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) = U x + 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 = |
|
where a^{2} + b^{2} = 1. A point x will be a fixed point of f if and only if U x + 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 = U x + v be a given orientation-reversing isometry without a fixed point. Since det U = -1, as in the argument for characterizing reflections,
U = |
|
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 U v <> -v. Let l be the line through the origin that is the axis of the reflection x -> U x. 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) = (U x + v') + v'', which shows that f is the composite obtained from translation by v'' following the reflection x -> U x + v' having axis parallel to the line l. Therefore f meets the condition characterizing glide reflections.
Let f be rotation about the point (1, 0) through the angle pi/4, and let g be rotation about the point (0, 1) through the angle pi/6. Show that g \circ f is a rotation, and find its center and its angle of rotation.
When is it the case that the composition of two rotations about different centers is a rotation?
What type of isometry is the composition of a reflection with the half turn, i.e., rotation through the angle pi, about a point not on the axis of the reflection?
If three lines l_{1}, l_{2}, l_{3} intersect so as to form a triangle, what type of isometry is the composition sigma_{3} \circ sigma_{2} \circ sigma_{1} of the reflections in those lines?