34

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 manifold of Euler characteristic $m \in \mathbb{Z}$ to get all rationals.

27

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 integers, and $\chi(M)$ is the characteristic of the surface
of the orbifold. It immediately follows that $\chi(O)\leq 2$ for all 2D orbiforlds, and as $1-1/q\geq 1/...

answered Apr 12 '17 at 23:33

Alexandre Eremenko

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I am not sure what kind of answer you are looking for. But if you have a stack $X$, then the inertia stack $IX$ is basically the gadget parametrizing pairs $(x,\sigma)$ where $x$ is a point of $X$ and $\sigma$ is in the isotropy group at $x$ (an automorphism of $x$). Informally, the locus where you have automorphism group $G$ becomes "doubled" $|G|$ times. ...

16

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 simplicity, Inst. Hautes Etudes Sci. Publ. Math. (1961), no. 9, 5 -
22
that a normal algebraic surface over $\mathbb{C}$, whose underlying topological space (in the ...

answered Mar 7 '16 at 22:25

Francesco Polizzi

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The integral cohomology ring of orbifolds does not satisfy Poincaré duality. Indeed, orbifolds do not even have finite cohomological dimension.

answered Jul 10 '13 at 19:59

André Henriques

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I'm not an expert and this might be wrong, but I think that Cerf theory should be impossible for orbifolds, and therefore all that comes from it, e.g. Kirby Calculus. Could somebody who knows please confirm this? I'm guessing that this comes not from orbifolds having singularities, but rather from orbifolds having built-in symmetries which mess up the ...

answered Jul 10 '13 at 8:54

Daniel Moskovich

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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 $n$ pairs $(i,j)$, and we have $\langle v_i, v_j \rangle \leq 0$ for all $i \neq j$.
But then
$$\left\langle \sum_{i=1}^n v_i, \sum_{i=1}^n v_i \right\rangle \leq ...

answered Aug 26 '14 at 19:33

David E Speyer

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11

A note before: the communities of low-dimensional topologists and categorically minded people have different opinions on what an
'orbifold' is. Let $G$ be a discrete group that acts properly and smoothly on the manifold $M$. As far as I understand,
low-dimensional topologists consider the quotient space $M/G$, together with the cleverly packaged ...

answered Jul 10 '13 at 18:47

Johannes Ebert

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11

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 $\tilde M_g$(and particular complex manifold) by a finite group $G$
Idea: Take $\Gamma(3)\subset \Gamma$ to be the preimage of congruence group $\{M\in Sp_{2g}(\...

10

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 has some nice slides on orbifolds.
Here is a link to Thurston's notes on the topic.
Here is a link to a book by Adem, Leida, Ruan which discusses, among many ...

10

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 they should be is not what they actually are.
To me, an orbifold should be to a manifold as a Deligne--Mumford stack is to a scheme. In particular, they should ...

9

Hi. Actually, a book by Bridson and Haefliger has much details: "Metric spaces of nonpositive curvature" published in Springer Verlag. Also, see my book MSJ memoirs Vol 27: "Geometric Structures on 2-Orbifolds: Exploration of Discrete Geometry" (http://www.academia.edu/2650286/Geometric_structures_on_2-orbifolds_exploration_of_discrete_symmetry or
available ...

9

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 points is the mapping class group $Mod_g$ of the surface $\Sigma_g$. To see this, recall that $\mathcal{M}_g$ is the quotient of Teichmuller space $\mathcal{T}_g$ ...

8

The orientable Euclidean orbifolds were described in section 7 of this paper of Dunbar.
To complete the list, note that every crystallographic group has an index 2 orientation-preserving subgroup. The corresponding orientable orbifold quotients will be realized as 2-fold orbifold covers of the non-orientable orbifolds. So one needs to look over Dunbar's ...

8

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 topological manifold if and only if $G$ is generated by pseudoreflections. Up to using the positive direction (and much harder direction) of Chevalley-Shephard-...

