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Being a happy beginner in algebra and groups, i would like to build a group that can systematically represent (in 2 dimensions):

  1. Rotation around origin.
  2. Translation along radius.

I have learned that complex numbers can represent rotation with some resolution $\theta$ : $$\bf M_{R(\theta)} = \begin{bmatrix}a&-b\\b&a\end{bmatrix}, \cases{a=\cos(\theta)\\b=\sin(\theta)}$$

And then translation with a resolution $k$ : $$\bf M_{T(k)}=\begin{bmatrix}1&0\\k&1\end{bmatrix}$$

Can I combine these somehow to achieve what I want? To build a general matrix with help of these matrices and their exponents $${\bf M(\theta,k)} = f({\bf M_{T(k)},M_{R(\theta)}},e_\theta,e_k)$$


Edit : An image to try and clarify : We want our group to take a given line to any other given line, we measure their distance to origo and their angle. The point of interest ( which i am talking about above ) will have a $\Delta$ in each of the dimensions: angle $\theta$ and radius translation $k$.

enter image description here

Example where the blue line is taken to the black line by concatenating the operations: $$\cases{r = r - 0\cdot 0.125\\\theta =\theta+ 2\cdot \frac{\pi}{8}}$$ The other tuples decide the numbers for lines of the other colors. We want to find a matrix representation of generators for a group which does this. The red helper circles are centered on origo and of radiuses $0.25,0.5$ to help see what the "translation" operation is supposed to do.

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  • $\begingroup$ I don't really understand what you mean by "translation along the radius" or "translation with resolution $k$"... Your matrix $\bf M_{T(k)}$ does not define a translation on $\Bbb R^2$, in fact no matrix defines a translation as translations are not linear. $\endgroup$ – Arnaud D. Feb 18 '17 at 11:23
  • $\begingroup$ It is only intended as a 1D translation in the radial direction ${\bf M_{T(k)}}^n = \begin{bmatrix}1&0\\kn&1 \end{bmatrix}$ is $n$ "steps" if k is 1 "step". The length in position $(2,1)$ of the matrix. And the inverses are intended to correspond to steps in opposite direction. $\endgroup$ – mathreadler Feb 18 '17 at 11:32
  • $\begingroup$ But it's not a translation, since it fixes the origin... $\endgroup$ – Arnaud D. Feb 18 '17 at 11:34
  • $\begingroup$ Yes, but it is not a scaling either. I don't know the word for it. I want to traverse back and forth radially on a linear scale and let the rotation steer the angle. $\endgroup$ – mathreadler Feb 18 '17 at 11:35
  • $\begingroup$ So if I understand correctly, if you have a point $P$, you want to map it to a point $Q$ that is on the same line through the origin $O$, and such that $d(O,Q)=k\cdot d(O,P)$? Or do you want something like $d(O,Q)=d(O,P)+k$? $\endgroup$ – Arnaud D. Feb 18 '17 at 12:21
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Assuming that we start with closest point at the x-axis, one way to calculate the group action on a point $(x,y) \in {\mathbb R}^2$:

$$\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}\underset{\text{Rotation generator}}{\underbrace{\begin{bmatrix}a&-b&0\\b&a&0\\0&0&0\\\end{bmatrix}}}^{\,\,m}\underset{\text{Translation generator}}{\underbrace{\begin{bmatrix}1&0&k\\0&1&0\\0&0&1\end{bmatrix}}}^n\begin{bmatrix}x\\y\\1\end{bmatrix}$$

Or in vector form with $\bf V$ having columns being points to act on and $\bf t$ being a translation vector:

$$\begin{bmatrix}\bf I_2&\bf 0\end{bmatrix}\begin{bmatrix}\bf R&\bf 0\\{\bf 0}^T&0\\\end{bmatrix}^m\begin{bmatrix}{\bf I_2}&\bf t\\\bf 0&1\end{bmatrix}^n\begin{bmatrix}\bf V\\{\bf 1}^T\end{bmatrix}$$

The lines above were generated using this recipe, so it has been verified to work.

An illustration for comparision to the fan beam scan we here instead imagine tangential tracks on a circular rail shooting parallell rays for each rotation of the gun, translation corresponds to relocating along the track, rotation is relocating along the circle: enter image description here

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