# Square of Pierce 1/2 elements

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?

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 $U_{p'} x = 0$.