# 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 $c,d\in A$,

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

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

$=$ | $a(b(cd))-a(b(cd))=0$ | (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 $A$.

Title | nucleus |
---|---|

Canonical name | Nucleus |

Date of creation | 2013-03-22 14:52:19 |

Last modified on | 2013-03-22 14:52:19 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 10 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 17A01 |

Defines | center of a nonassociative algebra |

Defines | nuclear |