I have two different, but related, questions about the type of geometry one can get on a knot complement.

Quickly some notation: $K$ will be a non-trivial smooth knot - living in $S^3$ - and $M$ will be the complement of a regular neighborhood of $K$.


How can $M$ ever admit a hyperbolic geometry (as in, how can hyperbolic knots exist)? $\pi_1(M)$ will contain a $\mathbb{Z}\times\mathbb{Z}$ subgroup (coming from the boundary of that regular neighborhood), and so $\pi_1(M)$ cannot be a hyperbolic group.

Most likely I am confusing details about hyperbolic structures and having a hyperbolic fundamental group.


This time let $K$ be the trefoil knot. A reliable source (who almost surely knows what he's talking about) told me, in this case, $M$ admits an $\widetilde{SL(2,\mathbb{R})}$ geometry. [I don't understand all the details of how he explained it, but I think it boiled down to viewing $M$ as $SL(2,\mathbb{R})$, quotient the action of the discrete group $SL(2,\mathbb{Z})$].

But wait! It is well-known in this case that $\pi_1(M)\cong B_3$, the braid group on $3$ strands. And $B_3$ has a subgroup of index $6$ which looks like $\mathbb{Z}\times F_2$, where $F_2$ is the free group on $2$ generators. This subgroup corresponds to a six-sheeted covering space $\hat{M}$; we also have $\hat{M}\sim N\times S^1$, with $N$ a compact surface and $\pi_1(N)\cong F_2$. Now this means $\hat{M}$ has an $\mathbb{H}^2\times\mathbb{R}$ geometry, and so $M$ must as well. [In fact, the underlying ideas here work for any torus knot.]

So what exactly went wrong?


For your first question, it is true that fundamental groups of closed hyperbolic manifolds cannot contain copies of $\mathbb{Z}^2$. However, a knot complement is not a closed manifold! The hyperbolic structure on the knot complement will be a complete hyperbolic manifold with finite volume, but with a cusp. The fundamental group of the cusp is $\mathbb{Z}^2$.

  • $\begingroup$ Thank you! I guess I was indeed confused about the relation between hyperbolic geometry and hyperbolic $\pi_1$. $\endgroup$ – user641 Oct 20 '11 at 0:18

One comment about your second question, which is a bit tedious to fit into a comment box:

The quotient $SL_2(\mathbb R)/SO(2)$ is isomorphic to $H^2$, and so $SL_2(\mathbb R)$ is a circle fibration over $H^2$. In fact, if we forget the group structure, there is a diffeomorphism $SL_2(\mathbb R) \cong H^2 \times SO(2).$ Thus, as a manifold, $\widetilde{SL_2(\mathbb R)} \cong \mathbb R \times H^2$.

  • $\begingroup$ Thanks! This makes me think the "problem" is that maybe a non-closed 3-manifold admits more than one geometry (for example, $\mathbb{R}^3$). $\endgroup$ – user641 Oct 20 '11 at 0:17

The hyperbolic structure on the complement of a trefoil knot is explained in Milnor's Introduction to Algebraic K-theory. He attributes the proof to Quillen.

Looking for the excerpt from Milnor's book online, one finds some similar arguments in one of John Baez' "this week..." postings:



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