# How should one understand orbifold fundamental groups?

I am studying orbifold fundamental group (or more generally orbifold homotopy groups). In a nutshell, my questions is: what are they intuitively? In what follows I give definitions and more precise questions. My definition of orbifold fundamental group is via classifying space of groupoid, which is explained in the next paragraph (so you may want to skip it if you know the definition).

Let $\mathcal{G}$ be a topological groupoid consisting a topological spaces $G_{0}$ of $objects$ and $G_{1}$ of $arrows$ together with usual continuous structure maps. Let $|\mathcal{G}|$ denote the associated topological space $G_{0}/G_{1}$. Let $G_{n}$ be the iterated fibered product $G_{n}=G_{1}\times_{s,t} G_{n-1}$. These $G_{n}$ have the structure of a simplicial topological space, called the $nerve$ of $\mathcal{G}$. Face operads $d_{i}:G_{n}\rightarrow G_{n-1}$ for $i=0,\dots,n$ are given by $$d_{i}(g_{1},\dots,g_{n})=(g_{1},\dots,g_{i}g_{i+1},\dots,g_{n})$$ for $i=1,\dots,n-1$ and $$d_{0}(g_{2},\dots,g_{n})=(g_{2},\dots,g_{n}), \ \ d_{n}(g_{1},\dots,g_{n})=(g_{2},\dots,g_{n-1}).$$ The classifying space $B\mathcal{G}$ of $\mathcal{G}$ is then defined as $$B\mathcal{G}=\bigsqcup_{n}(G_{n}\times \Delta^{n})((d_{i}(g),x)\sim(g,\delta_{i}(x))),$$ where $\Delta^{n}$ is the topological $n$-simplex and $\delta_{i}:\Delta^{n-1}\rightarrow \Delta^{n}$ is the standard facemap.\

The $n$-th orbifold homotopy group of $\mathcal{G}$ based at $x\in |\mathcal{G}|$ is defined to be $$\pi_{n}^{orb}(\mathcal{G},x)=\pi_{n}(B\mathcal{G},y),$$ where $y\in G_{0}$ maps to $x$ under the quotient map $G_{0}\rightarrow |\mathcal{G}|$.

The following are my questions:

1. Why is this a reasonable definition? Any manifold $M$ can be thought of topological groupoid via its chart i.e. $G_{0}=\bigsqcup_{i}U_{i}$ and $G_{1}=\bigsqcup_{i,j}U_{i}\times_{M} U_{j}$. It is not clear to me that the definition above reproduce $\pi_{n}(M)$.

2. I am aware of an explicit description the orbifold fundamental groups of the orbifold Riemann surface $\Sigma_{g,n,k}$ of genus $g$ and $n$ orbifold points $p_{i}$ of order $k_{i}$: $$\pi_{n}^{orb}(\Sigma_{g,n,k}) =\langle \alpha_{i},\beta_{i},\sigma_{j} \ (1\le i \le g,1\le j \le n)\ | \ \sigma_{1}\dots\sigma_{n}\prod_{i=1}^{g}[\alpha_{i},\beta_{i}]=1,,\sigma_{i}^{k_{i}}=1\rangle$$ Is it easy to see this explicit presentation by the definition above?

3. It seems there are several ways to define the fundamental group of an orbifold, such as covering space etc. How should one understand orbifold fundamental groups?

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I like the covering space approach. Locally constant sheaves on the corresponding etale groupoid will be the category of G-sets for a unique group (under semi-local continuity of the unit space) and so this group should be the fundamental group. – Benjamin Steinberg Aug 29 '12 at 20:24
("simplicial manifold" in paragraph 2 $\leadsto$ "simplicial topological space") – Theo Johnson-Freyd Aug 29 '12 at 20:37
+1 for writing up details! – Theo Johnson-Freyd Aug 29 '12 at 20:38
I organized the questions and corrected typos. – Michel Aug 30 '12 at 1:24
For a very nice discussion of 2-dimensional orbifolds, you should check out Peter Scott's article 'The geometries of 3-manifolds'. – HJRW Feb 19 '13 at 10:00

The definition of an orbifold in terms of a groupoid is flexible and technically useful and gives very clean definitions, but it's not so close to geometric intuition. Perhaps it is once you've mastered the art of thinking simplicially, which I probably haven't. I tend to think of orbifolds like this: the simplest orbifolds are the global quotients $[X/G]$. In this case the formalism has been cooked up so that geometry of $[X/G]$ is exactly the same as $G$-equivariant geometry on $X$ (whatever this means in a given context). Another catchphrase for this is that all group actions behave like free group actions. All other orbifolds can be obtained by gluing together ones of the above form. This leads to the definition in terms of coordinate patches of the form $[U/G]$.

The easiest example of an orbifold fundamental group is when your orbifold has the form $[X/G]$ with $X$ simply connected. Since every group action behaves like a free one, the map $X \to [X/G]$ is a covering map, which exhibits $X$ as the universal cover of the orbifold and $G$ as the group of deck transformations. So $\pi_1([X/G]) = G$.

