Transformation Geometry -- Math 331

March 1, 2004

Reflections and Glide Reflections

An isometry is orientation-reversing if and only if relative to Cartesian coordinates it has the form f(x) = U x + v where U is a reflection matrix, i.e., an orthogonal matrix of determinant -1.

Proposition. If U is a 2 \times 2 orthogonal matrix with determinant -1, then the linear transformation sigma(x) = Ux is the reflection in a line through the origin, and the following statements hold:

  1. 1 - U^{2} = (1 + U)(1 - U) = (1 - U)(1 + U) = 0.

  2. Any vector v of the form (1 + U)w for some vector w lies on the axis of sigma.

  3. Any vector v of the form (1 - U)w for some vector w is perpendicular to the axis of sigma.

Proof. It is elementary (see the assignment due February 20) that U has the form

 U  =  
(
a 
b
b
-a
)
 

where a^{2} + b^{2} = 1. Since both U^{-1} = transp(U) and transp(U) = U, clearly U^{2} = 1, and, therefore, (1 + U)(1 - U) = (1 - U)(1 + U) = 0. If v = (1 + U)w, then (1 - U)v = (1 - U)(1 + U)w = 0, hence, Uv = v, and, therefore, v lies on the axis of sigma. If, on the other hand v = (1 - U)w, then by similar reasoning Uv = -v, which characterizes vectors v perpendicular to the axis of sigma.

Proposition. Let f = Ux + v be an orientation-reversing isometry, and let

 v'   =  {1}/{2} 
(1 - U)
 v     and      v''  =  {1}/{2} 
(1 + U)
 v     . 

Then f is the composition of the isometry sigma(x) = Ux + v', which is a reflection with axis parallel to the axis of the reflection x -> Ux, followed by the translation tau(x) = x + v'' by the vector v'', which is parallel to the axis of sigma.

Proof. Clearly, v' + v'' = v, and, therefore, f = tau\circ sigma. Since v' is perpendicular to the axis of x -> Ux, translation by v' is the composition of the reflections in two lines parallel to the axis of x -> Ux, and one of those two lines may be chosen to be the axis of x -> Ux and then sigma is seen to be the reflection in the other of the two parallel lines. From this follows:

Theorem Let f = Ux + v be an orientation-reversing isometry. Then f is a reflection if and only if Uv = -v and is a glide reflection otherwise.

Exercises due Wednesday, March 3

  1. Let f be the affine transformation of the plane defined by

     f(x)  =  
    (
    3/5 
    4/5
    4/5
    -3/5
    		)
    (
    x_{1} 
    x_{2}
    		)
    		  + 
    (
    1 
    2
    		)
    		    . 

    1. What points x of the plane are “fixed” by f, i.e., satisfy f(x) = x ?

    2. What lines L in the plane are “stabilized” by f, i.e., satisfy the condition that f(x) is on L if x is on L ?

    3. Find a reflection sigma and a translation tau parallel to the axis of sigma such that f = tau\circ sigma.

  2. Explain briefly why the composition of any four reflections may always be written as the composition of two reflections.

  3. Two triangles ΔABC and ΔA'B'C' are, by definition, congruent if there is an isometry f for which the vertices of the one triangle are carried to the vertices of the other. If ΔABC and ΔA'B'C' are congruent, then how many isometries f of the first with the second are there when

    1. ΔABC is equilateral.

    2. The sides BC and CA have the same length, which is different from the length of AB.

    3. No two sides of ΔABC have the same length.

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