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Homenucleus

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# nucleus

Let $A$ be an algebra, not necessarily associative multiplicatively. The *nucleus* of $A$ is:

$\mathcal{N}(A):=\{a\in A\mid[a,A,A]=[A,a,A]=[A,A,a]=0\},$ |

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 *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 $c,d\in A$,

$\displaystyle[ab,c,d]$ | $\displaystyle=$ | $\displaystyle((ab)c)d-(ab)(cd)=(a(bc))d-(ab)(cd)$ | (1) | ||

$\displaystyle=$ | $\displaystyle a((bc)d)-(ab)(cd)=a(b(cd))-(ab)(cd)$ | (2) | |||

$\displaystyle=$ | $\displaystyle a(b(cd))-a(b(cd))=0$ | (3) |

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 *center of a nonassociative algebra* $A$ (which is slightly different from the definition of the center of an associative algebra):

$\mathcal{Z}(A):=\{a\in\mathcal{N}(A)\mid[a,A]=0\},$ |

where $[\ ,]$ is the commutator bracket.

Hence elements in $\mathcal{Z}(A)$ commute *as well as* associate with all elements of $A$. Like the nucleus, the center of $A$ is also a Jordan subalgebra of $A$.

## Mathematics Subject Classification

17A01*no label found*

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