从五边形到五角星
From Pentagons to Pentagrams

原始链接: https://johncarlosbaez.wordpress.com/2026/05/29/from-pentagons-to-pentagrams/

本文探讨了几何学与数论之间通过“伽罗瓦共轭”所建立的迷人联系。通过将“黄金域”中多面体坐标公式里的 $\sqrt{5}$ 替换为 $-\sqrt{5}$,可以将凸多面体转换为非凸的开普勒-潘索多面体。具体而言,该运算通过将黄金分割比与其负倒数互换,实现了从正五边形到正五角星的映射。 作者论证了,若正五边形的顶点位于黄金域内,则对这些坐标应用伽罗瓦映射即可得到一个正五角星。这种变换并非仅仅是数学奇观,它还与二十面体准晶体中准粒子(声子与相位子)的行为有关。其视觉效果显著:例如,对菱形二十十二面体应用此操作,会将其五边形面转换为五角星,而其正方形和等边三角形面则保持不变,因为这些面的边长在变换下保持恒定。最终,这凸显了一种将代数域论与星形多面体几何性质相连接的深邃而优雅的对称性。

```Hacker News 最新 | 过往 | 评论 | 提问 | 展示 | 招聘 | 提交 登录 从五角大楼到五角星 (johncarlosbaez.wordpress.com) 16 分,surprisetalk 发布于 5 小时前 | 隐藏 | 过往 | 收藏 | 3 条评论 | 帮助 fbu 1 小时前 | 下一条 [–] 大十二面体是我最喜欢的多面体,用纸就可以制作一个。有机会一定要试试! https://www.polyhedra.net/en/model.php?name-en=great-dodecah... 回复 red_trumpet 50 分钟前 | 上一条 | 下一条 [–] f(L^2) 为正这一事实无需检查坐标即可看出:毕竟它是 e_i' 的长度,因此为正。 回复 yepyoukno 4 小时前 | 上一条 [–] 关于星形多面体的精彩文章!标题含蓄且违反直觉,我差点就错过了! 回复 准则 | 常见问题 | 列表 | API | 安全 | 法律 | 申请 YC | 联系 搜索: ```
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原文

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 \sqrt{5} by -\sqrt{5} 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 \sqrt{5} by -\sqrt{5}, which is called Galois conjugation. This is mentioned near the end of this book:

• 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:

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 \mathbb{Q}(\sqrt{5}) be the golden field: that is, the set of all numbers

a + b \sqrt{5}

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

f \colon \mathbb{Q}(\sqrt{5}) \to \mathbb{Q}(\sqrt{5})

by

f(a + b \sqrt{5}) = a - b \sqrt{5}

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

\displaystyle{ \Phi = \frac{1 + \sqrt{5}}{2} = 1.6180339\dots }

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

\displaystyle{ -1/\Phi = \frac{1 - \sqrt{5}}{2} \approx -0.6180339\dots }

This suggests that Galois conjugation should somehow map regular pentagons to regular pentagrams! Why? Well, in a regular pentagon, each exterior turning angle is 2\pi/ 5 = 72^\circ:

while in a regular pentagram, each exterior turning angle is 4 \pi /5 = 144^\circ:

The cosine of the exterior turning angle for the pentagon is

\cos(2\pi/5) = 1/2 \Phi

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

\cos(4 \pi/5) = -\Phi/2

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 \mathbb{R}^n whose vertices lie in \mathbb{Q}(\sqrt{5})^n. Beware: such a pentagon is impossible in the plane!

Puzzle. Show this.

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

\begin{array}{ccl}  v_1 &=& (1,1,1)  \\  v_2 &=& (0, 1/\Phi,\Phi) \\  v_3 &=& (-1,1,1) \\  v_4 &=& (-1/\Phi,\Phi,0) \\  v_5 &=& (1/\Phi,\Phi,0)   \end{array}

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 \mathbb{Q}(\sqrt{5})^n coordinatewise; let’s abuse language and call this map

f \colon \mathbb{Q}(\sqrt{5})^n \to \mathbb{Q}(\sqrt{5})^n

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 v_1, \dots, v_5 \in \mathbb{Q}(\sqrt{5})^n be the vertices of a regular pentagon, listed in cyclic order. Then f(v_1), \dots, f(v_5), listed in the same cyclic order, are the vertices of a regular pentagram.

