PhyslibAlpha.ProbabilisticTheory.JordanOrderUnit.Power.Quadratic

Let $E$ be a Jordan algebra and $a \in E$. For any natural numbers $m, n \in \mathbb{N}$, the quadratic representation of the $m$-th power of $a$ acting on the $n$-th power of $a$ satisfies $U_{a^m}(a^n) = a^{2m+n}$.

In a Jordan algebra $E$, let $a \in E$ be an element and $n \in \mathbb{N}$ be a natural number. Let $a^n$ denote the $n$-th Jordan power of $a$ and $U_a$ denote the quadratic representation of $a$. Then, applying $U_a$ to $a^n$ raises the degree of the power by two: $U_a(a^n) = a^{n+2}$.

In a Jordan algebra $E$ with identity $1$, for any element $a \in E$ and any natural number $m \in \mathbb{N}$, the quadratic representation $U_{a^m}$ applied to the identity element $1$ satisfies $$U_{a^m}(1) = a^{2m}$$ where $a^k$ denotes the $k$-th Jordan power of $a$.

In a Jordan algebra $E$, let $U_x$ denote the quadratic representation of an element $x$, and let $a^n$ denote the $n$-th Jordan power of $a$. For any element $a \in E$ and natural numbers $m, n$, and $k$, the composition of quadratic representations acting on powers of $a$ satisfies $$U_{a^m}(U_{a^n}(a^k)) = a^{2m + 2n + k}.$$