## The affine orthogonal group and isometries

Last update: 06 March 2012

## The affine orthogonal group

The affine orthogonal group is The orthogonal group is The special orthogonal group is and $On(ℝ) SOn(ℝ) = {N,rN}$ where $N=SOn(ℝ) and r= -1 0 1 ⋱ 0 1 ,$ since $det: On(ℝ) → {±1} g ↦ det(g) and {±1}≃ ℤ 2ℤ .$ A rotation is an element of $N=S{O}_{n}\left(ℝ\right)$ and a reflection is an element of $rN,$ where $N=S{O}_{n}\left(ℝ\right)$ and $r= -1 0 1 ⋱ 0 1 .$

For $\mu \in {ℝ}^{n}$ and $g\in {O}_{n}\left(ℝ\right)$ let $Xμ = 0 0 0 0 0 1 0 μ 0 0 0 0 0 0 0 1 and g= 0 0 0 0 0 g 0 0 0 0 0 0 0 0 0 1 .$ Then $gXμg-1 = Xgμ and XμXν = Xμ+ν$ since $gXμ = 0 0 0 0 0 g 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 μ 0 0 0 0 0 0 0 1 = 0 0 0 0 0 g 0 gμ 0 0 0 0 0 0 0 1 and Xgμg = 0 0 0 0 0 1 0 gμ 0 0 0 0 0 0 0 1 0 0 0 0 0 g 0 0 0 0 0 0 0 0 0 1 = 0 0 0 0 0 g 0 gμ 0 0 0 0 0 0 0 1 .$ Let The group $A{O}_{n}\left(ℝ\right)$ acts on ${𝔼}^{n}$ by $g 0 x 0 1 = 0 gx 0 1 and Xμ 0 x 0 1 = 0 μ+x 0 1 .$ Note that, if $\mu \ne 0$ then $tμ: ℝn → ℝn x ↦ μ+x$ is not a linear transformation, in particular ${t}_{\mu }\left(0\right)\ne 0.$

Let $d:{𝔼}^{n}×{𝔼}^{n}\to {ℝ}_{\ge 0}$ be the metric on ${𝔼}^{n}$ given by $d(x,y) = |x-y| = (x1-y1)2 + (x2-y2)2 +⋯+ (xn-yn)2$ for $x= x1 x2 ⋮ xn 1 and y= y1 y2 ⋮ yn 1 .$

Let $⟨,⟩:{𝔼}^{n}×{𝔼}^{n}\to ℝ$ be the positive definite bilinear form given by $⟨x,y⟩ = x1y1+x2y2+⋯+xnyn$ for $x= x1 x2 ⋮ xn 1 and y= y1 y2 ⋮ yn 1 .$ Note that $d(x,y) = ⟨x-y,x-y⟩ and (AOG 1) ⟨x,y⟩ = 1 4 ( ⟨x+y,x+y⟩-⟨x-y,x-y⟩ ) = 1 4 ( d(x,-y)2-d(x,y)2 ) (AOG 2)$ An isometry of ${𝔼}^{n}$ is a function $f:{𝔼}^{n}\to {𝔼}^{n}$ such that

HW: Use (AOG 1) and (AOG 2) to show that if $f:{𝔼}^{n}\to {𝔼}^{n}$ is an isometry then $f$ satisfies

The group of isometries of ${𝔼}^{n}$ is with operation given by composition of functions.

Define $Φ: AOn(ℝ) → Isom(𝔼n) y ↦ fy$ where ${f}_{y}:{𝔼}^{n}\to {𝔼}^{n}$ is given by $fy( 0 x 0 1 ) = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 0 x 0 1 if y= 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 .$ Then $\Phi$ is a group isomorphism.

Let $\mu \in {ℝ}^{n}.$ Translation by $\mu$ is the function $tμ: 𝔼n → 𝔼n 0 x 0 1 ↦ 0 μ+x 0 1$ Note that ${t}_{\mu }$ is an isometry since $d(tμx,tμy) = ⟨(μ+x)-(μ+y),(μ+x)-(μ+y)⟩ = ⟨x-y,x-y⟩ = d(x,y).$

## Isometries in ${𝔼}^{2}$

Let $s= -1 0 0 1 and rθ= cosθ -sinθ sinθ cosθ$ and let $s$ is reflection in the $y-$axis ${L}_{y}$

${r}_{\theta }$ is rotation in an angle $\theta$ about $0$
${t}_{\gamma }$ is translation by $\gamma$
So $S{O}_{2}\left(ℝ\right)$ is the group of rotations about $0.$
1. Let $L$ be a line in ${ℝ}^{2}.$ Then there exist $c\in ℝ$ and $0\le \theta <\pi$ such that $L= rθt c 0 Ly.$ The reflection in the line $L$ is $sL = rθt c0 st -c0 r-θ.$
2. Let $p\in {ℝ}^{2}$ and $\theta \in \left[0,2\pi \right)$ Then rotation by $\theta$ around $p$ is $rθ,p = tprθt-p.$
3. The $d-$glide reflection in the line $L$ is: translate by a distance $d$ in a line parallel to $L$ and then reflect in $L.$

## Isometries

Let with $d(p,q) = (x1-x2)2+(y1-y2)2 if p= x1 y1 1 and x2 y2 1 .$ An isometry of ${𝔼}^{2}$ is a function $f:{E}^{2}\to {E}^{2}$ such that $d(fp,fq) = d(p,q).$ Note that

• a rotation fixes one point,
• a reflection fixes a line,
• a translation fixes no point.

