most recent 30 from http://mathoverflow.net 2013-05-20T05:20:20Z http://mathoverflow.net/feeds/question/48151 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48151/a-silly-technical-question-on-albert-algebras Pete L. Clark 2010-12-03T08:20:09Z 2010-12-03T09:04:50Z

<p>Apologies in advance for this spectacularly uninteresting question, but it has just come up in my work. (Okay, not in a truly important way, but I am trying to gauge the scope of a certain construction.) </p> <p>Let $K$ be a field and $A$ be a division <a href="http://en.wikipedia.org/wiki/Albert_algebra" rel="nofollow">Albert algebra</a> over $K$, i.e., a certain kind of $27$-dimensional commutative Jordan algebra over $K$. </p> <blockquote> <p>Is it true that for all nonzero elements $x,y \in A$, one has $(xy) y^{-1} = x$?</p> </blockquote> <p>Note that this is true in any finite-dimensional division algebra in which each subalgebra generated by two elements is associative, in particular in any composition algebra. But so far as I know (and by the way I know nothing about Albert algebras!), this "2-associativity" property does not hold here.</p>

http://mathoverflow.net/questions/48151/a-silly-technical-question-on-albert-algebras/48154#48154 Bugs Bunny 2010-12-03T09:04:50Z 2010-12-03T09:04:50Z

<p>No. In an associative algebra $(xy+yx)y^{-1} + y^{-1}(xy+yx)= 2x+ yxy^{-1} + y^{-1}xy \neq 4x$ unless $x$ and $y$ commute. Hence, it does not hold even in special Jordan algebras.</p>