Let $A$ and $B$ be two simple Euclidean Jordan Algebras. It is well known that $A\times B$ can be made into a Jordan Algebra in a unique way by defining the operation component-wise. Is it true that the product Jordan Algebra can be made Euclidean in a unique way? (the inner product on $A\times B$ is the sum of the corresponding inner products on $A$ and $B$)

  • 1
    $\begingroup$ To expect uniqueness, you probably want the requirement that it extends the original ones $\phi_A$ and $\phi_B$ on both $A\times\{0\}$ and $\{0\}\times B$. Otherwise you can take any positive linear combination of $\phi_A$ and $\phi_B$. $\endgroup$ – YCor Apr 17 '20 at 7:36
  • $\begingroup$ My guess is that as long you require the inner product of an atomic idempotent with itself to be normalized, that the inner product is unique. This probably follows in some way from the classification theorem for Euclidean Jordan algebras. $\endgroup$ – John Feb 23 at 13:10

Your Answer

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

Browse other questions tagged or ask your own question.