Thread:

1. 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
Attached Files
• P pp 462.pdf (537.3 KB, 12 views)
• P pp 463.pdf (732.4 KB, 8 views)

2. 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:

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

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

i'm convinced, how about you?

3. Re: Isometries of R2 and R3 - Specific example

Hi Deveno

Yes, worked through all the example you provided

CONVINCED!!

Thank you for your guidance and help! It has enabled me to pursue symmetries and symmetry groups further!

Peter
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