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Is there a simple algebraic proof?

Thanks!

UH1
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  • Well, if you compose a glide reflection with itself, what you get is a translation - which is an isometry.

    It's not hard to see this on euclidean spaces; even easier in the plane. But to do this with an algebraic proof?

    Try do proof this using only geometric arguments; if you have a glide reflection then it has a line in which it reflects; this line must divide the plane in two components, in such a way that when you apply the reflection again, the original geometric object which is being reflected is in the original "side" of the plane, but translated.

     – Marra Dec 12 '12 at 22:36
  • yeah i checked geometrically, i am trying to solve it algebraically as well by using the fact that Mk composed with Tab = Tab composed with Mk where AB is parallel to the line of reflection k. Tab is a translation and Mk is a reflection across line k AND Mk composed with Tab is glide reflection. – UH1 Dec 12 '12 at 23:07

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Let the glide reflection be $T$. Without loss of generality the reflection part of $T$ is reflection in the $y$-axis. Assume that the translation part is by the vector $(a,b)$.

The reflection takes $(x,y)$ to $(-x,y)$. The translation part now takes us to $(-x+a,y+b)$. So $T$ takes $(x,y)$ to $(-x+a,y+b)$.

Do it again. The reflection part takes $(-x+a,y+b)$ to $(x-a,y+b)$, and the translation part takes this to $(x,y+2b)$. So $T^2$ is translation parallel to the $y$-axis by $2b$.

André Nicolas
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  • this is more or less correct if I understand your idea correctly(unless you missed the word 'GLIDE REFLECTION') but I need to prove that composition of glide reflection with itself is a translation NOT that two reflections make a translation (that is indeed a very basic result) – UH1 Dec 12 '12 at 23:22
  • I was being sloppy, had vectors in mind but did not saay so. Have rewritten. A glide reflection is a reflection followed by a translation. The algebra above is that if $T$ is a glide reflection, then $T^2$ is a translation parallel to the axis of reflection. – André Nicolas Dec 12 '12 at 23:34
  • great i see now, thanks! – UH1 Dec 12 '12 at 23:41