Motivation

The notion that we can rigidly rotate the Euclidean plane about any point $Q$ and by any angle $\theta$ without breaking any geometric relations we value is deeply embedded in our experience. Perhaps this is less the case today than it was in the ancient world which did not use blackboards or projection screens to teach geometry. Instead, we can imagine students standing around the figures being drawn in the sand by the teacher. So there was no notion of "up" and "down" immediately associated with the figure. Clearly, up-down changes when we rotate the plane. But angles, distances, incidence of points on lines, parallelism and intersections, these do not change under rotations. Thus rotations should join translations and reflections as isometries we wish to study.

Definition of Rotation

Regardless how intuitive a rotation may seem, it is not easy to give an explicit formula for it. A rotation has a single fixed point. So, to give a rigorous definition we use property R3 for reflections we studied in the previous lesson.

Note the factor of 2, and the direction of the angle: $\angle k \ell = \pi - \angle \ell k$. Still, there is something arbitrary about the choice of mirrors. As we shall see presently, this is a good thing. We can easily change the mirrors if we prefer better ones than given.

For instance, given the rotation $\rho_{Q,\theta}$ and any desired mirror $m = (MQ)$ where M is some point other than $Q$, of course. We can make $m$ the bisector of an angle $\theta = \angle BQA = \angle ba$ in two ways. We are using the mnemonic "A,a" for "after" and "B,b" for "before".

• Draw $b$ so that $\angle bm = \frac{1}{2}\theta$. Then reflect $b$ in $m$ to get $a$.

• Define $a = \rho_{Q, \frac{1}{2}\theta}(m)$ and $\rho_{Q,-\frac{1}{2}\theta} (m) = b$. Note "half angle" and its sign.

 Twas brillig and the slithy toves did gyre and gimble in the wabe.
Question 1.

Why is $\rho \ne \sigma_m$ ?

Answer. [show] ▶

Apply both isometries to the point $M$ on $m$. The RHS leaves the point fixed. If the LHS would also fix $M$ then the rotation would have more than one fixed point. Indeed, you can see in the figure that $M^\rho = M'$ was moved.

Recalibration Theorem for Rotations

Proof: From the answer to question 1 we know that $\alpha := \sigma_m \rho$ is not the identity. So property R2 applies since $\alpha$ has two fixed points, $Q$ and $B$, namely that $\alpha = \sigma_{(BQ)$. Hence $\sigma_m \rho = \sigma_b \implies \rho = \sigma_m \sigma_b$. A similar argument for $\omega = \rho \sigma_m$ settles the "after mirror" cases. (Do it!)

$\square$

An immediate application of this theorem and the conjugacy theorem for reflections (CTR) is this extension of CTR to rotations.

Conjugacy Theorem for Rotations

Proof: Factor $\rho = \sigma_\ell \sigma_k$. Then

$\alpha \sigma_\ell \sigma_k \alpha^{-1} = \alpha \sigma_\ell \iota \sigma_k \alpha^{-1} = \alpha \sigma_\ell \alpha^{-1} \alpha \sigma_k \alpha^{-1} = \sigma_{\ell^\alpha} \sigma_{k^\alpha}$

where we applied CTR in the last equation. Note that the mirrors moved by $\alpha$ intersect at $Q^\alpha$. Since isometries preserve the dot product of vectors, they preserve the cosine of the angles. So, although the signed angle itself may not be preserved, its absolute value is.

Comment on $\pm \theta$

Consider a signed angle $\angle PQR$ under an isometry $\alpha$. It’s image must be $\angle P^\alpha Q^\alpha R^\alpha$. If the isometry reverses orientation, as for a reflection in a line, then the image angle has the opposite sense; cyclic goes to clockwise, and vice versa. This handedness of an isometry is determined by the parity of the number of mirrors it decomposes into; and the composition of an even number of mirrors yields an orientation preserving isometry, and the composition of an odd number of mirrors yields an orientation reversing isometry.

As we shall prove presently, translations and rotations are orientation preserving. Reflections and gliders are oriention reversing.

$\square$