Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 24 September 2014

Lecture 29: 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


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.

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.

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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