The affine orthogonal group and isometries

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

Last update: 06 March 2012

The affine orthogonal group

The affine orthogonal group is $A O_{n} (ℝ) = {(\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) | \binom{g \in O_{n} (ℝ)}{μ \in ℝ^{n}}}$ The orthogonal group is $O_{n} (ℝ) = {g \in M_{n \times n} (ℝ) | g g^{t} = 1} .$ The special orthogonal group is $S O_{n} (ℝ) = {g \in O_{n} (ℝ) | \det (g) = 1}$ and $\frac{O_{n} (ℝ)}{S O_{n} (ℝ)} = {N, r N}$ where $N = S O_{n} (ℝ) and r = (\begin{matrix} - 1 & 0 \\ 1 \\ ⋱ \\ 0 & 1 \end{matrix}),$ since $\begin{array}{rrcl} \det : & O_{n} (ℝ) & \to & {\pm 1} \\ g & \mapsto & \det (g) \end{array} and {\pm 1} ≃ \frac{ℤ}{2 ℤ} .$ A rotation is an element of $N = S O_{n} (ℝ)$ and a reflection is an element of $r N,$ where $N = S O_{n} (ℝ)$ and $r = (\begin{matrix} - 1 & 0 \\ 1 \\ ⋱ \\ 0 & 1 \end{matrix}) .$

For $μ \in ℝ^{n}$ and $g \in O_{n} (ℝ)$ let $X^{μ} = (\begin{matrix} 1 & μ \\ 0 & 1 \end{matrix}) and g = (\begin{matrix} g & 0 \\ 0 & 1 \end{matrix}) .$ Then $g X^{μ} g^{- 1} = X^{g μ} and X^{μ} X^{ν} = X^{μ + ν}$ since $\begin{array}{rcccl} g X^{μ} & = & (\begin{matrix} g & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & μ \\ 0 & 1 \end{matrix}) & = & (\begin{matrix} g & g μ \\ 0 & 1 \end{matrix}) and \\ X^{g μ} g & = & (\begin{matrix} 1 & g μ \\ 0 & 1 \end{matrix}) (\begin{matrix} g & 0 \\ 0 & 1 \end{matrix}) & = & (\begin{matrix} g & g μ \\ 0 & 1 \end{matrix}) . \end{array}$ Let $𝔼^{n} = {(\begin{matrix} x \\ 1 \end{matrix}) | x \in ℝ^{n}} = {(\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \\ 1 \end{matrix}) | x_{1}, x_{2}, ..., x_{n} \in ℝ} .$ The group $A O_{n} (ℝ)$ acts on $𝔼^{n}$ by $g (\begin{matrix} x \\ 1 \end{matrix}) = (\begin{matrix} g x \\ 1 \end{matrix}) and X^{μ} (\begin{matrix} x \\ 1 \end{matrix}) = (\begin{matrix} μ + x \\ 1 \end{matrix}) .$ Note that, if $μ \neq 0$ then $\begin{array}{rrcl} t_{μ} : & ℝ^{n} & \to & ℝ^{n} \\ x & \mapsto & μ + x \end{array}$ is not a linear transformation, in particular $t_{μ} (0) \neq 0 .$

Let $d : 𝔼^{n} \times 𝔼^{n} \to ℝ_{\geq 0}$ be the metric on $𝔼^{n}$ given by $d (x, y) = | x - y | = \sqrt{{(x_{1} - y_{1})}^{2} + {(x_{2} - y_{2})}^{2} + \dots + {(x_{n} - y_{n})}^{2}}$ for $x = (\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \\ 1 \end{matrix}) and y = (\begin{matrix} y_{1} \\ y_{2} \\ ⋮ \\ y_{n} \\ 1 \end{matrix}) .$

Let $⟨, ⟩ : 𝔼^{n} \times 𝔼^{n} \to ℝ$ be the positive definite bilinear form given by $⟨ x, y ⟩ = x_{1} y_{1} + x_{2} y_{2} + \dots + x_{n} y_{n}$ for $x = (\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \\ 1 \end{matrix}) and y = (\begin{matrix} y_{1} \\ y_{2} \\ ⋮ \\ y_{n} \\ 1 \end{matrix}) .$ Note that $\begin{array}{ll} d (x, y) = \sqrt{⟨ x - y, x - y ⟩} and & (AOG 1) \\ \begin{array}{rcl} ⟨ x, y ⟩ & = & \frac{1}{4} (⟨ x + y, x + y ⟩ - ⟨ x - y, x - y ⟩) \\ = & \frac{1}{4} (d {(x, - y)}^{2} - d {(x, y)}^{2}) \end{array} & (AOG 2) \end{array}$ An isometry of $𝔼^{n}$ is a function $f : 𝔼^{n} \to 𝔼^{n}$ such that $if x, y \in 𝔼^{n} then d (f x, f y) = d (x, y) .$

