I am giving a lecture on Braid Groups this month at a seminar and I am confused about how to understand the fundamental group of the configuration space of $n$ points, so I will define some terminology which I will be referring to and then ask a few questions.

The configuration space of $n$ points on the complex plane $\mathbb{C}$ is: $$C_{o,\hat{n}}(\mathbb{C})=\{(z_{1},...,z_{n}) \in \mathbb{C} \times \cdot \cdot \cdot \times \mathbb{C} \text{ | } z_{i} \neq z_{j} \text{ if } i \neq j\}$$

A point on $C_{o,\hat{n}}$ is denoted by a vector $\vec{z} = (z_{1},...,z_{n})$. The group action $S_{n} \times C_{o,\hat{n}}(\mathbb{C}) \rightarrow C_{o,\hat{n}}(\mathbb{C})$ has the property that $\forall \vec{z} \in C_{o,\hat{n}}(\mathbb{C})$, $\sigma\vec{z}=\vec{z}$ implies $\sigma = e$, where $\sigma \in S_{n}$ and $e$ is the identity permutation. Thus, the symmetric group acts freely on $C_{o,\hat{n}}$, permuting the coordinates in each $\vec{z} \in C_{o,\hat{n}}$. According to the notes I am reading, the orbit space $S_{n}(\vec{z})=\{\sigma\vec{z} \text{ | } \sigma \in S_{n}\}$ with $\vec{z} \in C_{o,\hat{n}}(\mathbb{C})$ is equivalent to the configuration space modulo the permutation group. That is,

$$C_{o,n} = S_{n}(\vec{z}) = C_{o,\hat{n}}(\mathbb{C})\Big/\Sigma_{n}$$

is the orbit space of the group action and we let $\tau: C_{o,\hat{n}} \rightarrow C_{o,n}$ be the orbit space projection. Then, letting $\vec{p}=(p_{1},...,p_{n})$ be a fixed base point, we define the pure braid group $\textbf{P}_{n}$ on $n$ strands and the braid group $\textbf{B}_{n}$ on $n$ strands to be the following fundamental groups, respectively.

$$\textbf{P}_{n} = \pi_{1}(C_{o,\hat{n}},\vec{p})$$ $$\textbf{B}_{n} = \pi_{1}(C_{o,n},\tau(\vec{p}))$$

EDIT: I have been able to answer some of my own questions recently by further thinking and research but I have one remaining question.

We place a base point $\vec{p} = (p_{1},...,p_{n})$ in our configuration space $C_{o,\hat{n}}$ and then we want to consider loops in configuration space. I can visualize how if we place our base point as a configuration of collinear points that we obtain a braid on $n$ strands, but I am not sure why there is a non-trivial loop. My understanding of homotopy is that if there is no "hole" in the space, then we can continuously retract our loop in $C_{o,\hat{n}}$ back to our base point $\vec{p}$. Why can we not do this in this case?

  • $\begingroup$ $C$ is the complement of a hyperplane arrangement. Why do you think that is has no holes? For example for $n=2$ the space $C$ is (expressed in specific coordinates) $\left(\mathbb{C} \setminus \{0\}\right) \times \mathbb{C}$. $\endgroup$
    – WimC
    Jul 19, 2012 at 6:33

1 Answer 1


Sorry, but my first answer was completely and utterly wrong. Since this question is one of the top search results for the query "braid group fundamental group configuration space" I think it's high time I've updated with a correct explanation! :-)

I am not sure why there is a non-trivial loop. My understanding of homotopy is that if there is no "hole" in the space, then we can continuously retract our loop back to our base point. Why can we not do this in this case?

Short answer. You're thinking of $B_n(\Bbb R)$ when you should be thinking of $B_n(\Bbb C)$.

Long answer. Let $X$ be a "nice" topological space (say, a manifold). Define $F_n(X)$ to be the subspace of $X^n$ comprised of tuples with distinct coordinates. The symmetric group $S_n$ acts on it freely, and we can form the $n$-configuration space as the quotient $SF_n(X):=F_n(X)/S_n$. Then we define the braid group as $B_n(X)=\pi_1(SF_n(X))$. (Of course, $SF_n(X)$ should be path-connected...)

If you take $X=\Bbb R$ then the connected components of $F_n(X)$ are blocks for the action of $S_n$. Given any two tuples $(x_1,\cdots,x_n)$ and $(y_1,\cdots,y_n)$ with $x_1<x_2<\cdots<x_n$ and $y_1<y_2<\cdots<y_n$, these two tuples will be path-connected: first shift all coordinates of $\vec{y}$ uniformly enough to the right so that $x_n<y_1$, then shift $y_1$ back until it's $x_1$, then shift $y_2$ back until it's $y_2$, and so on. The space of all tuples $(x_1,\cdots,x_n)$ with increasing coordinates is homeomorphic to $\Bbb R^n$ which is simply connected. Similarly for any other tuples whose coordinates are "ranked" in a given order.

