MIT 18.701 — Lecture 11

Point Groups, Lattices, Crystallographic Restriction, Group Actions

Last time, we classified all discrete subgroups of $O_2(\mathbb{R})$ . Our goal for today is to answer the following question:

Question. What do all the discrete symmetry groups $G \subseteq M_2$ of a shape $S \subseteq \mathbb{R}^2$ look like?

We'll set up today's discussion with the following theorem and definition.

Theorem. (Discrete Subgroups of $\mathbb{R}^2$ ) Every discrete subgroup of $(\mathbb{R}^2, +)$ is either of the form $\mathbb{Z}v$ or $\mathbb{Z}v + \mathbb{Z}w$ , where $v$ and $w$ are linearly independent vectors in $\mathbb{R}^2$ .

Proof: The proof is analysis-flavored, so we only provide a sketch here.

For any discrete subgroup $H \subseteq (\mathbb{R}^2, +)$ , we may take $v$ to be the nonzero vector of least magnitude in $H$ , then take $w$ to be the vector of least magnitude in $H \setminus \mathbb{Z}v$ .

If there is still some $x \in H \setminus (\mathbb{Z}v + \mathbb{Z}w)$ , then look for the “remainder” when $x$ is quotiented out by $\mathbb{Z}v + \mathbb{Z}w$ to contradict the minimality of $|v|$ and $|w|$ . $\blacksquare$

Definition (Lattice). A subset of $\mathbb{R}^2$ is a lattice if it is of the form $\mathbb{Z}v$ or $\mathbb{Z}v + \mathbb{Z}w$ for linearly independent $v$ and $w$ .

§ Point Groups and Lattices

Definition (Point Group). Consider the map $\pi: M_2 \to O_2(\mathbb{R})$ defined by $\pi: (x \mapsto Ax + b) \mapsto A$ . Intuitively, $\pi$ filters out the translations from the isometries in $M_2$ . Then for any discrete subgroup $G$ of $M_2$ :

The point group of $G$ is $\overline{G} := \pi(G)$ . In other words, $\overline{G}$ describes the isometries in $G$ “up to translation”.
The lattice of $G$ is $L := \ker(\pi) \cap G$ .

Why is $L := \ker(\pi) \cap G$ a lattice? Because an isometry $f(x) = Ax + b$ is in $\ker(\pi)$ if and only if $f$ is a translation. So $L$ is a discrete group of translations of $(\mathbb{R}^2, +)$ , so it's a lattice by the “Discrete Subgroups of $\mathbb{R}^2$ ” theorem.

Example. The tilings of $\mathbb{R}^2$ shown below are the square tiling, the pythagorean tiling, and the weaving tiling.

The point groups of their symmetry groups are the following:

The symmetry group of the square tiling has point group $D_4$ and lattice $\mathbb{Z}^2$ .
The symmetry group of the pythagorean tiling has point group $C_4$ and lattice shown by the dashed lines.
The symmetry group of the weaving tiling has point group $D_4$ . Some symmetries are glide reflections!

Remark. Importantly, the point group of a tiling is not the same as the group of isometries that fixes a single point. See the weaving tiling for an example of this.

§ Classifying Discrete Symmetry Groups: Mathematical Reasoning

Any discrete symmetry group $G \subseteq M_2$ is built out of its point group $\overline{G}$ and its lattice $L$ . What can we say about the relationship between $\overline{G}$ and $L$ ?

Theorem. (Point Groups are a Symmetry of the Lattice) Given a discrete subgroup $G \subseteq M_2$ , consider its corresponding point group $\overline{G} = \pi(G)$ and lattice $L = \ker(\pi) \cap G$ .

Then for any $A \in \overline{G}$ and any $x_0 \in L$ , we have $Ax_0 \in L$ . In other words, $\overline{G}(L) = L$ .

Proof: Since $A \in \overline{G}$ and $x_0 \in L$ , we know the following about $G$ :

The isometry $f_1: x \mapsto Ax + b$ is in $G$ for some translation $b \in \mathbb{R}^2$ .
The isometry $f_2: x \mapsto x + x_0$ is in $G$ .

The key idea is to consider the conjugation of $f_2$ by $f_1$ , which is in $G$ .

\begin{align*}f_1f_2f_1^{-1}: x \ \mapsto \ & A(A^{-1}x - A^{-1}b + x_0) + b \\ = \ & x + Ax_0.\end{align*}

Thus, $x \mapsto x + Ax_0$ is in $G$ , so $Ax_0$ is in $L$ , as desired. $\blacksquare$

What kind of point groups $\overline{G} \subseteq O_2(\mathbb{R})$ can preserve a lattice $L$ , then?

