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The group SO(3) acts naturally on $S^2$ and thus on $Conf(S^2, q)$, the configuration space of $q$ distinct points on the 2-sphere, via the diagonal action on $S^2 \times...\times S^2$. This is a question about $Conf(S^2, q)_{hSO(3)} = ESO(3) \times _{SO(3)} Conf(S^2, q)$, the $SO(3)$-equivariant homotopy type of this space.

Theorem: $Conf(S^2, q)_{hSO(3)}$ is aspherical for $q\geq 3$.

Proof (sketch): I will just write $C_q$ for $(Conf(S^2, q))_{hSO(3)}$. Projection to the first three coordinates $p:C_q \to C_3$ is a fibration (an equivariant version of the Fadell-Neuwirth fibration) with fiber $Conf(S^2-$ {3 points}, $q-3)$. But $C_3 \simeq *$, so the inclusion of the fiber $Conf(S^2-$ {3 points},$q-3)$ into $C_q$ is a homotopy equivalence. The fiber is a $K(\pi, 1)$ by an easy induction argument (using Fadell-Neuwirth again). QED.

Question: So $Conf(S^2, q)_{hSO(3)}$ is a $K(\pi, 1)$. What's $\pi?$

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Are you distinguishing the "SO(3) fundamental group" from the usual fundamental group? If so, would the former be defined as the fundamental group of the Borel construction? –  David Roberts Nov 19 '10 at 4:41
Right, the former; so I'm looking at $ESO(3) \times _{SO(S)} Conf(S^2, q)$. –  Romeo Nov 19 '10 at 4:44
Errr, $...\times _SO(3)...$, of course. –  Romeo Nov 19 '10 at 4:44
ugh, cant edit: $ESO(3) \times _{SO(3)} Conf(S^2, q)$. –  Romeo Nov 19 '10 at 4:46
We need to project down to the 1st 3 coordinates, so q$\geq3$. –  Romeo Nov 19 '10 at 16:53

1 Answer 1

$\pi$ is an iterated extension of free groups by free groups again using Fadell-Neuwirth. This is also called the pure braid group of the sphere minus three points on q-3 braids.

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$\pi_1(Conf(S^2, q)) is the "q-strand pure braids on $S^2$", but I'mquotienting by the SO(3) action here. –  Romeo Nov 19 '10 at 4:49
In your question I just noticed the symmetric group action is also present. Fundamental group of which space you are looking for...? Sorry I do not understand your notations. –  Roushon Nov 19 '10 at 5:19
I added the clarification from D Robert's comment to the main post. –  Romeo Nov 19 '10 at 5:40
I think Roushon's answer is correct (if you take ordered configurations). Proof: The fibre bundle $Conf^{q}(S^2) \to Conf^3 (S^2)$ is $SO(3)$-equivariant and the fibre is $Conf^{q-3} (C \setminus \{0,1\})$. So there is a fibre sequence $Conf^{q-3} (C \setminus \{0,1\}) \to Conf^q (S^2)_{hSO(3)} \to Conf^3(S^2)_{hSO(3)}$. The base is contractible (essentially because Möbius transformations on $S^2=CP^1$ act simply transitively on $Conf^3 (S^2)$. –  Johannes Ebert Nov 19 '10 at 16:41
Right, this is the argument in the post. So we are looking at $\pi_1$ of configurations of $q-3$ points in the thrice-punctured sphere. –  Romeo Nov 19 '10 at 17:16

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