# Knots in 3-manifolds

Consider a closed $3$-manifold $M$ and a knot $K$ in $M$.

Is it necessarily true that $\pi_2 (M \setminus K) = 0$?

If not, are there any conditions on $M$ and/or $K$ to ensure the above 2nd homotopy group of the knot complement is trivial?

Thanks!

(Note: This is, of course, true when $M$ is simply connected --> $S^3$)

## 1 Answer

EDIT - I've rewritten my previous answer in an attempt to remove everything except the answer to your question. All submanifolds are assumed to be smooth.

Suppose that $M$ is a closed, connected, oriented, irreducible three-manifold (and $M$ is not the three-sphere). Suppose that $K$ is a knot in $M$. Then $\pi_2(M - K)$ is non-trivial if and only if $K$ is contained in an embedded three-ball $B^3 \subset M$.

The proof is an exercise using Alexander's theorem (every embedded two-sphere in $S^3$ bounds balls on both sides), the sphere theorem, and the Poincaré conjecture. Hempel's book and Hatcher's notes on three-manifolds are standard references for the necessary background material.

• As I understand, by the Sphere Theorem if M is irreducible then $pi_2(M)=0$. But how does this also imply that $pi_2(M \setminus K) = 0$? – ali elgindi Dec 27 '14 at 17:56
• Regarding typesetting: \pi gives $\pi$ when surrounded by dollar signs. – Sam Nead Dec 27 '14 at 18:01
• That isn't making sense to me, maybe I don't understand your point. $K$ is cut out and $S$ bounds a ball in $M$, but the ball is now missing the knot. How is it still bounding a 3-ball? – ali elgindi Dec 27 '14 at 18:05
• I understand now, thank you. I will work on the proof (which seems direct). Thanks again! – ali elgindi Dec 27 '14 at 18:24