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This is exercise 1.9 in Lie Groups, Lie Algebras and Representations - Hall.

Suppose $a$ is an irrational real number. Show that the set $E_a$ of the numbers of the form $e^{2\pi i n a}$, $n \in \mathbb{Z}$, is dense in the unit circle $S_1$. Hint: Show that if we divided $S^1$ into $N$ equally sized "bins" of length $2\pi/N$, there is at least one bin that contains infinitely many elements of $E_a$. Then use the fact that $E_a$ is a subgroup of $S^1$.

My proof of this proposition is as follows. Since $a$ is irrational, you can determine that the set of rotations $E_a$ is infinite. Since $S^1$ is compact we can find two $r_1, r_2$ that are within $\epsilon$ of each other. Then $r_1^{-1}r_2$ is a small rotation of size $\epsilon$. Now, $r_1^{-1}r_2$ generates rotations that are within $\epsilon$ distance of any point of $S^1$.

That said, I don't believe that the hint suggested in the problem uses that technique. My knowledge of algebra is not all that strong so I was hoping someone could shed some light on what is being suggested there.

muaddib
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    How are you proving that such $r_1$ and $r_2$ exist by compactness? If you examine that argument, I suspect you'll find it's essentially the same as what the hint is saying. – Eric Wofsey Jul 26 '15 at 00:20
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    "Irrational rotations are dens in $S^1$" doesn't really capture the idea. It's not the set of all irrational rotations that is held to be dense. That much would be easy. Rather, it's the much smaller set of integer multiples of just one irrational rotation. That's a far stronger statement. One might said the closure under composition of an irrational rotation is dense. Or the set generated by an irrational rotation is dense. ${}\qquad{}$ – Michael Hardy Jul 26 '15 at 00:25
  • @MichaelHardy - Thank you for the suggestion. I modified the title. – muaddib Jul 26 '15 at 00:38

2 Answers2

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The hint is similar to your proof without directly using compactness. Instead it suggests you use the pigeonhole principle to show that we can choose arbitrarily small arcs that contain at least two elements of the group (in fact infinitely many), then proceed as in your argument.

Matt Samuel
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I think this should work: by Weirstrass' Theorem , every bounded infinite subset of the plane contains a limit point, say $p$. Then there is a sequence {$ka : k \in \mathbb N$}in $ E_a \cap S^1$ that converges to $p$. Now, for any other $q \neq p$, apply translation to construct a sequence that converges to $q$. I mean, scale the sequence in $E_a$ that converges to $p$ by a rotation by $\alpha -\theta$ , where $p=e^{i\theta} , q=e^{i \alpha} $ Then, if $e^{2\pi ka}\rightarrow p$ ,with some changes, $e^{i2\pi ka(\theta -\alpha)} \rightarrow q$. Now, obviously multiplying by $\theta- \alpha$ may push you out of $E_a$, but you can approximate the difference by a rational number.

EDIT: More clearly, as you pointed out, you can use this (Cauchy) sequence that converges to $p$ to construct an element $e_a$ in $E_a$ that gives you a rotation by any amount. Repeating this , i.e., the sequence {$ne_a$} gives you the dense subset of $S^1$. (you can construct a choice of $e_a$ for any $\epsilon$ you choose.)

Gary.
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