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)