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!

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