{1,4}-2{\beta }_1{\Delta }_{2,3}\right)$. Then,

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\alpha }_1{a}_{11},\hfill \\ \hfill {\beta }_2& =& -{\displaystyle \frac{1}{a_{11}}}\left({\alpha }_1{a}_{32}-{\alpha }_2{a}_{22}\right),\hfill \\ \hfill {\beta }_3& =& {\displaystyle \frac{1}{a_{11}}}\left(2{\alpha }_1{a}_{31}-2{\alpha }_2{a}_{21}+{\alpha }_3{a}_{11}\right).\hfill \end{array}$$

Let us consider the following cases:

• ${\alpha }_1\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_1}}& 0& 0& 0\\ 0& 1& 0& 0\\ -{\displaystyle \frac{{\alpha }_3}{2{\alpha }_1^2}}& {\displaystyle \frac{{\alpha }_2}{{\alpha }_1}}& {\displaystyle \frac{1}{{\alpha }_1^2}}& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_1}}\end{array}\right).$$
  Then, $\theta \ast \varphi =\left(0,{\Delta }_{1,2},0,{\Delta }_{1,4}-2{\Delta }_{2,3}\right)$. So, we obtain the algebra ${\mathcal{J}}_{4,31}$.
• ${\alpha }_1=0,{\alpha }_2\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}{\alpha }_2& 0& 0& 0\\ {\displaystyle \frac{1}{2}}{\alpha }_3& 1& 0& 0\\ 0& 0& {\alpha }_2^2& 0\\ 0& 0& {\alpha }_2{\alpha }_3& {\alpha }_2\end{array}\right).$$
  So, we obtain the representative $\theta =\left(0,0,{\Delta }_{1,2},0\right)$. Hence, we obtain the algebra ${\mathcal{J}}_{4,32}$.
• ${\alpha }_1={\alpha }_2=0$. Then, we have the representatives $\theta =\left(0,0,0,\alpha {\Delta }_{1,2}\right)$ and so we obtain the algebras ${\mathcal{J}}_{4,33}^{\alpha }$. Since for any $\varphi \in \mathrm{Aut}\left({\mathcal{J}}^{+}\right)$ we have ${\theta }^{\alpha }\ast \varphi ={\theta }^{\alpha }$, the algebras ${\mathcal{J}}_{4,33}^{\alpha },{\mathcal{J}}_{4,33}^{\beta }$ are isomorphic if and only if $\alpha =\beta$.

${\mathcal{J}}^{+}=A_{4,6}$. If $\theta =\left(B_1,B_2,B_3,B_4\right)$ is an arbitrary element of $Z^2({\mathcal{J}}^{+},{\mathcal{J}}^{+})$, then

$$\theta =(-{\alpha }_1{\Delta }_{1,3},{\alpha }_1{\Delta }_{2,3},{\alpha }_1{\Delta }_{1,2},{\alpha }_2{\Delta }_{1,2}+{\alpha }_3{\Delta }_{1,3}+{\alpha }_4{\Delta }_{2,3})$$

for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. Moreover, the automorphism group of ${\mathcal{J}}^{+}$, $\mathrm{Aut}\left({\mathcal{J}}^{+}\right)$, consists of the invertible matrices of the following form:

$$\varphi =\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& 0\\ a_{21}& a_{22}& a_{23}& 0\\ a_{31}& a_{32}& a_{33}& 0\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)$$

such that

$$\left(\begin{array}{ccc}{a}_{23}& a_{13}& a_{33}\\ a_{22}& a_{12}& a_{32}\\ a_{21}& a_{11}& a_{31}\end{array}\right)\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)=\left(\begin{array}{ccc}0& 0& a_{44}\\ a_{44}& 0& 0\\ 0& a_{44}& 0\end{array}\right)$$

(The algebra $A_{4,6}$ is denoted by $J_{4,5}$ in [10]). Let us consider the following cases:

• ${\alpha }_1\ne 0$. Choose $\varphi$ as follows:

  $$\varphi ={\displaystyle \frac{1}{{\alpha }_1^2}}\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& {\alpha }_1& 0& 0\\ 0& 0& {\alpha }_1& 0\\ -{\alpha }_3& {\alpha }_4& {\alpha }_2& 1\end{array}\right).$$
  Then, $\theta \ast \varphi =\left(-{\Delta }_{1,3},{\Delta }_{2,3},{\Delta }_{1,2},0\right)$. So, we obtain the algebra ${\mathcal{J}}_{4,34}$.
• ${\alpha }_1=0,{\alpha }_3\ne 0$. Set $\lambda ={\alpha }_2^2-2{\alpha }_3{\alpha }_4$. Let $\varphi$ be the first of the following matrices if $\lambda =0$ or the second if $\lambda \ne 0$:

