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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?

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(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.

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  • $\begingroup$ 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? $\endgroup$ – M. Alessandro Ferrari May 9 at 19:26
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    $\begingroup$ @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. $\endgroup$ – Kyle Miller May 9 at 19:56
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    $\begingroup$ To our notation, in "Knots and Links", pg. 218, Rolfsen wrote $D$ has a tubular neighborhood $D \times B^2$? $\endgroup$ – M. Alessandro Ferrari May 10 at 12:26
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    $\begingroup$ @AllessandroFerrari I didn't notice the typo: the 3 should have definitely been a 2. $\endgroup$ – Kyle Miller May 10 at 17:59

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