# Proof that there exist only 17 Wallpaper Groups (Tilings of the plane)

I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).

The punchline is

$$There \,\, are \,\, exactly \,\, 17 \,\, wallpaper \,\, groups \,\,(17 \,\, ways \,\, to \,\, tile \,\, the \,\, plane)$$

My question is $$2$$-fold:

1. Can someone sketch out the proof or at least give some high level ideas of why this may be true?

2. Can someone refer me to a website or a textbook that develops the proof in detail?

• Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof. – Don Thousand Mar 13 at 16:37
• That's why I suggested if anyone knows a good textbook/book then that would be good – NazimJ Mar 13 at 16:40
• Have you tried wikipedia... – Don Thousand Mar 13 at 16:53
• There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved. – Mark Bennet Mar 13 at 16:53

Sketch of the proof: Let $$\Gamma \le {\rm Iso}(\Bbb R^2)$$ be a wallpaper group. Then $$\Gamma$$ has a normal subgroup isomorphic to $$\Bbb Z^2$$ with finite quotient $$F$$. This finite group acts on the lattice $$\Bbb Z^2$$ by conjugation. We obtain a faithful representation $$F \hookrightarrow {\rm Aut}(\Bbb Z^2)\cong GL_2(\Bbb Z).$$ The group $$GL_2(\Bbb Z)$$ has exactly $$13$$ different conjugacy classes of finite subgroups, called arithmetic ornament classes: \begin{align*} C_1 & \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\rangle,\; C_2 \cong \left\langle \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle,\; C_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} \right\rangle, \\ C_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\rangle, \; C_6 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix} \right\rangle, \; D_1 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle, \\ D_1 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\rangle, \; D_2 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle, \\ D_2 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle, \; D_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}\right\rangle, \\ D_3 & \cong \left\langle \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} \right\rangle,\\ D_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\rangle, \; D_6\cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}\right\rangle. \end{align*}

This is an easy computation. Here $$C_1,C_2,C_3,C_4,C_6$$ are cyclic groups and $$D_1,D_2,D_3,D_4,D_6$$ are dihedral groups. We use here, that the order $$n$$ of a subgroup must satisfy $$\phi(n)=deg(\Phi_n)\mid 2$$, so that $$n=1,2,3,4,6$$. This is called the crystallographic condition. The wallpaper groups arise from these $$13$$ classes by equivalence classes of extensions $$1\rightarrow \Bbb Z^2\rightarrow \Gamma\rightarrow F\rightarrow 1,$$ determined by $$H^2(F,\Bbb Z^2)$$.

By computing $$H^2(F,\Bbb Z^2)$$ in each case we obtain $$18$$ inequivalent extensions, because in $$13$$ cases the cohomology is trivial, and in three cases we get $$C_2,C_2$$ and $$C_2\times C_2$$, i.e., $$5$$ additional possibilities, so that $$13+5=18$$. This yields $$17$$ different groups, because two of them turn out to be isomorphic.

• Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $\phi (n)$, $\Phi_n$, and $H^2 (F, \mathbb{R}^2 )$? – NazimJ Mar 13 at 21:50
• $\phi(n)$ is Euler's totient function and $\Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute. – Dietrich Burde Mar 13 at 22:40

The best I have, is this (and I admit it is not very good). The Euler characteristic of the infinite plane is 2.

The members of the wallpaper group have a notation:

632 or 4*2 or *2222

It uses some sequence of numbers, and the symbols $$*,\circ, \times$$

The numbers represent rotations, the $$*$$ represents the presence of reflection, the $$\times$$ represents a glide symmetry. The $$\circ$$ indicates translations without reflections or rotations.

This notation suggests an algebra. For each digit before the star we add $$\frac {n-1}{n}$$. The star adds 1. For each digit after the star we add $$\frac {n-1}{2n}$$ or one half of what you otherwise would add.

$$\times$$ adds 1, an $$\circ$$ adds 2.

This sum must equal 2.

For the groups above: $$\frac {5}{6}+\frac{2}{3} + \frac {1}{2} = 2$$ and $$\frac {3}{4} + 1 + \frac {1}{4} = 2$$ and $$1+\frac 14 + \frac 14 + \frac 14 + \frac 14=2$$

With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.

https://en.wikipedia.org/wiki/Orbifold_notation

However, I don't remember the proofs that associate this algebra to groups.