MIT 18.701 — Lecture 10

Isometries in $\mathbb{R}^2$ , Discrete Subgroups of $O_2(\mathbb{R})$

§ Isometries

Definition (Isometry). A function $f: \mathbb{R}^n \to \mathbb{R}^n$ is an isometry if $|f(x) - f(y)| = |x - y|$ for all $x, y \in \mathbb{R}^n$ .

Example. Every orthogonal matrix $M \in O_n(\mathbb{R})$ is an isometry. Translations $f: x \mapsto x + b$ are also isometries.

Theorem. (Isometry Classification) Every isometry $f: \mathbb{R}^n \to \mathbb{R}^n$ may be expressed as $f: x \mapsto Ax + b$ for some $A \in O_n(\mathbb{R})$ and $b \in \mathbb{R}^n$ . In other words, all isometries are an orthogonal matrix composed with a translation.

Proof: To eliminate the “up to translation” freedom, write $f(x) = g(x) + f(0)$ . So $g(x)$ is an isometry satisfying $g(0) = 0$ , and we would like to show $g(x)$ must look like $g: x \mapsto Ax$ for some $A \in O_n(\mathbb{R})$ .

Why is $g(0) = 0$ useful to know? Well, since $g$ is an isometry, it allows us to write:

|x| = |x - 0| = |g(x) - g(0)| = |g(x) - 0| = |g(x)|

Thus, $g$ preserves magnitudes. Unfortunately, we don't know that $g$ is a linear operator, so we can't use the “Properties of Orthogonal Matrices” theorem from last lecture. Let's fix that!

Observe that inner products $\langle x, y \rangle$ can be encoded in terms of magnitudes using the Polarization Identity.
$\langle x, y \rangle = \frac{|x|^2 + |y|^2 - |x - y|^2}{2}$
Thus, since $g$ preserves magnitudes and distances, $g$ must also preserve the inner product; that is, $\langle x, y \rangle = \langle g(x), g(y) \rangle$ .
Observe that additivity $z = x + y$ can be encoded in terms of inner products.
$z = x + y \iff |z - x - y|^2 = 0 \iff |x|^2 + |y|^2 + |z|^2 - 2 \langle x, z \rangle - 2 \langle y, z \rangle + 2 \langle x, y \rangle = 0.$
Thus, since $g$ preserves the inner product, $g$ must also preserve additivity; that is, $g(x + y) = g(x) + g(y)$ .
Observe that scaling $y = \lambda x$ can be encoded in terms of inner products.
$y = \lambda x \iff |y - \lambda x|^2 = 0 \iff |y|^2 - 2 \lambda \langle x, y \rangle + \lambda^2 |x|^2 = 0.$
Thus, since $g$ preserves the inner product, $g$ must also preserve scaling; that is, $g(\lambda x) = \lambda g(x)$ .

Thus, $g$ is a linear operator, so now the “Properties of Orthogonal Matrices” theorem finishes the job. $\blacksquare$

Definition (Isometry Group). The isometry group $M_n$ consists of all isometries of $\mathbb{R}^n$ under composition.

Some interesting properties of $M_n$ include:

Given the homomorphism $\pi: M_n \to O_n(\mathbb{R})$ via $\pi: (x \mapsto Ax + b) \mapsto A$ , then $\ker \pi = T_n$ is the set of translations.
However, we cannot write $M_n \cong T_n \times O_n(\mathbb{R})$ . This is because the group operation in $M_n$ looks like:
$\text{If } f_1 = A_1x + b_1 \text{ and } f_2 = A_2x + b_2, \text{ then } f_1 \circ f_2 = (A_1A_2)x + (A_1b_2 + b_1).$
Thus, $(A_1, b_1) \circ (A_2, b_2) = (A_1A_2, A_1b_2 + b_1)$ instead of the expected $(A_1, b_1) \circ (A_2, b_2) = (A_1A_2, b_1 + b_2)$ . Rather than writing $M_n \cong T_n \times O_n(\mathbb{R})$ , we instead write $M_n \cong T_n \rtimes O_n(\mathbb{R})$ , the semidirect product.