  $$\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_3}}& -{\alpha }_4& {\displaystyle \frac{{\alpha }_2}{{\alpha }_3}}& 0\\ 0& {\alpha }_3& 0& 0\\ 0& -{\alpha }_2& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}-{\displaystyle \frac{1}{{\alpha }_3^2}}\left({\alpha }_2^2-{\alpha }_3{\alpha }_4+\sqrt{\lambda }{\alpha }_2\right)& -{\displaystyle \frac{1}{2\lambda }}\left({\alpha }_3{\alpha }_4-{\alpha }_2^2+\sqrt{\lambda }{\alpha }_2\right)& {\displaystyle \frac{1}{\sqrt{\lambda }}}{\alpha }_4& 0\\ 1& -{\displaystyle \frac{1}{2\lambda }}{\alpha }_3^2& -{\displaystyle \frac{1}{\sqrt{\lambda }}}{\alpha }_3& 0\\ -{\displaystyle \frac{1}{{\alpha }_3}}\left({\alpha }_2+\sqrt{\lambda }\right)& -{\displaystyle \frac{1}{2{\lambda }^{\frac{3}{2}}}}{\alpha }_3\left(\lambda -\sqrt{\lambda }{\alpha }_2\right)& {\displaystyle \frac{1}{\sqrt{\lambda }}}{\alpha }_2& 0\\ 0& 0& 0& 1\end{array}\right).$$
  Then, $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,3}\right)$ if $\lambda =0$ or $\theta \ast \varphi =\left(0,0,0,\alpha {\Delta }_{1,2}\right)$ with $\alpha \ne 0$ if $\lambda \ne 0$. Thus, we obtain the algebras ${\mathcal{J}}_{4,35}$ and ${\mathcal{J}}_{4,36}^{\alpha \ne 0}$. Moreover, the algebras ${\mathcal{J}}_{4,36}^{\alpha \ne 0}$ and ${\mathcal{J}}_{4,36}^{\alpha \ne 0}$ are isomorphic if and only if ${\alpha }^2={\beta }^2$.
• ${\alpha }_1=0,{\alpha }_3=0$.

  –

  ${\alpha }_2\ne 0$. Choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& -{\displaystyle \frac{1}{2{\alpha }_2^2}}{\alpha }_4^2& {\displaystyle \frac{1}{{\alpha }_2}}{\alpha }_4& 0\\ 0& 1& 0& 0\\ 0& -{\displaystyle \frac{1}{{\alpha }_2}}{\alpha }_4& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,0,{\alpha }_2{\Delta }_{1,2}\right)$ and so we obtain again the algebras ${\mathcal{J}}_{4,36}^{\alpha \ne 0}$.

  –

  ${\alpha }_2=0$. If ${\alpha }_4=0$, then $\theta =0$ and we obtain the algebra ${\mathcal{J}}_{4,36}^{\alpha =0}$. If ${\alpha }_4\ne 0$, we choose $\varphi$ to be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}0& -{\alpha }_4& 0& 0\\ -{\displaystyle \frac{1}{{\alpha }_4}}& 0& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then, $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,3}\right)$ and we have again the algebra ${\mathcal{J}}_{4,35}$.

□

By some easy checking of identity (7) for all 4-dimensional general Jacobi–Jordan algebras in Theorem 6, we obtain the following result:

Theorem 7.

Let $\mathcal{A}$ be a complex noncommutative Jacobi–Jordan algebra of dimension 4. Then, $\mathcal{A}$ is isomorphic to one of the Lie algebras given in (Lemma 3 in [12]) or to one of the following algebras (as given in Theorem 6):

