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A continuous smooth ($C^{10}$-smooth, for example) map $f:S^1\to S^1$ of the circle is called expanding if $\inf_{x\in S^1} f'(x) > 1 $. Here $S^1 = [0,1]/\sim$, the segment with identified endpoints. It is well known that the dynamics of linear examples $f(x) = mx \pmod 1$ for natural $m\geq 2$ is conjugated (from the measure point of view) to the (full) shift on the space of one-sided sequences of digits $\{ 0, \ldots, m-1 \}$.

Question: is it true that any (non necessary linear) expanding map of the circle is conjugated to the one-sided full shift (with the number of letters equal to the topological degree of the map)?

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I think I know the answer now. Yes, the conjugation exists but the measure that is generated on the full shift by the coding is not necessary Bernoulli for nonlinear maps. The reason is simple -- the coding map is the same both for nonlinear map $f$ and for its linearisation (when branches are made linear conserving partitions of injectivity), the ergodic measure on the circle is no longer exactly Lebesgue measure (but something equivalent to it via a positive density).

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  • $\begingroup$ I know it's an old post but nevermind. Every expanding map on the circle or even on a compact set, topologically (semi-)conjugates with one one-sided shift, by considering a Markov partition of the map. I don't know if there is a better way but it's quite intuitive if one thinks about it a little bit. But I haven't thought what happens to the measures considered on the initial system, especially for the ones that we are extremely knowledgeable of, like the doubling map for example. very nice food for thought! $\endgroup$
    – alphaomega
    Jul 8, 2020 at 10:26

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