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Let $S^1$ be the unit circle and $T:S^1\to S^1$ be a continuous map. Suppose $\mu$ is a $T$-invariant Borel probability measure on $S^1$, that is, $\mu(T^{-1}A)=\mu(A)$ for every Borel subset $A$ of $S^1$. The Fourier coefficients $\{\hat{\mu}(n)\}_{n\in\mathbb{Z}}$ of $\mu$ is defined as $\hat{\mu}(n)=\int_{S^1} \bar{z}^n \,d\mu(z)$.

Question: is there any way to express the measurable entropy $h_\mu(T)$ of $T$ in terms of $\{\hat{\mu}(n)\}_{n\in\mathbb{Z}}$?

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    $\begingroup$ No. $\mu$ is invariant for both $T$ and $T^2$, but they have different entropies. $\endgroup$
    – Anthony Quas
    Commented Jan 10, 2014 at 17:29
  • $\begingroup$ But $h_\mu(T)$ may have a different expression from $h_\mu(T^2)$ in terms of Fourier coefficients of $\mu$. $\endgroup$
    – Huichi Huang
    Commented Jan 13, 2014 at 16:33
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    $\begingroup$ I cannot see how this can make sense. If you're allowing different expressions, then the entropy of $T$ is given by $h_\mu(T)\hat\mu(0)$. The question should be: can you determine the entropy of the transformation using the Fourier coefficients and nothing else. The answer is you can't because $T$ and $T^2$ have the same invariant measure; so the same Fourier coefficients but different entropies. $\endgroup$
    – Anthony Quas
    Commented Jan 13, 2014 at 19:22
  • $\begingroup$ I would interpret the question as follows: fix a transformation $T$, say the doubling map for concreteness. Let $\mathcal{M}$ be the space of invariant measures. Let $\mathcal{F}\colon \mathcal{M}\to \ell^\infty$ take $\mu$ to $\hat\mu$ and $h\colon \mathcal{M}\to [0,\infty)$ be the entropy function. Then $h\circ \mathcal{F}^{-1}$ takes as input the Fourier coefficients of a measure and gives as output that measure's entropy. This is a well-defined function on the space of sequences that are Fourier coefficients of an invariant measure, and it's reasonable to ask if it has a nice formula. $\endgroup$
    – Vaughn Climenhaga
    Commented Jan 13, 2014 at 20:03
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    $\begingroup$ Of course changing the transformation $T$ would yield a different function $h\circ \mathcal{F}^{-1}$, but one can still ask the question. $\endgroup$
    – Vaughn Climenhaga
    Commented Jan 13, 2014 at 20:09

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