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I'm having some difficulty wrapping my head around rigid motions in a plane. In particular, I'm trying to solve this following problem:

In a Euclidean plane, show that the product of two rotations around different points is equal to either a rotation around a third point or a translation. Hint: Show that it has at most one fixed point.

I working in a plane $\Pi$ over an Euclidean ordered field, so rotations are transformations defined by $$ \begin{cases} x'=cx-sy \\ y'=sx+cy \end{cases} $$ where $c^2+s^2=1$. I simply take two rotations $\psi$ and $\phi$ such that $$ \phi=\begin{cases} x'=cx-dy\\ y'=dx+cy \end{cases} $$ and $$ \psi=\begin{cases} x'=ex-fy\\ y'=fx+ey \end{cases} $$ where $c^2+d^2=1$ and $e^2+f^2=1$. Composing them, I see $$ \begin{align*} \psi\phi(x,y) &= (ecx-edy-fdx-fcy,fcx-fdy+edx+ecy) \\ &= ((ec-fd)x-(ed+fc)y,(fc+ed)x+(ec-fd)y) \end{align*} $$ but $$ \begin{align*} (ec-fd)^2+(ed+fc)^2 &= e^2c^2-2ecfd+f^2d^2+e^2d^2+2edfc+f^2c^2 \\ &= (e^2+f^2)c^2+(e^2+f^2)d^2 \\ &= c^2+d^2=1 \end{align*} $$ so $\psi\phi$ is a rotation. I feel I have done something wrong, since the problem seems to be prodding me in a different direction. What is the correct way to show this? Thanks.

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You have so far only considered rotations around the origin. – Alexander Thumm Mar 21 '11 at 8:02
@Alexander, thanks, I'll try to redo my argument. – yunone Mar 21 '11 at 8:17
up vote 2 down vote accepted

With no coordinate system, or at least, as long as possible without one.

An affine transformation of $\mathbb{R}^n$ consists of a linear transformation followed by a translation $x \mapsto A x+ b$, which we denote by $(A,b)$. The transformation $(A,b)$ is a rotation if and only if $A$ is an element of the special orthogonal group $SO(n)$ and $A$ is not identity. It is a translation (or the identity) if $A$ is identity.

The composition of two rotations $(A,b)$ and $(A',b')$ is $x\mapsto (A'A)x+(A'b+b')$. One knows that $A'A$ is an element of $SO(n)$. Hence, if $A'A$ is not the identity, the composition is a rotation and if $A'A$ is the identity, this is a translation.

In dimension $2$, any element $A$ of the group $SO(2)$ can be written as $$ A=\begin{pmatrix} \cos(u) & \sin(u)\\ -\sin(u) & \cos(u)\end{pmatrix} $$ with $u$ real, hence $A$ is identity if and only if $u$ is a multiple of $2\pi$. The product $A'A$, which you need to compute the composition of $(A,b)$ and $(A',b')$, is $$ A'A=\begin{pmatrix} \cos(u+u') & \sin(u+u')\\ -\sin(u+u') & \cos(u+u')\end{pmatrix} $$ (this is the addition formula for sines and cosines). The result is a proper rotation when $u+u'$ is not a multiple of $2\pi$, and a translation when it is.

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Thank you Didier, the vocabulary of this post is a little unfamiliar to me, but I think I can apply your explanation to my more elementary case. I worked it out with a coordinate system, (despite your warning not to!), and I get a transformation of the form $x\mapsto (A'A)x+(A'b+b')$, and it make more sense now. Thank you. – yunone Mar 21 '11 at 8:58
@yunone Hence the links to wikipedia pages in my post. If the post was useful in the end, everything is fine. – Did Mar 21 '11 at 10:23

All isometries (rigid transformations) of a plane can be expressed as a composition of 3 or fewer reflections.

  • A composite of two reflections over intersecting lines is a rotation about the point of intersection of the lines of reflection.
  • A composite of two reflections over parallel lines is a translation perpendicular to the lines of reflection.
  • A composite of three reflections is a "glide reflection" (which can be expressed as a reflection followed by a translation).

Isometries that can be expressed as a composite of an even number of reflections preserve orientation; those that can be expressed as a composite of an odd number of reflections reverse orientation. Since rotations preserve orientation, composing rotations also preserves orientation, which means that the result must be a rotation or a translation (the result must be an isometry and cannot be an orientation-reversing isometry, so cannot be a reflection or glide-reflection).

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Thank you for your answer Isaac, I was not as aware of how prevalent reflections are. The last paragraph of your post seems like it's hinting at the next exercise I want to try, so much thanks for that too! – yunone Mar 21 '11 at 9:00

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