## Group Theory and Linear Algebra

Last updated: 24 September 2014

## Lecture 28: The affine orthogonal group and isometries

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 $d(x,y)= ⟨x-y,x-y⟩ (a)$ and $⟨x,y⟩ = 1 4 ( ⟨x+y,x+y⟩-⟨x-y,x-y⟩ ) = 1 4 ( d(x,-y)2-d(x,y)2 ) (b)$ An isometry of ${𝔼}^{n}$ is a function $f:{𝔼}^{n}\to {𝔼}^{n}$ such that

HW: Use (a) and (b) 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 $d(tμx,tμy) = ⟨(μ+x)-(μ+y),(μ+x)-(μ+y)⟩ = ⟨x-y,x-y⟩ = d(x,y).$

## Notes and References

These are a typed copy of Lecture 28 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 11, 2011.