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Nov 20, 2019 at 20:04 comment added Mohammad Farajzadeh-Tehrani Hi John. Can't we realize every orbifold as a quotient of its Frame bundle by O(n)? see the book of Ruan et al (Orbifold and string topology page 13)
Nov 12, 2019 at 3:15 comment added John Pardon Your statement that every orbifold is a global Lie group quotient is correct, however as far as I know the first proof is given here arxiv.org/abs/1906.05816 which did not exist at the time your question was posted. Do you know of another proof of this result?
May 10, 2015 at 18:11 comment added Mohammad Farajzadeh-Tehrani @ Dylan: en.wikipedia.org/wiki/Branched_covering
May 10, 2015 at 1:48 comment added Dylan Thurston Can you specify exactly what you mean by a "branched covering map"?
May 10, 2015 at 1:46 comment added Dylan Thurston @AriyanJavanpeykar, there's a distinction between "locally a quotient by a finite group" and "locally a quotient by a compact group with finite stabilizers". I don't know how to prove the statement in the hint, but I do know an example: You can easily find actions of $S^1$ on $S^3$ so that the quotient is naturally one of the inadmissible orbifolds $\mathbb{P}^1_{mn}$ mentioned in the problem statement.
May 8, 2015 at 18:58 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 3.0
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May 8, 2015 at 18:56 comment added Mohammad Farajzadeh-Tehrani @ Ariyan: Think this way, if X is a non simply-connected manifold, instead of orbifold, then such M is simply a finite covering of X.
May 5, 2015 at 6:09 comment added Ariyan Javanpeykar In the algebraic setting every smooth orbifold is a global quotient stack; see Thm 2.18 in arxiv.org/pdf/math/9905049v3.pdf
May 5, 2015 at 6:08 comment added Ariyan Javanpeykar I'm a bit confused. You write "Not every orbifold is a global quotient", but then later you write "every orbifold is a global quotient $M/G$.
Apr 20, 2015 at 15:11 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 3.0
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Apr 20, 2015 at 0:57 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 3.0
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Apr 20, 2015 at 0:49 history asked Mohammad Farajzadeh-Tehrani CC BY-SA 3.0