The second simplest example is, I think, an (effective) orbifold Riemann surface. Fortunately you asked precisely about this one. Here you can really think concretely about loops on your surface. Fix an orbifold point $x$ of order $n$. The idea is that an orbifold point of order $n$ is $(1/n)$th of a point, so that an orbifold point is something inbetween an ordinary point and a puncture. The higher order stabilizer the point has, the closer it is to being an actual puncture. More concretely, what this means is that a loop on your surface that winds exactly $n$ times around $x$ can be shrunk across $x$. It's like the $n$ turns together add up to one whole point, which your loop can then slide across.

To see this slightly more formally, think in a chart centered on $x$, where your orbifold looks like $[D/\mu_n]$, with $D$ the unit disk and $\mu_n$ the group of $n$th roots of unity. Any loop in $D$ around the origin can be shrunk to a point, which should imply that the image of this loop in our orbifold is also homotopically trivial. But the image is just a loop that goes $n$ times around $x$. To see that no fewer than $n$ turns suffice, you need to think a bit about the definition of a loop on an orbifold. In any case, once we accept this fact we can obtain the presentation of the fundamental group that you gave in your question, in the same way as for the ordinary fundamental group of a punctured Riemann surface.

An example is $[S^2 / \mu_n]$ with $\mu_n$ acting by rotations along the equator. You have orbifold points at the north and south pole. Either of the two descriptions above immediately imply that the fundamental group is cyclic of order $n$: in terms of the second description, a generator is a simple loop around one of the poles, which becomes trivial when it is wound around itself $n$ times.

You might also find my recent question, and the answer by Jeffrey Giansiracusa, useful: Homotopy theory of topological stacks/orbifolds

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This is an extremely clear exposition, Dan! Now that orbifolds are not that scary and I even feel familiar with them. Having understood what orbifold fundamental groups should be, I now try to digest the definition and reproduce what you describe above by definition. Many thanks! – Michel Aug 30 '12 at 10:09
I'll throw in that this explanation extends as well to orbifolds with mirrors (boundary segments locally modelled on the quotient under a reflection in a line) and corner reflectors (locally modelled on finite dihedral group actions). A path that bounced off the mirror twice in a coordinate chart can be pulled off the orbifold locus. A path that bounces back and forth between the two mirrors adjacent to a $D_n$ corner reflector, with a total of $2n$ bounces, can also be pulled off the orbifold locus. Of course, these "bounces" must truly pass through the looking glass! – Lee Mosher Aug 30 '12 at 19:12

The short answer is, because I don't have time to write a full answer, is that the fundamental group is composed of equivalence classes of formal composites of paths and elements of the arrow space of the corresponding groupoid. If $p\to[0,1]$ is a cover by closed intervals only overlapping on boundaries (i.e. a partition), then such a formal composite is a functor $p\to G$. Geometric realisation makes this an element of your definition. Moerdijk and Mrcun (see e.g. this) do it this way, as does Hellen Colman (and so do I, see chapter 2). I gather the idea goes back to Haefliger.

Really orbifolds are objects of a bicategory (see Lerman's paper on this), and this bicategory can be described using anafunctors, and anafunctors from $[0,1]$ to an orbifold are equivalent to what I described above.

EDIT: To answer the second part of question 1, the geometric realisation of the groupoid associated to the open cover of a manifold is (at least) homeomorphic to the manifold, hence the fundamental group computed the two different ways are isomorphic. To see this via the description I give, one has to know that the 2-functor $\pi_1\colon OrbifoldGpd_* \to Grp$ factors through the bicategory of pointed orbifolds as constructed by e.g. Lerman (and many many others), i.e. sends Morita/weak equivalences to isomorphisms of groups. The open cover groupoid is weakly equivalent to the manifold thought of as a groupoid, and the description of the fundamental groupoid I give, in the case of a manifold (or space) is naturally isomorphic to the usual description.

To give a pointer for question 3, Moerdijk and Mrcun cover this in their chapter in this book, although possibly using sheaves instead of covering spaces.

I claim (in answer to the first part of question 1) that this (the definition given in the question) is a reasonable definition precisely because of its relation to the definition I give. I should probably discuss things like the homotopy type of topological stacks (see work by Noohi) and the relation between topological groupoids and topological stacks, various methods for computing with these and so on. But if the first paragraph is incomprehensible, then perhaps the long version will be as well :-)

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i downvote because you've answered precisely nothing of the OP's question. He did not ask "what is the most dense and incomprehensible interpretation of a orbifold and its homotopy groups"---which is what you've provided. – J. Martel Aug 30 '12 at 1:29
I explained why I gave a short answer, and I may expand on it. From this description one readily sees (in perhaps the sense that it takes 20 minutes thought) that $\pi_1$ of the groupoid associated to an open cover of a manifold is the $\pi_1$ of the manifold. It also gives a conceptual reason why the paths are defined as they are. And one can calculate with this description. – David Roberts Aug 30 '12 at 1:47
Although to be fair, one has to look up the references I gave to find the equivalence relation one puts on paths to get the fundamental group. – David Roberts Aug 30 '12 at 1:49
J. Martel, you may be right, but basically I appreciate any comments and any viewpoints. Thank you for the answer, David. – Michel Aug 30 '12 at 2:07
Second downvote! That's two for and two against. – David Roberts Aug 30 '12 at 2:26

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 for downloads from my homepage http://mathsci.kaist.ac.kr/~schoi/MSJbook2012.html). The aim of the book is to make an exposition of the elementary theory of orbifolds that an advanced undergraduate or MA students can understand.

