How many Fourier coefficients vanish?

Question

Let $G$ be a compact abelian connected metric group with Haar measure $\mu$ and let $f\colon G\to S^1$(=unit circle in $\mathbb{C}$) be a continuous function (not necessarily a group homomorphism) such that $\int_G fd\mu\neq0$. Can we say something about the set $\{k\in\mathbb{Z}\mid \int_G f^kd\mu\neq0\}$? Is this a question that has been studied in the literature?

Properties that I am potentially interested in would be: Does it always contain an arithmetic progression? Or does it even have finite complement in $\mathbb{Z}$?

Answer 1

By assumption, the set contains $1$. It's stable under negation since this complex conjugates the integral, so the set contains $-1$. Trivially, the set contains $0$.

This is sharp: $\{-1,0,1\}$ is a possibility.

Let $G$ be the unit circle. Fix $\delta<1$. There is a map from $S^1$ to $G$ that sends a point $e^{i \theta}$ to $ e^{i \theta + i\delta \sin \theta}$. Since $\delta<1$, this map has nowhere vanishing derivative and thus is invertible. Let $f$ be the inverse map.

Then change of variables gives, parameterizing $G$ with a variable $t$ and $S^1$ with $\theta$, $$\int_G f^k d\mu = \frac{1}{2\pi} \int_0^{2\pi} e^{ i k \theta(t)} dt = \frac{1}{2\pi} \int_0^{2\pi} e^{ i k \theta} \frac{d\theta}{ \frac{d\theta}{dt}}= \frac{1}{2\pi} \int_0^{2\pi} e^{ i k \theta} \frac{d\theta}{ (1 + \delta \cos \theta)^{-1} } $$ $$ = \frac{1}{2\pi} \int_0^{2\pi} e^{ i k \theta} (1+ \delta \cos \theta) d \theta$$

which is nonzero exactly for $k\in \{-1,0,1\}$.