4,5,6$. The solutions of are given by: $$\hat{\beta} \in \left\{ -\arctan\left(\frac{1}{2}\right), \; -\arctan\left(\frac{1}{2}\right)+\pi, \; \arctan(2), \; \arctan(2)-\pi \right\} =: S_{{\mathcal{T}}}.$$ Hence the number of Schur operators associated with ${\mathcal{T}}$ is finite; the elements in $S_{{\mathcal{T}}}$ provide all the possible boundary conditions for the analysis of transitions with ${\mathcal{T}}$-symmetry between icosahedral order. Specifically, since $S^1$ is connected, we can consider any path $\beta(t) : [0,1] \rightarrow S^1$ that connects $0$ with $\hat{\beta} \in S_{{\mathcal{T}}}$, i.e. $\beta(0) = 0$ and $\beta(1) = \hat{\beta}$. Then the corresponding Schur rotation $M_{{\mathcal{T}}}(t)$ is given by $M_{{\mathcal{T}}}(\beta) \circ \beta(t) = M_{{\mathcal{T}}}(\beta(t)) : [0,1] \rightarrow {\mathcal{Z}}({\widetilde{\mathcal{T}}}, {\mathbb{R}}) \cap SO(6)$.

  --------------------------------------------------------------------------------------------------
   Generator                Irrep $\Gamma_1$                            Irrep $\Gamma_2$
  ----------- --------------------------------------------- ----------------------------------------
     $g_2$       $\frac{1}{2} \left( \begin{array}{ccc}      $\frac{1}{2} \left( \begin{array}{ccc}
                          \tau-1 & 1 & \tau \\                       \tau-1 & -\tau & -1 \\
                          1 & -\tau & \tau-1 \\                      -\tau & -1 & \tau-1 \\
                           \tau & \tau-1 & -1                         -1 & \tau-1 & -\tau 
                          \end{array} \right)$                        \end{array} \right)$

   $g_{3d}$      $\frac{1}{2} \left( \begin{array}{ccc}      $\frac{1}{2} \left( \begin{array}{ccc}
                          1-\tau & 1 & \tau \\                       1-\tau & \tau & -1 \\
                          1 & \tau & 1-\tau \\                       -\tau & -1 & 1-\tau \\
                           -\tau & \tau-1 & -1                        -1 & \tau-1 & \tau 
                          \end{array} \right)$                        \end{array} 