# Reference request: The relationship between norm and trace forms on an Albert algebra

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!

• Perhaps, this link can be useful for you :Jordan operator algebra Aug 3, 2013 at 9:00
• @Sebastien: Thanks, I've certainly already made the effort to look at wikipedia.
– SMF
Aug 5, 2013 at 17:10
• 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... Aug 5, 2013 at 18:43
• 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).
– SMF
Aug 6, 2013 at 15:40
• 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. Aug 6, 2013 at 16:16