$$\begin{array}{lllll}• {\mathcal{A}}_{4,1}={\mathcal{J}}_{4,1}& • {\mathcal{A}}_{4,2}={\mathcal{J}}_{4,2}& • {\mathcal{A}}_{4,3}={\mathcal{J}}_{4,3}& • {\mathcal{A}}_{4,4}={\mathcal{J}}_{4,4}^{\alpha =-1}& • {\mathcal{A}}_{4,5}^{\alpha }={\mathcal{J}}_{4,5}^{\alpha }\\ • {\mathcal{A}}_{4,6}={\mathcal{J}}_{4,6}& • {\mathcal{A}}_{4,7}={\mathcal{J}}_{4,7}& • {\mathcal{A}}_{4,8}={\mathcal{J}}_{4,9}& • {\mathcal{A}}_{4,9}={\mathcal{J}}_{4,10}& • {\mathcal{A}}_{4,10}={\mathcal{J}}_{4,11}\\ • {\mathcal{A}}_{4,11}={\mathcal{J}}_{4,16}& • {\mathcal{A}}_{4,12}={\mathcal{J}}_{4,17}& • {\mathcal{A}}_{4,13}^{\alpha }={\mathcal{J}}_{4,19}^{\alpha }& • {\mathcal{A}}_{4,14}^{\alpha }={\mathcal{J}}_{4,21}^{\alpha }& • {\mathcal{A}}_{4,15}={\mathcal{J}}_{4,23}\\ • {\mathcal{A}}_{4,16}={\mathcal{J}}_{4,24}& • {\mathcal{A}}_{4,17}={\mathcal{J}}_{4,25}& • {\mathcal{A}}_{4,18}={\mathcal{J}}_{4,26}& • {\mathcal{A}}_{4,19}={\mathcal{J}}_{4,27}& • {\mathcal{A}}_{4,20}^{\alpha }={\mathcal{J}}_{4,28}^{\alpha }\\ • {\mathcal{A}}_{4,21}={\mathcal{J}}_{4,29}& • {\mathcal{A}}_{4,22}^{\alpha }={\mathcal{J}}_{4,30}^{\alpha }& • {\mathcal{A}}_{4,23}={\mathcal{J}}_{4,31}& • {\mathcal{A}}_{4,24}={\mathcal{J}}_{4,32}& • {\mathcal{A}}_{4,25}^{\alpha }={\mathcal{J}}_{4,33}^{\alpha }\\ • {\mathcal{A}}_{4,26}={\mathcal{J}}_{4,35}& • {\mathcal{A}}_{4,27}^{\alpha }={\mathcal{J}}_{4,36}^{\alpha }& & & \end{array}$$

The classification of all 4-dimensional nilpotent anticommutative algebras is given in [11]. And as a consequence of Theorem 6, we have the following classification of all 4-dimensional nilpotent complex general Jacobi–Jordan algebras:

Corollary 8.

Let $\mathcal{N}$ be a nilpotent complex general Jacobi–Jordan algebra of dimension 4. If $\mathcal{N}$ is anticommutative, then it is isomorphic to one of the following algebras:

• ${\mathcal{N}}_{4,1}:$ trivial algebra.
• ${\mathcal{N}}_{4,2}:e_1·e_2=e_3,e_2·e_1=-e_3.$
• ${\mathcal{N}}_{4,3}:e_1·e_2=e_3,e_2·e_1=-e_3,e_1·e_3=e_4,e_3·e_1=-e_4.$

Otherwise, it is isomorphic to one of the following algebras (as given in Theorem 6):

$$\begin{array}{lllll}• {\mathcal{N}}_{4,4}={\mathcal{J}}_{4,1}& • {\mathcal{N}}_{4,5}={\mathcal{J}}_{4,2}& • {\mathcal{N}}_{4,6}={\mathcal{J}}_{4,3}& • {\mathcal{N}}_{4,7}={\mathcal{J}}_{4,9}& • {\mathcal{N}}_{4,8}={\mathcal{J}}_{4,10}\\ • {\mathcal{N}}_{4,9}={\mathcal{J}}_{4,11}& • {\mathcal{N}}_{4,10}={\mathcal{J}}_{4,23}& • {\mathcal{N}}_{4,11}={\mathcal{J}}_{4,24}& • {\mathcal{N}}_{4,12}={\mathcal{J}}_{4,25}& • {\mathcal{N}}_{4,13}={\mathcal{J}}_{4,26}\\ • {\mathcal{N}}_{4,14}={\mathcal{J}}_{4,27}& • {\mathcal{N}}_{4,15}^{\alpha }={\mathcal{J}}_{4,28}^{\alpha }& • {\mathcal{N}}_{4,16}={\mathcal{J}}_{4,29}& • {\mathcal{N}}_{4,17}^{\alpha }={\mathcal{J}}_{4,30}^{\alpha }& • {\mathcal{N}}_{4,18}={\mathcal{J}}_{4,32}\\ • {\mathcal{N}}_{4,19}^{\alpha }={\mathcal{J}}_{4,33}^{\alpha }& • {\mathcal{N}}_{4,20}={\mathcal{J}}_{4,35}& • {\mathcal{N}}_{4,21}^{\alpha }={\mathcal{J}}_{4,36}^{\alpha }& & \end{array}$$

The next result is an immediate consequence of Theorem 7 and Corollary 8.

Corollary 9.

Any nilpotent complex general Jacobi–Jordan algebra of dimension 4 is a noncommutative Jacobi–Jordan algebra.

Remark 5.

It is noted that the nilpotent general Jacobi–Jordan algebras of dimensions $>4$ are not necessarily noncommutative Jacobi–Jordan algebras. For example, the 5-dimensional nilpotent non-Lie Malcev algebra with basis $\{e_1,\dots ,e_5\}$ and non-zero products

$$e_1·e_2=e_3,e_2·e_1=-e_3,e_3·e_4=e_5,e_4·e_3=-e_5,$$

is a nilpotent general Jacobi–Jordan algebra but not a noncommutative Jacobi–Jordan algebra.

