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Jordan algebra
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: $(a^{2}b)a=a^{2}(ba)$,
for any $a,b\in A$.
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.
If $A$ is a Jordan algebra, a subset $B\subseteq A$ is called a Jordan subalgebra if $BB\subseteq B$. Let $A$ and $B$ be two Jordan algebras. A Jordan algebra homomorphism, or simply Jordan homomorphism, from $A$ to $B$ is an algebra homomorphism that respects the above two laws. A Jordan algebra isomorphism is just a bijective Jordan algebra homomorphism.
Remarks.

If $A$ is a Jordan algebra such that $\operatorname{char}(A)\neq 2$, then $A$ is powerassociative.

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
$ab=\frac{1}{2}(a\cdot b+b\cdot a).$ (1) 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 $3\times 3$ Hermitian matrices over the octonions.
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