enough that $C_0$ and $D_0$ are two nearby clusters of galaxies, while the distances between $C_i$ and $D_i$ ($i=1,2$) are equal (within the observational error bounds) to the distance between $C_0$ and $D_0$.[^5] By parallel transporting the triangle $C_1D_1E_1$ along the geodesic segment $\overline{C_0C_1}$, one obtains two triangles with a common vertex, namely the triangle of nearby clusters $C_0D_0E_0$, and that of *transported* clusters of $C_1D_1E_1$. It is just a matter of elementary analytic geometry to determine the unique rotation that takes one triangle to the other. Note however that one can easily find also the unique reflection that takes one triangle to the other, if it exists. If the angle of rotation is different from $\pi$, $2 \pi/3$, $\pi/2$ or $\pi/3$, then the isometry that takes $C_0$ to $C_1$ is not a screw motion, but a reflection, and the Universe would be spatially non-orientable. On the contrary, if the angle of rotation is either $\pi$, $2 \pi/3$, $\pi/2$ or $\pi/3$, then one can think this is not by coincidence, so the Universe would be spatially orientable. In such a case, if the angle of rotation is different from $\pi$, it uniquely determines to which class the topology of the 