---- Let $M_{2}$ denote the [[group]] of [[euclidean isometry|euclidean isometries]] (rigid motions) of $\mathbb{R}^{2}$. Let $\text{SO}_{n}$ denote the [[special orthogonal group]]. > [!theorem] Theorem. ([[the structure of euclidean plane isometries]]) > [[euclidean isometry|Recall that]] $M_{2} = T \rtimes O_{2},$ > where $T$ denotes the [[subgroup]] of translations in $M_{2}$, and moreover that $\sigma t_{a} \sigma ^{-1}=t_{\sigma(a)}.$ The following properties hold: > 1. Every element of $O_{2}$ is represented by a [[matrix]] of the form $\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \text{ or } \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta. \end{bmatrix}$ > The first kind are (proper) rotations and the second kind of reflections over > 2. $O_{2} = \text{SO}_{2} \rtimes \langle r \rangle$, and if $\rho_{\theta} \in \text{SO}_{2}$ is rotation by $\theta$ then $r \rho_{\theta} r^{-1}=\rho_{-\theta}$. ([[dihedral group|compare]]) > 3. Every element of $M_{2}$ is of the form $t_{a} \rho _\theta$ (orientation-preserving) or $t_{a} \rho_{\theta}r$ (orientation-reversing). The following relations allow us to multiply any two elements in this form:$\begin{align} \rho_{\theta}t_{a}= & t_{a'}\rho_{\theta}, \text{ where } a':= \rho_{\theta}(a) \\ rt_{a} = & t_{a'}r, \ \ \text{ where } a' := r(a) \\ r \rho_{\theta} = & \rho_{-\theta}r \\ t_{a}t_{b}= & t_{a+b} \\ \rho_{\theta}\rho_{\theta'}= & \rho_{\theta + \theta'} \\ r^{2}= & e.\end{align}$ >4. Suppose a point $p$ has [[coordinate vector|coordinates]] $X=\begin{bmatrix}x \\ y\end{bmatrix}$ in some coordinate system. If we change coordinates, $p$ has new coordinates $X'=\eta ^{-1} X$ for some isometry $\eta$. Thus if $m \in M_{2}$ (given, say, in the standard coordinates), then the formula for $m$ in the new coordinates will be $\eta ^{-1} m \eta.$ >5. In the special case that said change of coordinates is given by a translation, $\text{pr}:M_{2} \to O_{2}$ commutes with it. >6. (Characterization of orientation-*preserving* isometries) Every element $m=t_{a} \rho_{\theta} \in M_{2}$ with $\theta \neq 0$ is a rotation by the (same) angle $\theta$ about some point $b$ in the plane. (If $\theta=0$ its just a translation.) >7. (Characterization of orientation-*reversing* isometries) Every element $m=t_{a}\rho_{\theta}r \in M_{2}$ is a (reflection or) [[glide reflection]]. >8. 6 and 7 along with [[multiplicativity of the determinant]] tell us that >- A composite of two rotations (OP $\cdot$ OP) is a rotation or translation (OP) >- A composite of a rotation (OP) and a glide (OR) is a glide >- A composite of two glides (OR $\cdot$ OR) is a rotation or translation (OP) > [!proof]- Proof. ([[the structure of euclidean plane isometries]]) > **1.** Let $L \in O_{2}$ have ([[orthogonal matrix|orthogonal]]) [[matrix]] $A$ wrt standard [[basis]] $e_{1},e_{2}$. [[isometry iff maps orthonormal lists to orthonormal lists|A preserves]] [[orthonormal basis|orthonormal bases]], hence $\|Ae_{1}\|_{2}=1$ so $Ae_{1} = \begin{bmatrix}\cos \theta \\ \sin \theta \end{bmatrix}$ for some $\theta \in [0, 2\pi)$ (must lie on the [[unit circle]]). Now, we see that two choices of $Ae_{2}$ preserve [[orthonormal|orthonormality]] with $Ae_{1}$; these are (draw picture) $\begin{bmatrix}-\sin \theta \\ \cos \theta\end{bmatrix}$ and $\begin{bmatrix}\cos \theta \\ -\sin \theta\end{bmatrix}$. So we have either $\begin{bmatrix} > \cos \theta & -\sin \theta \\ > \sin \theta & \cos \theta > \end{bmatrix} \text{ or } \begin{bmatrix} > \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta. > \end{bmatrix}$ > where we can see that LHS is a (proper) rotation and RHS is a reflection over the angle bisector of $e_{1}$ and $A_{1}$. > > **2.** $\text{SO}_{2}$ and $\langle r \rangle$ are [[subgroup]]s of $\text{O}_{2}$ so we will use [[internal semi-direct product]] characterization. First note that $\text{SO}_{2}$ has [[index of a subgroup|index]] $2$ in $O_{2}$: [[determinant|det]]$:\text{O}_{2} \to \mathbb{R}$ is a [[group homomorphism|homomorphism]] [[kernel iff normal subgroup|whose]] [[kernel]] is$\text{}$ $\text{SO}_{2}$, such that $\text{O}_{2} / \text{SO}_{2} \cong \{ \pm 1 \}.