Skip to main content
added 27 characters in body
Source Link
Thomas Rot
  • 7.6k
  • 2
  • 32
  • 54

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 $S^1$).

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 $S^1$).

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 $S^1$).

Source Link
Thomas Rot
  • 7.6k
  • 2
  • 32
  • 54

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 $S^1$).