6. Basic Concepts of Jordan Algebras

In this section, we state some basic notions and properties concerning Jordan algebras. We recall that a commutative algebra $(\mathcal{A},·)$ is a Jordan algebra if $\mathcal{A}$ satisfies the following Jordan identity:

$$x^2·(x·y)=x·(x^2·y),$$

(13)

for all $x,y\in \mathcal{A}$. The linearization of (13) is

$$(x,y,z·w)+(w,y,z·x)+(z,y,x·w)=0,$$

(14)

for $x,y,z,w\in \mathcal{A}$ (see ([8], p. 91) and ([7], p. 67)).

We say A is semisimple if it is a direct sum of simple algebras.

Proposition 10

([13]). Any finite-dimensional semisimple Jordan algebra has an identity element.

We recall that an element x of a Jordan algebra $\mathcal{A}$ is idempotent if satisfies $x^2=x$.

Proposition 11

([8,13]). Any finite-dimensional Jordan algebra which is not nilpotent contains a non-zero idempotent element.

The unique maximal nilpotent ideal of an algebra A is called the radical and denoted by $\mathcal{R}ad\left(A\right)$.

Theorem 8

([13]). Every finite-dimensional Jordan algebra $\mathcal{A}$ has a radical and the quotient $\mathcal{A}/\mathcal{R}ad\left(\mathcal{A}\right)$ is semisimple. Moreover, $\mathcal{A}=\mathcal{S}\oplus \mathcal{R}ad\left(\mathcal{A}\right)$ and $\mathcal{S}\cong \mathcal{A}/\mathcal{R}ad\left(\mathcal{A}\right)$, where ⊕ denotes the direct sum as vector spaces.

Definition 7.

An algebra A is decomposable if there are non-zero ideals $I_1$ and $I_2$ such that $A=I_1\oplus I_2$. Otherwise, it is indecomposable.

Let $\mathcal{A}$ be a Jordan algebra with idempotent element e. Then, the Peirce decomposition of $\mathcal{A}$ relative to the idempotent e is

$$\mathcal{A}={\mathcal{A}}_0\oplus {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}\oplus {\mathcal{A}}_1$$

where ${\mathcal{A}}_i={\mathcal{A}}_i\left(e\right):=\{x\in \mathcal{A}:x·e=ix\}$, with $i\in \{0,{\displaystyle \frac{1}{2}},1\}$.

Proposition 12

(see [13]). Let $\mathcal{A}={\mathcal{A}}_0\oplus {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}\oplus {\mathcal{A}}_1$ be the Peirce decomposition of the Jordan algebra $\mathcal{A}$ relative to e. Then, the multiplication table for the Peirce decomposition is

$$\begin{array}{ccc}{\mathcal{A}}_0^2\subset {\mathcal{A}}_0,& {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}^2\subset {\mathcal{A}}_0\oplus {\mathcal{A}}_1,& {\mathcal{A}}_1^2\subset {\mathcal{A}}_1,\\ {\mathcal{A}}_0·{\mathcal{A}}_{{\displaystyle \frac{1}{2}}}\subset {\mathcal{A}}_{{\displaystyle \frac{1}{2}}},& {\mathcal{A}}_0·{\mathcal{A}}_1=0,& {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}·{\mathcal{A}}_1\subset {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}.\end{array}$$

(15)

More generally, if $\mathcal{A}$ has unit $1={\sum }_{i=1}^n{e}_i$ which decomposes into a sum of pairwise orthogonal idempotents $e_i$, we have the Peirce decomposition of $\mathcal{A}$ relative to idempotents $\{e_1,\dots ,e_n\}$:

$$\mathcal{A}=\underset{1\le i\le j\le n}{⨁}{\mathcal{A}}_{ij}$$

(16)

where ${\mathcal{A}}_{ii}:=\{x\in \mathcal{A}:x·e_i=x\}$ and ${\mathcal{A}}_{ij}:=\{x\in \mathcal{A}:x·e_i=x·e_j={\displaystyle \frac{1}{2}}x\}$ for $i\ne j$. The multiplication table for the Peirce decomposition is

$$\begin{array}{c}{\mathcal{A}}_{ii}^2\subset {\mathcal{A}}_{ii},{\mathcal{A}}_{ij}·{\mathcal{A}}_{ii}\subset {\mathcal{A}}_{ij},{\mathcal{A}}_{ij}^2\subset {\mathcal{A}}_{ii}\oplus {\mathcal{A}}_{jj},\\ {\mathcal{A}}_{ij}·{\mathcal{A}}_{jk}\subset {\mathcal{A}}_{ik},{\mathcal{A}}_{ii}·{\mathcal{A}}_{jj}={\mathcal{A}}_{ii}·{\mathcal{A}}_{jk}={\mathcal{A}}_{ij}·{\mathcal{A}}_{kl}=0,\end{array}$$

(17)

where the indexes $i,j,k,l$ are all different (see [7,8,13]).

