TITLE: The differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$
QUESTION [8 upvotes]: The cosets of $\mathbb{Z}$ in $\mathbb{R}$ are all sets of the form $a+\mathbb{Z}$, with $0 ≤ a < 1$ a real number. Adding such cosets is done by adding the corresponding real numbers, and subtracting 1 if the result is greater than or equal to 1. -- Examples of Quotient Group, Wiki

I cannot figure out the differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$. Besides, "subtracting 1 if the result is greater than or equal to 1", what does "the result" mean here? Why do we need to subtract 1? I was wondering what is the background of $\mathbb{R}/ \mathbb{Z}$.

REPLY [13 votes]: I wrote some words about the algebraic structure on $\mathbf R/\mathbf Z$, but I think that's been treated very well in the other answers. By the tags I feel that we should also talk about the topological and analytic aspects of this group, so here's a start in that direction. I certainly wish that I knew more about this.
$\mathbf R/\mathbf Z$ is often called the circle group, because if we view $S^1$ as lying inside of the complex plane then the map $F\colon\mathbf R \to S^1$, $F(x) = e^{2\pi ix}$ induces a bijection $\mathbf R/\mathbf Z \to S^1$ which is both an isomorphism of groups and a homeomorphism of topological spaces (if we place the quotient topology on $\mathbf R/\mathbf Z$); in particular, $\mathbf R/\mathbf Z$ is naturally a compact space.
$\mathbf R/\mathbf Z$ is a nice home for periodic functions, since if $f\colon \mathbf R \to \mathbf R$ is such that $f(x + n) = f(x)$ for all $x \in \mathbf R$ and $n \in \mathbf Z$, then $f$ descends to a map $\mathbf R/\mathbf Z \to \mathbf R$. Moreover, Pontryagin duality says that Fourier analysis on the circle is related to functions on a certain "dual group", and the dual group of the circle is $\mathbf Z$, which is discrete! On the other hand, $\mathbf R$ is its own dual.
As for differential operators, that's a large subject and I can only sketch the situation and give Wiki references. One can view $\partial/\partial x$ as a vector field on $\mathbf R$. Each $a \in \mathbf R$ induces a translation diffeomorphism $L_a\colon \mathbf R \to \mathbf R$, $L_a(x) = x + a$, and the field $\partial/\partial x$ is called left-invariant if it is unchanged under the corresponding pushforwards: $L_{a*}(\partial/\partial x) = \partial/\partial x$.
In general, given a smooth map of manifolds $F\colon N \to M$ a vector field on $N$ may not correspond to a vector field on $M$. But you can check that in this case we can define a vector field on $S^1$, which people usually call $\partial/\partial\theta$, by $(\partial/\partial\theta)_{F(P)} = F_*(\partial/\partial x)_P$.