Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 24 September 2014

Lecture 28: The affine orthogonal group and isometries

The affine orthogonal group is $$A{O}_n\left(ℝ\right)=\left\{\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& \mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right) \right| \genfrac{}{}{0ex}{}{g\in O_n\left(ℝ\right)}{\mu \in {ℝ}^n}\}$$ The orthogonal group is $$O_n\left(ℝ\right)=\{g\in M_{n\times n}\left(ℝ\right) | g{g}^t=1\}.$$ The special orthogonal group is $$S{O}_n\left(ℝ\right)=\{g\in O_n\left(ℝ\right) | \det \left(g\right)=1\}$$ and $$\frac{O_n\left(ℝ\right)}{S{O}_n\left(ℝ\right)}=\{N,rN\}$$ where $$N=S{O}_n\left(ℝ\right)andr=\left(\begin{array}{cccc}-1& & & 0\\ & 1& & \\ & & \ddots & \\ 0& & & 1\end{array}\right),$$ since $$\begin{array}{rrcl}\det :& O_n\left(ℝ\right)& \to & \{\pm 1\}\\ & g& \mapsto & \det \left(g\right)\end{array}and\{\pm 1\}\simeq \frac{\mathbb{Z}}{2\mathbb{Z}}.$$ 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 $$r=\left(\begin{array}{cccc}-1& & & 0\\ & 1& & \\ & & \ddots & \\ 0& & & 1\end{array}\right).$$

For $\mu \in {ℝ}^n$ and $g\in O_n\left(ℝ\right)$ let $$X^{\mu }=\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 1& \phantom{0}& \mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)andg=\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& 0\\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right).$$ Then $$g{X}^{\mu }{g}^{-1}=X^{g\mu }and{X}^{\mu }{X}^{\nu }=X^{\mu +\nu }$$ since $$\begin{array}{rcccl}g{X}^{\mu }& =& \left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& 0\\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 1& \phantom{0}& \mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)& =& \left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& g\mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)and\\ X^{g\mu }g& =& \left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 1& \phantom{0}& g\mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& 0\\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)& =& \left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& g\mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right).\end{array}$$ Let $${𝔼}^n=\left\{\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right) \right| x\in {ℝ}^n\}=\left\{\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\\ 1\end{array}\right) \right| x_1,x_2,...,x_n\in ℝ\}.$$ The group $A{O}_n\left(ℝ\right)$ acts on ${𝔼}^n$ by $$g\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)=\left(\begin{array}{c}\phantom{0}\\ gx\\ \phantom{0}\\ 1\end{array}\right)and{X}^{\mu }\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)=\left(\begin{array}{c}\phantom{0}\\ \mu +x\\ \phantom{0}\\ 1\end{array}\right).$$ Note that, if $\mu \ne 0$ then $$\begin{array}{rrcl}{t}_{\mu }:& {ℝ}^n& \to & {ℝ}^n\\ & x& \mapsto & \mu +x\end{array}$$ is not a linear transformation, in particular $t_{\mu }\left(0\right)\ne 0.$

Let $d:{𝔼}^n\times {𝔼}^n\to {ℝ}_{\ge 0}$ be the metric on ${𝔼}^n$ given by $$d(x,y)=|x-y|=\sqrt{{(x_1-y_1)}^2+{(x_2-y_2)}^2+\cdots +{(x_n-y_n)}^2}$$ for $$x=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\\ 1\end{array}\right)andy=\left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_n\\ 1\end{array}\right).$$

Let $⟨,⟩:{𝔼}^n\times {𝔼}^n\to ℝ$ be the positive definite bilinear form given by $$⟨x,y⟩=x_1{y}_1+x_2{y}_2+\cdots +x_n{y}_n$$ for $$x=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\\ 1\end{array}\right)andy=\left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_n\\ 1\end{array}\right).$$ Note that $$\begin{array}{cc}d(x,y)=\sqrt{⟨x-y,x-y⟩}& \text{(a)}\end{array}$$ and $$\begin{array}{cc}\begin{array}{rcl}⟨x,y⟩& =& \frac{1}{4}(⟨x+y,x+y⟩-⟨x-y,x-y⟩)\\ & =& \frac{1}{4}(d{(x,-y)}^2-d{(x,y)}^2)\end{array}& \text{(b)}\end{array}$$ An isometry of ${𝔼}^n$ is a function $f:{𝔼}^n\to {𝔼}^n$ such that $$if\; x,y\in {𝔼}^n \; then\; d(fx,fy)=d(x,y).$$

HW: Use (a) and (b) to show that if $f:{𝔼}^n\to {𝔼}^n$ is an isometry then $f$ satisfies $$if\; x,y\in {𝔼}^n \; then\; ⟨fx,fy⟩=⟨x,y⟩.$$

The group of isometries of ${𝔼}^n$ is $$\mathrm{Isom}\left({𝔼}^n\right)=\{f:{𝔼}^n\to {𝔼}^n | f \; is\; an\; isometry\}$$ with operation given by composition of functions.

Define $$\begin{array}{rrcl}\Phi :& A{O}_n\left(ℝ\right)& \to & \mathrm{Isom}\left({𝔼}^n\right)\\ & y& \mapsto & f_y\end{array}$$ where $f_y:{𝔼}^n\to {𝔼}^n$ is given by $$f_y\left(\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)\right)=\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& \mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right)\left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)ify=\left(\begin{array}{cccc}\phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& g& \phantom{0}& \mu \\ \phantom{0}& \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& 0& \phantom{0}& 1\end{array}\right).$$ Then $\Phi$ is a group isomorphism.

Let $\mu \in {ℝ}^n.$ Translation by $\mu$ is the function $$\begin{array}{rrcl}{t}_{\mu }:& {𝔼}^n& \to & {𝔼}^n\\ & \left(\begin{array}{c}\phantom{0}\\ x\\ \phantom{0}\\ 1\end{array}\right)& \mapsto & \left(\begin{array}{c}\phantom{0}\\ \mu +x\\ \phantom{0}\\ 1\end{array}\right)\end{array}$$ Note that $t_{\mu }$ is an isometry since $$\begin{array}{rcl}d(t_{\mu }x,t_{\mu }y)& =& \sqrt{⟨(\mu +x)-(\mu +y),(\mu +x)-(\mu +y)⟩}\\ & =& \sqrt{⟨x-y,x-y⟩}\\ & =& d(x,y).\end{array}$$

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.

page history