Remark. Recall from a problem set that if $H$ and $K$ are normal subgroups of $G$ such that $H \cap K = \{1\}$ and $G = \{hk \mid h \in H, k \in K\}$ , then $G \cong H \times K$ . If we adjust this setup so that $K$ need not be normal, then we instead write $G \cong H \rtimes K$ , the semidirect product. Thus, since $T_n$ is a normal subgroup of $M_n$ , but $O_n(\mathbb{R})$ is not, we can write $M_n = T_n \rtimes O_n(\mathbb{R})$ .

§ Sidenote: Affine Stuffs

Definition (Affine). A function $T: \mathbb{R}^n \to \mathbb{R}^n$ is affine if $T(x) := Ax + b$ for some $A \in \mathbb{R}^{n \times n}$ and $b \in \mathbb{R}^n$ .

Example. All linear transformations are affine. All isometries are affine by the “Isometry Classification” theorem.

Definition (Affine Combinations). A linear combination $\sum_{i = 1}^k \lambda_i x_i$ of $\{x_1, \dots, x_k \} \subseteq \mathbb{R}^n$ is an affine combination if the coefficients sum to $1$ ; that is, $\sum_{i = 1}^k \lambda_i = 1$ .

So an affine combination of vectors is a weighted average of vectors (with possibly negative coefficients).

Definition (Affine Subspace). A subset $S \subseteq V$ is an affine subspace if $S = W + t$ for some subspace $W \subseteq V$ and vector $t \in V$ ; that is, $S$ is a translation of a subspace.

Theorem. (Affine Subspaces Make Sense) For any affine subspace $S \subseteq V$ , any affine combination of finitely many vectors in $S$ is also in $S$ .

Proof: Say the finitely many vectors in $S$ are $\{w_1 + t, \dots, w_k + t\}$ , where each $w_i \in W$ . Then an affine combination of these vectors is:

\sum_{i = 1}^k \lambda_i(w_i + t) = \left(\sum_{i = 1}^k \lambda_i w_i\right) + \left(\sum_{i = 1}^k \lambda_i\right)t = (\text{a linear combination of $\{w_i\}$}) + t \in W + t. ~ \blacksquare

You can generalize most vector space definitions to affine stuffs like this; we won't explore it further, though.

§ Symmetry Groups

Definition (Symmetry Group). For any subset $S \subseteq \mathbb{R}^n$ , the symmetry group of $S$ is the subgroup of $M_n$ consisting of all isometries $T \in M_n$ such that $TS = S$ .

Example. The symmetry group of $\mathbb{R}^n$ is $M_n$ . The symmetry group of any sphere centered at the origin is $O_n(\mathbb{R})$ .

Example. An equilateral triangle in $\mathbb{R}^2$ has six symmetries; its symmetry group is $D_3$ . However, if the edges of the equilateral triangle were oriented counterclockwise (say), its symmetry group would be only $C_3$ , since reflections would not preserve the counterclockwise orientation.

Example. The none pizza with left beef has two symmetries (assuming the left beef is placed symmetrically).

§ The Big Four: Isometries in $\mathbb{R}^2$

Here's the question we would like to answer.

Question. What do all of the two-dimensional isometries $f(x) = Ax + b \in M_2$ look like?

The key thing that makes $M_2$ nice is that all $A \in O_2$ fit in one of the following two cases:

Case X. Pure rotations counterclockwise by $\theta$ , via $A = \left[\begin{smallmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{smallmatrix}\right]$ .
Case Y. Pure reflections about a line at angle $\frac{\theta}{2}$ from the $x$ -axis, via $A = \left[\begin{smallmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{smallmatrix}\right]$ .

Case X. The trick is to try to rewrite $f(x) = Ax + b$ as $f(x) = A(x - x_0) + x_0$ , for some $x_0 \in \mathbb{R}^2$ . In this form, it is clear that $f(x)$ is a rotation by $\theta$ about some point $x_0 \in \mathbb{R}^2$ .

Remark. Equivalently, $f(x)$ is the conjugation of a rotation $x \mapsto Ax$ by a translation $x \mapsto x + x_0$ .

