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 AOn() = { 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1  |  gOn() μn } The orthogonal group is On() = {gMn×n()  |  ggt=1}. The special orthogonal group is SOn() = {gOn()  |  det(g)=1} and On() SOn() = {N,rN} where N=SOn() and r= -1 0 1 0 1 , since det: On() {±1} g det(g) and {±1} 2 . A rotation is an element of N=SOn() and a reflection is an element of rN, where N=SOn() and r= -1 0 1 0 1 .

For μn and gOn() 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 𝔼n = { 0 x 0 1  |  xn } = { x1 x2 xn 1  |  x1,x2,...,xn }. The group AOn() acts on 𝔼n by g 0 x 0 1 = 0 gx 0 1 and Xμ 0 x 0 1 = 0 μ+x 0 1 . Note that, if μ0 then tμ: n n x μ+x is not a linear transformation, in particular tμ(0) 0.

Let d:𝔼n×𝔼n 0 be the metric on 𝔼n 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 ,: 𝔼n×𝔼n 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 and (AOG 1) x,y = 1 4 ( x+y,x+y-x-y,x-y ) = 1 4 ( d(x,-y)2-d(x,y)2 ) (AOG 2) An isometry of 𝔼n is a function f:𝔼n𝔼n such that if   x,y𝔼n   then   d(fx,fy) = d(x,y).

HW: Use (AOG 1) and (AOG 2) to show that if f:𝔼n𝔼n is an isometry then f satisfies if   x,y𝔼n   then   fx,fy=x,y.

The group of isometries of 𝔼n is Isom(𝔼n) = { f:𝔼n𝔼n  |  f   is an isometry } with operation given by composition of functions.

Define Φ: AOn() Isom(𝔼n) y fy where fy:𝔼n𝔼n 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 Φ is a group isomorphism.

Let μn. Translation by μ is the function tμ: 𝔼n 𝔼n 0 x 0 1 0 μ+x 0 1 Note that tμ is an isometry since d(tμx,tμy) = (μ+x)-(μ+y),(μ+x)-(μ+y) = x-y,x-y = d(x,y).

Isometries in 𝔼2

Let s= -1 0 0 1 and rθ= cosθ -sinθ sinθ cosθ and let tγ: 2 2 x γ+x, for   γ= γ1 γ2   in   2. s is reflection in the y-axis Ly

Ly s sx x 0
rθ is rotation in an angle θ about 0
0 Ly Lx (0,1) (1,0) θ cosθ sinθ -sinθ cosθ
tγ is translation by γ
0 γ x tγx=x+γ
SO2() = {gM2×2()  |  ggt=1,  det(g)=1} = { a b c d  |  a,b,c,d, ad-bc=1 a b c d a c b d = 1 0 0 1 } = { a b c d  |  a,b,c,d,  a2+b2=1,  ac+bd=0 ca+db=0,  c2+d2=1,  ad-bc=1 } = { a b c d  |  a,b,c,d,  b=-c a2+b2=1,  a=d } = { a b -b a  |  a,b,c,d,  a2+b2=1} = { cosθ -sinθ sinθ cosθ  |  0θ<2π} = {rθ  |  0θ<2π}. So SO2() is the group of rotations about 0.
  1. Let L be a line in 2. Then there exist c and 0θ<π such that L= rθt c 0 Ly. The reflection in the line L is sL = rθt c0 st -c0 r-θ.
    0 { c θ L
  2. Let p2 and θ[0,2π) Then rotation by θ around p is rθ,p = tprθt-p.
  3. 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.
    0 { d L x gx

Isometries

Let 𝔼2 = { x y 1  |  x,y} with d(p,q) = (x1-x2)2+(y1-y2)2 if p= x1 y1 1 and x2 y2 1 . An isometry of 𝔼2 is a function f:E2E2 such that d(fp,fq) = d(p,q). Note that

Let f:𝔼2𝔼2 be an isometry. Suppose α,p are fixed points of f, fα=α and fβ=β

β α p { 0 0 0 R2 { 0 0 R1 C1 C2

Let p𝔼2. Since d(α,p) = d(fα,fp) = d(α,fp), fp   must lie on the circle   C1   of radius   R1=d(α,p)   centred at   α. Since d(β,p) = d(fβ,fp) = d(β,fp), fp   must lie on the circle   C2   of radius   R2=d(β,p)   centred at   β. So fpC1C2. If p is on the line Lαβ connecting α and β then C1C2={p}. So fp=p if pLαβ.

