Rotation (Euclidean Plane)

The set of rotations about a point in the Euclidean plane \(\mathbb{R}^{2}\) is an abelian subgroup of the set of all plane isometries.

Definition

The counter clockwise rotation of a point \(P\) about the centre of rotation \(C\) by an angle \(\theta\), denoted by \(\rho_{C, \theta}\), is the point \(P'\) such that \(\angle PCP'\) is \(\theta\) and the lengths \(CP\) and \(CP'\) are equal.

Form

To find the form of the function \(\rho_{C, \theta}\), we first consider \(\rho_{\boldsymbol{0}, \theta}\), the rotation about the origin.

To do this, consider the vector \(\begin{bmatrix} x \\ y \end{bmatrix}\) expressed in polar form as \(r\begin{bmatrix} \cos(\alpha) \\ \sin(\alpha) \end{bmatrix}\). To rotate this point by an angle of \(\theta\), we simply change the angle from \(\alpha\) to \(\alpha + \theta\).

\[\begin{align*} \rho_{\boldsymbol{0}, \theta}\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) &= \rho_{\boldsymbol{0}, \theta}\left(r\begin{bmatrix} \cos(\alpha) \\ \sin(\alpha) \end{bmatrix}\right)\\ &= r\begin{bmatrix} \cos(\alpha + \theta) \\ \sin(\alpha + \theta) \end{bmatrix} \\ &= r\begin{bmatrix} \cos(\alpha)\cos(\theta) - \sin(\alpha)\sin(\theta) \\ \sin(\alpha)\cos(\theta) + \cos(\alpha)\sin(\theta) \end{bmatrix} \\ &= r\begin{bmatrix} \cos(\alpha)\cos(\theta) - \sin(\alpha)\sin(\theta) \\ \cos(\alpha)\sin(\theta) + \sin(\alpha)\cos(\theta) \end{bmatrix} \\ &= r\cos(\alpha)\begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} + r\sin(\alpha)\begin{bmatrix} -\sin(\theta) \\ \cos(\theta) \end{bmatrix} \\ &= r\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\alpha) \\ \sin(\alpha) \end{bmatrix} \\ &= \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \left(r\begin{bmatrix} \cos(\alpha) \\ \sin(\alpha) \end{bmatrix}\right) \\ &= \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \\ \end{align*}\]

This means that counter clockwise rotation by an angle \(\theta\) around the origin corresponds with multiplication by the matrix:

\[\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}.\]

To generalise this to any rotation, we can first translate the point of rotation to the origin, conduct the rotation, and then translate back, that is:

\[ \rho_{C, \theta} = T_{\vec{c}} \circ \rho_{\boldsymbol{0}, \theta} \circ T_{-\vec{c}}\]

where \(c\) is the coordinate vector of the point \(C\).

In Terms of Reflections

Given that the set of plane isometries in \(\mathbb{R}^{2}\) is a group generated by reflections, we can express every rotation in terms of reflections.

Consider two intersection lines \(l\) and \(m\), with the directed angle from \(l\) to \(m\), \(\frac{\theta}{2} \in (-\frac{\pi}{2}, \frac{\pi}{2}]\), measured counter clockwise. Consider the points \(C\), at the intersection of \(m\) and \(l\), \(L\) which lies on \(l\) and \(M\) which lies on \(m\), with \(L \neq C\) and \(M \neq C\).

Now, we consider the reflection of the three points \(C\), \(L\) and \(M\) across the line \(l\) and \(m\). First note that \((\sigma_{m} \circ \sigma_{l})(C) = C\), since \(C\) lies on both lines.

Then, \(\sigma_{l}(L) = L\), since \(L \in l\), and we just need to calculate the reflection of \(L\) on line \(m\).

In order for \(L'\) to be the rotation of \(L\) around \(C\) by angle \(\theta\), \(\angle LCL' = \theta\) and the line segments \(LC\) and \(L'C\) need to be equal. This follows from a simple geometric argument. Note the orange line segment \(LL'\) below, which is such that \(m\) is its perpendicular bisector, be the definition of a reflection. Since both triangles share a side, we can then use the fact that the tangent of the green angles are the same, and thus the angles are the same. Additionally, by the pythagorean theorem the hypotenuses are the same, thus \(LC\) and \(L'C\) have the same length, as required.

A very similar argument follows for \(M\), however because the reflection is about the line \(l\) first, the above method needs to be applied twice.

The only problem with this argument is that the triangle used to justify the fact that the angles on either side of the reflecting line are equal is degenerate in the case where the two lines meet at \(\frac{\pi}{2}\).

However this case is easy to deal with separately, noting that the reflection of \(L\) about \(m\) lies on \(l\) because the lines are perpendicular, and this is also the point reflection (or halfturn) of \(L\) about \(C\). The same is true for the reflection of \(M\) about \(l\), noting that \(M'\) lies on \(m\) so \(\sigma_{m}(M') = M'\).

Since we have shown that the two isometries \(\sigma_{m} \circ \sigma_{l}\) and \(\rho_{C, \theta}\) agree for three non collinear points, we have shown that \((\rho_{C, \theta}^{-1} \circ \sigma_{m} \circ \sigma_{l})(\vec{x}) = \vec{x}\) for three non collinear points, and therefore

\[ \rho_{C, \theta}^{-1} \circ \sigma_{m} \circ \sigma_{l} = \mathrm{id} \implies \sigma_{m} \circ \sigma_{l} = \rho_{C, \theta}.\]

Point Reflection (Half-turn)

A half-turn is a special case of a rotation in which the angle of rotation is \(\pi\). This is also called a point reflection, as it is equivalent to taking the equivalent point on the "opposite" side of the centre of rotation. Note that this equivalent is a specific consequence of working in two dimensions.

A point reflection about the position vector \(\vec{c}\) takes the form \(\vec{x} \mapsto 2\vec{c} - \vec{x}\). This can be shown geometrically as below:

or by substituting \(\pi\) as the angle in a general rotation.

A shape which is invariant under a point reflection is called centrally symmetric.