However $(x_1,x_2,x_3,\cdots,x_n)$ will not be path-connected to $(x_2,x_1,x_3,\cdots,x_n)$ within $F_n(X)$. The difference between the first two coordinates would need to change from positive to negative, and hence by IVT must be zero at some point. In general, a path in $F_n(X)$ cannot change the "rank order" of the coordinates of a tuple. So there are no paths between points in any $S_n$-orbit in $F_n(X)$. Therefore, any based loop in $SF_n(X)$ when lifted back to $F_n(X)$ must also be a loop, hence must be nullhomotopic since $F_n(X)$'s connected components are simply connected, so $SF_n(X)$ is simply connected, so the braid group $B_n(\Bbb R)=\pi_1(SF_n(\Bbb R))$ is trivial.

Now consider $X=\Bbb C$ with $n=2$. We must delete the subspace $\{(z,z):z\in\Bbb C\}$ from $\Bbb C^2$. (Keep in mind for now that $C_2$ acts on the carved-out space by transposing coordinates.) This subspace is a plane inside Euclidean $4$-space, so it homeomorphic to $\Bbb R\times (\Bbb R^3- L)$ for a line $L\subset\Bbb R^3$. Better yet, consider the obvious Euclidean structure on the space and take the orthogonal complement $\{(z,-z):z\in\Bbb C\}$: there is an orthogonal projector given by $(z,w)\mapsto(z-w,w-z)/\sqrt{2}$ and then an isomorphism into the punctured plane given by $(u,-u)\mapsto u$. Thus, we have a deformation retract from $\Bbb C^2-{\rm diag}$ onto $\Bbb C^\times$, and we know $\pi_1(\Bbb C^\times)$ is infinite cyclic. (This is the pure braid group $P_2$.)

If one further deformation retracts $\Bbb C^\times\to S^1$ and has $C_2$ act by swapping antipodal points, then our deformation retract from $\Bbb C^2-{\rm diag}$ is equivariant. Thus we have a commutative diagram:

$$\begin{array}{ccc}\pi_1(\Bbb C^2-{\rm diag}) & \longrightarrow & \pi_1((\Bbb C^2-{\rm diag})/C_2) \\ \downarrow & & \downarrow \\ \pi_1(S^1) & \longrightarrow & \pi_1(S^1/C_2) \end{array} $$

The vertical maps are isomorphisms since they are induced from deformation retracts. As a result, we know that the inclusion of the pure braid group $P_2\hookrightarrow B_2$ is akin to $2\Bbb Z\hookrightarrow\Bbb Z$.

I don't think this kind of argument will generalize though.

So what about $n>2$? In configuration space (which has $2n$ real dimensions, so is hard to visualize) a single point, a "configuration," represents $n$ distinct points in a plane (which is easy to visualize). And a path in configuration space represents each of the $n$ points in the plane having a path in and out of it.

Thus, imagine a continuum (indexed by $[0,1]$) of copies of $\Bbb C$ (resting flat) piled on top of each other. If one lets the altitude represent time, then the paths traced out between the points represent strings, and if one looks at this picture from the side one sees braid diagrams! Example:

$\hskip 1.3in$ braid diagram

Since we can choose our basepoint for $\pi_1$ to be anything, without loss of generality we may assume it is $\{1,2,\cdots,n\}\subset\Bbb C$ for the purpose of visualization. Tuples in $\Bbb C^n$ with nondistinct coordinates represent two strings intersecting at the same point, which is why wemust delete this subspace from $\Bbb C^n$: to prevent collisions.

A path in $\Bbb C^n$ ending where it started means each colored string above would have to go back to its original point, and this defines a pure braid. If we quotient by the action of $S_n$, we essentially allow the path in configuration space to go to any of the permuted configurations, which means the strings in the braid diagram can connect different dots.

There is another way to visualize braids that is also very interesting: mapping classes of the closed unit disk with $n$ points inside deleted. I recently asked a question about generalizing this idea to generalized braid groups. Mappings can warp the unit disk like a sheet of rubber, but the rubber is attached to the boundary (the unit circle) which must remain fixed pointwise. When you delete $n$ points, that essentially means your mappings of the disk must restrict to a permutation of those $n$ points.

To visualize what such mappings look like, for $B_2$ imagine putting two fingers on the two points in the disk, then using your two fingers to warp the rubber disk by turning it one way or the other with your fingers. Remember the border of the rubber sheet is stuck in place, so you'll be twisting the inside of the rubber relative to the outside rather than lamely rotating it. In general for $n$ points, you can do the same thing by twisting the rubber around two points with two fingers.

There two ways to twist two marked points around each other with your fingers (clockwise or counterclockwise), corresponding to which string goes over/under which in the braid diagram. The paths that the marked points take throughout the twisting process essentially trace out a braid diagram. Intuitively, we should be able to "lift" any braid diagram into a composition of such twistings of the unit disk. More detail is given in the link.

  • $\begingroup$ You seem to cover most of the ground in your edit, but I was composing my answer before I saw it. $\endgroup$
    – anon
    Jul 19, 2012 at 6:23
  • $\begingroup$ I appreciate the response anyways and it did clear up a lot of my confusion. The main enlightening parts of your response for me are the last three paragraphs! $\endgroup$ Jul 19, 2012 at 6:32
  • $\begingroup$ @SamuelReid Thought I should let you know, I've just done a major correction and update of this answer. $\endgroup$
    – anon
    Jul 3, 2015 at 18:57

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