Transformation Geometry — Math 331

February 25, 2004

Discussion

• Theorem. Every orientation-preserving isometry of R^{2} without a fixed point is a translation.

Proof. Let f be a given orientation-preserving isometry of R^{2}. Since f is an affine transformation, f(x) = Ux + v for some invertible matrix U and some vector v. Since f is an isometry, U is an orthogonal matrix and since f is orientation-preserving, det U = 1 and, in fact, as in the argument for characterizing rotations,

U = ( \begin{matrix}a & −b \\ b & a \end{matrix} )

where a^{2} + b^{2} = 1. A point x will be a fixed point of f if and only if Ux + v = x, or, equivalently, x is a solution of the linear system (1 − U)x = v. So f certainly has a fixed point when the matrix 1 − U is invertible, which is the case when its determinant (1 − a)^{2} + b^{2} ≠ 0. Therefore, f can only fail to have a unique fixed point when a = 1 and b = 0, i.e., when U = 1 is the identity matrix and f, therefore, is a translation.

• Theorem. Every orientation-reversing isometry of R^{2} without a fixed point is a glide reflection.

Proof. Let f = Ux + v be a given orientation-reversing isometry without a fixed point. Since det U = −1, as in the argument for characterizing reflections,

U = ( \begin{matrix}a & b \\ b & −a \end{matrix} )

where a^{2}+b^{2} = 1. With this condition on U the matrix 1−U has determinant 0 and, in fact, has rank 1. Therefore, from the nature of U alone the fixed point equation (1 − U)x = v has a solution x if and only if v is in the one-dimensional column space of 1 − U. Since it is given that f has no fixed point, the vector v is not in the column space of 1−U, which means that Uv ≠ −v. Let l be the line through the origin that is the axis of the reflection x ↦ Ux. The condition obtained on v is equivalent to the statement that v is not perpendicular to l. Now decompose v = v′ + v′′ where v′ is perpendicular to l and v′′ is parallel to l. Then f(x) = (Ux + v′) + v′′, which shows that f is the composite obtained from translation by v′′ following the reflection x ↦ Ux + v′ having axis parallel to the line l. Therefore f meets the condition characterizing glide reflections.

Exercises due Friday, February 27

1. Let f be rotation about the point (1, 0) through the angle π/4, and let g be rotation about the point (0, 1) through the angle π/6. Show that g ∘ f is a rotation, and find its center and its angle of rotation.

2. When is it the case that the composition of two rotations about different centers is a rotation?

3. What type of isometry is the composition of a reflection with the half turn, i.e., rotation through the angle π, about a point not on the axis of the reflection?

4. If three lines l_{1}, l_{2}, l_{3} intersect so as to form a triangle, what type of isometry is the composition σ_{3} ∘ σ_{2} ∘ σ_{1} of the reflections in those lines?