# Ergodicity of non-homogeneous “rotations”

It is well known that a rotation $f(z)=e^{i\theta}z$ of the unit circle, is ergodic if and only if $e^{i\theta}$ is not a root of unity.

Now, what happens if we let $\theta$ depend on $z$ (say, continuously)?

To be more explicit: let $\theta:S^1\subset{\mathbb C}\rightarrow [0,2\pi)$ be continuous, and let $$f:S^1\rightarrow S^1,\qquad f(z)=e^{i\theta(z)}z.$$ Does there exist criteria (or some sort of classification) on whether $f$ is ergodic or not?

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I guess you mean that $\theta$ can depend on $z$? (Rather than on $\theta$.) –  Vaughn Climenhaga Feb 16 '12 at 1:01
Also your original rotation should be $f(z) = e^{i\theta} z$, and I assume that in your definition of a new map $f$ you want $f(z) = e^{i\theta(z)} z$. –  Vaughn Climenhaga Feb 16 '12 at 1:02
If I understand your definition, then this is an incredibly large class of systems, and there's essentially no hope of a complete answer. Two remarks jump out: (1) to talk about ergodicity you need an invariant measure. A more meaningful question might be whether or not $f$ is uniquely ergodic. (2) the maps you define are degree 1 maps of the circle. If you put a few more restrictions on $z$ so that they must be homeomorphisms, then there's a very rich theory of orientation preserving homeomorphisms of the circle. See en.wikipedia.org/wiki/Rotation_number, for instance. –  Vaughn Climenhaga Feb 16 '12 at 1:07
Also en.wikipedia.org/wiki/Circle_map gives some idea of the richness of behaviour that can happen. –  Vaughn Climenhaga Feb 16 '12 at 1:07
Vaughn, thanks. The wikipedia reference is very much in the direction I was looking for. I also edited as you observed. –  Federico Feb 16 '12 at 18:54