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).
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3$\begingroup$ 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. $\endgroup$– HJRWCommented Jun 25, 2019 at 8:56
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$\begingroup$ @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." $\endgroup$– Neil HoffmanCommented Jun 25, 2019 at 12:57
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2$\begingroup$ @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. $\endgroup$– HJRWCommented Jun 25, 2019 at 13:07
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1$\begingroup$ (I take the OP's term "ordinary orbifold" to mean "orbifold without boundary".) $\endgroup$– HJRWCommented Jun 25, 2019 at 13:09
1 Answer
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 $M$.
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2$\begingroup$ Orbifold atlases are quite messy (and many people have gotten muddled in the details, and thus ended up writing slightly wrong definitions). I recommend modelling orbifolds as Lie groupoids: that's a much safer way of developing the theory. Once you understand how Lie groupoids work, it's an easy step to go from there to groupoid objects in the category of manifolds with corners. $\endgroup$ Commented Jun 30, 2019 at 23:05
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1$\begingroup$ @AndréHenriques: OP asked to give a definition in terms of charts, which is what I am doing. As for which definition is the best, I think it is all in the eye of the beholder... $\endgroup$– MishaCommented Jul 1, 2019 at 1:35