Translation Symmetry

Question

It is a simple concept but I think I am making it harder than it is. I am having a hard time understanding the visual concept that polygons that are translated are not the same as the original object but lines can. I just think that if lines translated count as maintaining symmetry why do we not count a square being translated having translation symmetry?

Answer 1

Let's take euclidean real vector space $E=\Bbb R^2, \mathcal S=\{(\mp1,\mp1)\}, A=(1,3), u=(1,2), d_0=\Bbb R u, d=A+d_0$

$\mathcal S=\{B,C,D,E\}$, with $B=(-1,1), E=(-1,-1), C=(1,1), D=(1,-1)$

Let $t_u $ the translation of vector $u$, written mathematically, it means$$t_u:E\to E$$ $$M=(x,y)\mapsto M+u=(x,y)+(1,2)=(x+1,y+2)$$

Then :

• $t_u(A)=A+u=(2,5)\neq A,$ but $(2,5)\in d$;
• $t_{2u}(d)=\{A+tu+2u:t\in \Bbb R\}=\{A+(t+2)u:t\in \Bbb R\}=d$
• Let's go to the study of the square: $t_u(\color{cyan}B)=\color{cyan}B+(1,2)=(0,3)\notin \mathcal S$; on the other hand, if we define symmetry $\color{red}s$ with respect to $\color{red}{(EC)}$ $$s:(x,y)\mapsto (y,x)$$ $s(\color{red}E)=\color{red}E, s(\color{red}C)=\color{red}C, s(\color{cyan}B)=\color{green}D, s(\color{green}D)=\color{cyan}B, s(\mathcal S)=\mathcal S$

$s$ is one of the symetries of the square.