into $F$ the proof is finished.

\[thm3.2\] Let $F=F_0(X, Y)$. Let $A=M_3(F)$ and ${\sigma}=\operatorname{ad}_q$, where $$q={\langle}X,Y,XY{\rangle}.$$ The ${\sigma}$-symmetric element $XY$ is totally ${\sigma}$-positive, but is not a sum of hermitian squares in $(A,{\sigma})$.

It is clear that $XY\in\operatorname{Sym}(A,{\sigma})$ since $XY\in F$.

We first show that $XY$ is totally ${\sigma}$-positive. Since $T_{\sigma}\simeq q{\otimes}q$ (see [@Lew p. 227] or [@BOI 11.4]) we have $$\operatorname{sign}_P T_{\sigma}=(\operatorname{sign}_P q)^2 \in \{1,9\}$$ for any ordering $P\in X_F$. (Here $\operatorname{sign}_P T_{\sigma}$ denotes the signature of the quadratic form $T_{\sigma}$ with respect to the ordering $P$.) Hence, the set of ${\sigma}$-orderings on $F$ is not empty. It is exactly the set of $P\in X_F$ with $\operatorname{sign}_P T_{\sigma}=9$. (Note that $F$ has orderings for which both $X$ and $Y$, and thus $XY$, are positive so that the value $\operatorname{sign}_P T_{\sigma}=9$ can indeed be attained.)

Let $P$ be any ${\sigma}$-ordering on $F$. Then we have for any $a\in A$, $$\operatorname{Trd}({\sigma}(a)a)\geq_P 0$$ (by definition) and so for any $a\in A$, $$\operatorname{Trd}\bigl({\sigma}(a) XY a\bigr)=XY \operatorname{Trd}({\sigma}(a)a)\geq_P 0,$$ since $XY\geq_P 0$ (for otherwise $\operatorname{sign}_P T_{\sigma}=1$ and $P$ would not be a ${\sigma}$-ordering on $F$). Hence, $XY$ is totally ${\sigma}$-positive. An alternative argument showing that $XY$ is totally ${\sigma}$-positive can be given by observing that 