Regular polyhedral spaces

Question

By symmetrically gluing together opposite faces of a dodecahedron together, one of three spaces can be obtained, depending on the angle the faces are rotated by before twisting. In fact, this can be done with any of the regular polyhedra besides the tetrahedron, and in that case the stella octangula can be used instead. However, it seems only the case of the dodecahedron (or equivalently, its conjugate, the great stellated dodecahedron) has been investigated. What are the other spaces? Or rather, what are the spaces not already named by the following list?

• Stella octangula, 180 degrees (this is the 3-sphere)
• Stella octangula, 60 degrees
• Octahedron, 180 degrees (this is real projective space)
• Octahedron, 60 degrees
• Cube, 180 degrees (this is real projective space)
• Cube, 90 degrees
• Cube, 0 degrees (this is the 3-torus)
• Icosahedron, 180 degrees or great icosahedron, 180 degrees (this is real projective space)
• Icosahedron, 60 degrees or great icosahedron, 60 degrees
• Dodecahedron, 180 degrees or great stellated dodecahedron, 180 degrees (this is real projective space)
• Dodecahedron, 108 degrees or great stellated dodecahedron, 36 degrees (this is the Seifert-Weber space)
• Dodecahedron, 36 degrees or great stellated dodecahedron, 108 degrees (this is the Poincaré homology sphere)
• Great dodecahedron, 180 degrees or small stellated dodecahedron, 180 degrees
• Great dodecahedron, 108 degrees or small stellated dodecahedron, 36 degrees
• Great dodecahedron, 36 degrees or small stellated dodecahedron, 108 degrees

Answer 1

Identifying opposite faces of a cube via a 90 degree twist gives a spherical manifold whose fundamental group is the quaternion group $\{\pm1,\pm i,\pm j,\pm k\}$.

If one identifies opposite faces of an octahedron using a 60 degree twist, this also gives a spherical manifold. Its fundamental group has order 24 and is the binary tetrahedral group, the pre-image of the group of rotational symmetries of a tetrahedron under the two-to-one projection $S^3\to SO(3)$. There is a nice picture of the corresponding tessellation of $S^3$ by 24 octahedra in the classical book Geometry and the Imagination by Hilbert and Cohn-Vossen (on page 152 in the Chelsea edition).

If one uses the binary octahedral group instead of the binary tetrahedral group one obtains a spherical manifold that is not built from a regular polyhedron but from a truncated cube, identifying opposite octagonal faces by a 45 degree twist and opposite triangular faces by a 60 degree twist.