Cohen's idempotent theorem

Remarks

The first step is easy: $\widehat{\mu * \mu} = \widehat{\mu}^2$, so $\widehat{\mu}^2 = \widehat{\mu}$, and $\widehat{\mu}$ takes only the values $0$ and $1$. The hard part is the structure of the set $E = \set{\chi : \widehat{\mu}(\chi) = 1}$. Cohen's proof bounds the norm $\norm{\mu}$ of an idempotent measure from below in terms of the complexity of $E$. It uses this bound to show that $E$ is built from finitely many cosets.

Example: for $G = \mathbb{T}$ and an integer $m \ge 1$, let $\mu = \frac{1}{m}\sum_{k=0}^{m-1} \delta_{e^{2\pi i k/m}}$, the normalised Haar measure of the subgroup of $m$-th roots of unity. Then $\mu * \mu = \mu$, and $\widehat{\mu}(n) = \frac{1}{m}\sum_{k} e^{-2\pi i kn/m}$ is $1$ when $m$ divides $n$ and $0$ otherwise. So $E = m\Z$, a subgroup of $\widehat{\mathbb{T}} = \Z$.

A partial converse: if $E$ is in the ring of sets generated by the cosets of open subgroups of $\widehat{G}$, then $\mathbf{1}_E = \widehat{\mu}$ for an idempotent $\mu$. For an open subgroup $\Lambda$, the annihilator of $\Lambda$ is a compact subgroup of $G$ whose normalised Haar measure has transform $\mathbf{1}_\Lambda$. Translates, products and complements are then handled with the algebra structure of $M(G)$. When $G$ is compact, $\widehat{G}$ is discrete, every subgroup is open, and the theorem becomes a complete description of the idempotent measures. For $G = \mathbb{T}$ the result was first proved by Helson.

Going deeper

For a student

The theorem classifies the idempotents of the measure algebra $M(G)$ in Fourier terms. For $G = \mathbb{T}$ the coset ring of $\Z$ consists of the sets that differ by finitely many integers from a finite union of arithmetic progressions. So the idempotent measures on the circle are very rigid. The same work of Cohen gave the first lower bound in the Littlewood conjecture on the $L^1$ norms of exponential sums, with a related method. The theorem is a basic tool in the study of homomorphisms between group algebras.