# What is the physical implication(s) of the isomorphism between ${\rm SO}(2)$ and $\mathbb{R}/\mathbb{Z}$?

In the book Mathematical physics by V. Balakrishnan, he says (on page 329) that the isomorphism between ${\rm SO}(2)$ and $\mathbb{R}/\mathbb{Z}$, and the fact that $\mathbb{R}$ is the universal covering group of ${\rm SO}(2)$ has deep implications in two-dimensional systems in condensed matter physics and quantum field theory without any further explanation. Can someone elaborate on what he might have in mind?

• Aside: SU(2) is the universal and double cover of SO(3) and it's this, at least mathematically, that underlies the notion of spinors, and these are important in QM & QED. – Mozibur Ullah Jan 16 '18 at 18:29

In $d\ge 3$, the first homotopy group of $\mathrm{SO}^+(1,d)$ is $\mathbb Z_2$, which essentially leads to spin quantisation. In $d=2$, and due to $\mathrm{SO}(2)\sim\mathbb R/2\pi\mathbb Z$, we have $\pi_1(\mathrm{SO}^+(1,d))=\mathbb Z$, and therefore we don't have spin quantisation anymore. Particles are no longer classified into bosons vs. fermions, but they may have any statistics. We may find anyons, which lead to a very rich phenomenology (think fractional quantum Hall effect, etc.).
Recall that spin comes from the projective representations of the little group, to wit, $\mathrm{SO}(d)$. Unlike in higher dimensions, in $d=2$ we have that $\mathrm{Spin}(d)$ is not the universal cover of $\mathrm{SO}(d)$; indeed, $\widehat{\mathrm{SO}}(2)=\mathbb R$, which is non-compact. We thus no longer require $U(4\pi)=1$, so that spin is no longer a half-integer. Fun!
The distinction between the groups $\mathbb R$ and $\mathbb R/\mathbb Z\cong SO(2)\cong S^1$ is topological hence it is relevant in topological aspects of field theories.
A classic example of two (space) dimensional system where this stuff is important is the Abelian-Higgs model which appears in condensed matter physics (non relativistic) and high energy physics (relativistic). This consists basically in a gauge field theory with local symmetry algebra $\mathfrak u(1)$ and gauge symmetry spontaneously broken to the identity element $e$ by a complex scalar field (Higgs). If the gauge symmetry group is $\mathbb R/\mathbb Z\cong SO(2)\cong S^1$ then the static finite energy configurations provide maps $\phi:S^1\rightarrow S^1$ from the circle at infinity to the vacuum manifold $SO(2)/e\cong SO(2)\cong S^1$, which is another circle. The fundamental group of the second circle, $\pi_1(S^1)$, is non trivial: as we cover the first circle once, the image can cover the second circle $n$ times, having therefore winding number $n$. Maps with different winding number cannot be continuously deformed into each other, they are non homomorphic. All homomorphic maps are said to be in the same equivalence class and are associated to one element of the fundamental group. Therefore $\pi_1(S^1)\cong\mathbb Z$. The physical implication of all that is straithforward: since maps with different winding numbers are non homomorphic, the associated scalar field cannot decay into each other. This gives rise to stable configurations such as vortices which are experimentally verified in superconductivity, in the form of the Abrikosov vortex lines and are predicted in Grand Unified Theories as cosmic strings which may have cosmological implications such as in galaxies formation. It may also be relevant in the quark confinement problem, where the topological configurations are called flux tubes (Note that in all three examples, the configuration is extended in another space dimension but all topological relevance is in the two dimensional section). On the other hand if we just change the topology of the gauge group, saying that it is $\mathbb R$, then we shall look at configurations associated to $\phi:S^1\rightarrow \mathbb R$. But since all closed curves in $\mathbb R$ can be shrunk to point, all of them are homomorphic, i.e. $\phi(\mathbb R)= e$. Physically this means that all configurations can decay into the vacuum hence there is no stable vortices.