how to write down comatrix of the exceptional Jordan algebra

Question

Suppose we have exceptional Jordan algebra, which is a $3\times3$ matrix $X=\left(\begin{matrix}x_1&\phi_1&\phi_2\\\bar{\phi_1}&x_2&\phi_3\\\bar{\phi_2}&\bar{\phi_3}&x_3\\\end{matrix}\right)$, where $x_1,x_2,x_3$ are real number and $\phi_1,\phi_2,\phi_3$ are octonions. $\bar{\phi_1},\bar{\phi_2},\bar{\phi_3}$ are the conjugation of $\phi_1,\phi_2,\phi_3$.(conjugation means:$ a_0e_0+a_1e_1+a_2e_2+...+a_7e_7\rightarrow a_0e_0-a_1e_1-a_2e_2-...-a_7e_7$). And det(X) is defined by$\frac{1}{6}tr(X(XX))+\frac{1}{6}tr((XX)X)-\frac{1}{2}tr(X^2)tr(X)+\frac{1}{6}tr(X)^3$

My question is how to write down the comatrix of X,COM(X), which satisfies X COM(X)=det(X) I , here we assume det(X) is not zero.

Answer 1

The Jordan algebra is power associative, so the answer is $$ COM(X) = X^2 - tr(X) X + \sigma_2(X) I, $$ where $\sigma_2(X) = \tfrac12\bigl(tr(X)^2 - tr(X^2)\bigr)$.

From another point of view, this follows from the fact that, under the automorphism group $\mathrm{F}_4$ of the Jordan algebra $J$, every element is equivalent to a diagonal element. In the case of diagonal elements, this formula holds, so it's true in general.