matrix algebra with a *symplectic* involution, we will show in Theorem \[thm4.7\] below that the conjecture does hold.

Thus, we could ask if the PS Conjecture also holds for non-split central simple algebras with symplectic involution. The answer is “no”:

\[thm3.3\] Let $F=F_0(X, Y)$. Let $A=M_3(F){\otimes}_F H\cong M_3(H)$, where $H=(-1,-1)_F$ is Hamilton’s quaternion division algebra over $F$. Equip $A$ with the involution ${\sigma}=\operatorname{ad}_q{\otimes}\gamma$, where $\gamma$ is quaternion conjugation and ${\sigma}=\operatorname{ad}_q$ for $$q={\langle}X,Y,XY{\rangle}.$$ The algebra $A$ is central simple over $F$ of degree $6$ and the involution ${\sigma}$ is symplectic. The ${\sigma}$-symmetric element $XY$ is totally ${\sigma}$-positive, but is not a sum of hermitian squares in $(A,{\sigma})$.

The assertion about $(A,{\sigma})$ is clear, as is the fact that $XY\in\operatorname{Sym}(A,{\sigma})$ since $XY\in F$.

It is easy to verify that the involution trace form of $\gamma$, $T_\gamma$, is isometric to ${\langle}2{\rangle}{\otimes}N_H$, where $N_H={\langle}1,1,1,1{\rangle}$ is the norm form of $H$. Here $N_H(x):=\operatorname{Nrd}_H(x)$ for all $x\in H$, where $\operatorname{Nrd}_H$ denotes the reduced norm on $H$. Since $T_{\sigma}=T_{\operatorname{ad}_q{\otimes}\gamma}\simeq T_{\operatorname{ad}_q}{\otimes}T_\gamma$, we have $$\operatorname{sign}_P T_{\sigma}= (\operatorname{sign}_P T_{\operatorname{ad}_q})(\operatorname{sign}_P T_\gamma) = 4\operatorname{sign}_P T_{\operatorname{ad}_q}\in \{4, 36\}$$ for any ordering $P\in X_F$. Hence, the set of ${\sigma}$-orderings on $F$ is not empty. It is exactly the set of $P\in X_F$ with $\operatorname{sign}_P 