7. Malcev–Jordan Algebras

Definition 8.

An algebra $(\mathcal{A},·)$ is called a Malcev–Jordan algebra if it satisfies the following two identities:

$$x·y-y·x=0,$$

(18)

$$J(x,y,x·z)=J(x,y,z)·x,$$

(19)

for any $x,y,z\in \mathcal{A}$.

Another way to define a Malcev–Jordan algebra is as a commutative algebra which satisfies $J(x,y,x·z)=J(x,y,z)·x,$ for $x,y,z\in \mathcal{A}$. Note that any anticommutative algebra satisfying (19) is a Malcev algebra.

Lemma 12.

Any commutative associative algebra is a Malcev–Jordan algebra.

It is clear that the class of Malcev–Jordan algebras is significantly wider than that of Jacobi–Jordan algebras, and therefore, Malcev–Jordan algebras are not necessarily associative. Let $\mathcal{A}$ be a Malcev–Jordan algebra. If we consider $y=x$ in (19), we obtain $x^2·(x·z)=x·(x^2·z)$ for $x,y,z\in \mathcal{A}$. So, we have the following result.

Lemma 13.

Any Malcev–Jordan algebra is a Jordan algebra.

Remark 6.

In general, Jordan algebras are not necessarily Malcev–Jordan algebras. As a counterexample, let $(\mathcal{A},·)$ be a 3-dimensional Jordan algebra with basis $\{e_1,e_2,e_3\}$ and non-zero products

$$e_1^2=e_2^2=e_3^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3.$$

Then, $J(e_2,e_3,e_2·e_3)-J(e_2,e_3,e_3)·e_2=-e_1\ne 0$. Therefore, $\mathcal{A}$ is not a Malcev–Jordan algebra.

It is known that all Jacobi–Jordan algebras are nilpotent. This, in general, is not true for Malcev–Jordan algebras; they are not necessarily nilpotent.

Example 2.

Let $\mathcal{A}$ be a 2-dimensional semisimple commutative associative algebra with basis $\{e_1,e_2\}$ and non-zero products

$$e_1^2=e_1,e_2^2=e_2.$$

Clearly, any commutative associative algebra is a Malcev–Jordan algebra. Therefore, $\mathcal{A}$ is a Malcev–Jordan algebra.

The Structure of Malcev–Jordan Algebras

Proposition 13.

Let $(\mathcal{A},·)$ be a Malcev–Jordan algebra with a non-zero idempotent e. Then, $\mathcal{A}={\mathcal{A}}_0\oplus {\mathcal{A}}_1$ is a direct sum of ideals of $\mathcal{A}$, where ${\mathcal{A}}_i:=\{x\in \mathcal{A}:x·e=ix\}$, for $i\in \{0,1\}$.

Proof. 

Since $\mathcal{A}$ is a Jordan algebra as well as a Malcev–Jordan algebra, the Peirce decomposition of $\mathcal{A}$ relative to e is

$$\mathcal{A}={\mathcal{A}}_0\oplus {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}\oplus {\mathcal{A}}_1$$

where ${\mathcal{A}}_i:=\{x\in \mathcal{A}:x·e=ix\}$, for $i\in \{0,{\displaystyle \frac{1}{2}},1\}$. Now, for $x\in {\mathcal{A}}_{{\displaystyle \frac{1}{2}}}$,

$$x=2\left(J(e,x,e^2)-J(e,x,e)·e\right)=0,$$

so ${\mathcal{A}}_{{\displaystyle \frac{1}{2}}}=0$. Further, from (15), ${\mathcal{A}}_0$ and ${\mathcal{A}}_1$ are ideals of $\mathcal{A}$. □

Proposition 14.

Let $\mathcal{A}$ be a non-nilpotent Malcev–Jordan algebra. If $\mathcal{A}$ is indecomposable, then $\mathcal{A}$ is unital and has a unique idempotent element, that is, $e=1$.

Proof. 

By Proposition 11, there exists a non-zero idempotent $e\in \mathcal{A}$ and, by Proposition 13, it is $\mathcal{A}={\mathcal{A}}_0\oplus {\mathcal{A}}_1$, the direct sum of ideals of $\mathcal{A}$. Since $\mathcal{A}$ is indecomposable and $e\in {\mathcal{A}}_1\ne 0$, it follows that ${\mathcal{A}}_0=0$ and $\mathcal{A}={\mathcal{A}}_1$ so that $e=1$. Suppose that $e^{\prime }$ is an idempotent of $\mathcal{A}$. Then, ${\mathcal{A}}_1(e=1)=\mathcal{A}={\mathcal{A}}_1\left(e^{\prime }\right)$ and so $e^{\prime }=1$. □

Proposition 15.

