# Compressible torus in irreducible 3-manifold bounds a solid torus

In his 3-manifolds notes (page 11, item $$(4)$$), Hatcher shows that a 2-sided compressible torus $$T$$ in an irreducible 3-manifold $$M$$ either bounds a solid torus $$S^1 \times D^2$$ or is contained in a ball $$B \subset M$$.

Is it then correct to say that such a torus in a manifold meeting those conditions must always bound a solid torus? I believe (maybe I'm mistaken) that any torus inside a solid ball should bound a solid torus. Thanks!

## 1 Answer

You are indeed mistaken.

For a counterexample, start with any nontrivial knot $$K \subset \mathbb R^3 \subset S^3$$.

Let $$N \subset \mathbb R^3 \subset S^3$$ be a solid torus neighborhood of $$K$$, with boundary torus $$T$$.

Let $$B \subset \text{interior}(N)$$ be a solid ball embedded in the interior of $$N$$ and thus disjoint from $$T$$.

Then $$B^{c} = S^3 - \text{interior}(B)$$ is a solid ball into which $$T$$ is embedded, but $$T$$ does not bound a solid torus in $$B^c$$.

Examples like this are sometimes called "knotted holes" or "knotted hole balls". Here's a link depicting the knotted hole ball associated to a trefoil knot.

• Ahh, I see, thank you very much! May 17, 2022 at 20:07