Random thoughts on isometries on lines

Let \(L\) be a Euclidean line. This means \(L\) is an affine space over \(V\), where \(V\) is a vector space of dimension one with a euclidean norm, and \(L\) has the metric inherited from \(V\).

Now assume \(A\) is an isometry on \(L\). What could \(A\) be?

There are two cases:

• \(A\) has a fixed point.
• \(A\) does not have a fixed point.

Let's consider the case where \(A\) has a fixed point \(P\) first. Consider the mapping \[ B\colon V \ni v \mapsto (A(P + v) - P) \in V \] We have, since \(A\) is an isometry and \(A(P) = P \), \[ |B(v)| = |(A(P+v) - A(P)| = | (P + v) - P| = |v|. \] So \(B\) is an isometry.

There are only two such things! Either \(B\) is the identity, or \(B\) is \(v \mapsto -v\). In the first case, \(A\) is the identity on \(L\). In the second case, \(A\) is reflection in the point \(P\) on \(L\).

Now consider the case where \(A\) has no fixed point. Take any point \(P\) on \(L\), and let \(S\) be reflection in the point midway between \(P\) and \(A(P)\). The map \(S\circ A\) clearly has \(P\) as a fixed point. Call it \(A'\). According to the reasoning above, \(A'\) is either the identity or a reflection in some point. But \(A'\) cannot be the identity, because then \(A\) would be \(S\), and \(P\) would be a fixed point, but \(A\) had no fixed points. So \(A'\) must be a reflection in some other point. Call it \(S'\). We have \[ \begin{align} S\circ A = S' \\ A = S\circ S' \end{align} \] So \(A\) is the composition of two reflections. Some simple considerations will show that \(A\) is a translation on \(L\).

In conclusion, the kinds of isometries on a euclidean line are:

• The identity
• Reflections
• Compositions of two reflections (these are translations)