HW: Use (AOG 1) and (AOG 2) to show that if $f : 𝔼^{n} \to 𝔼^{n}$ is an isometry then $f$ satisfies $if x, y \in 𝔼^{n} then ⟨ f x, f y ⟩ = ⟨ x, y ⟩ .$

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

Define $\begin{array}{rrcl} Φ : & A O_{n} (ℝ) & \to & Isom (𝔼^{n}) \\ y & \mapsto & f_{y} \end{array}$ where $f_{y} : 𝔼^{n} \to 𝔼^{n}$ is given by $f_{y} ((\begin{matrix} x \\ 1 \end{matrix})) = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) (\begin{matrix} x \\ 1 \end{matrix}) if y = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) .$ Then $Φ$ is a group isomorphism.

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

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 $f = {id}_{𝔼^{2}} .$ $□$

Proof of the isometry/affine orthogonal group correspondence

Define $\begin{array}{rrcl} Φ : & A O_{n} (ℝ) & \to & Isom (𝔼^{n}) \\ y & \mapsto & f_{y} \end{array} where$ $f_{y} ((\begin{matrix} x \\ 1 \end{matrix})) = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) (\begin{matrix} x \\ 1 \end{matrix}) if y = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) .$ Then $Φ$ is a group isomorphism.

Proof.

To show:

$Φ$ is a function ( $Φ$ is well defined).
$Φ$ is a group homomorphism.
$Φ$ is a bijection.

(b) If

y, z \in A O_{n} (ℝ)

and

(\begin{matrix} x \\ 1 \end{matrix}) \in 𝔼^{n}

then

f_{y} f_{z} (\begin{matrix} x \\ 1 \end{matrix}) = y z (\begin{matrix} x \\ 1 \end{matrix}) = f_{y z} (\begin{matrix} x \\ 1 \end{matrix}) .

Φ

is a homomorphism.

(a) Assume

y = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) = X^{μ} g with μ \in ℝ^{n} and g \in O_{n} .

Then

f_{y} = t_{μ} g where \begin{array}{rrcl} t_{μ} : & 𝔼^{n} & \to & 𝔼^{n} \\ (\begin{matrix} x \\ 1 \end{matrix}) & \mapsto & (\begin{matrix} μ + x \\ 1 \end{matrix}) \end{array} is a translation,

and

\begin{array}{rrcl} g : & 𝔼^{n} & \to & 𝔼^{n} \\ (\begin{matrix} x \\ 1 \end{matrix}) & \mapsto & (\begin{matrix} g x \\ 1 \end{matrix}) \end{array} with g \in O_{n} (ℝ) so that g g^{t} = 1 .

x, z \in 𝔼^{n}

then

\begin{array}{rcl} d (t_{μ} x, t_{μ} z) & = & d (μ + x, μ + z) \\ = & \sqrt{⟨ (μ + x) - (μ + z), (μ + x) - (μ + z) ⟩} \\ = & \sqrt{⟨ x - z, x - z ⟩} \\ = & d (x, z) \end{array}

and

\begin{array}{rcl} ⟨ g x, g z ⟩ & = & (x_{1} \dots x_{n}) g^{t} g (\begin{array}{rcl} z_{1} \\ ⋮ \\ z_{n} \end{array}) \\ = & (x_{1} \dots x_{n}) (\begin{array}{rcl} z_{1} \\ ⋮ \\ z_{n} \end{array}) \\ = & ⟨ x, z ⟩ \end{array}

so that

\begin{array}{rcl} d (g x, g z) & = & \sqrt{⟨ g x - g z, g x - g z ⟩} \\ = & \sqrt{⟨ g (x - z), g (x - z) ⟩} \\ = & \sqrt{⟨ x - z, x - z ⟩} \\ = & d (x, z) . \end{array}

Thus

g, t_{μ}

and

f_{y} = t_{μ} g

are all isometries.

(c) To show: There is an inverse function to

Φ .

