TITLE: "The Galois group of $\pi$ is $\mathbb{Z}$."
QUESTION [19 upvotes]: Last year, in a talk of Michel Waldschmidt's, I remember hearing a statement along the lines of the title of this question, that is, "The Galois group of $\pi$ is $\mathbb{Z}$.". In what sense/framework is this true? What was meant exactly - and can this notion be made precise?

REPLY [17 votes]: I gather that the idea behind $\mathrm{Gal}(\pi)=\mathbb{Z}\backslash\{0\}$ (not $\mathbb{Z}$, $0$ is not a conjugate of $\pi$!) comes from Euler's formula:
$$\prod_{n\in \mathbb{Z\backslash\{0\}}}\bigg(1-\frac{x}{n\pi}\bigg)=\frac{\mathrm{sin}(x)}{x} \in\mathbb{Q}\{x\} $$
which can make you think of those $n\pi$ as the conjugates of $\pi$.
But in order for the Galois groups to act transitively on the conjugates you need all the non-zero rationals, so that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$.
For some actual evidence that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$ is the right answer (this is, consistent with the conjectural picture of periods and motives), see sections 3 and 5 of Galois theory, motives and transcendental numbers by Yves Andre, alredy mentioned in the comments.