Let $f: S^1 \to S^1$ be an homeomorphism. I'm trying to prove that if $R\circ f = f \circ R$ where $R$ is an irrational rotation in $S^1$ then $f$ is a rotation. So, using the definition in the $[0,1]$ interval, $R(x)=x+\alpha \mod 1$ where $\alpha$ is irrational. Then $f(x)+\alpha=f(x+\alpha)$. But I don't know what to do next. Thanks!
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1You may want to use Kronecker's theorem. It states that the multiples of an irrational number $\alpha$ reduced mod $1$ lie dense in $[0,1]$. – Zardo Jun 10 '15 at 22:07
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Consider that if $f$ commutes with $R$, then it commutes with $R^n$, the $n^{th}$ iterate of $R$. So, we can write: $$f(x+n\alpha)=f(x)+n\alpha$$ For integer $n$. However, given that $\alpha$ is irrational, the values of $n\alpha$ are dense mod $1$. You can, from there, use a continuity argument based on this fact to conclude that therefore $$f(x+\beta)=f(x)+\beta$$ for all $\beta$. Setting $x$ to zero shows $f$ to be a rotation.
Milo Brandt
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