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34 votes
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Is every rational realized as the Euler characteristic of some manifold or orbifold?

Products of 2-orbifolds with manifolds will do the trick. There are 2-orbifolds of Euler characteristic $1/n$ (take a quotient of $S^2$ by a rotation of order $2n$). Then take a product with a ...
Ian Agol's user avatar
  • 68.8k
27 votes

Is every rational realized as the Euler characteristic of some manifold or orbifold?

The answer for connected 2-dimensional orbifolds is no. Euler characteristic is $$\chi(O)=\chi(M)-\sum\left(1-\frac{1}{q}\right)-\frac{1}{2}\sum\left(1-\frac{1}{p}\right),$$ where $p,q\geq 2$ are ...
Alexandre Eremenko's user avatar
15 votes
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Is there a contractible hyperbolic 3-orbifold of finite volume?

Yes. For example, let $M$ be the figure-eight knot complement. So $M$ is a hyperbolic manifold with volume a bit more than 2. The manifold $M$ has a two-fold symmetry $\tau$ that fixes, pointwise, a ...
Sam Nead's user avatar
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14 votes
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Condensed / pyknotic approach to orbifolds?

Great question! I think the answer ought to be yes. But I must also make a disclaimer that I don't really know what orbifolds are, and the comments by David Roberts make be believe that what I thought ...
Peter Scholze's user avatar
13 votes

What is the official definition of $\mathcal{M}_g$ as an orbifold, and how much can I ignore it?

Since you seem to be mainly interested in $\mathcal{M_g}$, let me suggest a "quick and dirty" approach based on the following fact: $\mathcal{M}_g$ is quotient of a nonsingular algebraic variety $\...
Donu Arapura's user avatar
  • 35.2k
10 votes

What is the official definition of $\mathcal{M}_g$ as an orbifold, and how much can I ignore it?

Since in the comments you seem drawn to the "charts" approach to orbifolds, here are some resources which give definitions and more (there is no "official definition" as far as I know): Joan Porti ...
Sean Lawton's user avatar
  • 8,529
10 votes
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Do regular points of an orbifold form a connected set?

Part of the reason that people consider orbifolds whose singular points have codimension $\ge 2$ is because they are restricting their attention to oriented orbifolds. For example, if you are working ...
Lee Mosher's user avatar
  • 15.4k
9 votes
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Homotopy groups of smooth part of moduli space

I'm not sure why you're talking about codimension $1$ components of $S$ since the smooth locus $S$ is actually dense in $\mathcal{M}_g$. Anyway, the fundamental group of the locus $S$ of smooth ...
Andy Putman's user avatar
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9 votes
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When is a compact orbifold Riemann surface a global quotient of a Riemann surface

Let $M$ be a your compact orbifold Riemann surface and let $S\subset M$ be the finite set of orbifold singularities. Theorem: The following are equivalent: $M$ is the quotient of a closed Riemann ...
Nicolast's user avatar
  • 1,908
9 votes
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Suppose that $G$ is a finite subgroup of ${\rm SO}(3)$. Is there a *smooth* self-map of ${\bf R}^3$ whose fibers are precisely the orbits of $G$?

Here is a standard trick: Let $f: S^{n-1}\to S^{n-1}$ be a smooth map. We would like to extend it "radially" to a smooth map $F: \mathbb R^{n}\to \mathbb R^{n}$ such that if $f$ is $\rho$-...
Moishe Kohan's user avatar
  • 12.2k
8 votes

Is the Čech cohomology of an orbifold isomorphic to its singular cohomology?

As a reference I recommend this paper by K. Behrend: http://users.ictp.it/~pub_off/lectures/lns019/Behrend/Behrend.pdf Cech cohomology and De Rham cohomology can be defined for differentiable stacks ...
David C's user avatar
  • 9,870
7 votes
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Geodesic representatives in the orbifold fundamental group

First of all, I would use a different name for what you call geodesics on $X$, let's call them regular geodesics. The reason is that geodesics in Riemannian geometry are curves satisfying the equation ...
Misha's user avatar
  • 31.2k
7 votes
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Smooth rank one foliations with closed leaves

The result is not true without additional assumptions. See A counterexample to the periodic orbit conjecture by Sullivan. Added later. The paper by Sullivan linked above exhibits a foliation by ...
Jorge Vitório Pereira's user avatar
6 votes

Why study orbifolds?

