# Is there a “conceptual” proof that an Albert algebra is a Jordan algbera?

That is, that $x(yx^2) = (xy)x^2$ ?

In the original paper of A. Albert the proof is based on a direct calculation, so it is not even clear why $dim=3$ is important. References show that since then a rich theory with many generalizations has been developed - unfortunately too rich for a beginner to see the wood for the trees.

-
The question could benefit from some clarifications. Usually an Albert algebra is by definition a simple exceptional Jordan algebra of dimension $27$, so I think the question is really why the the $3 \times 3$ Hermitian matrices over an octonion algebra form a Jordan algebra. Also, I think the hardest thing is to show exceptionality. Something like what you are looking for can be found in McCrimmon's book where he talks about Jordan algebras defined from cubic forms. Also see chapter IX "Cubic Jordan Algebras" of "The Book of Involutions" by Knus, Merkurjev, Rost, and Tignol. –  Dan Fox Jan 1 '13 at 13:34

Why 3×3 Hermitian matrices over an octonion algebra form a Jordan algebra and what means its exceptionality is explained in a very nice book by Kevin McCrimmon "A taste of Jordan algebras" http://www.amazon.com/books/dp/0387954473 (I think it is possible to find an electronic version on the web). Much shorter review article by the same author "Jordan algebras and their applications": http://www.ams.org/journals/bull/1978-84-04/S0002-9904-1978-14503-0/ is also worth to read.

-