α β C2 p Lαβ C1
Thus, if f: 𝔼2𝔼2 is an isometry and α,β𝔼2 are such that αβ   and   fα=α   and   fβ=β then fp=p   for every   pLαβ where Lαβ is the line connecting α and β.

If f:𝔼2𝔼2 is an isometry and γ,α,β𝔼 are such that γLαβ and αβ and fα=α, fβ=β and fγ=γ then f fixes all of 𝔼2.

Proof.
f fixes Lαβ, Lαγ and Lβγ. If pLαβ and qLαγ then f fixes Lpq.
Lαβ Lβγ Lpq Lαγ p q α β γ x
Every point x𝔼2 is on some Lpq with pLαβ and qLαβ and so fx=x. So f=id𝔼2.

Proof of the isometry/affine orthogonal group correspondence

Define Φ: AOn() Isom(𝔼n) y fy where 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 Φ is a group isomorphism.

Proof.
To show:
  1. Φ is a function (Φ is well defined).
  2. Φ is a group homomorphism.
  3. Φ is a bijection.
(b) If y,zAOn() and 0 x 0 1 𝔼n then fyfz 0 x 0 1 = yz 0 x 0 1 = fyz 0 x 0 1 . So Φ is a homomorphism.

(a) Assume y= 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 =Xμg with   μn  and   gOn. Then fy = tμg where tμ: 𝔼n 𝔼n 0 x 0 1 0 μ+x 0 1 is a translation, and g: 𝔼n 𝔼n 0 x 0 1 0 gx 0 1 with   gOn()   so that   ggt=1. If x,z𝔼n then d(tμx,tμz) = d(μ+x,μ+z) = (μ+x)-(μ+z), (μ+x)-(μ+z) = x-z,x-z = d(x,z) and gx,gz = (x1  xn) gtg z1 zn = (x1  xn) z1 zn = x,z so that d(gx,gz) = gx-gz,gx-gz = g(x-z),g(x-z) = x-z,x-z = d(x,z). Thus g,tμ and fy=tμg are all isometries.

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

Define Ψ: Isom(𝔼n) AOn() f 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 where μ=f(0) and g= | | | g1 g2 gn | | | with | gj | = f 0 0 1 0 0  jth = f(ej) where ej= 0 0 1 0 0  jth. To show:
  1. Ψ is well defined.
  2. ΨΦ = idAOn and ΦΨ = idIsom.
(ii) Let 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 AOn. Then (ΨΦ) 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 = Ψ(fy) = 0 0 0 0 0 g' 0 μ' 0 0 0 0 0 0 0 1 where g'= | | | g1' g2' gn' | | | with gj' = fy(ej) = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 0 1 0 0 1 = | gj | 1 = | gj | and μ'=fy(0) with fy(0) = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 0 0 1 = 0 μ 0 1 =μ. So (ΨΦ) 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 = 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 . (i) Let fIsom. Then (ΦΨ)(f) = Φ 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 = fy where y= 0 0 0 0 0 g 0 μ 0 0 0 0 0 0 0 1 and g= | | g1 gn | | with f(ej) = | gj | = g 0 0 1 0 0 =gej, and μ=f(0).

To show: f 0 x 0 1 = fy 0 x 0 1 .

To show: t-μf = t-μfy.

Let g=t-μfy. Let h=t-μf.

To show: If x𝔼n then hx=gx.

We know hIsom(𝔼n) and h(0)=0. If x,z𝔼n then, since h(0)=0, hx,hz = 1 2 ( hx,hx+hz,hz-hx-hz,hx-hz ) = 1 2 ( d(hx,0)2 + d(hz,0)2 - d(hx,hz)2 ) = 1 2 ( d(hx,h0)2 + d(hz,h0)2 - d(hx,hz)2 ) = 1 2 ( d(x,0)2 + d(z,0)2 - d(x,z)2 ) = 1 2 ( x,x + z,z - x-z,x-z ) = x,z. Assume x= x1 xn 1 𝔼n. Since hei=gei, jth entry of   hx = hx,ej = h(x1e1++xnen),ej = x1e1++xnen,h-1ej = x1e1,h-1ej ++ xnen,h-1ej = x1he1,ej ++ xnhen,ej = x1gj1 ++ xngjn = jth  entry of   g x1 xn . So hx=gx.

Notes and References

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References

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