# $S^3 \setminus S^1$ doesn't have hyperbolic structure

I need to prove that $M = S^3 \setminus S^1$ doesn't admit any metric of constantly negative sectional curvature s.t. $M$ is complete respect for this metric. I know that it is consequence of famous Thurston Theorem, but it is quite uncomfortable to use it, so, maybe there is exist direct argument?

• You need to assume that the $S^1$ is unknotted. Then consider the limit set of the underlying Kleinian group of $S^3-S^1$. It will contain at most 2 points, and so couldn't be a finite covolume Kleinian group. Jun 6, 2018 at 22:45
• I am not sure why people offer complicated answers. Your $M$ does admits a complete hyperbolic metric, namely, the quotient of the upper half space $\mathbf{H^3}=\{(z,t): z\in\mathbb C, t>0\}$ by the cyclic group generated by the dilation $(z,t)\to (2z, 2t)$. Jun 7, 2018 at 20:07

You probably mean $M$ does not admit complete hyperbolic metrics of finite volume.

Since $M$ is topologically the interior of a solid torus, a complete hyperbolic structure just identifies $M$ as the quotient of $\mathbb{H}^3$ by a single loxodromic or parabolic isometry, but such quotients have infinite volume.

• Thank you! Does exist such simple explanation in case of infinite volume too? Jun 5, 2018 at 5:24

The following contribution comes from conversations with Bill Goldman, any mistakes however are mine alone.

Any (geodesically complete) geometric 3-manifold $N=\mathbb{M}/G$ with infinite order elements in its fundamental group (isomorphic to $G$) is covered by an open solid torus admitting the same geometric structure (having model geometry $\mathbb{M}$). Note that $G$ is a subgroup of the isometry group of $\mathbb{M}$ and arises from a discrete faithful homomorphism $\pi_1(N)\to Isom(\mathbb{M})$.

To see this, let $T\in G$ be of infinite order. Then the covering space of $N$ given by $\mathbb{M}/\langle T\rangle$ is an open solid torus (coverings of manifolds admitting geometric structures admit geometric structures themselves). Although I have not verified it, I am pretty sure that $\mathbb{M}$ can be any of Thurston's eight geometries excepting only $\mathbb{S}^3$.

Here is an example construction (when $\mathbb{M}$ is hyperbolic 3-space $\mathbb{H}^3$). Take a hyperbolic line $L$ (complete geodesic) in $\mathbb{H}^3$ and a hyperbolic translation (loxodromic element) along that line. Call the translation $T$. Consider the cyclic group $\langle T\rangle.$ Then $T$ leaves the line $L$ invariant.

Take any unit disk $D$ orthogonal to $L$. Translate $D$ by $T$ along $L$ to obtain the disk $T(D)$. The two disks $D$ and $T(D)$ bound a fundamental domain for $\langle T\rangle$. Fundamental Domain in Ball Model; Image made by Marvin Castellon

The quotient of the fundamental domain by $\langle T\rangle$ is homeomorphic to a solid torus (one can do the same thing when $T$ is parabolic). The resulting manifold is geodesically complete.

The volume of such a torus must be infinite. For otherwise, by Mostow Rigidity, there would be a unique hyperbolic structure. However, from the construction above we see there are infinitely many distinct structures (pick any non-conjugate loxodromic element).

So in your case the manifold is topologically an open solid torus (assuming $S^1$ is unknotted) and so it admits many different complete geometric structures (including many hyperbolic structures).