Let $\hat{D} := D \backslash \{0\}$ be a ball in $R^n$ with the origin $\{0\}$ removed. Assume that $\hat{D}$ has a structure as an orbifold (may be distinct from its standard manifold structure). Is it possible to extend the orbifold structure from $\hat{D}$ to $D$ ?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
No. This is not true in dimension three (nor in any higher dimension).
The three-dimensional orbifold structures allowed at a point are controlled by the list of finite subgroups of $\mathrm{SO}(3)$. In particular, there are at most three one-dimensional loci that converge at any point.
Let $S^2(2^m)$ be the two-orbifold which is topologically a two-sphere, decorated with $m$ orbifold points of order two. Suppose that $m > 3$. We form $\hat{D} = S^2(2^m) \times (0, 1]$. This is the desired example.
Another way to think about this: the link of a point $x$ in an $n$-dimensional orbifold must be a spherical orbifold in dimension $n-1$. In the example above, (for $m > 4$) we found a particular hyperbolic orbifold structure on $S^2$. However any euclidean or hyperbolic orbifold structure on $S^2$ would have served as well.
-
1$\begingroup$ Do you mean $\{0\}$ is the limit of $m$ one-dimension loci correponding to $m$ orbifold points, so it is a contradiction? Can you explain why, for any point in $D$, there are at most four one-dimensional loci that converge to it? $\endgroup$– Hao YuCommented Mar 16, 2022 at 13:38
-
$\begingroup$ A1: Yes. A2: I made a mistake earlier, the bound is three, not four. This follows from the classification of finite subgroups of SO(3). Each subgroup (trivial, cyclic, dihedral or platonic) gives an orbifold quotient which has either zero, two, three, or three (respectively) one-dimensional loci (that is, the link has that many orbifold points). $\endgroup$– Sam NeadCommented Mar 16, 2022 at 20:39
-
1$\begingroup$ What if we first specify an orbifold structure on $S^{n-1}$, then ask if there is an extension of it to an orbifold with boundary $\bar{D}$, so that $\partial{\bar{D}} = S^{n-1}$ (as an orbifold)? $\endgroup$– Hao YuCommented Mar 17, 2022 at 7:32
-
$\begingroup$ That seems like an interesting problem... I don't know an answer off the top of my head. Do you want to additionally require that the topological space "underlying" $\bar{D}$ is an $n$-ball? That is more restrictive but perhaps more interesting. Also, there is the general problem of "which orbifolds bound?" which is less restrictive. $\endgroup$– Sam NeadCommented Mar 17, 2022 at 9:33
-
$\begingroup$ If my answer addresses your original question, it would be polite to accept it. You should then ask a your new question in another post - it is not the "done thing" to edit an old question into a new form. $\endgroup$– Sam NeadCommented Mar 17, 2022 at 9:34