34 votes
Accepted

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 ...
  • 62.2k
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 ...
16 votes
Accepted

What is an example of an orbifold which is not a topological manifold?

It is quite easy to give an example in real dimension $4$. In fact, it was shown by D. Mumford in the paper The topology of normal singularities of an algebraic surface and a criterion for ...
13 votes
Accepted

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 ...
12 votes

why most of the angles are right

Let the root system be $v_1$, …, $v_n$ with all elements normalized to be length $1$. So $\langle v_i, v_i \rangle =1$, we have $\langle v_i, v_j \rangle \leq - \cos (\pi/3) = -1/2$ for at least $n$ ...
12 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 $\...
  • 32.1k
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 ...
  • 8,094
9 votes
Accepted

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 ...
  • 40.4k
9 votes
Accepted

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 ...
  • 14.7k
8 votes
Accepted

Is there a topological Chevalley-Shephard-Todd Theorem?

Since the bounty is now ended, I will post my comment above as an answer. When the dimension $n$ equals $2$, the quotient of $\mathbb{C}^n$ by a finite group $G$ of holomorphic automorphisms is a ...
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 ...
  • 9,500
8 votes
Accepted

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 ...
7 votes
Accepted

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 ...
  • 30.2k
6 votes

Conditions for underlying space of an orbifold $\Bbb T^n/\Gamma$ to be a sphere?

Here is a theorem due to Armstrong: A group $G$ of homeomorphisms of a topological space $X $is called discontinuous if (1) the stabilizer of each point of $X$ is finite, and (2) each point $\chi \...
6 votes
Accepted

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 $RP^{d-1}$....
  • 5,121
6 votes
Accepted

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. ...
  • 37.5k
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 ...
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 $\...
  • 6,132
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 ...
  • 51.1k
5 votes
Accepted

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 ...
5 votes
Accepted

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 ...
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 ...
  • 30.2k
5 votes
Accepted

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 ...
  • 30.2k
5 votes
Accepted

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 ...
4 votes

Orbifold fundamental group in terms of loops?

Here are two more references where the fundamental group of an orbifold is defined in terms of loops: The chapter by Haefliger on "orbi-espaces" in this book: E. Ghys et P. de la Harpe. Sur les ...
4 votes

Conditions for underlying space of an orbifold $\Bbb T^n/\Gamma$ to be a sphere?

Let me clarify my comment. $S_n$ acts on $T^n$ in the obvious way, and preserves the multiplication map $T^n \to T$, so acts on the fiber $T^{n-1}$ in a natural way. I think the quotient by this ...
  • 119k
4 votes
Accepted

Is there a Riemann existence theorem for orbifolds?

The answer is yes. See Theorem 20.4 on p. 80 in Behrang Noohi's http://arxiv.org/pdf/math/0503247v1.pdf
4 votes

Conditions for underlying space of an orbifold $\Bbb T^n/\Gamma$ to be a sphere?

I am just expanding abx's comment: it is possible to realize this branched cover as a group quotient. If abx prefers to explain this, I am happy to delete this answer. Denote by $\mu_2$ the $2$-...
4 votes
Accepted

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.
  • 6,508
4 votes
Accepted

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 $...
  • 1,587

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