1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you for your excellent answer! Thank you for helping me always! $\endgroup$ – user42804 Sep 21 '15 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.