Question

This question is inspired by an answer of Tim Porter.

Ronnie Brown pioneered a framework for homotopy theory in which one may consider multiple basepoints. These ideas are accessibly presented in his book Topology and Groupoids. The idea of the fundamental groupoid, put forward as a multi-basepoint alternative to the fundamental group, is the highlight of the theory. The headline result seems to be that the van-Kampen Theorem looks more natural in the groupoid context.
I don't know whether I find this headline result compelling- the extra baggage of groupoids and pushouts makes me question whether the payoff is worth the effort, all the more so because I am a geometric topology person, rather than a homotopy theorist.

Do you have examples in geometric topology (3-manifolds, 4-manifolds, tangles, braids, knots and links...) where the concept of the fundamental groupoid has been useful, in the sense that it has led to new theorems or to substantially simplified treatment of known topics?

One place that I can imagine (but, for lack of evidence, only imagine) that fundamental groupoids might be useful (at least to simplify exposition) is in knot theory, where we're constantly switching between (at least) three different "natural" choices of basepoint- on the knot itself, on the boundary of a tubular neighbourhood, and in the knot complement. This change-of-basepoint adds a nasty bit of technical complexity which I have struggled with when writing papers. A recent proof (Proposition 8 of my paper with Kricker) which would have been a few lines if we hadn't had to worry about basepoints, became 3 pages. In another direction, what about fundamental groupoids of braids?
Have the ideas of fundamental groupoids been explored in geometric topological contexts? Conversely, if not, then why not?

Answer 1

Here is an interesting example where groupoids are useful. The mapping class group $\Gamma_{g,n}$ is the group of isotopy classes of orientation preserving diffeomorphisms of a surface of genus $g$ with $n$ distinct marked points (labelled 1 through n). The classifying space $B\Gamma_{g,n}$ is rational homology equivalent to the (coarse) moduli space $\mathcal{M}_{g,n}$ of complex curves of genus $g$ with $n$ marked points (and if you are willing to talk about the moduli orbifold or stack, then it is actually a homotopy equivalence)

The symmetric group $\Sigma_n$ acts on $\mathcal{M}_{g,n}$ by permuting the labels of the marked points.

Question: How do we describe the corresponding action of the symmetric group on the classifying space $B\Gamma_{g,n}$?

It is possible to see $\Sigma_n$ as acting by outer automorphisms on the mapping class group. I suppose that one could probably build an action on the classifying space directly from this, but here is a much nicer way to handle the problem.

The group $\Gamma_{g,n}$ can be identified with the orbifold fundamental group of the moduli space. Let's replace it with a fundamental groupoid. Fix a surface $S$ with $n$ distinguished points, and take the groupoid where objects are labellings of the distinguished points by 1 through n, and morphisms are isotopy classes of diffeomorphisms that respect the labellings (i.e., sending the point labelled $i$ in the first labelling to the point labelled $i$ in the second labelling).

Clearly this groupoid is equivalent to the original mapping class group, so its classifying space is homotopy equivalent. But now we have an honest action of the symmetric group by permuting the labels on the distinguished points of $S$.

Answer 2

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

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Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The isotropy group is the alternating group $A_n$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$A_n$ at point -1" with "$A_n$ at point +1" via any element of $S_n$ that sends -1 to +1. A choice of such an element yields an automorphism of $A_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(A_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

Answer 3

A short example.

The family of pure braid groups does not possess a symmetric operad structure.

But the fundamental groupoid of the little 2-discs operad is naturally a symmetric operad.

Although the fundamental groups of the little 2-discs operad are the pure braid groups, there is no way to choose basepoints consistent with the operad structure.

The moral is that groupoids are not naturally pointed, whilst groups are. If you're working with fundamental groups you should really be working with pointed spaces. Ofcourse you can ignore this and you'll only run into trouble if your mathematics doesn't work with pointed spaces, see the example above.

Answer 4

Steenrod defined a local coefficient system as a functor from the path-groupoid of your space to a category. It is tricky to define homology/cohomology with local coefficients by just picking base points since the identifications given by the paths are needed as well. Calculations of induced maps are especially prone to error. Perhaps my favorite example is the following.

Let $\tau\colon S^{2n} \to S^{2n}$ be the antipodal map with quotient $RP^{2n}$ and quotient map $\pi\colon S^{2n} \to RP^{2n}$. There is a twisted coefficient system $Z^w$ on $RP^{2n}$ so that $H_{2n}(RP^{2n};Z^w) \cong Z$ which is used in the non-orientable Poincare duality theorem.There is a natural notion of the pull-back coefficient system and the system $\pi^\ast Z^w$ is equivalent to the usual trivial system. Hence $H_{2n}(S^{2n};\pi^\ast Z^w) \cong Z$ but the equivalence involves a choice.

