Necessary condition for the product of two rotations of the plane to be a rotation

Question

I know that any rotation of $\mathbb{R}^2$ can be expressed as the product of two reflections in non-parallel lines, and hence the product of two rotations can be written as $R_{L_4}R_{L_3}R_{L_2}R_{L_1}$, where $R_{L_i}$ denotes reflection in the line $L_i$ for $1 \leq i \leq 4$. It is then clear that a sufficient condition for this product of two rotations to yield another rotation is for $L_2 = L_3$ (in which case the product $R_{L_3}R_{L_2}$ reduces to the identity) and for $L_1$ and $L_4$ to be non-parallel. My question is whether this is also a necessary condition for the product of two rotations of the plane to yield another rotation - I am inclined to believe that it is, and it would be greatly appreciated if someone here can quickly confirm this for me.

Answer 1

I want to present another approach to the question.

$\newcommand\rg{\rm rg}$ $\newcommand\arho[2]{{\raise.5ex\rho}_{#1,\,#2}}$ $\newcommand\vrho[1]{{\raise.5ex\rho}_{#1}}$Let us consider two affine rotations of the affine plane $\mathcal A=\mathbb R^2$, say

$$ \arho{O_1}{\theta_1} \; : \; P\; \to \; O_1 + \vrho{\theta_1}(P-O_1) $$ $$ \arho{O_2}{\theta_2} \; : \; Q\; \to \; O_2 + \vrho{\theta_2}(Q-O_2) $$

around $O_1$ of angle $\theta_1$ the first, and around $O_2$ of angle $\theta_2$ the second, where $\vrho {\theta_1}$ and $\vrho {\theta_2}$ are the correspondig rotations of the vector space $E=\mathbb R^2$.
Their product of composition is given by \begin{align} \arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}(P) &= \arho{O_2}{\theta_2}\big(O_1+\vrho{\theta_1}(P-O_1)\big) =\\[1.5ex] &= O_2 + \vrho{\theta_2}\big(O_1-O_2+\vrho{\theta_1}(P-O_1)\big)=\tag{*}\label{*}\\[1.5ex] &= O_2+\vrho{\theta_2}(O_1-O_2)+\vrho{\theta_1+\theta_2}(P-O_1). \end{align}

A fixed point $\;C\;$ of $\;\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}$ must satisfy the equation

$$ O_2+\vrho{\theta_2}(O_1-O_2)+\vrho{\theta_1+\theta_2}(C-O_1) = C $$

so

$$ \vrho{\theta_2}(O_1-O_2)+\vrho{\theta_1+\theta_2}(C-O_1) = (C-O_1)+(O_1-O_2) $$

so

$$ (\vrho{\theta_1+\theta_2}-1_E)(C-O_1) = (\vrho{\theta_2}-1_E)(O_2-O_1), \tag{**}\label{**} $$

where $\;1_E\;$ is the identity function of the vector space $E=\mathbb R^2$.

Now, IF the linear function $\;\vrho{\theta_1+\theta_2}-1_E\;$ is invertible, we deduce that there is a unique fixed point $C$ for $\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}$, given by

$$ C = O_1 + (\vrho{\theta_1+\theta_2}-1_E)^{-1}\circ(\vrho{\theta_2}-1_E)(O_2-O_1) $$

and we can verify that $\;\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}\;$ is a rotation centered on $C$ and of angle $\;\theta_1+\theta_2$. In fact, for all $P\in \mathcal A$, we have, from \eqref{*}

\begin{align} &\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}(P)-C = \arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}(P)-\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}(C) =\\[1.5ex] &= \big(O_2+\vrho{\theta_2}(O_1-O_2)+\vrho{\theta_1+\theta_2}(P-O_1)\big)-\big(O_2+\vrho{\theta_2}(O_1-O_2)+\vrho{\theta_1+\theta_2}(C-O)\big)= \\[1.5ex] &= \vrho{\theta_1+\theta_2}\big((P-O_1)-(C-O_1)\big) = \vrho{\theta_1+\theta_2}(P-C). \qquad\blacksquare \end{align}

The (sufficient) condition seen above is equivalent to $\;\det(\vrho{\theta_1+\theta_2}-1_E)\neq0$, that is, with respect to the standard orthonormal basis of $E$,

$$\det\begin{bmatrix} \cos(\theta_1+\theta_2)-1 & \quad-\sin(\theta_1+\theta_2) \\[2ex] \sin(\theta_1+\theta_2) & \quad\cos(\theta_1+\theta_2)-1 \end{bmatrix}\neq0 $$

i.e.

