Jordan identity

Remarks

Proof idea: $x \circ x = x^2$, so both sides are symmetrized products with $x^2$. Expanding gives $(x \circ y) \circ x^2 = \frac{1}{4}(xyx^2 + yx^3 + x^3 y + x^2 y x)$. Expanding the right side gives $\frac{1}{4}(xyx^2 + x^3 y + yx^3 + x^2 y x)$, the same four terms. The product $\circ$ is commutative, so $A^+$ satisfies both axioms of a Jordan algebra.

Example: for real $2 \times 2$ matrices, $x = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has $x^2 = 0$, so both sides vanish for every $y$.

Equivalent forms: write $L_a$ for the map $b \mapsto a \circ b$. In a commutative algebra the Jordan identity says that $L_x$ and $L_{x \circ x}$ commute for every $x$. The product $\circ$ is not associative in general. For the $2 \times 2$ matrix units $e_{ij}$, one has $(e_{12} \circ e_{21}) \circ e_{11} = \frac{1}{2} e_{11}$ but $e_{12} \circ (e_{21} \circ e_{11}) = \frac{1}{4}(e_{11} + e_{22})$.

Going deeper

For a student

The algebras $A^+$ and their subalgebras are the special Jordan algebras, and they are the main source of examples. The Jordan identity implies that a Jordan algebra is power-associative, so powers $x^n$ are well defined. Linearizing the identity, that is, replacing $x$ by $x + t z$ and comparing coefficients of $t$, gives multilinear identities that hold in every Jordan algebra over a field of characteristic $0$. Not every Jordan algebra is of the form $A^+$ or a subalgebra of one: the Albert algebra of $3 \times 3$ Hermitian octonion matrices is exceptional.