Define

\begin{array}{rrcl} Ψ : & Isom (𝔼^{n}) & \to & A O_{n} (ℝ) \\ f & \mapsto & (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) \end{array}

where

μ = f (0)

and

g = (\begin{matrix} | & | & | \\ g_{1} & g_{2} & \dots & g_{n} \\ | & | & | \end{matrix}) with (\begin{matrix} | \\ g_{j} \\ | \end{matrix}) = f (\begin{matrix} 0 \\ ⋮ \\ 0 \\ 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}) j^{th} = f (e_{j})

where

e_{j} = (\begin{matrix} 0 \\ ⋮ \\ 0 \\ 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}) j^{th} .

To show:

$Ψ$ is well defined.
$Ψ \circ Φ = {id}_{A O_{n}}$ and $Φ \circ Ψ = {id}_{Isom} .$

(ii) Let

(\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) \in A O_{n} .

Then

(Ψ \circ Φ) (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) = Ψ (f_{y}) = (\begin{matrix} g' & μ' \\ 0 & 1 \end{matrix})

where

g' = (\begin{matrix} | & | & | \\ g_{1}' & g_{2}' & \dots & g_{n}' \\ | & | & | \end{matrix}) with g_{j}' = f_{y} (e_{j}) = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ ⋮ \\ 1 \\ 0 \\ ⋮ \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} | \\ g_{j} \\ | \\ 1 \end{matrix}) = (\begin{matrix} | \\ g_{j} \\ | \end{matrix})

and

μ' = f_{y} (0) with f_{y} (0) = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ ⋮ \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} μ \\ 1 \end{matrix}) = μ .

(Ψ \circ Φ) (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) .

(i) Let

f \in Isom .

Then

(Φ \circ Ψ) (f) = Φ (\begin{matrix} g & μ \\ 0 & 1 \end{matrix}) = f_{y} where y = (\begin{matrix} g & μ \\ 0 & 1 \end{matrix})

and

g = (\begin{matrix} | & | \\ g_{1} & \dots & g_{n} \\ | & | \end{matrix}) with f (e_{j}) = (\begin{matrix} | \\ g_{j} \\ | \end{matrix}) = g (\begin{matrix} 0 \\ ⋮ \\ 0 \\ 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}) = g e_{j},

and

μ = f (0) .

To show:

f (\begin{matrix} x \\ 1 \end{matrix}) = f_{y} (\begin{matrix} x \\ 1 \end{matrix}) .

To show:

t_{- μ} f = t_{- μ} f_{y} .

Let

g = t_{- μ} f_{y} .

Let

h = t_{- μ} f .

To show: If

x \in 𝔼^{n}

then

h x = g x .

We know

h \in Isom (𝔼^{n})

and

h (0) = 0 .

x, z \in 𝔼^{n}

then, since

h (0) = 0,

\begin{array}{rcl} ⟨ h x, h z ⟩ & = & \frac{1}{2} (⟨ h x, h x ⟩ + ⟨ h z, h z ⟩ - ⟨ h x - h z, h x - h z ⟩) \\ = & \frac{1}{2} (d {(h x, 0)}^{2} + d {(h z, 0)}^{2} - d {(h x, h z)}^{2}) \\ = & \frac{1}{2} (d {(h x, h 0)}^{2} + d {(h z, h 0)}^{2} - d {(h x, h z)}^{2}) \\ = & \frac{1}{2} (d {(x, 0)}^{2} + d {(z, 0)}^{2} - d {(x, z)}^{2}) \\ = & \frac{1}{2} (⟨ x, x ⟩ + ⟨ z, z ⟩ - ⟨ x - z, x - z ⟩) \\ = & ⟨ x, z ⟩ . \end{array}

Assume

x = (\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \\ 1 \end{array}) \in 𝔼^{n} .

Since

h e_{i} = g e_{i},

\begin{array}{rcl} j^{th} entry of h x & = & ⟨ h x, e_{j} ⟩ \\ = & ⟨ h (x_{1} e_{1} + \dots + x_{n} e_{n}), e_{j} ⟩ \\ = & ⟨ x_{1} e_{1} + \dots + x_{n} e_{n}, h^{- 1} e_{j} ⟩ \\ = & x_{1} ⟨ e_{1}, h^{- 1} e_{j} ⟩ + \dots + x_{n} ⟨ e_{n}, h^{- 1} e_{j} ⟩ \\ = & x_{1} ⟨ h e_{1}, e_{j} ⟩ + \dots + x_{n} ⟨ h e_{n}, e_{j} ⟩ \\ = & x_{1} g_{j_{1}} + \dots + x_{n} g_{j_{n}} \\ = & j^{th} entry of g (\begin{array}{c} x_{1} \\ ⋮ \\ x_{n} \end{array}) . \end{array}

h x = g x .

$□$

Notes and References

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References

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