Dos Equis (Foldables 3)










The Fractal (Foldables 2)

The second bifoldable object Jiangmei showed me was this:

Fractal 1






Fractal 2

And, just for fun, the generation 10 fractal:

 Fractal 10 colorgradient


The Butterfly (Foldables 1)












Unpacking the Hypercube

Last week we learned how Rototiler moves can unpack a cube.  As a warmup, below are the moves for a 2x2x2 cube projected parallel along a cube diagonal onto hexagons:


We start at the left, remove the frontmost cube, and keep going. The solution is far from being unique, but not too complicated. Today, we do the same with a  hypercube. The projection of a 1x1x1x1 into 3-space along a main diagonal is a rhombic dodecahedron, tiled by four rhomboids. These rhombic dodecahedra have 8 obtuse, 3-valent vertices at the corners of a cube, and 6 acute, 4-valent vertices at the corners of an octahedron.


There are two ways to tile the rhombic dodecahedron with these rhomboids, and changing one to the other corresponds to a rototiler move in space. Let’s do this with the 2x2x2x2 hypercube, whose projection is a rhombic dodecahedron tiled by 32 rhomboids.


At first it seems as if there is no swappable rhombic dodecahedron available, but if we remove three rhomboids and  look inside (which is the direction of the fourth dimension, after all), we can see it. After swapping it, we also remove the frontmost rhomboid of the  swapped dodecahedron.




We then see that the four removed rhomboids together make up another swappable dodecahedron. We replace it by its swap. The same can be done at three other places.


The next thing to do is to swap 6 more dodecahedra. One of them is the one which shows yellow and purple rhomboids in the right figure above, sitting between the red and blue “vertex”. All these dodecahedra correspond to the edges of the tetrahedron whose vertices are the already swapped four peripheral dodecahedra. Doing these six swaps leads to a tiling very much like the one above to the right, where now the other four obtuse vertices mark swappable dodecahedra. Swapping these and finally the hidden central dodecahedron  completely unpacks the hypercube. It took us 1+4+6+4+6+1 = 32 swaps, as expected.


Next week we’ll see what this is good for…












Vertical Flux (The pleasures of Failure IV)

One of the standard examples of a failed minimal surface construction is the Finite Riemann surface, named in reference to the rather successful infinite Riemann minimal surface




The attempt was to make an embedded minimal sphere with 3 ends. Two ends is easy: the catenoid will do. If you want more, the ends can only be catenoidal or planar, and the catenoidal ends need to point in the same direction, i.e. have the same limiting normal. Such a sphere will always have what is called vertical flux, which is disastrous in its usefulness.


Whenever you have vertical flux, you can deform the minimal surface in a very simple way, keeping the catenoidal ends vertical but tilting non-horizontal planar ends like above. This is nice but also disastrous because it allows to prove that embedded surface of this sort cannot exist.


This is a famous theorem by Francisco José Lópe and Antonio Ros. The proof consists of three steps: If the surface is embedded, its stays embedded for all parameter values of the deformation. Secondly, if such a surface has either a planar end or a finite point with vertical normal, it cannot stay embedded. These two steps imply that such a surface can only have catenoidal ends and no points with vertical normal. Being a sphere then implies that there can only two ends, and we have the catenoid.



It is amusing to make more examples of surfaces with vertical flux. Above are single periodic versions with vertical catenoidal and annular ends. The first one feels a little bit embedded because the only intersections that happen are those of a catenoidal end with a translational copy, and I am inclined not to count that.



Dihedralization (The Pleasures of Failure II)

W1 2Up above is Meinhard Wohlgemuth’s first surface. It is very similar to Costa’s surface, but has one more end and genus 2. In Mathematics, like in every experimental science, you gain intuition about truth through observation. For instance, every minimal surface person I know will probably believe in dihedralization

W1 5

Whenever you have a minimal surface that has two vertical symmetry planes making an angle π/2, there should be more symmetric version where the planes make the angle π/n. Like up above for n=5, observed for Wohlgemuth’s first surface.

W2 2

Then there is Wohlgemuth’s second surface. It has genus 3, squeezing in a little suspicious handle between the two planar ends, so that it looks a little bit like two copies of the ill-fated Horgan surface stacked together. But this time it works, one (Meinhard) can prove this surface does indeed exist. So, let’s make it more symmetric: W2 3

This is what I get when I run the experiment. The parameter values are so extreme that I don’t think this actually exists. So, do we trust our intuition, or the experiment? Neither. The only way out is to dig deeper. Let’s cut Wohlgemuth’s surface open (in half, to be acurate).

W2 2 cut

We can now apply the dihedralization continously, slowly increasing n=2 to n=3, thereby tracking the numerical solutions. They show a clear deformation,up to approximately n=2.9 (below).  After that, the periods don’t close accurately enough anymore to make me feel confident about the surface.W2 2 9 cut

So I tend to think that this is the first case where dihedralization fails.