Proof. Define the edge vectors

e_i \;=\; v_{i+1} - v_i \;\in\; \mathbb{Q}(\sqrt{5})^n, \qquad i = 1, \dots, 5 \pmod 5

All these have the same squared length

e_i \cdot e_i = L^2 \in \mathbb{Q}(\sqrt{5})

The exterior turning angle at each vertex of the pentagon is 2\pi/5, so this is the angle between the consecutive edge vectors e_i and e_{i+1}. Since

\cos(2\pi/5) = 1/2\Phi

we have

e_i \cdot e_{i+1} \;=\; \cos(2\pi/5) L^2  \;=\; (1/2\Phi) \, L^2

Now let v_i' = f(v_i) and e_i' = f(e_i) = v_{i+1}' - v_i'. It is easy to check that the usual dot product of v, w \in \mathbb{Q}(\sqrt{5})^n obeys

f(v) \cdot f(w) = f(v \cdot w)

so

e_i' \cdot e_i' \;=\; f(L^2)

and

e_i' \cdot e_{i+1}' \;=\;  f(1/2\Phi) \, f(L^2) \;=\; -(\Phi/2) \, f(L^2)

Therefore the cosine of the angle between consecutive edge vectors e_i' is

\displaystyle{  \frac{e_i' \cdot e_{i+1}'}{\sqrt{(e_i' \cdot e_i')(e_{i+1}' \cdot e_{i+1}')}}     \;=\;  \frac{-(\Phi/2) f(L^2)}{\sqrt{f(L^2)^2}}  \; = \; -\Phi/2 }

Since

\cos(4\pi/5)   \; = \; -\Phi /2

it follows that the angle between consecutive edge vectors is 4\pi/5.

But wait! The above calculation secretly assumed f(L^2) is positive, because we claimed that the usual positive square root of f(L^2)^2 equals f(L^2), and we also felt free to divide by f(L^2). Why is f(L^2) positive? Writing

e_i = (c_1, \dots, c_n)

with c_j \in \mathbb{Q}(\sqrt{5}), we have

L^2 = \sum_j c_j^2

and thus

f(L^2) = \sum_j f(c_j)^2

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

The five points v_i' are coplanar, since coplanarity amounts to the vanishing of certain 3 × 3 minors in the matrix of coordinate differences, which is a polynomial condition over \mathbb{Q}(\Phi), hence preserved by f. These points are also distinct, since f is injective. The edges e'_i thus form a planar closed 5-gon with equal edge lengths \sqrt{f(L^2)} and constant exterior turning angle 4\pi/5 at each vertex. This is a regular pentagram.   █

The same style of argument shows that applying Galois conjugation to a regular pentagram with vertices in \mathbb{Q}(\sqrt{5})^n, we get back a regular pentagon. And if you’re worried about what happened to the plane, fear not! We can draw regular pentagons in the plane whose vertices have coordinates in a certain quadratic extension of \mathbb{Q}(\sqrt{5}). This larger field again has an automorphism that carries regular pentagons to regular pentagrams.

We can also play similar games with heptagons and the like, using different fields.

Acknowledgments and addenda

The red pictures of polyhedra were made using were created using Robert Webb’s Stella software and placed on Wikicommons. The diagram of Kepler–Poinsot polyhedra was created by Tilman Piesk and placed on Wikicommons.

For a description of the small stellated dodecahedron and great dodecahedron as Riemann surfaces—branched coverings of the sphere—try this:

• John Baez, Small stellated dodecahedron, Visual Insight, 15 June 2016.

I do not know how these branched coverings are related to the Galois theory perspective given here!

On Mastodon, J. M. animated an icosahedron morphing into a great icosahedron:

and a dodecahedron morphing to a great stellated dodecahedron:

Here’s another fun example. If you take a rhombicosidodecahedron with vertex coordinates all in the golden field:

and apply the Galois transformation \sqrt{5} \mapsto -\sqrt{5}, the pentagons turn into pentagrams, while the squares stay squares and the equilateral triangles stay equilateral triangles. It gets messy if we draw everything, but if we draw just the pentagrams it’s beautiful:

Interestingly the squares and triangles, not drawn, stay the same size—because the squared lengths of their edges are rational! But the squared lengths of the pentagon edges involve \sqrt{5}, so they change as they become pentagram edges.

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