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.

  • $\begingroup$ $B_3$ is the fundamental group of the configuration space of unordered triplets of points in $\mathbb{C}$. This space might be homotopy equivalent to $S^3 \setminus K$...? $\endgroup$ – Qiaochu Yuan May 11 '12 at 0:59
  • $\begingroup$ This may be useful to you: mathoverflow.net/questions/96423/… $\endgroup$ – Samuel Reid May 11 '12 at 1:22
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    $\begingroup$ Speaking of John Baez, have you seen this?: math.ucr.edu/home/baez/week261.html It explains it very clearly! $\endgroup$ – user641 May 11 '12 at 4:39
  • $\begingroup$ 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. $\endgroup$ – Dan Rust May 27 '13 at 12:53
  • $\begingroup$ 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. $\endgroup$ – Dan Rust May 27 '13 at 12:54

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.


The 3-strand braid group is the fundamental group of $UC_3(\mathbb{C}) = \{\{z_1,z_2,z_3\} \subset \mathbb{C} \mid z_i \not= z_j\}$, the unordered configuration space of distinct triples in the complex plane. By translating, this space deformation retracts to the subspace of triples where $z_1+z_2+z_3=0$. By positive real scaling, this deformation retracts further so the space where $|z_1|^2+|z_2|^2+|z_3|^2=1$. (Note that these conditions are invariant up to permutation of the $z_i$.) Call this new space $X$.

Now we see that $X$ is a 3-manifold. That it is a manifold should be clear. As for the dimension, you can heuristically think of there being $2$ degrees of freedom for where to pick a first point in the plane. The normalization tells you that you can pick one more point (very nearly) freely on a circle, so $1$ degree of freedom, and then the last point is uniquely determined.

Not only is $X$ a 3-manifold, it's also a Seifert fiber space: there is a circle action $S^1 \curvearrowright X$ given by rotation, or multiplication by $e^{i\theta}$ if you prefer. This action is almost free. There are just two exceptional orbits, given by the triples of points $\{\pm 1/\sqrt{2},0\}$ (collinear) and $\{1/\sqrt{3},e^{\pm 2\pi i/3}/\sqrt{3}\}$ (equilateral triangle). From here, you see that $X$ has exceptional fibers of order 2 and order 3. There is also one "missing" generic fiber, corresponding to the collapse of two points. (Our normalization prevents the collapse of three points, so need need to worry about that). Maybe this is enough to make you start to suspect that you're looking at the trefoil knot complement.

There's another explanation along the same lines where you view the points in $\mathbb{C}$ as roots of complex polynomials. Steve Trettel has a nice exposition of this at his website.


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