# Isometries of R2 and R3 - Specific example

• Jan 8th 2012, 10:47 PM
Bernhard
Isometries of R2 and R3 - Specific example
I am reading Papantonopoulou: Algebra Ch 14 Symmetries. I am seeking to fully understand Theorem 14.21 (see attached Papantonopoulou pp 462 -463)

On page 462, Papantonopoulou defines translations, rotations and reflections for R2 and R3 (see attached). Note that the rotations are defined as about the origin and the reflections are about the X-axis or $e_1$ axis.

Then on Page 463 he states Theorem 14.21 as follows:

14.21 Theorem An isometry S of $\mathbb{R}^2$ or $\mathbb{R}^3$ can be uniquely expressed as $S = t_b \circ \rho_{\theta} \circ r^i$ where i = 0 or 1

I would like to use Theorem 14.21 to specify S for the isometry of $\mathbb{R}^2$ that maps the line y = x to the line y = 1 - 2x? [ ie what is $r^i$ , $\rho_{\theta}$ , $t_b$ in this case?]

Peter
• Jan 9th 2012, 02:25 AM
Deveno
Re: Isometries of R2 and R3 - Specific example
there is more than one isometry that does this. we get two seperate rotations that take y = x to y = -2x (one through some angle θ, and one through π+θ).

on the other hand, if we let i = 1, this takes the line y = x, to the line y = -x (well, actually, the line -y = x, but so what?), and we have two different rotations that map y = -x to y = -2x.

after each 4 of these elements of O(2), we can translate using (0,1) to get the desired affine isometry.

explicitly, here is one such mapping:

let i = 0. this is the "identity" reflection (not reflecting at all). note that:

$r^i = \begin{bmatrix}1&0\\0&(-1)^i\end{bmatrix}$

the next step is to compute θ. let's rotate counter-clockwise. the angle we want is:

$\theta = \frac{3\pi}{4} - \text{arctan}(2)$

then we have:

$\rho_{\theta} = \begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$

if we want to obtain a specific matrix for this rotation, we can rotate by the two angles (3π/4 and -arctan(2)) separately:

$\rho_{3\pi/4} = \begin{bmatrix}\frac{-\sqrt{2}}{2}&\frac{-\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&\frac{-\sqrt{2}}{2}\end{bmatrix}$

while:

$\rho_{-\text{arctan}(2)} = \begin{bmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5 }}\\ \frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{bmatrix}$

so that:

$\rho_{\theta} = \rho_{3\pi/4} \circ \rho_{-\text{arctan}(2)} = \begin{bmatrix}\frac{\sqrt{10}}{10}&\frac{-3\sqrt{10}}{10}\\ \frac{3\sqrt{10}}{10}&\frac{\sqrt{10}}{10} \end{bmatrix}$

(if i did my arithmetic and geometry right, i dislike calculation like this :()

finally, we translate by (0,1), giving the transformation:

$S:\begin{bmatrix}x\\y\end{bmatrix} \to \left(\begin{bmatrix}\frac{\sqrt{10}}{10}&\frac{-3\sqrt{10}}{10}\\ \frac{3\sqrt{10}}{10}&\frac{\sqrt{10}}{10} \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix}\right)$

so let's see if we map points on y = x to points on y = 1 - 2x, and preserve the distance.

for our 2 test points, i'll pick (0,0) and (4,4). clearly S takes (0,0) to (0,1) and 1 = 1 - 2(0), so that is on our target line. a brief (for you, maybe) calculation shows that S takes (4,4) to the point (-4√10/5, (8√10 + 5)/5), and again:

(8√10 + 5)/5 = 1 - 2(-4√10/5), so that, too, is on our line.

now the distance between (0,0) and (4,4) is 4√2. let's see what the distance between our 2 image points are:

$\sqrt{\left(\frac{8\sqrt{10}+5}{5} - 1\right)^2 + \left(\frac{-4\sqrt{10}}{5}\right)^2}$

$= \sqrt{\frac{640}{25} + \frac{160}{25}} = \sqrt{\frac{800}{25}} = \sqrt{32} = 4\sqrt{2}$