PhyslibAlpha.AlgebraicFramework.Algebra.NuclearInvolution

An element $x$ in an algebra $D$ is nuclear if it associates trivially in every slot. Specifically, for all $y, z \in D$, the associator $[a, b, c] = (ab)c - a(bc)$ satisfies $[x, y, z] = 0$, $[y, x, z] = 0$, and $[y, z, x] = 0$.

In an algebra $D$, if $n \in D$ is a nuclear element—meaning that for all $a, b \in D$, the associator $[a, b, c] = (ab)c - a(bc)$ satisfies $[n, a, b] = [a, n, b] = [a, b, n] = 0$—then for any elements $x, y, z \in D$, the associator satisfies the identity $[nx, y, z] = n[x, y, z]$.

Let $D$ be an algebra. An element $n \in D$ is said to be nuclear if it associates trivially in every position, meaning that for all $a, b \in D$, the associator $[x, y, z] = (xy)z - x(yz)$ satisfies $[n, a, b] = 0$, $[a, n, b] = 0$, and $[a, b, n] = 0$. For any such nuclear element $n$ and any elements $x, y, z \in D$, the identity $[x, ny, z] = n[x, y, z]$ holds.

Let $D$ be an algebra. Suppose $n \in D$ is a nuclear element, meaning it associates trivially in every slot (i.e., $[n, a, b] = [a, n, b] = [a, b, n] = 0$ for all $a, b \in D$). For any elements $x, y, z \in D$, the associator satisfies the identity $[x, y, n z] = n [x, y, z]$, where $[x, y, z] = (xy)z - x(yz)$.

Let $D$ be an algebra. If $n \in D$ is a nuclear element—meaning that for all $a, b \in D$, the associator $[a, b, c] = (ab)c - a(bc)$ satisfies $[n, a, b] = [a, n, b] = [a, b, n] = 0$—then for any elements $x, y, z \in D$, the associator satisfies the identity $[x, y, z \cdot n] = [x, y, z] \cdot n$.

Let $D$ be an alternative algebra equipped with a star involution denoted by $x \mapsto x^*$. For any element $x \in D$, the sum $x + x^*$ is nuclear. An element $n \in D$ is defined as nuclear if it associates trivially with all elements $y, z \in D$ in the sense that the associator $[a, b, c] = (ab)c - a(bc)$ satisfies $[n, y, z] = 0$, $[y, n, z] = 0$, and $[y, z, n] = 0$.

In an alternative algebra $D$ equipped with a star involution, let $[x, y, z] = (xy)z - x(yz)$ denote the associator of elements $x, y, z \in D$. For any elements $x, y, z \in D$, the associator of the star of $x$ with $y$ and $z$ is the negative of the associator of $x, y, z$: $$[x^*, y, z] = -[x, y, z],$$ where $x^*$ denotes the star involution applied to $x$.

Let $D$ be a $\ast$-alternative algebra. For any elements $x, y, z \in D$, the associator $(x, y, z) = (xy)z - x(yz)$ satisfies the identity $(x, y^*, z) = -(x, y, z)$, where $y^*$ denotes the star involution of $y$.

In an alternative algebra $D$ equipped with a star involution $x \mapsto x^*$, for any elements $x, y, z \in D$, the associator $[x, y, z] = (xy)z - x(yz)$ satisfies the identity $[x, y, z^*] = -[x, y, z]$.

Let $D$ be an algebra and $n \in D$ be a nuclear element, meaning that for all $y, z \in D$, the associator $[a, b, c] = (ab)c - a(bc)$ satisfies $[n, y, z] = 0$, $[y, n, z] = 0$, and $[y, z, n] = 0$. Then for any $x, y, z \in D$, the element $n$ commutes with the associator of $x, y, z$, such that $n \cdot [x, y, z] = [x, y, z] \cdot n$.