every smooth manifold can be triangulated, is it true for orbifold? Is it a known result? If yes, is there any reference?
reply to the comment : G does not need to be any subgroup of Sn , any finite group is fine
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2$\begingroup$ By orbisimplex, do you mean some spaces like $\Delta^n/G$, where $G\subset S_{n+1}$ is a subgroup? Then how do you realise orbifold singularities in dimension $n$ whose isotropy group is too large to fit into $S_{n+1}$? $\endgroup$– Sebastian GoetteCommented Mar 8, 2017 at 15:10
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$\begingroup$ @Sebastian Goette, I just want a similar result with "smooth manifold has a triangulation" for orbifold setting. I dont know what the right definition of orbi simplex is, (or if there is one in the literature), perhaps G doesn't have to act on a simplex globally or maybe orbihedron fits better. $\endgroup$– haoyuCommented Mar 8, 2017 at 23:21
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$\begingroup$ @SebastianGoette, a good example to think about is the standard triangulation (with two 2-simplices) of the sphere with three cone points. $\endgroup$– HJRWCommented Mar 9, 2017 at 15:08
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$\begingroup$ @HJRW In that case I would suspect that one can triangulate by ordinary simplices such that each closed stratum is a subcomplex (but I know neither a proof nor a reference for that). I thought the OP wanted to have a particular structure at the orbifold points (hence, orbisimplex), maybe such that simplices are transversal to the strata. $\endgroup$– Sebastian GoetteCommented Mar 9, 2017 at 16:16
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You have to be a bit more specific about the meaning of a "triangulation" for orbifolds. Assuming that you just want to triangulate the underlying space, the claim follows from
C. T. Yang, "The triangulability of the orbit space of a differentiable transformation group", Bull. of Amer. Math. Soc. 69 (1963), 405-408.
In order to use Yang's result, note that each smooth $n$-dimensional orbifold $O$ is the quotient of a smooth manifold $FO$ by the smooth action of $O(n)$, where $FO$ is the orthonormal frame bundle of $O$ (equipped with a Riemannian metric). If you just want a reference, you can quote Proposition 1.2.1 in
I. Moerdijk and D.A. Pronk, "Simplicial Cohomology of Orbifolds" Indagationes Mathematicae, Vol. 10, Issue 2 (1999) 269-293.
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$\begingroup$ @haoyu: I do not understand what you are asking. $\endgroup$– MishaCommented Mar 12, 2017 at 3:22
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$\begingroup$ ,I mean the interior of a triangulable space must be a topological manifold, right? (by the definition of triangulation), but if an orbifold can be triangulated by ordinary simplex, is the interior of any orbifold always a topological manifold? $\endgroup$– haoyuCommented Mar 12, 2017 at 4:20
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$\begingroup$ i feel orbifold should be triangulated by orbi simplex, but the paper proves that ordinary simplex is enough. surprised $\endgroup$– haoyuCommented Mar 12, 2017 at 4:33
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$\begingroup$ @haoyu: What is the "interior of a triangulable space"? What do you mean by "interior of an orbifold"? $\endgroup$– MishaCommented Mar 12, 2017 at 15:37
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1$\begingroup$ No, a triangulable space need not be a manifold with corners (just take a look at the 1-dimensional case). $\endgroup$– MishaCommented Mar 14, 2017 at 15:38