T_{\sigma}=36$. (Note again that this value can indeed be attained since there are orderings on $F$ for which both $X$ and $Y$, and thus $XY$, are positive.) Arguing similarly as in the proof of Theorem \[thm3.2\] we can verify that $XY$ is totally ${\sigma}$-positive.

Before proceeding, note that the involution $\gamma$ is adjoint to the hermitian form ${\langle}1{\rangle}_\gamma$ over $(H,\gamma)$. Hence, ${\sigma}$ is adjoint to the hermitian form $h=q{\otimes}{\langle}1{\rangle}_\gamma ={\langle}X,Y,XY{\rangle}_\gamma$ over $(H,\gamma)$. Thus $$h(x,y)= \gamma(x_1) X y_1 +\gamma(x_2) Y y_2 + \gamma(x_3)XY y_3$$ for vectors $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ in the right $H$-vector space $H^3$.

Next we show that $XY$ is not a sum of hermitian squares in $(A,{\sigma})=(M_3(H), \operatorname{ad}_h)$. We identify $XY$ with $XYI_3$ in $M_3(H)$, where $I_3$ denotes the $3{\times}3$ identity matrix. Assume for the sake of contradiction that $XYI_3$ is a sum of elements of the form ${\sigma}(a)a$ with $a=[a_{ij}]_{1\leq i,j\leq 3} \in M_3(H)$. Recall that $${\sigma}(a)a =\operatorname{ad}_h(a)a=
\left[\begin{smallmatrix}
X & & \\
& Y & \\
& & XY
\end{smallmatrix}\right] \cdot \gamma(a)^t\cdot 
\left[\begin{smallmatrix}
X & & \\
& Y & \\
& & XY
\end{smallmatrix}\right]^{-1}\cdot a,$$ where $\gamma(a)=\big[\gamma(a_{ij})\big]_{1\leq i,j\leq 3}$. The $(3,3)$-entry of ${\sigma}(a)a$ is equal to $$\gamma(a_{13})Ya_{13}+\gamma(a_{23})X a_{23} +\gamma(a_{33}) a_{33}=YN_H(a_{13})+XN_H(a_{23})+N_H(a_{33}).$$ Since $N_H={\langle}1,1,1,1{\rangle}$, each of $N_H(a_{13})$, $N_H(a_{23})$, 