What does it take to do this? Well, we need to pick $x_0$ so that $b = x_0 - Ax_0$ , or so that $(I_2 - A)x_0 = b$ .

This is doable as long as $I_2 - A$ is invertible. Well, $I_2 - A$ is noninvertible if and only if $A$ has an eigenvalue of $1$ . But rotations $A = \left[\begin{smallmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{smallmatrix}\right]$ have eigenvalues $e^{\pm i \theta} \neq 1$ as long as $A \neq I_2$ .

— Summary of Case X —

If $A \neq I_2$ , then $f(x) = Ax + b$ looks like $f(x) = A(x - x_0) + x_0$ , which is a rotation about $x_0$ by $\theta$ .

If $A = I_2$ , then $f(x) = Ax + b$ looks like $f(x) = x + b$ , which is a translation in the direction of $b$ .

Case Y. Let's try to use the same trick as in Case X. Is writing $f(x) = A(x - x_0) + x_0$ still possible?

Unfortunately, not always; we need $b = (I_2 - A)x_0$ , but $I_2 - A$ is never invertible because reflections $A$ always have an eigenvalue of $1$ . Maybe we can still be lucky, though; which $b$ are in the image of $I_2 - A$ ?

b = (I_2 - A)x_0 ~ \implies ~ Ab = Ax_0 - A^2 x_0 = Ax_0 - x_0 = (A - I_2)x_0 = -b.

Thus, we can write $f(x) = Ax + b$ in the form $f(x) = A(x - x_0) + x_0$ only if $Ab = -b$ , in which case $f(x) = A(x - x_0) + x_0$ performs a reflection across the line with angle $\frac{\theta}{2}$ passing through $x_0$ .

If $b$ is not in the image of $I_2 - A$ , we're not so fortunate. But we can still cleverly rewrite $f(x) = Ax + b$ as:

f(x) = \underbrace{\left(Ax + \dfrac{b - Ab}{2}\right)}_{f'(x)} \ \underbrace{+ \left(\dfrac{Ab + b}{2}\right)}_{T(x)}.

Then $f'(x) = Ax + \frac{b - Ab}{2}$ can be written in $f'(x) = A(x - x_0) + x_0$ form, since $A\left(\frac{b - Ab}{2}\right) = -\left(\frac{b - Ab}{2}\right)$ . Thus, $f'(x)$ corresponds to the reflection about the line with angle $\frac{\theta}{2}$ through $x_0$ .

Furthermore, we claim the translation $T: x \mapsto x + \frac{Ab + b}{2}$ is along the direction of $\frac{\theta}{2}$ . The reason why is by geometry; the midpoint of any vector $b$ and its reflection $Ab$ about a line $\ell$ must lie on $\ell$ .

— Summary of Case Y —

If $f(x) = Ax + b$ satisfies $Ab = -b$ , then we may write $f(x) = A(x - x_0) + x_0$ , which is a reflection:
specifically, a reflection about a line with angle $\frac{\theta}{2}$ through $x_0$ .

Otherwise, we write $f(x) = \left(Ax + \frac{b - Ab}{2}\right) + \left(\frac{Ab + b}{2}\right)$ , which is a glide reflection:
a reflection about a line $\ell$ with angle $\frac{\theta}{2}$ through $x_0$ and a translation along $\ell$ by $\frac{Ab + b}{2}$ .

In summary, the isometries in $\mathbb{R}^2$ are the rotation, translation, reflection, and glide reflection.

§ Defining Discreteness

We've classified all of the isometries. Naturally, it makes sense to try to answer the following bigger question:

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

Unfortunately, this question is very hard. The answer isn't “every subgroup of $M_2$ ” because, say, the group $SO_2(\mathbb{R})$ of all rotations about the origin is not a symmetry group: any set it preserves is a union of circles centered at the origin, which the reflections preserve too.

Some symmetry groups are also very stupid; for example, all of $M_2$ is the symmetry group of $\emptyset$ . We introduce the following definition to filter out subgroups of $M_2$ that feel overly “dense”.

