Defining a map into $S^1$ as an "angle" in a non simply connected domain

Question

Suppose that ambient space is $\mathbb R^2$, and $\Omega \subset \mathbb{R}^2 $ is a smooth domain, non simply connected domain. To fix ideas,we can assume $$\Omega = \{(x_1,x_2) : 1< x_1^2+x_2^2 <4\}.$$

Given $p\geq2$, if $V$ is simply connected, given a function $\phi\in W^{1,p}(V;S^1)$ there exists $\theta \in W^{1,p}(V,\mathbb R)$ such that $$ \phi = (\cos(\theta),\sin(\theta)). $$

This is well known, see comments below (which date from an earlier formulation of this post).

It isn't the case here ($\phi = x/|x|$ for example) . But what can be said? That there is a smooth lift taking values in, say $\mathbb{R}/2\pi\mathbb{Z}$ ? Does such a result exist ?

What I have found are results saying that if the domain is not simply connected, it isn't as usual. But they do not dwell on what happens then.

To clarify, I am aware that if we make a cut somewhere, then we can define an angle. But this unfolding of the domain does not appeal to me, as I am interested in a variational problems and it would require me to specify some sort of boundary condition on the cut, which I don't want to do.

Given a map $\phi$ taking values in $S^1$, since $\phi^T\phi =1$, what you can always say is that $$ D\phi^T\phi = 0 \text{ and } \phi^T D\phi =0, $$ and therefore the gradient is at most of rank one. Writing $J=\begin{pmatrix} 0& 1 \\ -1& 0 \end{pmatrix}$, since $\phi,J\phi$ are orthogonal, this means that writing $ z =D\phi^T J\phi$, one has $$ D\phi = J\phi z^T. $$ My question is: is $z$ the gradient of something?

Answer 1

Let $\Omega$ be connected. There is a lifting criterion: A map $f:\Omega\rightarrow S^1$ lifts to the universal cover $\mathbb R\rightarrow S^1$ iff $f_*(\pi_1(\Omega))=0$, where $f_*$ is the induced map $f_*:\pi_1(\Omega)\rightarrow \pi_1(S^1)$. Your space $W^{1,p}(\Omega,S^1)$ has different connected components: There is one connected component (namely those for which the criterion before is satisfied) for which the lifts are possible.

In general you cannot lift a map to the universal cover. However subgroups of the fundamental group of index $k$ correspond to $k$ fold covers. For the circle the cover is again a circle, only wrapped around the base circle $k$ times.

Over the other components of $W^{1,p}(\Omega,S^1)$, where $f_*(\pi_1(\Omega))$ is the index $k$ subgroup, you can lift to the $k$ fold cover of $S^1$ (which is still $S^1$).