Let $\mathcal{A}$ be a Malcev–Jordan algebra. If $\mathcal{A}$ has unit $1={\sum }_{i=1}^n{e}_i$ where all $e_i$ are orthogonal idempotent pairwise, then $\mathcal{A}={⨁}_{1\le i\le n}{\mathcal{A}}_{ii}$ is a direct sum of ideals of $\mathcal{A}$, where ${\mathcal{A}}_{ii}=\{x\in \mathcal{A}\mid x·e_i=x\}\ne 0$, for $i\in \{1,\dots ,n\}$.

Lemma 14.

Let $(\mathcal{A},·)$ be a commutative algebra. Consider the following identities: Then, $(\mathcal{A},·)$ satisfies

$$J(x,y,z)·x=J(x,y,x·z),$$

(20)

if and only if it satisfies

$$J(x,y,z)·t=J(x,y,t·z),$$

(21)

for $x,y,z,t\in \mathcal{A}$.

Proof. 

Let us suppose that $\mathcal{A}$ satisfies (20). The linearization of (20) is

$$J(x,y,z)·t+J(t,y,z)·x=J(x,y,t·z)+J(t,y,x·z),$$

for $x,y,z,t\in \mathcal{A}$. This is equivalent to

$$y·\left(t·(x·z)\right)+y·\left(x·(t·z)\right)+(t·z)·(x·y)+(x·z)·(t·y)=x·\left(t·(y·z)\right)+x·\left(z·(t·y)\right)+t·\left(x·(y·z)\right)+t·\left(z·(x·y)\right).$$

By (1), it can be rewritten as

$$y·\left(t·(x·z)\right)+y·\left(x·(t·z)\right)+(t,z,x·y)+(x,z,t·y)=x·\left(t·(y·z)\right)+t·\left(x·(y·z)\right).$$

(22)

Since $\mathcal{A}$ is a Jordan algebra as well, it satisfies the linearized Jordan identity (14). Thus, we have

$$(x,z,t·y)+(t,z,x·y)+(y,z,x·t)=0.$$

(23)

From (22) and (23), we have

$$y·\left(t·(x·z)\right)+y·\left(x·(t·z)\right)=x·\left(t·(y·z)\right)+t·\left(x·(y·z)\right)+(y,z,x·t).$$

Again, by (1) it can be rewritten as

$$y·\left(t·(x·z)\right)+y·\left(x·(t·z)\right)+y·\left(z·(x·t)\right)=x·\left(t·(y·z)\right)+t·\left(x·(y·z)\right)+(y·z)·(x·t).$$

This is equivalent to

$$J(t,x,z)·y=J(t,x,y·z).$$

Permuting $t\to x\to y\to t$ and leaving z fixed, we obtain (21) as desired. Conversely, by taking $t=x$ in (21), we obtain (20). □

Proposition 16.

Any unital Malcev–Jordan algebra is associative.

Proof. 

Let $\mathcal{A}$ be a unital Malcev–Jordan algebra with identity element 1. By considering $x=1$ in (21), we obtain $3t·(y·z)=3y·(t·z)$ for $y,z,t\in \mathcal{A}$. Since $\mathcal{A}$ is commutative, it follows that $\mathcal{A}$ is associative. □

From here, if $(\mathcal{A},·)$ is a non-associative Malcev–Jordan algebra, then its unital hull $\widehat{\mathcal{A}}:=\mathbb{F}1\oplus \mathcal{A}$ with product

$$(\alpha 1+x)·(\beta 1+y):=\alpha \beta 1+(\alpha y+\beta x+x·y),$$

for $\alpha ,\beta \in \mathbb{F}$ and $x,y\in \mathcal{A},$ is not a Malcev–Jordan algebra.

Proposition 17.

Let $\mathcal{A}$ be a Malcev–Jordan algebra. If $\mathcal{A}$ is simple, then $\mathcal{A}\cong \mathbb{F}$.

Proof. 

By Proposition 10 (also Proposition 14), $\mathcal{A}$ is a unital algebra. Moreover, by Proposition 16, $\mathcal{A}$ is a simple commutative associative algebra and so $\mathcal{A}\cong \mathbb{F}$. □

Proposition 18.

Let $\mathcal{A}$ be a Malcev–Jordan algebra. If $\mathcal{A}$ is indecomposable, then either of the following holds:

i. 

$\mathcal{A}$ is an indecomposable nilpotent Malcev–Jordan algebra;

ii. 

