Relations between translation and rotations

Question

Can every translation be written as a product of two non-involutory rotations? why or why not

I feel like it can be written as two rotations of right angles as there are no other restrictions on the choosing the rotations and rotations of 90 degrees are indeed non-involutory.

Is this the right proof?

Answer 1

Let's suppose you're dealing with arbitrary rotations. Choose a coordinate system where the first point you want to rotate about is the origin. To rotate through an angle of $\theta$, we map any point $\begin{bmatrix}x_1\\ x_2\end{bmatrix}$ to $$ \begin{bmatrix} x_1'\\ x_2' \end{bmatrix} = \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}. $$ Now, to rotate about our second angle, we need to translate our coordinate system so that our second point, let's call it $\begin{bmatrix}y_1\\ y_2\end{bmatrix}$, is now the origin, i.e. we need to map $\begin{bmatrix}x_1'\\ x_2'\end{bmatrix}$ to $$ \begin{bmatrix} x_1'\\ x_2' \end{bmatrix} - \begin{bmatrix} y_1\\ y_2 \end{bmatrix}, $$ then rotate and translate back. Thus to rotate around $\begin{bmatrix}y_1\\ y_2\end{bmatrix}$ through an angle of $\phi$, we map $\begin{bmatrix}x_1'\\ x_2'\end{bmatrix}$ to $$ \begin{align} \begin{bmatrix} x_1''\\ x_2'' \end{bmatrix} &= \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \left( \begin{bmatrix} x_1'\\ x_2' \end{bmatrix} - \begin{bmatrix} y_1\\ y_2 \end{bmatrix} \right) + \begin{bmatrix} y_1\\ y_2 \end{bmatrix} \\ &= \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \left( \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} - \begin{bmatrix} y_1\\ y_2 \end{bmatrix} \right) + \begin{bmatrix} y_1\\ y_2 \end{bmatrix} \\ &= \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} - \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \begin{bmatrix} y_1\\ y_2 \end{bmatrix} + \begin{bmatrix} y_1\\ y_2 \end{bmatrix} \\ &= \begin{bmatrix} \cos(\theta+\phi)&-\sin(\theta+\phi)\\ \sin(\theta+\phi)&\cos(\theta+\phi) \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} - \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \begin{bmatrix} y_1\\ y_2 \end{bmatrix} + \begin{bmatrix} y_1\\ y_2 \end{bmatrix} \\ &= \begin{bmatrix} \cos(\theta+\phi)&-\sin(\theta+\phi)\\ \sin(\theta+\phi)&\cos(\theta+\phi) \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} + \begin{bmatrix} y_1'\\ y_2' \end{bmatrix}, \end{align} $$ where $$ \begin{bmatrix} y_1'\\ y_2' \end{bmatrix} = - \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \begin{bmatrix} y_1\\ y_2 \end{bmatrix} + \begin{bmatrix} y_1\\ y_2 \end{bmatrix}. $$ This will be a translation only if $$ \begin{bmatrix} \cos(\theta+\phi)&-\sin(\theta+\phi)\\ \sin(\theta+\phi)&\cos(\theta+\phi) \end{bmatrix} = \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}, $$ i.e. only if $\theta+\phi=2k\pi$ for some integer $k$.

So, it would seem that yes, you can describe a translation $\begin{bmatrix}x_1\\ x_2\end{bmatrix}\mapsto\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\begin{bmatrix}y_1'\\ y_2'\end{bmatrix}$ by successive rotations about arbitrary points, provided you can find $\phi$ and $\begin{bmatrix}y_1\\ y_2\end{bmatrix}$ such that $$ \begin{bmatrix} y_1'\\ y_2' \end{bmatrix} = - \begin{bmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{bmatrix} \begin{bmatrix} y_1\\ y_2 \end{bmatrix} + \begin{bmatrix} y_1\\ y_2 \end{bmatrix} $$ (which you can actually do for any $\phi$ which is not coterminal with $0$), but your rotations have to amount to some integer amount of full turns, and the angles are determined by the translation.

Answer 2

You can indeed obtain any translation of the plane as the composite of two $90^\circ$ rotations. To see this, let me first consider a very special (and easy to understand) pair of $90^\circ$ rotations: First rotate counterclockwise by $90^\circ$ about the origin $(0,0)$; then rotate clockwise by $90^\circ$ about the point $(0,1)$. I claim that this composite operation amounts to a translation by $(-1,1)$. To verify the claim, look first at where $(0,0)$ gets moved; the first rotation leaves it fixed, and then the second rotation sends it to $(-1,1)$. So the point $(0,0)$ gets translated as I claimed. Second, look at where $(1,0)$ gets moved; the first rotation sends it to $(0,1)$, which is then fixed by the second rotation. So $(1,0)$ gets sent to $(0,1)$, agreeing with my claimed translation. But an orientation-preserving rigid motion of the plane is determined by where it sends two points, so in fact the composite of the two rotations agrees with the claimed translation everywhere.

That takes care of one translation, by the vector $(-1,1)$.What about translations by all sorts of other vectors $\vec v$? No problem! Given any non-zero $\vec v$ (the case of $\vec 0$ is even easier), introduce a new coordinate system in which $\vec v$ has the coordinates $(-1,1)$, and then proceed as in the preceding paragraph in this new coordinate system.