# Rigid motions of $\mathbf R^2$ form a group

I am trying to show that the rigid motions of $\mathbf R^2$ form a group. The rigid motions are distance preserving transformations.

But how to find all the group elements? The following are distance preserving transformations: rotations, translations and reflections. Are there any other? Once I have them, how do I show that these are all?Thank you.

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Note that to show that elements with a certain property form a group, you do not need to find another way to describe them. Just show that the compositions of two such maps is again such a map, and that the inverse is also such a map. – Tobias Kildetoft Jan 16 '13 at 16:36
The question of listing the elements of this group is much harder than showing that it indeed is a group (see Tobias comment). – Siméon Jan 16 '13 at 16:38
@Tobias Thank you. If it is possible to determine all elements I am interested in knowing how to do it. – Anna Jan 16 '13 at 16:41
@Anna A rigid motion is determined uniquely by how three noncollinear points are mapped. – Hagen von Eitzen Jan 16 '13 at 16:50
One "geometric" way to distinguish the different types that you have enumerated is that translations have no fixed points, rotations have one fixed point, reflections have a line as the fixed points, and the identity has the entire plane for fixed points. – Thomas Andrews Jan 16 '13 at 17:00

I'll do this for arbitrary dimension. Just replace even $n$ by a $2$ is you prefer. You have to split this problem up. We can consider rotations separately from translations. Consider the first case, i.e. no translations.

The rotations preserve distances and angles, and so preserve the scalar product. Given column vectors $v$ and $w$, the scalar product can be written as $\langle v,w\rangle = v^{\top}w$. Now, consider a transformation $f$ with matrix $M$: $f(v) = Mv$ and $f(w)=Mw$. It follows that for all vectors $v,w$

$$\langle v, w \rangle = \langle f(v), f(w) \rangle \iff v^{\top}w = (Mv)^{\top}(Mw) \iff v^{\top}w = v^{\top}M^{\top}Mw \iff M^{\top}M = I$$

where $I$ is the identity matrix. So the transformations that preserve distance and angle have matrices with the property that $M^{\top}M = I$. Such matrices are called orthogonal, and the group of $n \times n$ orthogonal matrices is denoted by $O(n)$. However this also included reflection. Notice that:

$$M^{\top}M=I \implies \det(M^{\top}M) = \det(I) \implies \det(M)^2 = 1 \implies \det(M) = \pm 1 \, .$$

The $-1$ case involved reflections, so we assume that $\det(M)=+1$. We call this group the special orthogonal group and write $\text{SO}(n)$.

Now, here's the trick. Let $M \in SO(n)$ be a special orthogonal matrix and let $t \in \mathbb{R}^n$ be a translation. (We can associate a translation with the vector starting at $0$ and ending at the image of $0$.) As a set, the group of Euclidean transformations of $\mathbb{R}^n$ looks like $SO(n) \times \mathbb{R}^n$, with an element being $(M,t)$. We can actually embed the group of Euclidean transformation, as a Lie group, into the space of $(n+1) \times (n+1)$ matrices: $\text{Mat}_{n+1}\mathbb{R}$. What we do is this:

$$(M,t) \mapsto \left( \begin{array}{c|c} M & t \\ \hline 0 & 1 \end{array}\right) .$$

The expression on the right is a block matrix, where $M$ is an $n\times n$ matrix, $t$ is an $n \times 1$ column vector, $0$ is a $1 \times n$ zero vector and $1$ is the $1 \times 1$ matrix $(1)$. Composition is given by:

$$(N,\tau) * (M,t) = \left( \begin{array}{c|c} N & \tau \\ \hline 0 & 1 \end{array}\right)\left( \begin{array}{c|c} M & t \\ \hline 0 & 1 \end{array}\right) = \left( \begin{array}{c|c} NM & Nt+\tau \\ \hline 0 & 1 \end{array}\right).$$

Thus, define an operation $*$ on $SO(n) \times \mathbb{R}^n$ by $(N,\tau) * (M,t) = (NM,Nt+\tau).$

Let's prove that we do actually have a group structure here:

1. $(N,\tau) * (M,t) = (NM,Nt+\tau).$ We need to check that $NM \in SO(n)$ and $Nt + \tau \in \mathbb{R}^n$. Well, if $M^{\top}M = I$ and $N^{\top}N = I$ then $(NM)^{\top}(NM) = M^{\top}N^{\top}NM = M^{\top}IM=M^{\top}M = I$. Clearly $Nt + \tau \in \mathbb{R}^n$.
2. Since matrix multiplication is associative, so is our operation $*$.
3. The identity is $(I,0)$. Checking: $(M,t) * (I,0) = (MI,M0+t) = (M,t).$
4. The inverse of $(M,t)$ is $(M^{-1},-M^{-1}t).$ Checking: $(M,t)*(M^{-1},-M^{-1}t) = (MM^{-1},-MM^{-1}t+t) = (I,0).$

This shows how a previous translation is distorted by a following linear transformation (we don't just add the translations). It's tempting to think that the inverse is just $(M^{-1},-t)$, but we have to "unpick" $M$'s action on the translation. As a Lie group, the group of rigid transformations is a semi-direct product: $\text{Euc}(n) = \text{SO}(n) \ltimes \mathbb{R}^n$.

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To prove it's a group (under composition), you must show the following (it shouldn't be difficult):

(i) There is a rigid motion that acts as an identity element.

(ii) For each rigid motion, there is a rigid motion that acts as an inverse.

(iii) A composition of rigid motions is a rigid motion.

Ordinarily, we'd need to prove that the given operation is associative, but composition of functions $\Bbb R^2\to\Bbb R^2$ satisfies this.

In answer to your other question, there is yet a fourth type of rigid motion called a "glide reflection", which is a reflection over a line, followed (or preceded) by a translation along that line. This is just a combination of a reflection and a translation, though.

It turns out that all rigid motions of the plane can be obtained as a composition of some subcollection of {a translation, a rotation, a reflections}--this subset may be empty, in the case of the identity transformation. In particular, if we pick any line in the plane (call it $\ell$), and any point on that line (call it $p$), then we can obtain any rigid motion of the plane by a composition of a subcollection of {a translation, a rotation about $p$, reflection about $\ell$}.

It's simplest to choose $\ell$ as the $x$-axis and $p$ as the origin, but it isn't strictly necessary to do so. If you're interested full treatment of the classification of rigid motions of the plane can be found in Artin's "Algebra", Chapter 5, Section 2.

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From a different perspective, all rigid motions of the plane can also be obtained as the composition of reflections; a translation by some vector $\vec{v}$ is the composition of two reflections through planes orthogonal to $\vec{v}$ and spaced $|\vec{v}|/2$ apart from each other, and similarly a rotation through some point $\vec{p}$ can be expressed as the composition of reflections through two suitably-chosen planes passing through $\vec{p}$. – Steven Stadnicki Jan 16 '13 at 17:21