# Conjecture Regarding rotation of a set by a sequence of rational angles.

1. Aug 2, 2012

### mehr1methanol

Conjecture Regarding Rotation of a Set by a Sequence of Angles.

Consider the following sequence, where the elements are rational numbers mulriplied by $\pi$:
$(\alpha_{i}) = \hspace{2 mm}\pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/16,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/32,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/16,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/64,\hspace{2 mm} \pi/4,\hspace{2 mm} \cdots$

Let $K \subset ℝ^{2}$ be a compact set. Also let $R_{\alpha_{i}}$ denote the rotation by $\alpha_{i}$.

Suppose $R_{\alpha_{i}}K = \hspace{2 mm} K$ for each $\alpha_{i} \in (\alpha_{i})$.

Question: Is it true that for all $\theta \in [0, 2\pi)$ $R_{\theta}K = \hspace{2 mm} K$.

Note:
If instead we had the sequence $(n\alpha)$ where $\alpha$ is an irrational number, it is trivial that the conjecture holds. This is trivial due to the following fact from the study of continued fractions:
Given any real number on a circle, it can be approximated arbitrarily close by multiples of an irrational number.
But if $\alpha$ is a rational number this doesn't hold since after a finite number of rotations you will get back to where you started from. However in the question above we don't have rotations by a fixed rational number and the answer is not immediate!

Last edited: Aug 2, 2012
2. Aug 2, 2012

### haruspex

There are some things not making sense to me in this question.
Most obviously, the sequence clearly consists of irrationals, so I guess you mean rational multiples of pi. Secondly, I don't see any significance in showing it as a sequence, including repeats. It seems to be used only as a set - so why the repeats?
That said, if I've understood the question...
K is closed under rotations by 3nπ.2-m, for all positive integers m, n.
Given θ, let xi be the bits of the binary fraction expressing θ/3π. From this you can construct a sequence of rotations converging on θ.

3. Aug 3, 2012

### mehr1methanol

That's exactly right. Thank you for pointing that out. I corrected the question.

The reason for the repeats is the following:
I'm preforming a symmetrization on the set K and the algorithm is such that it produces the above sequence.

I got confused myself because once I got the sequence, I thought the rule is that I must follow the sequence to get arbitrary close to a θ. But you're absolutely right. Once I show $R_{\alpha_{i}}K = K$ for each $\alpha_{i}$, I'm done. This is because of the role that "n" is playing in your solution.

When I get a little too excited I need someone to check I'm not doing something stupid. Thanks for the comment!