$\mathcal{A}$ is the unital hull of a nilpotent commutative associative algebra B (i.e., $\mathcal{A}=\widehat{B}=\mathbb{F}1\oplus B$, so $\mathcal{A}$ is a local unital commutative associative algebra).

Proof. 

By Theorem 8, $\mathcal{A}=\mathcal{S}\oplus \mathcal{R}ad\left(\mathcal{A}\right)$. If $S=0$ then $\mathcal{A}=\mathcal{R}ad\left(\mathcal{A}\right)$ is nilpotent. Otherwise, $\mathcal{A}$ is not nilpotent. By Propositions 14 and 16, $\mathcal{A}$ is a unital commutative associative algebra with a unique idempotent $e=1$. If $\mathcal{R}ad\left(\mathcal{A}\right)$ is non-associative, then so is $\mathcal{A}$. Thus, $\mathcal{R}ad\left(\mathcal{A}\right)$ is a nilpotent commutative associative algebra. By Proposition 17 and since $\mathcal{A}$ has a unique idempotent $e=1$, $\mathcal{S}=\mathbb{F}1$ and hence $\mathcal{A}=\mathbb{F}1\oplus \mathcal{R}ad\left(\mathcal{A}\right)$. □

From here, it is clear that the classification of nilpotent Malcev–Jordan algebras is sufficient for the classification of all other Malcev–Jordan algebras. In [10,14], the classification of (associative and non-associative) nilpotent Jordan algebras up to dimension 5 is given. This along with Proposition 18 yields, in particular, a classification of Malcev–Jordan algebras up to dimension 5.

Proposition 19.

All nilpotent Jordan algebras of dimensions $n\le 5$ are Malcev–Jordan algebras.

Proof. 

It is enough to show that any nilpotent Jordan algebra of dimension 5 is a Malcev–Jordan algebra. Thus, let $\mathcal{A}$ be a 5-dimensional nilpotent Jordan algebra. If $\mathcal{A}$ is associative, then it is a Malcev–Jordan algebra. If $\mathcal{A}$ is non-associative, then from the results of [10], it is easy to see that ${\mathcal{A}}^4=0$. Hence, $J(x,y,z)x=J(x,y,xz)=0$ for $x,y,z$ in $\mathcal{A}$. Therefore, $\mathcal{A}$ is a Malcev–Jordan algebra. □

Note that the classes of nilpotent Jordan algebras and Jordan nilalgebras are the same; see [8]. The following example shows that the nilpotent Jordan algebras of dimensions $>5$ are not necessarily Malcev–Jordan algebras.

Example 3.

Let $(\mathcal{A},·)$ be a 6-dimensional Jordan algebra with basis $\{u_1,\dots ,u_6\}$ and non-zero products

$$u_1^2=u_3,u_2^2=u_4,u_6^2=-{\displaystyle \frac{1}{2}}{u}_5,u_1·u_2=u_6,u_3·u_4=u_5.$$

This algebra is the unique, up to isomorphism, Jordan nilalgebra with a nilindex of 4 and dimension 6 such that ${\mathcal{A}}^{\langle 4\rangle }=0$ and ${\mathcal{A}}^{\langle 2\rangle }·{\mathcal{A}}^{\langle 2\rangle }\ne 0$ (see [15]). So, we have

$$J(u_1,u_1,u_2)·u_2=0$$

and

$$J(u_1,u_1,u_2^2)=u_1^2·u_2^2=u_3·u_4=u_5\ne 0.$$

Thus, $\mathcal{A}$ is not a Malcev–Jordan algebra.

From here, it is a natural problem to describe those nilpotent Jordan algebras of dimension 6 which lie beyond the limits of the narrower variety of Malcev–Jordan algebras. Let $\mathcal{A}$ be a 6-dimensional Jordan algebra such that $\mathcal{A}$ is not Malcev–Jordan algebra. If $\mathcal{A}$ is a nilalgebra with a nilindex of at most 3, then $\mathcal{A}$ is a Jacobi–Jordan algebra and hence a Malcev–Jordan algebra (see [5]). Also, if $\mathcal{A}$ is a nilalgebra with a nilindex of 7, then $\mathcal{A}$ is associative and so a Malcev–Jordan algebra. Furthermore, from the results of [15], we have

Lemma 15.

Let $(\mathcal{A},·)$ be a 6-dimensional Jordan nilalgebra. If $\mathcal{A}$ is not a Malcev–Jordan algebra, then $\mathcal{A}$ is a nilalgebra with a nilindex of 4 or 5. Moreover, if $\mathcal{A}$ is a nilalgebra with a nilindex of 4, then there exists a basis $\{u_1,u_2,u_3,u_4,u_5,u_6\}$ of $\mathcal{A}$ with non-zero products

$$u_1^2=u_3,u_2^2=u_4,u_6^2=-{\displaystyle \frac{1}{2}}{u}_5,u_1·u_2=u_6,u_3·u_4=u_5.$$

If $\mathcal{A}$ is a nilalgebra with a nilindex of 5, then exists a basis $\{u_1,u_2,u_3,u_4,u_5,u_6\}$ of $\mathcal{A}$ such that the products are given by one of the following list:

Type 1.

