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 30: Matching the affine orthogonal group with isometries

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

Proof.

To show:

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

(b)

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)

To show: If

y \in A O_{n} (ℝ)

then

f_{y}

is an isometry.
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} & ⟶ & 𝔼^{n} \\ (\begin{matrix} x \\ 1 \end{matrix}) & ⟼ & (\begin{matrix} μ + x \\ 1 \end{matrix}) \end{array} is a translation,

and

\begin{array}{rrcl} g : & 𝔼^{n} & ⟶ & 𝔼^{n} \\ (\begin{matrix} x \\ 1 \end{matrix}) & ⟼ & (\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}) & ⟶ & A O_{n} (ℝ) \\ f & ⟼ & (\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:

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

(cb)

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}) .

(ca)

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

These are a typed copy of Lecture 30 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 14, 2011.

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