Let $R$ be a commutative ring with $1\neq 0$ . An $R$ -algebra $A$ with multiplication not assumed to be associative is called a (commutative) Jordan algebra if

1. $A$ is commutative: $ab=ba$ , and
2. $A$ satisfies the Jordan identity: $(a^2b)a=a^2(ba)$ ,

The above can be restated as

1. $[A,A]=0$ , where $[\ , ]$ is the commutator bracket, and
2. for any $a\in A$ , $[a^2,A,a]=0$ , where $[\ , , ]$ is the associator bracket.

Remarks.

• If $A$ is a Jordan algebra such that $\operatorname{char}(A)\neq2$ , then $A$ is power-associative.
• If in addition $2=1+1\neq\operatorname{char}(A)$ , then by replacing $a$ with $a+1$ in the Jordan identity and simplifying, $A$ is flexible.
• Given any associative algebra $A$ , we can define a Jordan algebra $A^{+}$ . To see this, let $A$ be an associative algebra with associative multiplication $\cdot$ and suppose $2=1+1$ is invertible in $R$ . Define a new multiplication given by \begin{equation} ab=\frac{1}{2}(a\cdot b+b\cdot a). \end{equation}It is readily checked that this new multiplication satisifies both the commutative law and the Jordan identity. Thus $A$ with the new multiplication is a Jordan algebra and we denote it by $A^{+}$ . However, unlike Lie algebras, not every Jordan algebra is embeddable in an associative algebra. Any Jordan algebra that is isomorphic to a Jordan subalgebra of $A^{+}$ for some associative algebra $A$ is called a special Jordan algebra. Otherwise, it is called an exceptional Jordan algebra. As a side note, the right hand side of Equation (1) is called the Jordan product.
• An example of an exceptional Jordan algebra is $H_3(\mathbb{O})$ , the algebra of $3\times3$ Hermitian matrices over the octonions.