Let $p$ be a nontrivial idempotent in a JB-algebra $A$ with Pierce decomposition $A = A\_1 \oplus A\_{1/2} \oplus A\_0$. Then the projection onto $A\_1$ (resp. $A\_0$) is given by $U\_p$ (resp. $U\_{p'}$). Let $x \in A\_{1/2}$ be a nonzero element. Then $x^2 \in A\_1 \oplus A\_0$.


The statement I am interested in is that both $U\_p x^2$ as well as $U\_{p'} x^2$ are nonzero. This is true, since by the Shirshov-Cohn Theorem the algebra generated by $x$ and $p$ is special, so one can view this algebra as an algebra of operators on a Hilbert space. Then $x$ is of the form $\left( \begin{array}{cc} 0 & B \\ B^\* & 0 \end{array} \right)$, and so $x^2$ is of the form $\left( \begin{array}{cc} BB^\* & 0 \\ 0 & B^\*B \end{array} \right)$.


This proof is far from elementary. Is there a much easier proof of this statement?



Square of Pierce 1/2 elements
After some more research I found a much easier proof: it follows immediately from Corollary 2.10 in "A Gelfand-Neumark Theorem for Jordan Algebras" by Alfsen, Schultz and Stormer. This corollary states that for positive $x \in A$, $x = U\_p x$ if and only if $U\_{p'} x = 0$.


