From Pentagons to Pentagrams | Azimuth

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From Pentagons to Pentagrams

I recently showed you that if you take the regular icosahedron:

considered in a coordinate system based on the golden ratio, and then replace √5 by -√5 in all your formulas, you get the great icosahedron:

But this fact isn’t an isolated one-off! If we do the same for the regular dodecahedron:

we get the great stellated dodecahedron:

There’s also a star polyhedron called the great dodecahedron:

and if we play the same game, replacing by in all the formulas, we get the small stellated dodecahedron:

These six polyhedra form a family; the four nonconvex ones are called the Kepler–Poinsot polyhedra. I never understood what was so great about them, though of course they look ravishingly attractive. So it was nice to learn that if we include the convex ones, they come in three pairs related by the operation of replacing by which is called Galois conjugation . This is mentioned near the end of this book:

&bull; John Horton Conway, Heidi Burgiel and Chaim Goodman-Strauss, The Symmetries of Things, A K Peters, Natick, Massachusetts, 2008.

These authors spend more energy describing three other relations among this family of polyhedra:

But I’m more interested in Galois conjugation, which carries each polyhedron in this picture to the one at the opposite corner of the hexagon. I got interested in Galois conjugation because it interchanges two kinds of quasiparticles that propagate in icosahedral quasicrystals, called phonons and phasons:

&bull; Phasons in Quasicrystals.

But there’s some simple geometry behind it, which I’d like to discuss here.

You’ll notice that in all the examples I gave, Galois conjugation takes regular pentagons to regular pentagrams. And that turns out to be a general fact!

Let be the golden field : that is, the set of all numbers

with rational, equipped with the usual addition, multiplication, subtraction and division. Define Galois conjugation

by

This map preserves all the field operations, and if you apply it twice you get back where you started. Thus, it’s like complex conjugation in many respects.

The golden field gets its name because it contains the golden ratio

If we apply Galois conjugation to the golden ratio, we get its negative reciprocal:

This suggests that Galois conjugation should somehow map regular pentagons to regular pentagrams! Why? Well, in a regular pentagon, each exterior turning angle is :

while in a regular pentagram, each exterior turning angle is

The cosine of the exterior turning angle for the pentagon is

and we apply Galois conjugation to this, we get the cosine of the exterior turning angle for the pentagram!

This is not quite a proof that Galois conjugation turns regular pentagons into regular pentagrams—indeed, we have to clarify what we even mean by that claim. But it’s a key ingredient of the proof.

To be more precise, let’s consider a regular pentagon in whose vertices lie in Beware: such a pentagon is impossible in the plane!

Puzzle. Show this.

But it’s possible in 3 or more dimensions. For example:

taken in cyclic order, are the vertices of a regular pentagon in 3 dimensions. And once you can get one, you can get plenty, by translations and rotations. It may take a bit of thought to dream up rotation matrices with entries in the golden field, but there are lots: even rotation matrices with rational entries are dense among all rotations.

Now, Galois conjugation acts on coordinatewise; let’s abuse language and call this map

If we take a regular pentagon and apply this map to its vertices and edges, what do we get? A regular pentagram, I claim!

We can simplify the proof by noticing that only the cyclic ordering on the vertices is needed to distinguish a regular pentagon and a regular pentagram.

Theorem. Let be the vertices of a regular pentagon, listed in cyclic order. Then listed in the same cyclic order, are the vertices of a regular pentagram.

Proof. Define the edge vectors

All these have the same squared length

The exterior turning angle at each vertex of the pentagon is so this is the angle between the consecutive edge vectors and Since

we have

Now let and It is easy to check that the usual dot product of obeys

so

and

Therefore the cosine of the angle between consecutive edge vectors is

Since

it follows that the angle between consecutive edge vectors is

But wait! The above calculation secretly assumed is positive, because we claimed that the usual positive square root of equals and we also felt free to divide by Why is positive? Writing

with we have

and thus

This is a sum of squares of real numbers, hence nonnegative. It is strictly positive because is injective, so the are not all zero.

The five points are coplanar, since coplanarity amounts to the vanishing of certain 3 &times; 3 minors in the matrix of coordinate differences, which is a polynomial condition over hence preserved by These points are also distinct, since is...