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Version 4 |
| \PMlinkescapeword{associate} |
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| Let $A$ be an algebra, not necessarily associative multiplicatively. The \emph{nucleus} of $A$ is: |
Let $A$ be an algebra, not necessarily associative multiplicatively. The \emph{nucleus} of $A$ is: |
| $$\mathcal{N}(A):=\lbrace a\in A\mid [a,A,A]=[A,a,A]=[A,A,a]=0 \rbrace,$$ |
$$\mathcal{N}(A):=\lbrace a\in A\mid [a,A,A]=[A,a,A]=[A,A,a]=0 \rbrace,$$ |
| where $[\ , , ]$ is the associator bracket. In other words, the nucleus is the set of elements that multiplicatively associate with all elements of $A$. |
where $[\ , , ]$ is the associator bracket. In other words, the nucleus is the set of elements that multiplicatively associate with all elements of $A$. |
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| $\mathcal{N}(A)$ is a subalgebra of $A$. To see this, let $a,b\in \mathcal{N}(A)$. Then for any $c,d\in A$, |
$\mathcal{N}(A)$ is a subalgebra of $A$. To see this, let $a,b\in \mathcal{N}(A)$. Then for any $c,d\in A$, |
| \begin{eqnarray} |
\begin{eqnarray} |
| [ab,c,d] &=& ((ab)c)d-(ab)(cd) = (a(bc))d-(ab)(cd) \\ |
[ab,c,d] &=& ((ab)c)d-(ab)(cd) = (a(bc))d-(ab)(cd) \\ |
| &=& a((bc)d)-(ab)(cd) = a(b(cd))-(ab)(cd) \\ |
&=& a((bc)d)-(ab)(cd) = a(b(cd))-(ab)(cd) \\ |
| &=& a(b(cd))-a(b(cd)) = 0 |
&=& a(b(cd))-a(b(cd)) = 0 |
| \end{eqnarray} |
\end{eqnarray} |
| Similarly, $[c,ab,d]=[c,d,ab]=0$ and so $ab\in\mathcal{N}(A)$. |
Similarly, $[c,ab,d]=[c,d,ab]=0$ and so $ab\in\mathcal{N}(A)$. |
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| 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): |
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): |
| $$\mathcal{Z}(A):=\lbrace a\in \mathcal{N}(A)\mid [a,A]=0 \rbrace,$$ |
$$\mathcal{Z}(A):=\lbrace a\in \mathcal{N}(A)\mid [a,A]=0 \rbrace,$$ |
| where $[\ , ]$ is the commutator bracket. |
where $[\ , ]$ is the commutator bracket. |
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| 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 subalgebra of $A$. |
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 subalgebra of $A$. |
