The hyperbolic metric on $S^3 \setminus S^1$

Question

We remove a circle $S^1$ from the sphere $S^3$ and $M = S^3 \setminus S^1$. By the Thurston Theorem $M$ doesn't admit any metric of constant negative sectional curvature $K = -1$ such that $M$ is complete respect dor this metric. I need to prove this statement, but you can not use the Thurston theorem. I think that there is some direct argument

Answer 1

No. A famous theorem of Thurston states that $S^3 - K$, where $K$ is a knot (a tamely embedded circle), is hyperbolic precisely when $K$ is neither a satellite of a non-trivial knot nor a torus knot. Here are two surveys of possible interest. Maybe others can recommend the best place to start reading about this subject.

Hyperbolic Knots by Adams: https://arxiv.org/pdf/math/0309466.pdf

The Hyperbolic Revolution by Bonahon: http://www-bcf.usc.edu/~fbonahon/Research/Preprints/GeomTopMAA.pdf