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In a previous answer see here by Samuel Reid, I read the following:

"The configuration space of $n$ points in a topological space $X$ is usually defined to be,

$$C_{\hat{n}}(X) = \{(z_{1},...,z_{n}) \in X^{n} \; | \; z_{i} \neq z_{j} \; \text{if} \; i\neq j \}$$

[...]We can define the orbit space $C_{n}(X) := C_{\hat{n}}(X) / \Sigma_{n}$ as configuration space modded out by the symmetric group on $n$ elements $\Sigma_{n}$. Then it can be shown that $$\pi_{1}(C_{\hat{n}},\vec{p}) = PB_{n}, \; \text{and} \; \pi_{1}(C_{n},\tau(\vec{p})) = B_{n}$$ Where, $\tau: C_{\hat{n}}(X) \rightarrow C_{n}(X)$ is an $n!$-sheeted covering map called the orbit space projection, and $PB_{n}$ and $B_{n}$ are the pure braid group and braid group on $n$ strands, respectively."

Here is my question: it is well known that $\pi_{1}(C_{n}[\mathbb{R}^2])=B_n$ but also that $ \pi_{1}(C_{n}[\mathbb{R}^d])=\Sigma_n$ for $d>2$. So how is this compatible with Samuel Reid's statement (which I must have misunderstood) ? Also does Samuel Reid's statement mean that for any connected manifold $X$, then $ \pi_{1}(C_{n}[X])$ is either trivial, $B_n$ or $\Sigma_n$ or, at the opposite, can I come up with a connected manifold $X$ for which $ \pi_{1}(C_{n}[X])$ is yet another group?

Thank you for your answers!

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PLG
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  • Hint: Consider $C_{\hat{1}}(X)$. – Baby Dragon Mar 19 '14 at 18:53
  • Better to not use personal names in the title (and minimize the number of uses in the question and comments). – zyx Mar 19 '14 at 18:58
  • Right, so its clear that $C_1(X)=C_{\hat{1}}(X)$ and hence $\pi_1(C_1(X))=\pi_1(X)$ can be whatever, e.g. $\mathbb{Z}$. I don't understand how Samuel Reid's comment can be correct then, is the comment right and I misundertand it? Because it seems to me that it implies $\pi_1(C_n(X))=B_n$ for any $X$. – PLG Mar 19 '14 at 23:19

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