# What are orbifolds with corners?

What is the geometric definition of orbifolds with corners? Here “geometric" means that there is a definition in chapter 8 of the draft of Dominic Joyce's book D-manifolds and d-orbifolds: a theory of derived differential geometry (book website, direct pdf link), but that is too technical and abstract, and so I am trying to find a definition parallel to the definition of orbifolds in the normal way (locally it is an open set acted by a smooth group, etc.) or more intuitive than the definition in that book. If it is defined as an open set with boundary (its closure) acted by a group keeping the boundary invariant seems does not work, as the boundary point of a manifold with boundary is itself a point in an orbifold (a manifold with boundary is an ordinary orbifold).

• I haven't looked closely at Joyce's definition, but the statement "a manifold with boundary is an ordinary orbifold" does not agree with the definition of orbifold I usually use (cf. Scott's BLMS article "The geometries of 3-manifolds"). It is true that an orbifold may have a manifold with boundary as its underlying space. But, without specifying additional (necessarily non-trivial) isotropy subgroups, a manifold with boundary is not an orbifold. – HJRW Jun 25 '19 at 8:56
• @HJRW I often find it convenient to discuss the set of all (3-)orbifolds which for me includes (3-)manifolds and (3-)orbifolds, but I understand your convention that if something is an manifold, you would have to have a good reason for calling it a "orbifold with trivial isotropy groups." – Neil Hoffman Jun 25 '19 at 12:57
• @NeilHoffman -- I think our conventions agree: a manifold is the same thing as an orbifold with trivial isotropy groups; in particular, all manifolds are orbifolds. But I'm trying to make a different point. There are orbifolds without boundary whose underlying spaces are manifolds with boundary (the easiest example is the quotient of the circle by a reflection) but, conversely, a manifold with boundary and trivial isotropy groups is not an orbifold without boundary. I think this contradicts the OP's final assertion in parentheses. – HJRW Jun 25 '19 at 13:07
• (I take the OP's term "ordinary orbifold" to mean "orbifold without boundary".) – HJRW Jun 25 '19 at 13:09

Orbifolds with corners are defined by the same axioms as manifolds with corners and ordinary orbifolds: A topological $$n$$-dimensional orbifold with corners is a topological space $$X$$ (2nd countable and Hausdorff) equipped with a (maximal) "orbifold atlas" $$\{U_i \ldots : i\in I\}$$ consisting of open subsets $$U_i\subset X$$, open subsets $$V_i\subset [0,\infty)^n$$ and finite affine groups $$\Gamma_i$$ preserving $$V_i$$'s, together with homeomorphisms $$\phi_i: V_i/\Gamma_i\to U_i$$, satisfying a long list of compatibility conditions which are identical to the ones for ordinary orbifolds with one important addition: Gluing maps $$\psi_{ij}: V_i\to V_j$$ preserve the boundary stratifications of $$V_i$$ and $$V_j$$ given by the product structure of $$[0,\infty)^n$$ (just as in the case of manifolds with corners).
As a special case, a good orbifold with corners is the quotient $$M/\Gamma$$ of a topological manifold with corners $$M$$ by a properly discontinuous group action $$\Gamma\times M\to M$$, which is locally linearizable and is by automorphisms of the manifold with corners $$M$$.