Three tetrahedra are called desmic if one can join each
vertex of a tetrahedron to pairs of vertices of the other two tetrahedra by
means of a straight line. The
picture below (made with Cabri) shows a 3-dimensional projective space RP3 in which you see three desmic
tetrahedra. The red and blue one make a stella octangula that has its 8 vertices
on those of a cube (in grey). The third one is the green coordinate tetrahedron.
This one has three of its edges on the 3 coordinate axes and 3 edges in the
plane that becomes the plane at infinity if we want the grey cube to get the
familiar affine shape with parallel edges. The 16 yellow lines (12 of them are
partly grey to accentuate the cube) are the straight lines that join the
vertices of the 3 tetrahedra in the desired way to make them desmic. Twelve of
them are the extrapolations of the edges of the cube and four of them are the
body diagonals of the cube.
There is a nice correspondence between a projective space and an elliptic space. A model for an elliptic space is the 4-dimensional (hyper-)sphere S3 on which an elliptic point is represented by a pair of diametrically opposite points. Lines in this model are great circles (like the equator) of the hypersphere. By means of a stereographic projection we can return to the 3-dim space we live in. The same configuration of desmic tetrahedra then becomes in an elliptic space:
The 12 vertices of the 3 tetrahedra are now doubled to 12 pairs of vertices. You can count only 23 points because the 24th vertex, that is the partner of the central vertex of the cube, is at the end of the 3 straight green lines. It is the point of the hypersphere that serves as the centre of (the stereographic) projection. Besides a doubling the points, there is also a doubling of the tetrahedra. The big red and blue one are perhaps not so easy to visualize, but it might help your imagination if you know that their centre is the 24th vertex again (the centre of projection).
A pair of 2 tetrahedra can in an obvious way be completed
with another 14 tetrahedra to become the 16 tetrahedral cells of a spherical
16-cell that lives on the hypersphere. The compact green cell is just one octant
of the inner space of the sphere and the non-compact one octant outside the
The 3 circles make an elliptic triangle that looks the
same as a spherical octahedron. The 3 axes complete the 3 circles to the
elliptic tetrahedron that looks the same as a spherical 16-cell.
The other 2
spherical 16-cells (elliptic tetrahedra) are:
The nice thing about the 24 vertices is that they are the
vertices of a 24-cell. The 16 yellow lines are the (6x16=) 96 edges of the
24-cell. On each yellow line lie 6 vertices that divide each line in 6 segments
that are 6 edges of the 24-cell.
To see this, the 16 lines are shown without the
tetrahedra and they are suitably colored to accentuate the edges:
We recognize the grey cube as a pair of cubes in red,
that has another 6 cubic cells to make a 4-dimensional (hyper-)cube with 8 cubes
as 3-dim. cells. Three of the 24 octahedral cells made of 8 yellow lines and 4
grey edges of the cube were already visible in the projective space. Now we see
6 of them made of edges of the small red cube and green and blue lines. Another
6 are visible attached to the large red cube. With some more effort you can see
the remaining 12 octahedral cells on the 12 edges of the small red cube that
join edges of the big red cube
making a square diagonal plane of an octahedral cell.
Furthermore you can discern 96 triangular faces: 24 cells have 24x8=192 triangular faces but each face is contained in 2 cells so we have 192/2=96 triangular faces.
Let us return to the projective 3-space to see what
happens when we join the 6 intersection points of the edges of the red and blue
tetrahedron. The 12 brown lines make the edges of an octahedron.
By extrapolation of the brown edges we find the 12 blue lines. They intersect the plane at infinity in 6 points, because the 12 edges are parallel in pairs. You can see only 3 of them, making a triangle. To see the other 3 and the configuration they make at infinity, we leave out the tetrahedra and the yellow diagonals and zoom out a little bit.
Now we see that the 4 new dark green lines at infinity make a complete quadrilateral. The 4 dark green lines and the 12 light blue (and brown) lines make a configuration of 16 lines that is completely equivalent to the 16 yellow lines in the first picture. These 16 lines make the (16x6=) 96 edges of another 24-cell. The relation between the 24-cells is comparable to the relation between the cube and the octahedron. They are dual to each other with respect to a sphere. The 6 square faces of the cube correspond to the 6 vertices of the octahedron. The 8 vertices of the cube correspond to the 8 triangular faces of the octahedron. The 12 edges to 12 edges… . The 2 24-cells are dual to each other with respect to a hypersphere. The 24 vertices correspond to 24 octahedral cells. The 96 edges correspond to 96 triangular faces. A 24-cell is self-dual, just like the 3-dimensional tetrahedron!
How does the 24-cell appear with the blue and green lines? Like this:
Just as before in changing to an elliptic model, the 6 vertices of the octahedron are doubled to make an elliptic octahedron (a pair of spherical octahedra). The brown octahedron changed color into blue and its partner, whose edges are lying on the same light blue lines, has 12 red edges. The plane at infinity contains (2x6 =) 12 vertices of a cuboctahedron. The 4 dark green lines have changed to 4 light green circles, making a cuboctahedron. A cuboctahedron can thus be seen as the elliptic equivalent of the complete quadrilateral in projective space. The 4 light green lines in the plane at infinity in the projective case are the (4x4=)16 diagonals of the squares of the cuboctahedron. I did not draw these 4 circles.
In a similar way one can treat the other 2 platonic
polyhedra: the dodecahedron and the icosahedron.
Some pictures of a desmic surface, starting with a picture that looks like a present and is actually one of the 24 cells of the 24-cell that is covered by the desmic surface:
Then unwrap the present to see the other 7 cells of the 24-cell that are covered by the desmic surface:
And a detail of the picture above that lies on one hemisphere of the hypersphere:
Now select another set of 8 covered cells of the 24-cell:
Peeling off a little bit:
The same picture rotated over 90°.
On one hemisphere of the hypersphere:
Now a desmic surface on the dual 24-cell:
And another set of 8 covered cells: