Is Fürstenberg's topology useful? - MathOverflow

Is Fürstenberg's topology useful?

2010-10-18T06:32:29Z

It's hard not to be amused and perhaps even amazed when first encountering Fürstenberg's clever "topological" proof that there are infinitely many primes. Closer inspection, however, reveals the disappointing truth that there really isn't anything topological going on there, as pointed out by BCnrd in a comment to this answer.

Nevertheless, the topology on $\mathbb{Z}$ introduced in the proof, where an open set is defined as any union of arithmetic sequences, does seem both natural and interesting.

My question is this: Can anything useful be done with this topology? Useful would include a new theorem, a simplification to a proof of a known result, or even fresh insight into standard material.

Answer by S. Carnahan for Is Fürstenberg's topology useful?

2010-10-18T07:59:49Z

The answer to your question is yes, but it is a stretch to claim that the topology is due to Furstenburg. There is an extended discussion on Furstenburg's proof in the comments to this answer. The short version is as Chandan Singh Dalawat said in the comments above: this topology on the integers is the profinite topology, and people had been studying profinite topologies long before Furstenburg.

The topology is useful in the sense that profinite completions are useful. In particular, you may argue that it is a natural topology on the fundamental group of a circle (or the punctured complex affine line), since its profinite completion is the geometric fundamental group of the multiplicative group $\mathbb{G}_m$ . It also appears in some form whenever one uses the ring of adeles $\mathbb{A}_\mathbb{Q}$ , which you may encounter when studying Tate's thesis or automorphic representations.

Answer by Greg Kuperberg for Is Fürstenberg's topology useful?

2010-10-19T11:15:42Z

A more rigorous version of Scott's answer: If a topology on a group $G$ is translation-invariant, then it also defines a uniformity on $G$, by definition a distinguished set of neighborhoods of the diagonal $G \times G$ that is analogous to a metric. Actually, in the present example with $G = \mathbb{Z}$, the uniformity comes from a metric. Like the metric spaces that they generalize, uniform spaces have completions. The completion of $\mathbb{Z}$ with respect to the uniformity cited by Furstenberg is exactly the adelic profinite completion of $\mathbb{Z}$. Or if $G$ is any group, there is a similar topology generated by finite-index subgroups, and a uniformity, and the completion is the profinite completion.