# The knot $K\subset \Bbb S^3$ is smoothly slice, but the disc $D\subset \Bbb D^4$ is only locally flat. Can $D$ be smoothed?

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 $$K$$.

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.
• 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? Apr 5, 2022 at 21:34
• 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. Apr 5, 2022 at 22:39