Roman surface

ROMAN SURFACE

Surface studied by Steiner in 1844.
This surface was named this way by Steiner because he discovered during a stay in Rome.

Cartesian equation as a tetrahedral surface of Kümmer:

, where

,
so that f = 0 is the equation of the sphere with center O and radius a and pqrs = 0 is the equation of a regular tetrahedron centered on O and the edges of which are at distance a from O.
Cartesian equation in a frame turned by 45° around Oz:

.
Cartesian parametrization:

with

;
i.e., by taking

so that

.
i.e. by taking

One-sided quartic surface, special case of Steiner surface.

The Roman surface is the image of the quotient of the sphere(center O O, radius 1) by the antipodal relation (in other words the real projective plane) by the map: .
It is historically the first representation of the projective plane as a surface in . It has three segment lines of self-intersection that form a right trihedron each of them ended by two cuspidal points and intersecting at their center in a triple point (here O).

The above definition as a special case of Kummer surface shows that it has the symmetries of the regular tetrahedron.
With the second equation, the vertices of the tetrahedron are the points , with an even number of minus signs.

View showing a partition into 4 isometric parts (containing the vertices of the tetrahedron),
and a view showing a partition into 6 isometric parts (containing the edges of the tetrahedron).

The Roman surface is, in three ways, the union of ellipses (with the second equation above, they are the sections by the planes containing the axes); we could say that the Roman surface, in the manner of the cross-cap, is a triply "ellipsed" surface.
Opposite, the family of ellipses located in the planes passing by the vertical double line.

The sections by the planes perpendicular to the axes of rotation of order 3 form the family of hypotrochoids with 3 branches.
In particular, the section by the central plane is the regular trifolium.

The sections by the planes perpendicular to the double lines are the lemniscates of Booth.

Finally, here is a polyhedral version of the Roman surface: an another one, where the cubes were truncated in orthocentric tetrahedra:

Find the 3 double segment lines ended by the 6 pinch points, the 4 vertices, and the triple point!

Attention 1: if you see hollow cubes, you are victim of an illusion!
Attention 2: it is not a true polyhedron: the double edges are common to 4 faces. Here neither, it is not a polyhedron but the central faces are united into 4 squares, we get a crossed polyhedron, that is even semiregular, called tetrahemihexahedron.

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The Roman surface is, in three ways, the union of ellipses (with the second equation above, they are the sections by the planes containing the axes); we could say that the Roman surface, in the manner of the cross-cap, is a triply "ellipsed" surface. Opposite, the family of ellipses located in the planes passing by the vertical double line.
The sections by the planes perpendicular to the axes of rotation of order 3 form the family of hypotrochoids with 3 branches. In particular, the section by the central plane is the regular trifolium.
The sections by the planes perpendicular to the double lines are the lemniscates of Booth.

Finally, here is a polyhedral version of the Roman surface:	an another one, where the cubes were truncated in orthocentric tetrahedra:
Find the 3 double segment lines ended by the 6 pinch points, the 4 vertices, and the triple point!
Attention 1: if you see hollow cubes, you are victim of an illusion! Attention 2: it is not a true polyhedron: the double edges are common to 4 faces.	Here neither, it is not a polyhedron but the central faces are united into 4 squares, we get a crossed polyhedron, that is even semiregular, called tetrahemihexahedron.