$ > Hence $\text{SO}_{2}$ is a [[normal subgroup]] of $\text{O}_{2}$. Since $\det r=-1$ so $\text{SO}_{2} \cap \langle r \rangle=(e)$; because $|\langle r \rangle|=2$ we conclude that the [[Frobenius product of subgroups|frobenius product]] $\text{SO}_{2} \langle r \rangle$ equals $\text{O}_{2}$. The conditions are satisfied. > > **3.** So far we have seen > ![[CleanShot 2023-10-27 at [email protected]|200]], > [[internal semi-direct product|which tells us that]] (casing on $\{ e,r \}=\langle r \rangle$) an element $m \in M_{2}$ may be uniquely written as a product $t_{a} \rho _\theta$ or $t_{a} \rho_{\theta}r$. Now, the relation $\rho_{\theta}t_{a}=t_{\rho_{\theta}(a) }\rho _\theta$ follows from the [[conjugate|conjugation]] property given in the problem statement. That $rt_{a}r^{-1}=t_{r(a)}$ follows from the same property. We can verify that $r \rho_{\theta}=\rho_{-\theta}r$ follows by just [[matrix product|multiplying matrices]]:![[CleanShot 2023-10-27 at [email protected]|300]]. > Clearly $t_{a}t_{b}(x)=t_{a}(b+x)=a+b+x = t_{a+b}(x)$. Similarly, it is easy to verify with matrix multiplication and trig identities that $\rho_{\theta}\rho_{\theta'}=\rho_{\theta+ \theta'}$. Finally $r^{2}=e$ is immediate by matrix multiplication. > > The image below illustrates the relationship between translations and rotations. ![[IMG_BA0F92C8B07D-1.jpeg]] > > **4.** Suppose $p$ has coordinates $X=\begin{bmatrix} x \\ y\end{bmatrix}$ in the original coordinates can coordinates $X'=\begin{bmatrix} x' \\ y'\end{bmatrix}$ in the new coordinates. The new coordinate system has same angles/distances as the original, hence differs from the original by a [[euclidean isometry]] $\eta$. That is, the new coordinate system expressed from the perspective of the old one, is $\eta(\mathbb{R}^{2})$. So $\eta X$ gives the new coordinates of $X$ expressed in the old coordinate system. So to express the new coordinates of $X$ in the new coordinate system, we need to reverse the isometry $\eta$. So $X'=\eta ^{-1}X$. (It always helps to draw a picture with this idea.) Hence the formula for $m$ in the new coordinates will be $\eta ^{-1}m \eta$. > > **5.** We show that $\text{pr}:M_{2} \to O_{2}$ commutes with a change of coordinates corresponding to [[euclidean isometry|isometry]] $\eta$ if $\eta=t_{a}$ for some $a \in \mathbb{R}^{2}$. Let $m=t_{b}\rho_{\theta} \in M_{2}$ be arbitrary. We have $\begin{align} > \text{pr}(\eta ^{-1} m \eta)= & \text{pr}(t_{-a}m t_{a}) \\ > = & \text{pr}(t_{-a}t_{b}\rho_{\theta}t_{a}) \\ > = & \text{pr}(t_{-a}t_{b} t_{\rho_{\theta}(a)}\rho_{\theta}) \\ > = & \rho_{\theta} > \end{align}$ > $\eta ^{-1}\text{pr}(m)\eta=\eta ^{-1} \rho_{\theta}\eta=t_{-a}\rho_{\theta}t_{a}$ ?? > > **6.** Assuming the result is true, changing coordinates by a translation to the center of rotation $b$ would result in a formula for the transformation which visibly shows it is a rotation. So we'll try to find $b$ such that $t_{b}mt_{b}^{-1}$ is in $\text{SO}_{2}$— that is, we want $\begin{align} > t_{b} t_{a} \rho_{\theta} t_{-b}= & t_{a+b - \rho_{\theta}(b)} \rho_{\theta} > \end{align}$ > to 'look like a rotation'. That would entail $a+b-\rho_{\theta}(b)=0$, i.e., $\big( \begin{bmatrix} > \cos \theta & -\sin \theta \\ > \sin \theta & \cos \theta > \end{bmatrix} - \begin{bmatrix} > 1 & 0 \\ > 0 & 1 > \end{bmatrix} \big) b = a.$ > The [[determinant of a matrix|determinant]] is $2(1-\cos \theta)$ which is nonzero because $\theta \neq 0$. Thus we can solve for $b$. > > **7.** Set $m=t_{a}\rho _\theta r$. $\rho_{\theta}r$ is a [[reflection]] across some [[line]] $\ell$. We change coordinates (keeping the same origin) so that $\ell$ becomes the $x$-axis. In these new coordinates $m$ has the form $t_{a'}r$ for some $a'$, $\begin{bmatrix} > 1 & 0 \\ 0 & -1 > \end{bmatrix} \begin{bmatrix} > x \\ y > \end{bmatrix} + \begin{bmatrix} > a_{1}' \\ a_{2} ' > \end{bmatrix} =\begin{bmatrix} > x + a_{1}' \\ -y + a_{2}' > \end{bmatrix}= \begin{bmatrix} > x \\ -y + a_{2}' > \end{bmatrix} + \begin{bmatrix} > a_{1}' \\ 0 > \end{bmatrix}$ > which we recognize as a reflection over the line $x_2=\frac{a_{2}'}{2}$ (since the line is $x_{2}=\text{midpoint}(y, -y + a_{2}')=\frac{y+(-y+a_{2}')}{2}=\frac{a_{2}'}{2}$, we need) followed by a translation by the vector $a_{1}'e_{1}$ parallel to that line. > > > > ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```