In my July diary I posed the following brain-teaser.
Imagine a tetrahedron that is hollow, its surface made of rubber. Pump it full of air: it turns into a sphere, the four vertices and six edges of the tetrahedron just visible as four dots joined by six lines on the surface of the sphere. Take a movie of this process. Now run the movie backwards, the sphere deflating to a tetrahedron. The sphere is "equivalent" to a single tetrahedron.
So far, so good. Now, in all topological enquiries, after you've done with the sphere, you turn your attention to the torus. Obviously (I hope!) a torus is not "equivalent" to a single tetrahedron, the way the sphere was. Is it, though, equivalent to some number of tetrahedra glued together by their faces?
To ask a different way: Is there some pattern of dots and lines I could draw on the surface of a rubber torus so that, on deflating the torus, I end up with some number of tetrahedra — not necessarily regular ones — glued together facewise? If so, what is the smallest number of tetrahedra for which I can do this?
(Note that kaleidocycles aren't allowed here, as they are glued together along edges, not faces. The enclosed air, when you try to pump up a kaleidocycle into a torus, won't be able to get past those glued-together edges.)
The object of the exercise is to "collapse" a torus (think donut, or tire inner tube) into some number — the smallest possible number —of tetrahedra glued together by their faces, without losing the essential torus-ness of the torus. That is, if all the faces are made of rubber, and the "internal" (i.e. glued-together) faces are totally porous to air, you can inflate the flat-faced, straight-edged figure of your solution into a regular torus by using a bicycle pump.
One of my readers described the problem very imaginatively as that of "diamondizing" the torus. I hadn't heard the topic expressed in this way before, and was so taken by this word I am going to refer to this problem as "diamondizing the torus." Yes: we want to turn the smoothly-rounded surface of the torus into a thing of flat facets, each one triangular — we want to diamondize it. The facets will be joined by straight edges, which meet at vertices — two facets meeting at each edge, and at least three edges (if you think about it a minute) at each vertex. Then we want to chop this diamondized torus up into tetrahedra, and count how many tetrahedra there are. Then we want to figure out the minimum number of tetrahedra we could get from this process.
The main point of torus-ness about a torus is, it has a donut hole going through it. If you "deflate" your torus to something with flat faces and straight edges, and you still want that hole, what's the simplest shape the hole could have? Well, it will be a triangle.
So let's start with a triangle, and take the three sides of the triangle to be edges of three tetrahedra. Figure 1 gives the idea.
(Note: Regular tetrahedra — all four faces of each tetrahedron congruent equilateral triangles — don't work here. the angles are all wrong. If you used three regular tetrahedra, the two triangular faces of two adjacent tetrahedra "looking at each other" across one apex of the triangular donut hole have an angle between them of more than 180 degrees--more than 190, in fact. To get a nice construction, you need to "stretch" your tetrahedra away from the hole, as I did in Figure 1, bringing the angle between these faces to less than 180 degrees.)
Now, to get properly solid torus-ness, we just have to fill the three gaps between these tetrahedra with some more tetrahedra. How many shall we need?
Well, if you look carefully at one of the gaps, it has five vertices, one of them a corner of the "donut hole" triangle. It helps to cheat a bit here and think of this gap-shape as a "square" pyramid (think Egypt), whose apex is that triangle-corner. I put "square" in quotes because the "base" of this pyramid is not, in general, square. It is merely a quadrilateral — and not even necessarily a plane one!
Figure 2 shows one of these gap-filling "square" pyramids, ready to be moved down to slot into its gap. One edge of the base quadrilateral (dotted) is hidden.
But this base quadrilateral, even though it is not, in general, plane, can be resolved into two triangles simply by drawing a diagonal. And then the gap-filling "square" pyramid can be Cheopped — oops, sorry, I mean "chopped" — into two tetrahedra by slicing it with the plane defined by that diagonal and the apex of the pyramid.
So to fill a gap, you need two tetrahedra.
To fill all three gaps, you need six tetrahedra.
Added to the three you started with, you need nine tetrahedra altogether.
Note that the diamondized torus, as well as being composed of nine tetrahedra, has:
- 9 vertices, since adding the three "square" pyramids, once everything is glued together, doesn't add any vertices. Every vertex of a "square" pyramid has a vertex of the original three-tetrahedra configuration to go to.
- 18 facets (I am not, of course, counting the "internal," glued-together faces as facets): there are 6 from the original three tetrahedra (the other 6 become "internal") plus 6 more from the three quadrilateral pyramid bases, each base divided into two triangular facets by drawing a diagonal, plus 6 more from the non-internal faces of the pyramids.
- 27 edges: the 18 you see in Figure 1 (well, three of them are hidden so you don't actually see them — use your imagination!) plus those three base diagonals of the "square" pyramids, plus six more from the pyramid-bases (the other 6 coincide with the outer-most edges of the original three-tetrahedra figure).
The Poincaré formula requires that for a torus, whose genus (number of donut holes) is 1, the number of vertices, edges, and faces satisfy the equation V − E + F = 0. Our diamondized torus, with V = 9, E = 27, F = 18, does indeed satisfy this equation. Phew!
(The above construction is not rigorous, since I have shown "sufficient" but not "necessary." As my own math professors used to say: "I leave you to fill in the details.")