---- > [!definition] Definition. ([[euclidean isometry]]) > An **[[Lipschitz continuous|isometry]] of $\mathbb{R}^{n}$** endowed with [[metric]] $d$ is a distance preserving transformation $f: \mathbb{R}^{n} \to \mathbb{R}^{n} \text{ s.t. } d\big(f(x), f(y)\big)=d(x,y) \ \forall x,y \in \mathbb{R}^{n}.$ > > [!warning] > In these notes, the unadorned term 'isometry' sometimes refers to a [[linear isometry]]— a distance-preserving map on an [[inner product space|inner product space]] [[inner product space]]. When $V=\mathbb{R}^{n}$, the [[linear isometry|linear isometries]] are precisely the euclidean isometries that fix the origin. > > > The [[group]] of [[euclidean isometry|isometries]] of $\mathbb{R}^{n}$ is denoted **$M_{n}$**. As a [[group]] it looks like $T \rtimes O_{n}$: [[general orthogonal group|orthogonal transformations]] and translations, where the [[conjugate|conjugation]] [[group action|action]] of $O_{n}$ on $T$ is given by $L \circ T_{x} \circ L^{-1}=T_{L(x)}.$ \ Abstractly, $T \cong (\mathbb{R}^{n}, +)$ (via the identification of $T_x$ with $x$) and the [[group homomorphism|homomorphism]] $O_{n} \to GL_{n}(\mathbb{R}) \subset \text{Aut}(T)$ is just the inclusion. \ As a consequence, there is a natural homomorphism whose [[kernel]] is $T$, $\pi:M_{n} \to O_{n}$ given by dropping $T$ — i.e., [[internal semi-direct product|(uniquely)]] writing $T_{x}O \in M_{n}$ we define $\pi(T_{x}O):=O$. > [!justification] > Let's show these properties of $M_{n}$. > **Fact 1. $T$ is a [[normal subgroup]] of $M_{n}$.** > **Proof.** Let $x \in \mathbb{R}^{n}$ and $L \in O_{n}$. Then $L \circ T_x = T_{L(x)} \circ L,$ > since $(L \circ T_{x})(y)=L(x+y)=L(x)+L(y)=T_{L(x)}L(y)=T_{L(x) \circ L}(y).$ > So, we get $L T_{x} L^{-1} = T_{L(x)},$ > and this is suffices to show the result. > **Fact 2.** $M_{n} \cong T \rtimes O_{n}$. Clearly $T \cap O_{n}=(e)$. Now we've satisfied [[external semi-direct product]] characterization. > [!basicexample] Example. (Isometries of $\mathbb{R}^{1}$) > We can write down some obvious isometries of the real line. >- Translations: $T_{x}$, defined by $T_{x}(y)=x+y$; >- Reflections: for any $x \in \mathbb{R}^{1}$, let $r_x$ denote reflection about the point $x$. \ **Claim.** These are in fact all the elements of $M_{1}$. **Proof.** Let $f \in M_{1}$ and let $x=f(0)$. Then $T_{-x} \circ f(0)=0$ (fixes $0$). Clearly, an element of $M_{1}$ that fixes $0$ must either be the identity or $r_{0}$. If $T_{-x} \circ f = e$, then $f=T_{x}$. Else $T_{-x} \circ f =r_{0}$, in which case $f= T_{x} \circ r_{0}=r_{\frac{x}{2}}$. Thus every element in $M_{1}$ is either a translation or a reflection. \ It follows that $M_{1}$ is a [[group]]. Now we describe its [[group]] structure. Let $T$ be the [[subgroup]] of translations. We have just seen that $M=T \langle r_{0} \rangle$ where $\langle r_{0} \rangle=\langle e, r_{0} \rangle \cong C_{2}$ denotes the [[cyclic group]] [[generating set of a group|generated by]] $r_{0}$. Observe that $r_{0} T_{x} r_{0}^{-1}(y) = r_{0}T_{x} (-y)=r_{0}(x-y)=-x+y=T_{-x}(y),$ and clearly $T \cap \langle r_{0} \rangle=\{ e \}$ so conclude that $M_{1}=T \rtimes_{\phi} C_{2}=\mathbb{R} \rtimes_{\phi} C_{2}$, where the map $\phi: C_{2} \to \text{Aut }(\mathbb{R})$ giving the [[external semi-direct product|semi-direct product]] is the one sending the [[generating set of a group|generator]] of $C_{2}$ to the [[automorphism]] $x \mapsto -x$. \ **Fact.** $M_{1}=T \rtimes \langle r_{x} \rangle$ for any $x$, and the [[subgroup|subgroups]] $\langle r_{x} \rangle$ as $x$ varies are all [[conjugate]] in $M_{1}$. \ **Proof of Fact.** $\langle r_{x} \rangle \cong C_{2}$ for any $x$, so first part seems obvious? \ Next notice that the formula to reflect over a point $x$ is to 'center' at origin (blue), negate (green), and shift back (pink). For example, the formula to reflect $5$ over the point $3$ is $\textcolor{LimeGreen}{-}(5\textcolor{Skyblue}{-3})\textcolor{Thistle}{+3}$. More generally, $r_{x}(y)=-(y-x)+x=\textcolor{Thistle}{T_{x}}\textcolor{LimeGreen}{r_{0}}\textcolor{Skyblue}{T_{-x}}$. This is exactly the [[conjugate|conjugation]] of $r_{0}$ by $T_{x}$. \ **Fact.** The set of [[continuous]] [[automorphism|automorphisms]] of $(\mathbb{R},+)$, $\text{Aut}_{\text{cont}}(\mathbb{R})$, is [[group isomorphism|isomorphic]] to $\mathbb{R}^{\times}$. (Compare to [[automorphism group of cyclic group]]?) \ **Proof of Fact.** An [[automorphism]] of $(\mathbb{R}, +)$ is a function $f$ s.t. $f(a+b)=f(a)+f(b)$. Additionally, $f(0)=0$. [[TODO]] ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY 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 > ```