PhyslibAlpha.ProbabilisticTheory.JordanOrderUnit.Compatibility

Let $E$ be a Jordan algebra. Two elements $a, b \in E$ are Jordan-compatible if their corresponding left multiplication operators $L_a$ and $L_b$ commute, such that $L_a \circ L_b = L_b \circ L_a$, where $L_x(y) = x \cdot y$. This serves as the generalization of commutativity for non-associative Jordan structures.

Let $E$ be a Jordan algebra. For any element $a \in E$, $a$ is Jordan-compatible with itself.

Let $E$ be a Jordan algebra. For any elements $a, b \in E$, if $a$ and $b$ are Jordan-compatible, then $b$ and $a$ are also Jordan-compatible. (Recall that $a$ and $b$ are Jordan-compatible if their left multiplication operators $L_a$ and $L_b$ commute, i.e., $L_a \circ L_b = L_b \circ L_a$).

In a Jordan algebra $E$ with unit element $1$, the left multiplication operator $L_1$ (defined such that $L_1(x) = 1 \cdot x$ for all $x \in E$) is equal to the identity linear map $\text{id}_E$.

In a Jordan algebra $E$ with unit $1$, for any element $a \in E$, the unit element $1$ is Jordan-compatible with $a$. This means that the left multiplication operator $L_1$ (where $L_1(x) = 1 \cdot x$) commutes with the left multiplication operator $L_a$ (where $L_a(x) = a \cdot x$), such that $L_1 \circ L_a = L_a \circ L_1$. Since $L_1$ is the identity map on $E$, this condition is satisfied for all $a$.

In a Jordan algebra $E$ with unit element $1$, every element $a \in E$ is Jordan-compatible with $1$. Specifically, the left multiplication operator $L_a$ and the left multiplication operator $L_1$ commute, such that $L_a \circ L_1 = L_1 \circ L_a$.