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 29: Isometries in $𝔼^{2}$

Let $s = (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) and r_{θ} = (\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix})$ and let $\begin{array}{rrcl} t_{γ} : & ℝ^{2} & \to & ℝ^{2} \\ x & \mapsto & γ + x, \end{array} for γ = (\begin{matrix} γ_{1} \\ γ_{2} \end{matrix}) in ℝ^{2} .$ $s$ is reflection in the $y -$ axis $L_{y}$

r_{θ}

is rotation in an angle

θ

about

0

t_{γ}

is translation by

γ

\begin{array}{rcl} S O_{2} (ℝ) & = & {g \in M_{2 \times 2} (ℝ) | g g^{t} = 1, \det (g) = 1} \\ = & {(\begin{matrix} a & b \\ c & d \end{matrix}) | \binom{a, b, c, d \in ℝ, a d - b c = 1}{(\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} a & c \\ b & d \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})}} \\ = & {(\begin{matrix} a & b \\ c & d \end{matrix}) | \binom{a, b, c, d \in ℝ, a^{2} + b^{2} = 1, a c + b d = 0}{c a + d b = 0, c^{2} + d^{2} = 1, a d - b c = 1}} \\ = & {(\begin{matrix} a & b \\ c & d \end{matrix}) | \binom{a, b, c, d \in ℝ, b = - c}{a^{2} + b^{2} = 1, a = d}} \\ = & {(\begin{matrix} a & b \\ - b & a \end{matrix}) | a, b, c, d \in ℝ, a^{2} + b^{2} = 1} \\ = & {(\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}) | 0 \leq θ < 2 π} \\ = & {r_{θ} | 0 \leq θ < 2 π} . \end{array}

S O_{2} (ℝ)

is the group of rotations about

0 .

Let $L$ be a line in $ℝ^{2} .$ Then there exist $c \in ℝ$ and $0 \leq θ < π$ such that $L = r_{θ} t_{(\binom{c}{0})} L_{y} .$ The reflection in the line $L$ is $s_{L} = r_{θ} t_{(\binom{c}{0})} s t_{(\binom{- c}{0})} r_{- θ} .$
Let $p \in ℝ^{2}$ and $θ \in [0, 2 π)$ Then rotation by $θ$ around $p$ is $r_{θ, p} = t_{p} r_{θ} t_{- p} .$
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 $𝔼^{2} = {(\begin{matrix} x \\ y \\ 1 \end{matrix}) | x, y \in ℝ}$ with $d (p, q) = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} if p = (\begin{matrix} x_{1} \\ y_{1} \\ 1 \end{matrix}) and (\begin{matrix} x_{2} \\ y_{2} \\ 1 \end{matrix}) .$ An isometry of $𝔼^{2}$ is a function $f : E^{2} \to E^{2}$ such that $d (f p, f q) = d (p, q) .$ 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 $α, p$ are fixed points of $f,$ $f α = α and f β = β$

Let $p \in 𝔼^{2} .$ $\begin{array}{ll} Since & d (α, p) = d (f α, f p) = d (α, f p), \\ f p must lie on the circle C_{1} of radius R_{1} = d (α, p) centred at α . \\ Since & d (β, p) = d (f β, f p) = d (β, f p), \\ f p must lie on the circle C_{2} of radius R_{2} = d (β, p) centred at β . \\ So & f p \in C_{1} \cap C_{2} . \end{array}$ If $p$ is on the line $L_{α β}$ connecting $α$ and $β$ then $C_{1} \cap C_{2} = {p} .$ So $f p = p$ if $p \in L_{α β} .$

Thus, if

f : 𝔼^{2} \to 𝔼^{2}

is an isometry and

\begin{aligned} α, β \in 𝔼^{2} & are such that \\ α \neq β and f α = α and f β = β \\ then & f p = p for every p \in L_{α β} \end{aligned}

where

L_{α β}

is the line connecting

α

and

β .

If $f : 𝔼^{2} \to 𝔼^{2}$ is an isometry and $γ, α, β \in 𝔼$ are such that $γ \notin L_{α β}$ and $α \neq β$ and $f α = α, f β = β and f γ = γ$ then $f$ fixes all of $𝔼^{2} .$

		Proof.
	$f$ fixes $L_{α β}, L_{α γ}$ and $L_{β γ} .$ If $p \in L_{α β}$ and $q \in L_{α γ}$ then $f$ fixes $L_{p q} .$ Every point $x \in 𝔼^{2}$ is on some $L_{p q}$ with $p \in L_{α β}$ and $q \in L_{α β}$ and so $f x = x .$ So $f = {id}_{𝔼^{2}} .$ $□$

Notes and References

These are a typed copy of Lecture 29 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 12, 2011.

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