MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\pi_{1}: \text{Top}^* \rightarrow \text{Grp}$ denote the fundamental group functor and let $H_{1}: \text{Top}^* \rightarrow \text{Grp}$ denote the first homology group functor. We can then define a natural transformation $\eta: \pi_{1} \rightarrow H_{1}$ with a component $\eta_{X}: \pi_{1}(X) \rightarrow H_{1}(X)$. We define the configuration space on $n$ points by, $$C_{\hat{n}}(\mathbb{R}) = \{(z_{1},...,z_{n}) \in \mathbb{R}^n \; | \; z_{i} \neq z_{j}\}$$ and the unordered configuration space on $n$ points by $C_{n}(\mathbb{R}) = C_{\hat{n}}(\mathbb{R}) \big/ \Sigma_{n}$, where $\Sigma_{n}$ is the symmetric group. We then have for $\vec{p} \in C_{\hat{n}}(\mathbb{R})$ that $\pi_{1}(C_{\hat{n}}(\mathbb{R}),\vec{p}) \cong PB_{n}$ and $\pi_{1}(C_{n}(\mathbb{R}),\tau(\vec{p})) \cong B_{n}$, where $\tau : C_{\hat{n}}(\mathbb{R}) \rightarrow C_{n}(\mathbb{R})$ is the orbit space projection, $PB_{n}$ is the pure braid group on $n$-strands, and $B_{n}$ is the braid group on $n$-strands.

Consider the components of the natural transformation between the fundamental group functor and the first homology group functor given by,

$$\eta_{C_{\hat{n}}(\mathbb{R})} : \pi_{1}(C_{\hat{n}}(\mathbb{R})) \rightarrow H_{1}(C_{\hat{n}}(\mathbb{R}))$$ $$\eta_{C_{n}(\mathbb{R})} : \pi_{1}(C_{n}(\mathbb{R})) \rightarrow H_{1}(C_{n}(\mathbb{R}))$$

What objects in $\text{Grp}$ will these components of $\eta: \pi_{1} \rightarrow H_{1}$ associate with $PB_{n}$ and $B_{n}$, and how do they relate to $C_{\hat{n}}(\mathbb{R})$ and $C_{n}(\mathbb{R})$?

I am interested in this question because I am wondering if $H_{1}(C_{\hat{n}}(\mathbb{R}))$ and $H_{1}(C_{n}(\mathbb{R}))$ have any unnoticed connections to knot theory since every knot is the closure of a braid.

share|cite|improve this question
$H_{1}(C_{n}(\mathbb{R})) = H_1(B_n) = \mathbb{Z}$ and $H_{1}(C_{\hat{n}}(\mathbb{R})) = H_1(PB_n) = \mathbb{Z}^{n(n-1)/2}$, as can be seen by abelianizing. This can be seen using the usual presentations ( – Richard Kent Dec 1 '12 at 3:41
You probably want to do configurations in $\mathbb R^2$; otherwise your configuration spaces are contractible. – Mariano Suárez-Alvarez Dec 1 '12 at 3:44
Mariano: the configuration space of $n$ points in $\mathbb R^m$ is not contractible for any $n > 1$. I think you meant to write "simply connected". – Dan Petersen Dec 1 '12 at 6:14
@Dan, isn't the configuration space of $n$ unordered points in the line a cell? (Each component of the configuration space of $n$ ordered points in the line is an intersection of half-spaces in $\mathbb{R}^n$, which means it's convex. The symmetric group just permutes these.) – Richard Kent Dec 1 '12 at 14:48
@Richard Kent, $C_{n}([0,1]) = \Delta^{n}$ and $C_{\hat{n}}([0,1]) = \coprod_{i=1}^{n!} \Delta^{n}$. – Samuel Reid Dec 1 '12 at 20:40
up vote 1 down vote accepted

In line with Richard Kents answer, what you obtain by passing from the fundamental group to the first homology group of the ordered configuration spaces should be generated by letting any two strands in the pure braid group tangle together - and allow for no further tangling of three and higher strands. You should be able to make this clear by considering the morphism $\gamma_n \colon C_{\hat{n}}(\mathbb{R}^2) \to \prod_{n \choose 2} C_{\hat{2}}(\mathbb{R}^2)$ picking any two configurations of points in the domain and mapping them onto the configuration indexed by this choice of two points.

The morphisms $\gamma_{n-1}$ and $\gamma_n$ fits as a morphism of vector bundles between the bundle

$$ \bigvee_{n-1} S^1 \to C_{\hat{n}}(\mathbb{R^2}) \to C_{\hat{n-1}}(\mathbb{R^2})\\ $$

and the trivial bundle

$$ \prod_{n-1} C_{\hat{2}}(\mathbb{R}^2) \to \prod_{{n} \choose 2} C_{\hat{2}}(\mathbb{R}^2) \to \prod_{{n-1} \choose 2} C_{\hat{2}}(\mathbb{R}^2) $$

(Pascals triangle tells you that the indices in the fibers indeed match up) - and using the homotopy equivalence $C_{\hat{2}}(\mathbb{R}^2) \simeq S^1$, you can see the fibers of the upper bundle as the 1-skeleton of the $(n-1)$-tori in the fiber of the lower bundle, inductively meaning that $\gamma_n$ becomes an $H_1$-isomorphism for all $n$.

share|cite|improve this answer

Fred Cohen and Sam Gitler have written a paper "Loop spaces of configuration spaces, braid like groups and knots" available on Fred Cohen's homepage. They study deep and interesting relationships between the loop space homology of configuration spaces and knot theory. For example they show how to recover universal Yang-Baxter lie algebras from the study of this loop space homology. In section 7 of this paper they study the closure of braids, and in the two next sections are devoted to the study of embedding spaces and Vassiliev's theory.

Maybe I do not answer your question but I want to focus on the fact that there are some very deep relationships between homology of configuration spaces (in fact the homology of the based loop space of configuration spaces) and knot theory. Taking care of the first homology group of configuration spaces is not enough.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.