# Gluing of orbifolds

Suppose that $$P$$ and $$Q$$ are $$n$$-dimensional orbifolds, with boundaries. Suppose also that there is an isomorphism $$f \colon \partial P \rightarrow \partial Q$$ (as orbifolds). Is there a way to glue $$P$$ and $$Q$$ using $$f$$ to form a new closed orbifold? This is an analogy with manifolds with boundary.

• It seems there is a clear strategy to prove this. Namely try to equivariantly glue two orbifold charts with boundary around every boundary point to give standard orbifold charts around each point. Does anything go wrong with that? Jun 26, 2022 at 13:53
• @NickL I think the OP was asking how to prove that the gluing of $2$ boundary orbifold points yields an interior orbifold point. In the manifold case, $2$ half Euclidean spaces are naturally glued to become a whole Euclidean spaces, but in the orbifold case we have to consider the change of group action. Jun 26, 2022 at 17:42
• @Zerox. I think that this issue is covered by the fact the orbifold boundaries are isomorphic. This means there is an equivariant homeomorphism of orbifold charts locally along the boundary, which are in turn compatible with the charts of each manifold with boundary. It is true the isotropy group may be a subgroup of isotropy group of the associated chart of the manifold with boundary, but that doesn't seem to create a problem. Jun 26, 2022 at 19:17
• @NickL can you provide more specific construction ? what is the finite group acting on open set of $R^n$ of the glued chart? Jun 27, 2022 at 1:11

Lets assume that the two charts about our boundary point are of the form $$[0,\infty) \times \mathbb{R}^{n-1}$$ with finite groups $$G_{i}$$ acting, each preserving $$\{0\} \times \mathbb{R}^{n-1}$$.
We may assume that $$G_{i}$$ acts effectively and linearly on each chart, respectively. This implies that the induced action on $$\{0\} \times \mathbb{R}^{n-1}$$ is effective (since the chart has boundary it is not possible to do a reflection in the boundary hyperplane).
By assumption the two orbifolds $$(\{0\} \times \mathbb{R}^{n-1},G_{i})$$ are isomorphic. That is we have $$G_{1} = G_{2}$$ (because both of the actions are effective) and that there is a homeomorphism of $$(\{0\} \times \mathbb{R}^{n-1})$$ respecting the $$G_i$$ actions. This allows us to glue the two half spaces equivariantly creating an orbifold chart $$(\mathbb{R}^{n}, G_{1} = G_{2})$$, as required.