$$ 2\big(1-\cos(\theta_1+\theta_2)\big)\neq0 $$

so that a sufficient condition for $\;\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}\;$ to be a rotation is that

$$ \theta_1+\theta_2 \neq 0. \tag{***}\label{***}$$

Conversely, suppose that $\;\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}$ is a rotation around the fixed point $C$. Let us consider two cases:

• $\rg(\vrho{\theta_1+\theta_2}-1_E)=0\qquad\text{[rg denote the rank]}$

We have $\;\vrho{\theta_1+\theta_2} = 1_E,\;$ so $\;$[by $\;$\eqref{*}]

$$ \arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}(P) = O_2+\vrho{\theta_2}(O_1-O_2) + (P-O_1) = P+ (O_2-O_1) + \vrho{\theta_2}(O_1-O_2) $$

i.e. a (effective) translation or the identity $\;1_{\mathcal A}$, depending on whether

$\hskip4cm\;(O_2-O_1) + \vrho{\theta_2}(O_1-O_2)\neq 0\;$ or $\;=0$.

• $\rg(\vrho{\theta_1+\theta_2}-1_E)=1$

The equation of fixed points $\;$\eqref{**}$\,$ admits infinite solutions $\;C-O_1\;$ if it is compatible, or no solutions otherwise. For each solution $\;C-O_1$, the point $\;C\;$ is a fixed point for $\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}$ and vice versa. Indeed, by $\eqref{*}$ and $\eqref{**}$, we have

\begin{align} \arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}(C) &= O_2+\vrho{\theta_2}(O_1-O_2)+\vrho{\theta_1+\theta_2}(C-O_1) =\\ &= O_2 + \vrho{\theta_2}(O_1-O_2) + C-O_1 + (\vrho{\theta_1+\theta_2}-1_E)(C-O_1) =\\ &= O_2 + \vrho{\theta_2}(O_1-O_2) + C-O_1 +\vrho{\theta_2}(O_2-O_1) + (O_1-O_2) =\\ &= O_2 + (C-O_1) + (O_1-O_2) = C. \end{align}

Similarly for vice versa.
It follows that, if the rotation $\;\arho{O_2}{\theta_2}\circ\arho{O_1}{\theta_1}\;$ is not the identity, then $\rg(\vrho{\theta_1+\theta_2}-1_E)=2$, and therefore, as we have already seen, that the inequality $\;\eqref{***}\;$ holds.

Answer 2

After doing a quick search on Mathematics Meta, I see that it is generally considered acceptable to post an answer to one's own question if no complete answers have been provided by other users, so here is what I have managed to work out.

Let $a$ and $b$ be points in the plane. A rotation about $a$ by angle $\theta \in [0,2\pi)$ can be expressed as a pair of reflections in any lines $L_1$ and $L_2$ that form an angle of $\frac{\theta}{2}$ at their point of intersection $a$. Likewise, a rotation about $b$ by angle $\phi \in [0,2\pi)$ can be expressed as a pair of reflections in any lines $L_3$ and $L_4$ that form an angle of $\frac{\phi}{2}$ at their point of intersection $b$. Insofar as we can always construct lines $L_1, L_2, L_3, L_4$ such that $L_2 = L_3$ passes through both points $a$ and $b$ and forms an angle of $\frac{\theta}{2}$ and $\frac{\phi}{2}$ with $L_1$ and $L_4$ at $a$ and $b$, respectively, this "condition" as stated in my earlier question will always be satisfied. That is, the pair of rotations is always expressible as the product of reflections $R_{L_4}R_{L_3}R_{L_2}R_{L_1}$ = $R_{L_4}R_{L_1}$, with $L_2 = L_3$.

The necessary condition is merely that $\theta + \phi \neq 2\pi$, since this is what is required for $L_1$ and $L_4$ to intersect at some point $c$. The product of rotations is then a rotation about $c$ by angle $\theta + \phi$. On the other hand, if $\theta + \phi = 2\pi$, then $L_1$ and $L_4$ will be parallel, and the product of reflections $R_{L_4}R_{L_1}$ will be a translation by twice the distance between $L_1$ and $L_4$ in the direction perpendicular to $L_1$.