I'm self studying with Artins Algebra and reached chapter 5, section 2. There, he defines four types of symmetry (orientation-preserving: rotation and translation; orientation-reversing: reflection and glide-reflection) and goes on to show that they can all be represented as the composition of a reflection on the x1-axis (if it's orientation-reversing), a rotation around the origin and a translation. In one of the exercises, you are then asked to "prove that a conjugate of a reflection or a glide-reflection is a motion of the same type", but I don't understand how I can tell reflection and glide reflection apart on an algebraic level since a reflection around a line that doesn't intersect the origin also includes translations, and I haven't found anything in the chapter about it.

Edit: using $f(f(x))=x$ for a reflection $f(x)=t_a(\rho_\theta(r(x)))$, I got the result that, for a translation vector $a$ and a rotation angle $\theta$, we would have

$$\begin{bmatrix}a_1\\a_2 \end{bmatrix}=\begin{bmatrix}-\cos(\theta)a_1-\sin(\theta)a_2\\ -\sin(\theta)a_1+\cos(\theta)a_2 \end{bmatrix}$$ if and only if $f$ is a reflection. Is this correct?

  • 3
    $\begingroup$ A reflection $f$ will satisfy $f(f(x)) = x$. For a glide reflection, $f(f(x))$ is a translation. $\endgroup$ Aug 15 '20 at 9:24
  • $\begingroup$ Ah okay that helps a lot. Thanks! $\endgroup$ Aug 15 '20 at 9:29
  • $\begingroup$ I edited my post with what I got as an answer for an equation a reflection has to satisfy. Is this correct? (Also sorry if this breaks any etiquette I'm not aware of) $\endgroup$ Aug 15 '20 at 9:56
  • $\begingroup$ I don't understand your notation. What are $\rho_{\theta}$ and $r(x)$? $\endgroup$ Aug 15 '20 at 10:14
  • 1
    $\begingroup$ Note that there is a proof strategy that avoids these explicit coordinates: first, prove that the conugate of a translation is also a translation. Then, using the fact that $f$ is a glide reflection iff $f^2$ is a translation, we see that any element $gfg^{-1}$ conjugate to $f$ satisfies $$ (gfg^{-1})^{2} = gf^{2}g^{-1}, $$ which means that $(gfg^{-1})^{2}$ is the conjugate of a translation, which means that $(gfg^{-1})^{2}$ is a translation, which means that $gfg^{-1}$ is a glide reflection. $\endgroup$ Aug 15 '20 at 10:32

Given that $f(x) = t_a\circ \rho_\theta\circ r$, we see that $f$ is a reflection if and only if $f(f(0)) = 0$. We note that $$ \begin{align} f(f(0)) & = f(a) = a + \rho_\theta(r(a)) = a + \pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta } \pmatrix{a_1\\ -a_2} \\ & = \pmatrix{a_1\\a_2} + \pmatrix{\cos \theta a_1 + \sin \theta a_2\\ \sin \theta a_1 - \cos \theta a_2}. \end{align} $$ So indeed, your statement is correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.