# What`s the algebraic difference between a reflection and a glide reflection?

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?

• A reflection $f$ will satisfy $f(f(x)) = x$. For a glide reflection, $f(f(x))$ is a translation. Aug 15 '20 at 9:24
• Ah okay that helps a lot. Thanks! Aug 15 '20 at 9:29
• 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) Aug 15 '20 at 9:56
• I don't understand your notation. What are $\rho_{\theta}$ and $r(x)$? Aug 15 '20 at 10:14
• 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. Aug 15 '20 at 10:32

## 1 Answer

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.