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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?

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$C$ 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) $\left(\mathbb{C} \setminus \{0\}\right) \times \mathbb{C}$. –  WimC Jul 19 '12 at 6:33
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1 Answer

up vote 4 down vote accepted

I don't think it's a good idea to begin understanding the algebraic topology treatment of braid groups by visualizing $2n$-dimensional space directly; even the smallest $n=2$ case can't be drawn on paper or easily animated because it's going to be four-dimensional! Instead, the easiest way for me to visualize this fundamental group is through projections.

First of all, the base point is indeed in $C_{o,\hat{n}}$, but its location doesn't matter. In any path-connected space, the fundamental group $\pi_1(X,a)$ is isomorphic to $\pi_1(X,b)$ simply because for any loops attached to $a$, we can shrink them sufficiently close and then transport them in such a way that $a$ is taken to $b$ and we then have loops around $b$, and vice-versa. We therefore disregard its location.

Consider a path $\gamma:[0,1]\to C_{o,\hat{n}}$ with $p=\gamma(0)=\gamma(1)$. Since $p$ can be chosen anywhere, let's assume that it is such that its individual coordinates, taken as $n$ points in $\Bbb C$, are all collinear. Note the $k$th coordinate of $\gamma$ traces out a path in $\Bbb C$ from $p_k$ to itself, the only stipulation is that no two cross simultaneously, i.e. at the same $t\in[0,1]$ value. A drawing:

$~~~~$ braid

On the left the three coordinates of $p$ are three black dots in the plane, and the three coordinate paths are traced out in green, purple and orange (the color of dry-erase markers I have on hand).

I've inserted a small blue circle and a large pink perimeter as well as some gray demarking lines; "cut" along the blue and pink and the main vertical gray line, and then warp the resulting sheet until it becomes the figure on the right. This establishes the correspondence between paths and the usual depiction of braids. Note that on each crossing, one coordinate path crosses over the other: since no two coordinates can be the same at the same time, inevitably one path "got there first" and so we can put it overtop.

Continuous deformation of $\gamma$ is realized as simultaneous deformation of the coordinate curves; the "crossings" will move around but the order of which goes over/under the other will not change, for that would mean that at some point in the deformation we created a $\gamma$ where two coordinates were the same for some $t$. This is why we cannot shrink each loop to a point; they are braided together!

This manifests the pure braid group $P$. The same idea works with $B$, but we lose the restriction that each coordinate path ends at the same black dot it started on - they can instead be permuted. (This also illustrates why $P$ is the kernel of the obvious homomorphism $B_n\to S_n$.)

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You seem to cover most of the ground in your edit, but I was composing my answer before I saw it. –  anon Jul 19 '12 at 6:23
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! –  Samuel Reid Jul 19 '12 at 6:32
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