I would like to propose an answer to this question, since 15 year ago I was asking it to myself and was thinking that orbiolds are useless. I read your question (maybe wrongly) as a question in ...
6 votes
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Does the torus $T^d$ 2-fold cover an orbifold $Q^d$ with underlying space $S^d$?

Given the specific quotient described in the question, the answer is no. Under this quotient there is a fixed point at the origin and the neighborhood of this point quotients to a cone over $RP^{d-1}$....
Neil Hoffman's user avatar
  • 5,259
6 votes
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Maps to the universal punctured elliptic curve

It is indeed true that a holomorphic map $\mathbb C \to \mathcal E'$ is constant. This is because it must factor through the universal cover of $\mathcal E'$, since $\mathbb C$ is simply connected. ...
Dan Petersen's user avatar
  • 40.2k
6 votes

Is there a contractible hyperbolic 3-orbifold of finite volume?

As pointed out by Moishe Kohan in the comments below, the following doesn't answer the question as asked, because my group $\Gamma$ is not contained in $SO(3,1)$. Anyway, here is an easy description ...
IJL's user avatar
  • 3,451
5 votes

Is the Čech cohomology of an orbifold isomorphic to its singular cohomology?

David C's answer gets this nicely, but another way to intuitively think of it is the following. Choose a fine enough open cover $\{U_i\}_i$ of $\mathcal{O}$ such that each $U_i$ is isomorphic to $\...
Simon Rose's user avatar
  • 6,290
5 votes

Stable homotopy theory of orbifolds

Since Mike asked this question almost 10 years ago, Schwede's work on global equivariant homotopy theory, alluded to by Tyler in the comments, has become a classic. I'm personally not aware of the ...
Tim Campion's user avatar
  • 63.9k
5 votes
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Does geometrization of Alexandrov 3-spaces follow from that of 3-orbifolds?

In principle, yes. Note, however, that topologically singular Alexandrov 3-spaces are homeomorphic to non-orientable orbifolds. We could not find an appropriate reference for the geometrization in the ...
Fernando Galaz-García's user avatar
5 votes
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Is a free and discrete group action on the plane a covering space action?

There are examples of free actions on $\mathbb{R}^2$ where every orbit is discrete and closed but the action is not properly discontinuous and the quotient is non-Hausdorff. The example is rather ...
Misha's user avatar
  • 31.2k
5 votes

What are orbifolds with corners?

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 ...
Misha's user avatar
  • 31.2k
5 votes
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Why study orbifolds?

I can not say why one studies orbifolds (or e.g. why one studies math at all). However, I can try the approach which might convince your funding agency: There are tons of interesting examples of how ...
Alexander Schmeding's user avatar
5 votes

Smooth rank one foliations with closed leaves

The question was already answered by Jorge Vitório Pereira, but let me add here what I have already found. Recall that a foliation on a Riemannian manifold is called "Riemannian foliation" ...
Misha Verbitsky's user avatar
5 votes
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Non compact Seifert manifolds

This is copied, in part, from the comments. So I've made it community wiki. Please feel free to edit and improve. The consensus in the comments is that "the classification in the non-compact ...
4 votes
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Can we perturb a surface away from an orbifold point?

The answer to the question is no. In general the intersection $[f(S)] \cdot [f(S)]$ can lie in $\mathbb{Q} \setminus \mathbb{Z}$, then of course in this case there is no such two cycle.
Nick L's user avatar
  • 6,995
4 votes
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Absolute and relative tilings of the hyperbolic plane

Let $T$ be the pictured tiling, where we forget about the colouring. Then its symmetry group is the maximal discrete subgroup of isometries of $\mathbb{H}^2$ which fixes $T$. Call that symmetry group $...
Josh Howie's user avatar
  • 1,617
4 votes
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Almost free Lie group action

I think the answer to the first question is yes and the answer to the second one is no: Yes, the quotient is an orbifold. The action of the finite group $G_x$ in a neighbourhood of $x$ can be ...
Nicolast's user avatar
  • 1,908
4 votes
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In what sense is Bass-Serre theory the one-dimensional version of orbifold theory

A graph is a $1$-dimensional manifold with singularities and a graph of groups is, roughly, a $1$-dimensional orbifold with singularities. Every graph of groups has a Bass-Serre covering tree which is ...
Qiaochu Yuan's user avatar
4 votes

References for orbifold curves

The answer to your question is that there is no good reference. I am finishing up my PhD thesis on moduli of sheaves on stacky curves. If you want I can send you some drafts, that will hopefully be ...
Lisanne's user avatar
  • 41

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