Suppose I am given a smoothly slice knot $K\subset\Bbb S^3$. But I am only given a locally flat disc $D\subset \Bbb D^4$ with boundary $K$.

Question: Is there a smooth disc $D'\subset\Bbb D^4$ with boundary $K$ that is $\epsilon$-close to $D$ and approaches $D$ towards the boundary?

More precisely, I want $D'$ to be in $D_\epsilon\cap X$, where $D_\epsilon$ is an $\epsilon$-thickening of $D$ and

$$X:=\bigcup_{x\in D} B_x(d(x,K)),$$

$B_x(r)$ being a ball around $x$ of radius $r$, and $d(x,K)$ being the distance of $x$ from $K$.


1 Answer 1


Nice question! Here's an example to think about. Let J be a knot that is topologically but not smoothly slice, and let $D_J$ be a locally flat disc that is smooth near the boundary, with boundary J. Note that $-J$ (the usual concordance inverse) is also slice, with disk $-D_J$. Let $D = D_J\natural(-D_J)$, where the $\natural$ denotes boundary connected sum done along a band near the boundary. Let $K = J \# (-J)$; then it is smoothly slice by a standard argument.

At first glance, it looks hard for any slice disk $D'$ for $K$ to be close to $D$ without splitting into smooth slice disks for $J$ and $-J$. But the proof is not obvious because you are approximating in the $C^0$ topology whereas one would presumably need a stronger sense of approximation near the boundary.

As I said, something to think about.

  • $\begingroup$ Thanks Danny, nice example. Does "the proof is not obvious" mean that there is a proof or does it mean that you are not sure? Since this is fairly new terrain for me, can you say more about why $K$ is smoothly slice? $\endgroup$
    – M. Winter
    Apr 5, 2022 at 21:34
  • 3
    $\begingroup$ Sorry, that wasn't clear: I meant that I don't know how to turn that into a proof. The standard argument for sliceness of $J\# -J$ is to view J as an arc with endpoints on a plane in $R^3$, and then spin that arc in 4-space. See for instance Rolfsen, Chapter 8, Lemma E.11. $\endgroup$ Apr 5, 2022 at 22:39

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