8

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 using a double complex associated to the underlying Lie groupoid.
These two cohomologies are isomorphic (see remark 10 p. 262).
Singular cohomology is also ...

8

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 with $n$-orbifolds over the complex numbers $\mathbb{C}$ then each element of each isotropy group is a holomorphic transformation, and every holomorphic ...

7

There are 10 orbifolds that can be covered by the torus i.e. 10 compact Euclidean 2 orbifolds. However, only 7 of them are quotients of the torus by a cyclic group or Abelian product of cyclic groups. The intuition that Euler characteristics are zero is correct. Formulas for the orbifold Euler characteristic appear throughout the literature. I like Chapter ...

7

Yes. One can extend the Gauss-Bonnet theorem to 2-orbifolds since but area an orbifold Euler characteristic behave well under the covering map.
To find all hyperbolic 2-orbifolds with area smaller than $\pi/2$, we need to get a list of orbifolds $\{O_i\}$ $0>\chi(O_i)>-1/4$.
Thankfully, this is not a horrible task. Background and pictures for this ...

7

The answer to question 1, which asks whether this definition recovers homotopy groups of a manifold, is yes. As in the question, we'll take $M$ to be manifold, $G_0 = \bigsqcup_i U_i$ the disjoint union of the sets of an open cover of $M$ and $G_1 = G_0 \times_M G_0 = \bigsqcup_{i,j} U_i \cap U_j$. It is easy to see that $G_2 = \bigsqcup_{i,j,k} U_i \cap U_j ...

answered Apr 3 '13 at 4:06

Omar Antolín-Camarena

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7

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 $\nabla_{c'} c'=0$. For an $n$-dimensional Riemannian orbifold $X$ this equation makes sense not on $X$ itself but in its local coordinate orbifold charts $U$, ...

6

Here are a couple papers on oriented orbifold cobordism. The first gives rational invariants and generators and also shows that rationally odd dimensional orbifolds bound.
The second paper develops machinery for handling the torsion (all dimensions) and applies that to show that every oriented three-orbifold bounds.
K.S. Druschel. Oriented Orbifold ...

6

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 $\chi \in X$ has a neighborhood $U$ such that any element of $G$ not in the stabilizer of $\chi$ maps $U$ outside itself. Theorem: Let $G$ be a discontinuous group of ...

answered Aug 7 '15 at 22:42

Igor Rivin

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6

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 $RP^{d-1}$. For $d=2$, $RP^1=S^1$ so the cone has underlying space a disk, however for $d>2$, the underlying cone will not be homeomorphic a $d$-ball and so it is not ...

6

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. But the universal cover is a product of two copies of the complex unit disk, so the result follows by the maximum principle.
Nevertheless I don't think that this ...

5

As far as (co)homology is concerned, you may replace the orbifold quotient of a manifold $M$ by a finite group $G$ by the so-called Borel construction $(M\times EG)/G$. The "tangent bundle" of the orbifold is then a usual vector bundle over the space $(M\times EG)/G$, and you can define all the usual characteristic classes.
I should say:
- $EG$ is a ...

answered Jul 15 '13 at 21:15

André Henriques

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5

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 connection between orbifolds and equivariant homotopy theory having been deeply explored, but maybe it has (and maybe somebody will chime in on this question to ...

5

Indeed, you always get a fibration $M\to \hat B$, where $\hat B$ is the Haefliger's classifying space of the orbifold (the space whose cohomology is the orbifold cohomology of $B$). The fiber of this fibration is the principal leaf of your Seifert fibration.
One can write the corresponding spectral sequence. And indeed, if the orbifold is not a manifold ...

5

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 $\mathbb{R}^N$ and such that the local groups are all sitting in $O(N)$. Then each $U_i$ and $U_i/G_i$ are contractible, and so all of the terms in your Čech ...

5

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 orbifolds arise in "applications" (= mathematics):
The quotient spaces appearing in symplectic reduction are not always manifolds. If they fail to be manifolds,...

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