$u_1^2=4\beta u_2-4{\beta }^2{u}_4+{\gamma }_3{u}_5+{\gamma }_4{u}_6,u_1·u_2=-2{\beta }^2{u}_5+{\gamma }_0{u}_6,u_1·u_3=u_2,u_1·u_5=-2\beta u_6,$

$u_2·u_3=\beta u_5+\gamma u_6,u_3^2=u_4,u_3·u_4=u_5,u_3·u_5=u_6,u_4^2=u_6,\beta \ne 0.$

Type 2.

$u_1^2=-4\beta u_2-4{\beta }^2{u}_4+{\gamma }_3{u}_5+{\gamma }_4{u}_6,u_1·u_2=-6{\beta }^2{u}_5+{\gamma }_0{u}_6,u_1·u_3=u_2,u_1·u_5=-2\beta u_6,$

$u_2^2=-4{\beta }^2{u}_6,u_2·u_3=\beta u_5+\gamma u_6,u_3^2=u_4,u_3·u_4=u_5,u_3·u_5=u_6,u_4^2=u_6,\beta \ne 0.$

Type 3.

$u_1·u_4=u_2,u_1^2=\lambda u_2+\delta u_4+\gamma u_5+\epsilon u_6,u_1·u_2=\delta u_6,$

$u_3^2=u_4,u_3·u_4=u_5,u_3·u_5=u_6,u_4^2=u_6,\delta \ne 0.$

In the description of Lemma 15, the isomorphisms inside the class of Jordan nilalgebras with a nilindex of 5 are unknown, and there are three families of algebras depending either on 4 or 5 parameters. So, we have the following result:

Theorem 9.

Let $\mathcal{J}$ be a nilpotent Jordan algebra of dimension 6. Then, either $\mathcal{J}$ is a Malcev–Jordan algebra or it is isomorphic to one of the following algebras:

• ${\mathcal{J}}_{6,0}:a^2=c,b^2=d,e^2=-{\displaystyle \frac{1}{2}}f,a\circ b=e,c\circ d=f.$
• ${\mathcal{J}}_{6,1}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{1}{2}}e,d^2=f,b\circ e=2f.$
• ${\mathcal{J}}_{6,2}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{1}{2}}e,a\circ c=f,d^2=f,b\circ e=2f.$
• ${\mathcal{J}}_{6,3}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{1}{2}}e,a\circ d=f,d^2=f,b\circ e=2f.$
• ${\mathcal{J}}_{6,4}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{1}{2}}e,a\circ c=f,a\circ d=f,d^2=f,b\circ e=2f.$
• ${\mathcal{J}}_{6,5}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{3}{2}}e,c\circ d=f,a\circ e=f.$
• ${\mathcal{J}}_{6,6}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{3}{2}}e,b^2=f,c\circ d=f,a\circ e=f.$
• ${\mathcal{J}}_{6,7}:a^2=c,a\circ b=d,b\circ c=e,a\circ d={\displaystyle \frac{3}{2}}e,b\circ d=f,c\circ d=f,a\circ e=f.$
• ${\mathcal{J}}_{6,8}:a\circ b=c,a\circ c=d,b\circ c=e,a\circ e=f,b\circ d=f,c^2=f.$
• ${\mathcal{J}}_{6,9}:a\circ b=c,a\circ c=d,a^2=e,b\circ c=e,a\circ e=f,b\circ d=f,c^2=f.$
• ${\mathcal{J}}_{6,10}:a\circ b=c,b^2=d,a\circ c=d,a^2=e,b\circ c=e,a\circ e=f,b\circ d=f,c^2=f.$

Proof. 

Suppose that $\mathcal{J}$ is not a Malcev–Jordan algebra. Then, by Lemma 15, $\mathcal{J}$ is a nilalgebra with a nilindex of 4 or 5. If $\mathcal{J}$ is a nilalgebra with a nilindex of 4, then we have the algebra ${\mathcal{J}}_{6,0}$. On the other hand, if $\mathcal{J}$ is a nilalgebra with a nilindex of 5, then from Lemma 15, we can easily see that $\mathcal{J}/Ann\left(\mathcal{J}\right)$ is non-associative. All 6-dimensional nilpotent Jordan algebras $\mathcal{J}$ such that $\mathcal{J}/Ann\left(\mathcal{J}\right)$ is non-associative were classified in [16]. Therefore, we have the algebras ${\mathcal{J}}_{6,1},\dots ,{\mathcal{J}}_{6,10}$. □