Ttgjyuk

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?

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:
beginalign*
C_1 & cong leftlangle beginpmatrix 1 & 0 \ 0 & 1 endpmatrix rightrangle,;
C_2 cong leftlangle beginpmatrix -1 & 0 \ 0 & -1 endpmatrix rightrangle,;
C_3 cong leftlangle beginpmatrix 0 & 1 \ -1 & -1 endpmatrix rightrangle, \
C_4 & cong leftlangle beginpmatrix 0 & 1 \ -1 & 0 endpmatrix rightrangle, ;
C_6 cong leftlangle beginpmatrix 0 & 1 \ -1 & 1 endpmatrix rightrangle, ;
D_1 cong leftlangle beginpmatrix 1 & 0 \ 0 & -1 endpmatrix rightrangle, \
D_1 & cong leftlangle beginpmatrix 0 & 1 \ 1 & 0 endpmatrix rightrangle, ;
D_2 cong leftlangle beginpmatrix 1 & 0 \ 0 & -1 endpmatrix,beginpmatrix -1 & 0 \ 0 & -1 endpmatrix
rightrangle, \
D_2 & cong leftlangle beginpmatrix 0 & 1 \ 1 & 0 endpmatrix,beginpmatrix -1 & 0 \ 0 & -1 endpmatrix
rightrangle, ; D_3 cong leftlangle beginpmatrix 0 & 1 \ 1 & 0 endpmatrix,
beginpmatrix 0 & 1 \ -1 & -1 endpmatrixrightrangle, \
D_3 & cong leftlangle beginpmatrix 0 & -1 \ -1 & 0 endpmatrix,beginpmatrix 0 & 1 \ -1 & -1 endpmatrix
rightrangle,\
D_4 & cong leftlangle beginpmatrix 0 & 1 \ 1 & 0 endpmatrix,beginpmatrix 0 & 1 \ -1 & 0 endpmatrix
rightrangle, ; D_6cong leftlangle beginpmatrix 0 & 1 \ 1 & 0 endpmatrix,
beginpmatrix 0 & 1 \ -1 & 1 endpmatrixrightrangle.
endalign*

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
$$
1rightarrow Bbb Z^2rightarrow Gammarightarrow Frightarrow 1,
$$
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 $C_2times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
This yields $17$ different groups, because two of them turn out to be isomorphic.

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 $frac n-1n$. The star adds 1. For each digit after the star we add $frac n-12n$ 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: $frac 56+frac23 + frac 12 = 2$ and $frac 34 + 1 + frac 14 = 2$ 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.