The complete understanding of the homotopy theory of orbifolds should be fully understandable only from a categorical point of view. This is in the book "Orbifolds and stringy topology" by Adem et al. This is also a very nice book to read once one gets used to the elementary category theory here.

Please note that Haefliger pioneered the fundamental group theory of orbifolds from the point of view of Ehresmann on foliations. These objects are really very abstract but when viewed as leaf spaces we topologists become more comfortable. The main obstacle here is French, which is very desirable to learn in this field.

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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 $G_2 = \bigsqcup_{i,j,k} U_i \cap U_j \cap U_k$, and generally, $G_n$ is the disjoint union of the $(n+1)$-fold intersections of the $U_i$. To prove that $B\mathcal{G}$ has the same homotopy groups as $M$, we'll show in fact that these two spaces are homotopy equivalent.

First of all there is a clear map $B \mathcal{G} \to M$, that sends any point $(x,p) \in G_n \times \Delta^n$ to just $x$. This map is well defined since (1) $G_n$ is just a disjoint union of some open subsets of $M$, so any point of $G_n$ can be thought of as a point of $M$, and (2) all the face and degeneracy maps in $\mathcal{G}$ just perform bookkeeping: they change which multiple intersection of $U_i$'s a point is regarded as being in, but don't actually change the point.

Now, we'll define a section $M \to B \mathcal{G}$. For this, take a locally finite partition of unity $\phi_i$ subordinate to the cover $\{U_i\}_i$. Given $x \in M$, it'll belong to the support of finitely many $\phi_i$, say $\phi_{i_0}, \ldots, \phi_{i_k}$ and we can send $x$ to the equivalence class of $(x, (\phi_{i_0}(x), \ldots, \phi_{i_k}(x))) \in G_k \times \Delta^k$. Here, $x$ is thought of as belonging to the $U_{i_0} \cap \cdots \cap U_{i_k}$ term of $G_k$, and we regard $\Delta^k$ as the set of points $(t_0,\ldots,t_k) \in \mathbb{R}^{k+1}$ such that $t_i \ge 0$ and $t_0 + \cdots + t_k = 1$. It's straightforward to check that the definition above does give a function $M \to B\mathcal{G}$ (for example, adding extra $\phi_{i_j}$ with $\phi_{i_j}(x)=0$, does not change the image of $x$ in $B\mathcal{G}$), and it is clear this map is a section of the map $B\mathcal{G}\to X$ above.

Finally, the composite $B\mathcal{G} \to M \to B\mathcal{G}$ is homotopic to the identity by a straight line homotopy within the $\Delta^k$'s (that is, via homotopy that is the identity on the $G_k$ coordinate and moves in a straight line segment in the $\Delta^k$ coordinate).

This argument has probably been discovered many times and is well-known, I fairly recently found out that it does appear in print at least in Segal's Classifying Spaces and Spectral Sequences, proposition 4.1. It doesn't use that $M$ is a manifold, just that the open cover has a continuous subordinate partition of unity, so for example it works for any cover of a paracompact space $M$. Even without subordinate partitions of unity, for any open cover $U_i$ of any any space $M$ the homotopy groups of $M$ and $B \mathcal{G}$ are the same: these two spaces might not be homotopy equivalent in this general setting, but they are still weakly homotopy equivalent. See Dugger and Isaksen’s Hypercovers in Topology, theorem 2.1. This simplicial space $\mathcal{G}$ is called a Čech cover, and Dugger and Isaksen's paper also deals with the more general notion of a hypercover.

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I think in your last paragraph you mean "...with a cover $U_i$ possessing A subordinate partition of unity..." – Dan Petersen Apr 3 '13 at 7:24
No Dan, the argument I gave works whenever you have a continuous partition of unity subordinate to the cover (I.e.,for paracompact spaces M even if M is not a manifold) and gives you a homotopy equivalence. What I'm saying in the last paragraph is that without a subordinate partition of unity you don't get a homotopy equivalence anymore, but you still get a weak equivalence. – Omar Antolín-Camarena Apr 3 '13 at 10:45
I've reworded the last paragraph to make it clearer, since Dan Petersen's question made me see I wasn't being very clear. – Omar Antolín-Camarena Apr 3 '13 at 10:56
Oh, I understand now. Thanks! – Dan Petersen Apr 3 '13 at 12:11

A reference for the basics in the topology of orbifolds is http://kaist.academia.edu/SuhyoungChoi/Papers/236402/Geometric_Structures_on_Orbifolds_and_Holonomy_Representations

It is a Theorem of Thurston (Theorem 8 on Page 18 of the above notes) that every connected orbifold has a universal covering and that the orbifold fundamental group is the same as the group of deck transformations of the universal covering.

Of course this helps only for the fundamental group, not the higher homotopy groups....

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