12
$\begingroup$

Forgive me if this is a basic question. I'm just learning about orbifolds, and covering spaces are my happy place for thinking about group actions.

If $M$ is a manifold and $G$ is a group acting properly discontinuously on $M$, then $M/G$ is an orbifold. It seems to me like $M$ should also be a branched cover of $M/G$. Is this always true? If not, is there an illuminating example?

Right now when I think of orbifold, I'm secretly thinking of quotients of manifolds by group actions. And when I think of quotients of manifolds by group actions, I'm secretly thinking about covering spaces. If anyone could help me out by showing me some spots where either of these bridges break down I'd be very appreciative.

$\endgroup$
  • 3
    $\begingroup$ You have to be careful with what you mean when you say "branched cover". For instance, the group $\mathbb{Z}/2$ acts on $\mathbb{R}^2$ via the map that takes $(x,y)$ to $(-x,y)$. The quotient is a half-space $X$ which is an orbifold whose "orbifold points" are the entire boundary. But usually when you define branched covers of surfaces, the branch locus is discrete. $\endgroup$ – Andy Putman Oct 15 '13 at 23:59
  • 3
    $\begingroup$ (but it is possible to define branched cover so that the maps $M \rightarrow M/G$ are branched covers whenever $G$ is a group acting on the manifold $M$ properly discontinuously). $\endgroup$ – Andy Putman Oct 16 '13 at 0:00
5
$\begingroup$

I think, what you are curious about is the difference between good orbifolds (which have a manifold cover) and bad ones (having no manifold cover). There are standard examples of a "teardrop" and a "spindle" as bad orbifolds. The best way to reach them (with explanations), is to read Thurston's notes (or google out some other lecture notes on orbifolds, which are now numerous).

$\endgroup$
  • 2
    $\begingroup$ Another excellent source, at least for 2-dimensional orbifolds, is Peter Scott's article 'The geometries of 3-manifolds'. $\endgroup$ – HJRW Oct 16 '13 at 12:01
2
$\begingroup$

I will not presume that you asked about what a good cover is. Thinking about orbifolds the way you described is useful: a group $G$ acts on $M$ properly and discontinuously but because the action is not free the topological quotient $X = G\backslash M$ may be missing some information. In particular the map $M \twoheadrightarrow G\backslash M=X$ is not a covering map.

One definition of an orbifold is the space $X$ equipped with an atlas $\mathcal A$ such that each chart contains some extra information (i.e. a local group action.) This is the extra information that is lost, for example, in passing from $M$ to $X=G\backslash M$. With this information, you can recover $M$ from $X$ and $G$, should $M$ actually exist (this has to do with "goodness"). So in this sense an orbifold is a quotient of a proper discontinuous group action.

A covering space of $Y$ can be obtained by taking a open cover $\mathcal{U}=\{U_i\}$ of $X$ and glueing together copies of the $U_i$, one of the criteria for being a covering space though is that two distinct lifts of some $U_i \in \mathcal U$ are disjoint.

If you want to make an orbifold cover, i.e. get $M$ from $G\backslash M$, the problem is that distinct lifts of the same chart in $\mathcal A$ can not be made to be disjoint.

Suppose first that the non-trivial fixed sets of the $G$ action on $M$ are points (and that $M$ is a surface). Let $X = G\backslash M$ and let $\mathcal A$ be the orbifold atlas on $X$. Then $M$ can be obtained by taking copies of the charts $V_i$ in $\mathcal A$ (which are all discs), but sometimes you will need to slice some of the $V_i$ open (to get something that looks like a slice of watermelon), you can then get $M$ by glueing together possibly sliced charts from $\mathcal A$ in a manner analogous to building a covering space. Moreover, these "slices" in charts are exactly the cuts in branched covers. So in this case the orbifold cover is a branched cover.

However, as Andy Putman commented, not all actions of $G$ on $M$ have fixed sets that are points. For example a reflection will fix an arc, and the example that Andy gave ($\mathbb R^2$ modulo reflection) will not correspond to a branched cover. So to answer your question, branched covers are a way to construct special (but important) cases of orbifold covers. The special case, as Andy commented, is when all the non trivial $G$-fixed sets in $M$ are points (provided $M$ is a surface).

$\endgroup$
  • 3
    $\begingroup$ Someday, I dream that mathematicians will learn how to spell my last name... $\endgroup$ – Andy Putman Oct 16 '13 at 16:25

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.