Im new to the topic so this could be basic nonsense to you.
Any Albert-Algebra $A$ has a trace map $T:A \rightarrow k$ and thus one can assign a quadratic form $q_A$ of rank $27$ by setting $q_A(x) = T(x^2)/2$.
Do we have: $A$ is split iff $q_A$ is isotropic ?
In Garibaldi,Merkurjev,Serre - Cohomological invariants in Galois Cohomology p.50 it says:
$q_A = <2,2,2> \oplus (q_3 \otimes <-a_1,-a_2,a_1a_2>)$ for reduced $A$ and certain $q_3,a_1,a_2$.
Why might this be wrong if $A$ is not reduced?
- Do you have an example of some $q_A$ which is a subform of a rank $32$ Pfisterform (which is a $5$-Pfisterform).
Edit: There is no obvious reason to ask question 3 but im interessted in the motivic decomposition of $q_A$ and if the answer to 3. is "Yes" the decomposition can be obtained by some easy formula of Rost.