Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\pi?$

share|improve this question
    
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
1  
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.

share|improve this answer
    
$\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
2  
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
1  
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

Your Answer

 
discard

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.