Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I apologise in advance for the vagueness of this question but I have not been able to find very much info on the topic and have made very little progress on my own.

I am trying to understand why the knot group $\pi_1 (S^3 - K)$ of the trefoil is isomorphic to Artin's 3-strand braid group $B_3$. I know that the Wirtinger presentation for $\pi_1 (S^3 - K)$ gives Artin's presentation for $B_3$ directly but I was hoping someone could paint a more topological picture which takes homotopy classes directly to braids (or vice-versa) without using group presentations as the middle man. Thanks in advance.

Edit: Thanks for the replies guys. I should have stated that $S^3-K$ is diffeomorphic to the space $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$ (John Baez says so in his blog). It is possible that the disk with 3 holes is a deformation retract of $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$ but I don't know much about this space. I'll update if I find the answer myself.

share|improve this question
    
$B_3$ is the fundamental group of the configuration space of unordered triplets of points in $\mathbb{C}$. This space might be homotopy equivalent to $S^3 \setminus K$...? –  Qiaochu Yuan May 11 '12 at 0:59
    
This may be useful to you: mathoverflow.net/questions/96423/… –  Samuel Reid May 11 '12 at 1:22
1  
Speaking of John Baez, have you seen this?: math.ucr.edu/home/baez/week261.html It explains it very clearly! –  user641 May 11 '12 at 4:39
    
To answer the point made by @QiaochuYuan , it's well known that knot complements are aspherical spaces, and so $S^3\setminus K$ is a $K(B_3,1)$. So is the configuration space of unordered triples in $\mathbb{C}$ - by considering the long exact sequence in homotopy of the Fadell-Neuwirth fibration on the ordered space of points in $\mathbb{C}$ with $m$ puncture points, and then taking a covering space projection on to the space of unordered points. Both of these spaces are homotopy equivalent to CW-complexes, and so are weakly homotopy equivalent by the classification of $K(G,n)$s. –  Daniel Rust May 27 '13 at 12:53
    
Obviously, this uses the fact that $S^3\setminus K$ and the configuration space of unordered triples have isomorphic fundamental group, so doesn't answer the OP's question, but it does suggest looking for an explicit homotopy equivalence between the spaces. –  Daniel Rust May 27 '13 at 12:54
add comment

1 Answer

The universal cover of the trefoil complement is also the universal cover of a disk with three holes in it, cross an interval. Both are obtained from tilings of $\mathbb{H}^2\times\mathbb{R}$ by prisms which are ideal triangles cross intervals. The only difference is the way the two groups act on this tiling. Each takes two adjacent prisms as its fundamental domain. So the two are very closely related.

share|improve this answer
add comment

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.