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Define the configuration space of $n$ points in a general manifold $M$, where $\dim M=m$, as $K=(M^n-D)/S_n$ where $S_n$ is the permutation group and $D=\{(x_1,\cdots,x_n)| \exists i,j\ s.t. x_i=x_j \}$.

Then my question is

(1) How to prove $M=\mathbb{R}^m$ then $\pi_1(K)=S_n$ for $m>2$ and $\pi_1(K)=B_n$ for $m=2$ where $B_n$ is the Braid group. And what is the $\pi_1(K)$ when $m=1$?

(2) In general when $M$ is non-simply connected then what is $\pi_1(K)$ and how to calculate it? For example, $M=T^2$.

same question: https://math.stackexchange.com/questions/1748136/how-to-calculate-the-fundamental-group-of-general-configuration-space

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    $\begingroup$ The first question is a basic fact about the braid groups. This is where their name comes from too. $\endgroup$ – Dimitri Chikhladze Apr 19 '16 at 12:02
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    $\begingroup$ These are called braid groups of manifolds. Key to calculations are the so-called Fadell-Neuwirth fibrations. You could probably find an answer for the torus by searching using these terms. $\endgroup$ – Mark Grant Apr 19 '16 at 12:30
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(1a) Assume $n\geq 2$ since otherwise the quotient is trivial. Since the action of $S_n$ is free, the quotient map is a covering map. Since $m\geq 3$, and $M$ is simply-connected, $M^n-D$ is simply-connected by transversality since the codimension of $D$ is greater than or equal to 3 (so the inclusion map is 2-connected). From covering space theory, $\pi_1(K)=S_n$.

(1 b and c) See the comments here

(2) For non-simply connected $M$, use the idea in (1a) to get a quotient of $\pi_1(K)$ by the image of the fundamental group of $M^n-D$ via the covering map. Use that to analyze $\pi_1(K)$. See here for examples.

Note: here,here and here are some great introductory slides on this topic by Fred Cohen, who has written many papers about the topology of configuration spaces.

Note: By the Dold-Puppe Theorem, the fundamental group of the singular space $M^n/S_n$ is simply connected when $M$ is simply connected even though $(M^n-D)/S_n$ is not simply connected.

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  • $\begingroup$ What's the Dold-Puppe Theorem? $\endgroup$ – Dan Ramras Apr 22 '16 at 4:26
  • $\begingroup$ ams.org/mathscinet/search/… $\endgroup$ – Sean Lawton Apr 22 '16 at 5:19
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    $\begingroup$ Ah, cool. This is sort of an unstable version of the Dold-Thom theorem. For those who can't follow the above link, the paper is Ann. Inst. Fourier Grenoble 11 1961 201–312. $\endgroup$ – Dan Ramras Apr 22 '16 at 5:40

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