I am interested in either a nice reference, or some clarification.

Overview: I am considering $J_3(\mathbb{O})$, the Jordan algebra of $3\times 3$ self adjoint octonionic matrices. This algebra is a Jordan Banach algebra, which means it is equipped with a cubic norm $||.||_n$ which satisfies: $$ ||ab||_n\leq||a||_n.||b||_n,\hspace{0.2cm}||a^2||_n = ||a||_n^2,\hspace{0.2cm}||a^2||_n\leq ||a^2 + b^2||_n. $$ for $a,b\in J_3(\mathbb{O})$. This algebra is also equipped with a symmetric, positive definite, bilinear, nondegenerate trace form $$ Tr(a,b) = Tr(ab) $$

Question: If I represent $J_3(\mathbb{O})$ on itself in the natural way (ie using the associative multiplication algebra of left and right acting algebra elements), then it seems (at least naively) that I can use the trace form to define an inner product structure on the vector space. It then seems as though I can use this inner product to define an operator norm. $$ ||A||_o := sup\{||Av||_t:||v||_t=1\} $$ where the trace norm $||v||_t := \sqrt{Tr(v,v)}$ is defined on elements of the vector space. If everything I have said above is correct, I want to know how this operator norm $||.||_o$ on the 'representation' of $J_3(\mathbb{O})$ on itself relates (if at all) to the Norm form $||.||_n$ defined on $J_3(\mathbb{O})$. Is it possible for example that these two norms define the same topology?

Or perhaps stated another way, is it possible to define the norm form on $J_3(\mathbb{O})$ such that $$ ||a||_n = ||L_a||_o $$

I hope this question is clear!

  • $\begingroup$ Perhaps, this link can be useful for you :Jordan operator algebra $\endgroup$ Aug 3, 2013 at 9:00
  • $\begingroup$ @Sebastien: Thanks, I've certainly already made the effort to look at wikipedia. $\endgroup$
    – SMF
    Aug 5, 2013 at 17:10
  • $\begingroup$ Ok, so if your question is not "trivial for someone" here, you can also open a talk on wikipedia with the contributors of the page Albert algebra, in order to update this page by an answer of your question, if you will... $\endgroup$ Aug 5, 2013 at 18:43
  • $\begingroup$ My apologies, I am not a regular on the stack exchange. Is my question or the way I have asked it inappropriate (honest question)? In any case I have edited the language a little bit to reflect your response.$$ $$ The page could certainly be updated. They don't even mention the trace form (for example). $\endgroup$
    – SMF
    Aug 6, 2013 at 15:40
  • $\begingroup$ There is no problem for me. But my point of view is not "representative" of what is correct on mathoverflow, because I'm active on, only since 2 months. What I note is that the subject "non-associative algebras" is very few developed on mathoverflow (there are very few followers and questions on this subject). Perhaps this will changed, because the non-associative algebras are more and more fashionable. In the meantime, as I said, if you want, you can try to open a talk with some experts on Wikipedia. I'm not at all an expert but I know some of them. $\endgroup$ Aug 6, 2013 at 16:16


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