<?xml version="1.0" encoding="UTF-8"?>

<record version="8" id="6351">
 <title>flexible algebra</title>
 <name>FlexibleAlgebra</name>
 <created>2004-10-10 14:34:05</created>
 <modified>2008-10-07 19:46:25</modified>
 <type>Definition</type>
 <creator id="3771" name="CWoo"/>
 <author id="3771" name="CWoo"/>
 <classification>
	<category scheme="msc" code="17A20"/>
 </classification>
 <defines>
	<concept>left power</concept>
	<concept>right power</concept>
	<concept>flexible</concept>
 </defines>
 <related>
	<object name="Associator"/>
	<object name="AlternativeAlgebra"/>
 </related>
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 <content>A non-associative algebra $A$ is \emph{flexible} if $[\ a,b,a\ ]=0$ for all $a,b\in A$, where $[\ , , ]$ is the associator on $A$.  In other words, we have $(ab)a=a(ba)$ for all $a,b\in A$.  Any associative algebra is clearly flexible.  Furthermore, any alternative algebra with characteristic $\neq 2$ is flexible.

Given an element $a$ in a flexible algebra $A$, define the \emph{left power} of $a$ iteratively as follows:
\begin{enumerate}
\item $L^1(a)=a$,
\item $L^n(a)=a\cdot L^{n-1}(a)$.
\end{enumerate}
Similarly, we can define the \emph{right power} of $a$ as:
\begin{enumerate}
\item $R^1(a)=a$,
\item $R^n(a)=R^{n-1}(a)\cdot a$.
\end{enumerate}
Then, we can show that $L^{n}(a)=R^{n}(a)$ for all positive integers $n$.  As a result, in a flexible algebra, one can define the (multiplicative) power of an element $a$ as $a^n$ unambiguously.</content>
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