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The {${\cal T }_{20}$} commutators vanish by inspection since all operators commute
{$$[3I_zS_z-\mathbf I\cdot \mathbf S,[3I_zS_z-\mathbf I\cdot \mathbf S,I_z]] = 0$$}
The trace of the product of the {${\cal T }_{2\pm1}$} commutators with {$\rho(0)=Iz$} yield
{$$\begin{align*}
Tr\,[I_zS_\pm+I_\pm S_z,\,[I_zS_\mp\alpha+I_\mp S_z\beta,I_z]]\,I_z &= Tr\,[I_zS_\pm+I_\pm S_z,\,[I_\mp,\,I_z]S_z\beta]\,Iz \\
&= \pm Tr\,[I_zS_\pm+I_\pm S_z,\,I_\mp S_z\beta]\,I_z\\
&= \pm\beta\, Tr\,([I_z,\,I_\mp]I_z \,S_\pm S_z + I_zI_\mp \,[S_\pm,\,S_z]+[I_\pm ,\,I_\mp]\, I_zS_z^2\\
& = \beta\, Tr I_z^2S_z^2\\
& = \beta\, \frac {I(I+1)} 3 \, \frac {S(S+1)} 3
\end{align*}
$$}
with {$\alpha,\beta = \exp(\mp i \omega_{S,I}\tau)$}. We used commutators
{$$ \begin{align*}
[I_\pm,I_\mp] &= -i[I_x,I,y] + i [I_y,I_x] = \pm 2I_z\\
[I_z,I_\pm] &= \pm I_\pm
\end{align*}
$$}
and {$Tr I_\pm I_z =0$} together with their {$S$} equivalent.
Finally, the trace of the product of the {${\cal T }_{2\pm2}$} commutators with {$\rho(0)=Iz$} yield
{$$\begin{align*}
Tr\,[I_\pm S_\pm,\,[I_\mp S_\mp,I_z]]\,I_z &= Tr\,[I_\pm S_\pm,\,[I_\mp,\,I_z]S_\mp]\,Iz \\
&= \pm Tr\,[I_\pm S_\pm,\,I_\mp S_\mp]\, I_z\\
&= \pm Tr\,([I_\pm,\,I_\mp]I_z\, S_\pm \,S_\mp + I_\pm I_z \,[S_\pm,\,S_\mp]\\
& = 4 Tr I_z^2\,S_z^2\\
& = 4 \frac {I(I+1)} 3 \, \frac {S(S+1)} 3
\end{align*}
$$}
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