Translation (Euclidean Plane)

The set of translations by some vector in \(\mathbb{R}^{2}\) is a subgroup of the isometry group on \(\mathbb{R}^{2}\).

Definition

The translation \(T_{\vec{b}}\) is defined to be the mapping which takes each point \(P\) and maps it to a point \(P'\) such that \(\overrightarrow{PP'} = \vec{b}\)

Form

The translation \(T_{\vec{b}}\) is given by the formula:

\[ \vec{x} \mapsto \vec{x} + \vec{b}.\]

This satisfies the definition above, since:

\[\begin{align*} \overrightarrow{PP'} &= \vec{p}' - \vec{p} \\ &= T_{\vec{b}}(\vec{p}) - \vec{p} \\ &= \vec{p} + \vec{b} - \vec{p} \\ &= \vec{b}. \end{align*}\]

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 translation in terms of reflections.

To translate by a vector \(\vec{b}\), we select two parallel lines, both perpendicular to \(\vec{b}\) with a perpendicular separating distance of \(\left|\frac{\vec{b}}{2}\right|\), and apply the reflections in the order that matches the direction of the translation.

Consider the diagram below showing three points \(A\), \(B\), and \(C\), the first two of which lie on a line \(l\) while the third lies on a line \(m\), with \(l\) parallel to \(m\). Additionally, \(A\) and \(C\) are such that the line segment \(AC\) is perpendicular to both \(l\) and \(m\).

Now we let \(\vec{b} = 2\overrightarrow{AC}\), and consider the result of reflecting \(A\), \(B\) and \(C\) about the line \(l\) and then the line \(m\), that is, applying the transformation \(\sigma_{m} \circ \sigma_{l}\).

First lets consider what happens to \(A\). Since the reflection fixes the line of reflection \(\sigma_{l}(A) = A\) given that \(A \in l\). Then, let \(A' = \sigma_{m}(A)\). By the definition of a reflection, \(m\) must be the perpendicular bisector of \(AA'\). Since \(\vec{b}\) is perpendicular to \(m\), and adding \(\frac{\vec{b}}{2}\) maps \(A\) to be on \(m\), we simply must add \(\vec{b}\) to give the required construction.

Hence \(A' = T_{\vec{b}}(A)\).

A very similar argument follows for \(B\).

Then for \(C\), we reflect about \(l\) first, and therefore end up with the point \(C'\) after translating by \(-\vec{b}\). This is because of the same rationale used to reflect \(A\) about \(m\), where in this case we need \(l\) to be the perpendicular bisector of \(CC'\).

Then, the vector \(\vec{b}\) takes \(C'\) to be on line \(m\) by the shortest (perpendicular) distance. Thus for \(m\) to be the perpendicular bisector of \(C'C''\) where \(C''\) is the result of the reflection, we must translate by \(2\vec{b}\). This means the overall translation is \(T_{-\vec{b}} \circ T_{2\vec{b}} = T_{\vec{b}}\).

This means, for three non-collinear points, the result of the two reflections is the same as the translation by \(\vec{b}\), \(T_{\vec{b}}(\vec{x}) = (\sigma_{m} \circ \sigma_{l})(\vec{x})\) or alternatively \((T_{-\vec{b}} \circ \sigma_{m} \circ \sigma_{l})(\vec{x}) = \vec{x}\). Any plane isometry which fixes three points must be the identity, and therefore:

\[ T_{-\vec{b}} \circ \sigma_{m} \circ \sigma_{l} = \mathrm{id} \implies T_{\vec{b}} = \sigma_{m} \circ \sigma_{l}.\]