Nucleus 2004-12-09 16:57:12 2004-12-15 13:00:39 Definition center of a nonassociative algebra nuclear % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \PMlinkescapeword{associate} Let $A$ be an algebra, not necessarily associative multiplicatively. The \emph{nucleus} of $A$ is: $$\mathcal{N}(A):=\lbrace a\in A\mid [a,A,A]=[A,a,A]=[A,A,a]=0 \rbrace,$$ where $[\ , , ]$ is the associator bracket. In other words, the nucleus is the set of elements that multiplicatively associate with all elements of $A$. An element $a\in A$ is \emph{nuclear} if $a\in\mathcal{N}(A)$. $\mathcal{N}(A)$ is a Jordan subalgebra of $A$. To see this, let $a,b\in \mathcal{N}(A)$. Then for any $c,d\in A$, \begin{eqnarray} [ab,c,d] &=& ((ab)c)d-(ab)(cd) = (a(bc))d-(ab)(cd) \\ &=& a((bc)d)-(ab)(cd) = a(b(cd))-(ab)(cd) \\ &=& a(b(cd))-a(b(cd)) = 0 \end{eqnarray} Similarly, $[c,ab,d]=[c,d,ab]=0$ and so $ab\in\mathcal{N}(A)$. Accompanying the concept of a nucleus is that of the \emph{center of a nonassociative algebra} $A$ (which is slightly different from the definition of the center of an associative algebra): $$\mathcal{Z}(A):=\lbrace a\in \mathcal{N}(A)\mid [a,A]=0 \rbrace,$$ where $[\ , ]$ is the commutator bracket. Hence elements in $\mathcal{Z}(A)$ commute \emph{as well as} associate with all elements of $A$. Like the nucleus, the center of $A$ is also a Jordan subalgebra of $A$.