Open tubular neighborhood of $2$-disk in $4$-ball and its exterior

I try to be familiar with the notion of smoothly sliceness of knots and disks.

A $$2$$-disk $$D$$ is said to be a slice disk if it a smoothly and properly embedded in $$B^4$$. The boundary of $$D$$ in $$S^3$$ is also called a slice knot.

Let $$\nu(.)$$ denote the open tubular neighborhood in the ambient space. It is a well-known fact that $$\nu(K) \approx S^1 \times D^2$$ and the knot exterior $$\overline{S^3 - \nu(K)}$$ is a compact $$3$$-manifold with the boundary $$S^1 \times S^1$$.

My questions are as follows:

1) If $$K$$ is a slice knot, then $$\overline{S^3 - \nu(K)} \approx S^1 \times D^2$$? (if $$K$$ is unknot, it is true.)

2) For the slice disk $$D$$ in $$B^4$$, are $$\nu(D)$$ and $$\overline{B^4 - \nu(D)}$$ homeomorphic to explicit manifolds?

(I'm going to use $$B^n$$ to denote an $$n$$-dimensional open ball and $$D^n$$ to denote a closed one.)

Theorem. Let $$K$$ be a knot. $$S^3-\nu(K)\approx S^1\times D^2$$ if and only if $$K$$ is the unknot.

Proof. We will show the stronger statement that $$\pi_1(S^3-\nu(K))\approx\mathbb{Z}$$ if and only if $$K$$ is the unknot. Consider a minimal genus Seifert surface $$\Sigma$$ for $$K$$, and let $$\Sigma'=\Sigma-\nu(K)$$. The induced map $$\pi_1(\Sigma')\to \pi_1(S^3-\nu(K))$$ must be injective, since otherwise by Kneser's lemma there would be a way to compress $$\Sigma'$$ and reduce its genus. If the knot genus is greater than $$0$$, then $$\pi_1(S^3-\nu(K))$$ contains a free group on at least two generators. The rest follows from the fact that the unknot is the unique genus-$$0$$ knot. $$\square$$

In fact, Gordon and Luecke proved that if there is an orientation-preserving homeomorphism $$S^3-\nu(K)\approx S^3-\nu(K')$$, then $$K$$ and $$K'$$ are equivalent knots.

Since disks are contractible, the tubular neighborhood $$\nu(D)$$ of a disk $$D$$ in $$D^4$$, regarded as the embedded normal bundle, must be a trivial $$\mathbb{R}^2$$ bundle over $$D$$. So, $$\nu(D)\approx D\times B^2$$. Its closure is $$\overline{\nu(D)}\approx D^4$$.

The complement $$D^4-\nu(D)$$ can be rather complicated. While I don't have any examples on hand, it seems that $$\pi_1(D^4-\nu(D))\neq \mathbb{Z}$$ if $$D$$ is a ribbon disk for a non-trivial ribbon knot by the van Kampen theorem.

In contrast to the Gordon and Luecke result, I just found a paper by Abe and Tange, "Ribbon disks with the same exterior", where they give a family of inequivalent slice knots such the the complements of a slice disk for each are all diffeomorphic.

• Thanks a lot. I have one question. What do you mean when saying trivial $\mathbb{R}^3$ bundle over $D$? It seems that it is a trivial fiber bundle over $D$ with $\mathbb{R}^3$ as fiber. In any case, I don't understand well why $\nu(D) \approx D \times B^3$. Is there any good reference to read something about it? – M. Alessandro Ferrari May 9 at 19:26
• @AlessandroFerrari In Lee's book on differential topology there is the homeomorphism between a tubular neighborhood and the normal vector bundle. In any case, it is a trivial fiber bundle, and "trivial" means it is a product. – Kyle Miller May 9 at 19:56
• To our notation, in "Knots and Links", pg. 218, Rolfsen wrote $D$ has a tubular neighborhood $D \times B^2$? – M. Alessandro Ferrari May 10 at 12:26
• @AllessandroFerrari I didn't notice the typo: the 3 should have definitely been a 2. – Kyle Miller May 10 at 17:59