Theorem. (Crystallographic Restriction) Consider some point group $\overline{G} \subseteq O_2(\mathbb{R})$ , which must look like $C_n$ or $D_n$ for some $n$ . Then if $L$ is a nontrivial lattice preserved by $\overline{G}$ , it must in fact be the case that $n \in \{1, 2, 3, 4, 6\}$ .

Proof: The key idea is to consider the nonzero vector $x$ of minimal length in $L$ .

If $n > 6$ , then $\overline{G}$ contains some rotation $R$ by an angle less than $\frac{\pi}{3}$ . But then $|Rx - x| < |x|$ by geometry, contradicting the minimality of $|x|$ .
Similarly, if $n = 5$ , then $\overline{G}$ contains the rotation $R$ by angle $\frac{2\pi}{5}$ . But then $|R^2x + x| < |x|$ by geometry, contradicting the minimality of $|x|$ .

So the above implies $n \in \{1, 2, 3, 4, 6\}$ , and we're done. $\blacksquare$

Thus, our answer to the big question for today is the following:

Answer. If $G \subseteq M_2$ is a discrete symmetry group with a nontrivial lattice, then $\overline{G} = \pi(G)$
must look like $C_n$ or $D_n$ for some $n \in \{1, 2, 3, 4, 6\}$ .

The original question asked for all the discrete symmetry groups of some shape $S \subseteq \mathbb{R}^2$ . But our analysis so far has completely ignored the shape $S$ , so what remains is just shape-inspection for each candidate point group $\pi(G)$ .

This shape-inspection, unfortunately, is mostly tedious. So we'll just skip to the results.

§ Classifying Discrete Symmetry Groups: The Results

We categorize our results based on the rank of our lattice.

Example (Rank 0 Lattices). If the lattice $L$ has rank zero, then $L = \{0\}$ , and every single point group $C_n$ or $D_n$ works (even if $n \not \in \{1, 2, 3, 4, 6\}$ ). This is uninteresting.

Example (Rank 1 Lattices: Frieze Groups). Consider the lattice $L = \mathbb{Z}v$ . Then it turns out $\overline{G} \in \{C_1, D_1, C_2, D_2\}$ , achievable in $7$ different ways. These are called Frieze Groups.

Example (Rank 2 Lattices: Wallpaper Groups). For $L= \mathbb{Z}v + \mathbb{Z}w$ , there are $17$ different Wallpaper Groups.

And these are all of the discrete symmetry groups of some shape $S \subseteq \mathbb{R}^2$ . The end.

Remark. These are not meant to be memorized, unless you're a crystallographer or something.

§ Group Actions and Cayley's Theorem

Now for something completely different.

Definition (Action). Let $G$ be a group and $S$ be a set. Then an action of $G$ on $S$ is a map $G \times S \to S$ satisfying the following properties:

If $1$ is the identity of $G$ , then $(1, s) \mapsto s$ for all $s \in S$ .
For any $g, h \in G$ and any $s \in S$ , both $(gh, s)$ and $(g, (h, s))$ correspond to the same element of $S$ .

For ease of writing, we will often denote the image of $(g, s)$ by $g \cdot s$ .

Example. Some common examples of group actions are:

The permutation group $G = S_n$ acts on $S = \{1, 2, \dots, n\}$ .
The group of linear transformations $G = GL_n(\mathbb{R})$ acts on $\mathbb{R}^n$ .
Every group $G$ acts on the set of its own elements.

Importantly, the third point in the example above provides the intuition behind the following theorem.

Theorem. (Cayley's Theorem) Every group $G$ is isomorphic to a subgroup of $S_n$ , where $n = |G|$ .

Proof: Think of $G$ as a group action on the set of its own elements. Expressed more formally, there exists a map $f: G \to \mathrm{Perm}(G)$ , where $f(g)$ is the map $f(g): s \mapsto gs$ for all $s \in G$ .

Intuitively, $f$ is an injective homomorphism. Here's the formal mathematical jargon that justifies this intuition.

To show $f$ is a homomorphism, just write $f(gh)(s) = (gh) \cdot s = g \cdot (h \cdot s) = g \cdot (f(h)(s)) = (f(g) \circ f(h))(s)$ .
To show $f$ is injective, note that if $f(g) = f(h)$ , then $gs = hs$ , meaning $g = h$ .

Thus, $H = f(G) \subseteq S_n$ is a subgroup of $S_n$ for which $f: G \to H$ is an isomorphism, meaning $G \cong H$ and win. $\blacksquare$

Remark. This isomorphism from groups $G$ to subgroups of $S_n$ is very “space-inefficient”. For example, the encoding of $S_8$ as a group action on its own elements would yield a representation of $S_8$ as a subgroup of $S_{40320}$ .