The heptahedron consists of six vertices with
(x, y, z) coordinates
A(1,0,0),
B(0,1,0),
C(0,0,1),
D(1,0,0),
E(0,1,0),
F(0,0,1),
twelve edges
AB, BD, DE, EA, AC, BC, DC, EC, AF, BF, DF, EF and
seven faces
ABC, CDE, AEF, BDF, BCEF, ACDF, ABDE as shown in Figure
I. The three square faces intersect in three mutually perpendicular straight
line segments that are not edges and a point in the center that is not
a vertex. Exactly two faces meet at every edge and we can travel from any
face to any other face by crossing edges. The result is a polyhedron called
the heptahedron which was originally constructed by David Hilbert.

Figure I. Hilbert's heptahedron

We shall first show that the heptahedron is topologically equivalent
to a Möbius band joined along its boundary to the boundary of a disc.
To visualize this, we cut a small portion of the surface along the zaxis
and along the edges of the square face ABDE above and below the
xyplane
as shown in Figure II. A simple band without any twists and a disc with
a hole are now visible in Figure II.

Figure II. A simple band and a disc with a hole

Closing the hole of the disc recovers the square face ABDE, shown in
Figure III.

Figure III. The square face

Rejoining the simple band along the inside portion of its boundary yields
a Möbius band, shown in Figure IV. If we now rejoin the boundary of
the disc along the edge of the Möbius band, we can recover the original
heptahedron of Figure I.

Figure IV. The Möbius band

Next, referring to Figure II, we deform the simple band in the
Heptahedron to the simple band given by Equation I:
z = (xy/(2x^{2}+2y^{2}))(1(14x^{2}4y^{2})1/2)
for ε ≤ x^{2}+y^{2} ≤ 1/4  ε and we deform
the disc with the hole in the heptahedron to the disc with a hole given
by Equation II:
z = (xy/(2x^{2}+2y^{2}))(1+(14x^{2}4y^{2})1/2)
for ε ≤ x^{2}+y^{2} ≤ 1/4  ε where ε>0
is a small constant. The result is shown in Figure V.

Figure V. The simple band and disc with a hole

Now let ε approach zero in Figure V. We obtain a disc given by
Equation II for 0 ≤ x^{2}+y^{2} ≤ 1/4 shown in Figure
VI.

Figure VI. The disc obtained by closing the hole

We also obtain a Möbius band given by Equation I for 0 ≤ x^{2}+y^{2} ≤ 1/4
shown in Figure VII.

Figure VII. The Möbius band obtained by closing
the hole

Finally, to complete the deformation, the boundaries of the Möbius
band and disc are joined along the curve given by z = 2xy for x
and y on the circle x^{2}+y^{2 }= 1/4. The
resulting closed surface shown in Figure VIII is called the Roman surface
which was first discovered by Jakob Steiner.

Figure VIII. Steiner's Roman surface

The equation for the Roman surface may be obtained as follows.
For any fixed x and y, Equation I and Equation II are roots
of the quadratic equation in z given by Equation III:
(x^{2}+y^{2})z^{2}+(xy)z+(x^{2}y^{2})
= 0.
Thus from Equation III, the Roman surface is a surface of the fourth
order given by Equation IV:
x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+xyz
= 0.
Note that all surfaces given by the equation
x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+2kxyz
= 0
where k is nonzero, are called Roman surfaces. We have used k
= 1/2 to make the above argument.
The heptahedron and Roman surface are both nonorientable surfaces since
they contain a Möbius band. Being equivalent to a Möbius band
joined along its boundary to the boundary of a disc, both the heptahedron
and Roman surface are topological models of the projective plane.