Definition (Discrete Subgroups). A subgroup $G \subseteq M_2$ or $G \subseteq M_1$ is discrete if there is some $\varepsilon > 0$ such that no two elements of $G$ are less than $\varepsilon$ apart. (A discrete subgroup of $M_n$ has no translation or rotation with magnitude less than $\varepsilon$ .)

This motivates the following alternative question.

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

§ Classifying Discrete Subgroups of $O_2(\mathbb{R})$

To end today's lecture, we'll answer the following simplified question.

Question (Simplified). What do all the discrete subgroups of $O_2(\mathbb{R})$ look like?

In other words, only consider the symmetry groups without translations. The answer is the following:

Theorem. (Discrete Subgroups of $O_2(\mathbb{R})$ ) Every discrete subgroup of $O_2(\mathbb{R})$ looks like one of the following:

\begin{cases} C_n = \{1, x, \dots, x^{n - 1} \} \text{ where } x \text{ is a rotation by } \frac{2\pi}{n}. \\ D_n = \{1, x, \dots, x^{n - 1}, y, yx, \dots, yx^{n - 1} \} \text{ where } y \text{ is a reflection, and } x \text{ is a rotation by } \frac{2\pi}{n}. \end{cases}

To prove this, we'll need the following helper theorem.

Theorem. (Discrete Subgroups of $\mathbb{R}$ ) Every discrete subgroup of $(\mathbb{R}, +)$ is of the form $\alpha \mathbb{Z}$ for some $\alpha \in \mathbb{R}$ .

Proof: Let $S$ be a discrete subgroup of $(\mathbb{R}, +)$ ; if $S = \{0\}$ , then $S = 0 \mathbb{Z}$ , so assume otherwise. Then there is some smallest positive element $\alpha \in S$ .

By additive closure, every integer multiple of $\alpha$ is in $S$ , so $\alpha \mathbb{Z} \subseteq S$ .
For any real number $t \not \in \alpha \mathbb{Z}$ , let $n$ be the unique integer such that $t \in (n \cdot \alpha, (n + 1) \cdot \alpha)$ . Then if $t$ were in $S$ , so would $t - n \cdot \alpha$ , contradicting the minimality of $\alpha$ . Thus, for all $t \not \in \alpha \mathbb{Z}$ , we also have $t \not \in S$ .

This proves $S = \alpha \mathbb{Z}$ , as desired. $\blacksquare$

Furthermore, we'll assume the following analysis-flavored theorem for free.

Theorem. (Compactness of $O_2(\mathbb{R})$ ) Any subgroup of $O_2(\mathbb{R})$ is discrete if and only if it is finite.

We're now ready to provide the proof to the “Discrete Subgroups of $O_2(\mathbb{R})$ ” theorem.

Proof: From the “Compactness of $O_2(\mathbb{R})$ ” theorem, it is equivalent to consider finite subgroups $G \subseteq O_2(\mathbb{R})$ .

Suppose for now that $G \subseteq SO_2(\mathbb{R})$ . Then $G$ consists only of rotations, so we may define the isomorphism:

\varphi: SO_2(\mathbb{R}) \to \mathbb{R}/2\pi \mathbb{Z} ~~~ \text{ defined so that } ~~~ \varphi: \left[\begin{smallmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{smallmatrix}\right] \mapsto \theta + 2 \pi \mathbb{Z}.

Then the preimage of $\varphi(G)$ in $\mathbb{R}$ is a discrete subgroup of $\mathbb{R}$ containing $2 \pi \mathbb{Z}$ , meaning it is $\alpha \mathbb{Z}$ with $\alpha = \frac{2\pi}{n}$ for some positive integer $n$ . So $G$ must look like $C_n$ .

We now confront the $G \not \subseteq SO_2(\mathbb{R})$ case. By the reasoning above, we know $H = G \cap SO_2(\mathbb{R})$ is isomorphic to $C_n$ . Thus, $G = H \cup y \, H$ for any $y \in G \setminus H$ . By inspection, the only possibility then is that $y$ refers to a reflection, meaning $G \cong D_n$ . $\blacksquare$