Since $\pi\circ \tau = \pi$, there is a ``natural'' identification of $(\pi\circ\tau)^\ast Z^w$ with $\pi^\ast Z^w$. With these choices, the antipodal map has degree 1. If you make your choices so that the antipodal map has degree $-1$, then the two maps induced by $\pi$ have different degrees.

Answer 5

I came into groupoids by trying to find a new proof of the fundamental group of the circle. It turned out that one could do this using the fundamental groupoid on two base points. Writing the 1968 edition of my book now called `Topology and Groupoids' (T&G) (available on amazon.com and e-version from www.kagi.com) convinced me that all of 1-dimensional homotopy theory was better expressed in terms of groupoids rather than groups, in that one obtained more powerful theorems with simpler proofs. Later results on the fundamental groupoid of orbit spaces (Chapter 11 of T&G) are more awkward to express in terms of groups; this elaborates on the point by Dustin Clausen.

Henry Whitehead answered the question of "Why not restrict to CW-complexes with just one vertex?" by considering covering spaces. Philip Higgins gave a considerable generalisation of Grusko's theorem by considering covering morphisms of groupoids, see his 1971 book `Categories and groupoids' available as a TAC Reprint, 2005.

In 1966 I thought about prospective uses of groupoids in higher homotopy theory, and this led over many years to higher dimensional Seifert-van Kampen Theorems, with a range of new nonabelian calculations of second relative homotopy groups and triad homotopy groups (for the latter, see the "nonabelian tensor product of groups"). That sounds relevant to geometric topology!

So one answer to the original question is that the use of groupoids opens new worlds of possibilities.

Actually the idea of `change of base point for the fundamental group' is a bit bizarre: one does not describe a railway timetable in terms of return journeys and change of starting point for these.

In the end, an aesthetic viewpoint implies more power!

Thanks to those above who give me additional examples.

More information on my page "http://www.bangor.ac.uk/r.brown/gpdsweb.html" From groups to groupoids.

Ronnie Brown

Answer 6

My knowledge of algebraic topology is limited, but I have found one particular example where groupoids are almost imposing themselves: a purely algebraic proof that every subgroup of a free group is free.

The usual geometric proof that a subgroup $H$ of the free group $F_2$ on two generators is free goes like this: one represents $F_2$ as the fundamental group $F_2 = \pi_1(S^1\vee S^1)$. Then, $H$ is the fundamental group of a covering of $p : X \to S^1\vee S^1$. But $X$ is a graph and the fundamental group of a graph is always free.

Of course, it should be possible to translate the above into a purely algebraic proof, but this is very difficult if you don't introduce groupoids or a related notion. I haven't done it in detail, but it appears to me that one has to treat the group $F_2$ as a groupoid "inside" the subgroup $H$. The objects of this groupoid are equivalence classes

$$ g \sim g' \iff \exists h \in H. gh = g' $$

and the morphisms are given by $\lbrace a : [g] \to [ga] \rbrace \cup \lbrace b : [g] \to [gb] \rbrace$ where $a$ and $b$ are the generators of $F_2$. This groupoid is a purely algebraic model of the covering space $X$. The group $H$ corresponds to the subgroupoid of morphisms of the form $\lbrace h : [e] \to [e] \rbrace$. It remains to be shown that $X$ can be contracted along a spanning tree while $H$ is left unchanged.

I think the point is that this last contraction rips apart the group structure of $F_2$.

Answer 7

Crisp and Paris's proof of Tits's conjecture that the subgroup of an Artin group generated by the squares of the generators is itself a right-angled Artin group uses groupoids in an essential way. For each small type Artin group, they construct a surface with boundary by gluing annuli along squares and mark each square with a base point, and then then study the action of the Artin group on the fundamental groupoid (with respect to a basepoints which correspond to the gluing squares) of the obvious graph which is a deformation retract of this surface. In this way, they construct a representation of such an Artin group into automorphisms of the fundamental groupoid of a graph. Here is the reference:

Crisp and Paris, ``The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group'', Invent. Math. 145 (2001).

Answer 8

If you study the set of ideal triangulations of a fixed punctured surface you find out that:

• ("generators") by using sequences of flips you can relate any pair of triangulations,
• ("relators") two such sequences are related by a well-understood set of moves (the most important one is the pentagon relation)

The object you get with these "generators" and "relators" is not really a group, it is just a groupoid, called the Ptolemy groupoid. See for instance the paper of Chekhov and Fock introducing quantum Teichmuller space.