Let $f:{𝔼}^{2}\to {𝔼}^{2}$ be an isometry. Suppose $\alpha ,p$ are fixed points of $f,$ $fα=α and fβ=β$

Let $p\in {𝔼}^{2}.$ If $p$ is on the line ${L}_{\alpha \beta }$ connecting $\alpha$ and $\beta$ then ${C}_{1}\cap {C}_{2}=\left\{p\right\}.$ So $fp=p$ if $p\in {L}_{\alpha \beta }.$

Thus, if $f:{𝔼}^{2}\to {𝔼}^{2}$ is an isometry and where ${L}_{\alpha \beta }$ is the line connecting $\alpha$ and $\beta .$

If $f:{𝔼}^{2}\to {𝔼}^{2}$ is an isometry and $\gamma ,\alpha ,\beta \in 𝔼$ are such that $\gamma \notin {L}_{\alpha \beta }$ and $\alpha \ne \beta$ and $fα=α, fβ=β and fγ=γ$ then $f$ fixes all of ${𝔼}^{2}.$

 Proof. $f$ fixes and ${L}_{\beta \gamma }.$ If $p\in {L}_{\alpha \beta }$ and $q\in {L}_{\alpha \gamma }$ then $f$ fixes ${L}_{pq}.$ $Lαβ$ $Lβγ$ $Lpq$ $Lαγ$ $p$ $q$ $α$ $β$ $γ$ $x$ Every point $x\in {𝔼}^{2}$ is on some ${L}_{pq}$ with $p\in {L}_{\alpha \beta }$ and $q\in {L}_{\alpha \beta }$ and so $fx=x.$ So $f={\mathrm{id}}_{{𝔼}^{2}}.$ $\square$

## Proof of the isometry/affine orthogonal group correspondence

Define $Φ: AOn(ℝ) → Isom(𝔼n) y ↦ fy where$ $fy( 0 x 0 1 ) = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 0 x 0 1 if y= 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 .$ Then $\Phi$ is a group isomorphism.

 Proof. To show: $\Phi$ is a function ($\Phi$ is well defined). $\Phi$ is a group homomorphism. $\Phi$ is a bijection. (b) If $y,z\in A{O}_{n}\left(ℝ\right)$ and $\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)\in {𝔼}^{n}$ then $fyfz 0 x 0 1 = yz 0 x 0 1 = fyz 0 x 0 1 .$ So $\Phi$ is a homomorphism. (a) Assume Then $fy = tμg where tμ: 𝔼n → 𝔼n 0 x 0 1 ↦ 0 μ+x 0 1 is a translation,$ and If $x,z\in {𝔼}^{n}$ then $d(tμx,tμz) = d(μ+x,μ+z) = ⟨(μ+x)-(μ+z), (μ+x)-(μ+z)⟩ = ⟨x-z,x-z⟩ = d(x,z)$ and so that $d(gx,gz) = ⟨gx-gz,gx-gz⟩ = ⟨g(x-z),g(x-z)⟩ = ⟨x-z,x-z⟩ = d(x,z).$ Thus $g,{t}_{\mu }$ and ${f}_{y}={t}_{\mu }g$ are all isometries. (c) To show: There is an inverse function to $\Phi .$ Define $Ψ: Isom(𝔼n) → AOn(ℝ) f ↦ 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1$ where $μ=f(0)$ and where To show: $\Psi$ is well defined. $\Psi \circ \Phi ={\mathrm{id}}_{A{O}_{n}}$ and $\Phi \circ \Psi ={\mathrm{id}}_{\mathrm{Isom}}.$ (ii) Let $\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& \mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)\in A{O}_{n}.$ Then $(Ψ∘Φ) 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 = Ψ(fy) = 0 0 0 0 0 g' 0 μ' 0 0 0 0 0 0 0 1$ where $g'= | | | g1' g2' ⋯ gn' | | | with gj' = fy(ej) = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 0 ⋮ 1 0 ⋮ 0 1 = | gj | 1 = | gj |$ and $μ'=fy(0) with fy(0) = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 0 ⋮ 0 1 = 0 μ 0 1 =μ.$ So $(Ψ∘Φ) 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 .$ (i) Let $f\in \mathrm{Isom}.$ Then $(Φ∘Ψ)(f) = Φ 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 = fy where y= 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1$ and $g= | | g1 ⋯ gn | | with f(ej) = | gj | = g 0 ⋮ 0 1 0 ⋮ 0 =gej,$ and $\mu =f\left(0\right).$ To show: $f\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)={f}_{y}\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right).$ To show: ${t}_{-\mu }f={t}_{-\mu }{f}_{y}.$ Let $g={t}_{-\mu }{f}_{y}.$ Let $h={t}_{-\mu }f.$ To show: If $x\in {𝔼}^{n}$ then $hx=gx.$ We know $h\in \mathrm{Isom}\left({𝔼}^{n}\right)$ and $h\left(0\right)=0.$ If $x,z\in {𝔼}^{n}$ then, since $h\left(0\right)=0,$ $⟨hx,hz⟩ = 1 2 ( ⟨hx,hx⟩+⟨hz,hz⟩-⟨hx-hz,hx-hz⟩ ) = 1 2 ( d(hx,0)2 + d(hz,0)2 - d(hx,hz)2 ) = 1 2 ( d(hx,h0)2 + d(hz,h0)2 - d(hx,hz)2 ) = 1 2 ( d(x,0)2 + d(z,0)2 - d(x,z)2 ) = 1 2 ( ⟨x,x⟩ + ⟨z,z⟩ - ⟨x-z,x-z⟩ ) = ⟨x,z⟩.$ Assume $x=\left(\begin{array}{c}{x}_{1}\\ ⋮\\ {x}_{n}\\ 1\end{array}\right)\in {𝔼}^{n}.$ Since $h{e}_{i}=g{e}_{i},$ So $hx=gx.$ $\square$

## Notes and References

Where are these from?

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