we can choose an arbitrary orientation of the polygon corresponding to the cyclic symmetry group. For simplification, we choose the face of the regular polygon to be perpendicular to the $z$-axis in the Bloch sphere. In other words, the $z$-axis is the common axis of the rotations. For instance, the $5$-sided regular polygon is shown in Figure \[Figur cyclic vectors\].

The cyclic symmetry group of the polygon is generated by the $2\pi/m$ rotation about the $z$-axis. In the Hilbert space, this rotation of the Bloch sphere corresponds to the matrix ${\rm diag}(1,\omega) \in
{\mathbb C}^{2 \times 2}$ where $\omega := {\rm exp}(- 2 \pi i/m)$ is an $m$th complex root of unity. The diagonal form of the matrix is the reason for choosing the $z$-axis as common rotation axis. If we choose the vector $(1,1)^T \in {\mathbb C}^2$ and consider the orbit under the symmetry group then we get the vectors $(1, \omega^{j})^T
\in {\mathbb C}^2$ for $j \in \{0, \ldots, m-1\}$. Other vectors of the Bloch sphere do not lead to POVMs or to POVMs that correspond to a polygon with another rotation. The latter case is discussed at the end of the previous section. Due to the identity 