Are slice disks with the same boundary isotopic?

Question

Given two smoothly properly embedded disks $f,g : D^2 \to D^4$ with $f|_{S^1} = g|_{S^1}$, does it follow that $f$ and $g$ are isotopic?

Answer 1

If you mean proper isotopies (An isotopy of $D^2$ in $B^4$ which restricts to an isotopy of $S^1$ inside $S^3$), the answer is no. Take for instance the flat disk bounding the unknot (or more generally any doubly-slice disk) and connect sum any non-trivial 2-knot (an embedded $S^2$ in $S^4$ not smoothly isotopic to a longitudinal $S^2$). If this disk was isotopic to the flat disk, then it's easy to construct an isotopy from this 2-knot to the standard longitudinal $S^2$. The first example of such a 2-knot is due to E. Artin, and I am having trouble finding a relatively accessible survey focusing on 2-knots. Maybe Hillman's book is a decent place to start (though it is certainly a large amount of material) https://arxiv.org/pdf/math/0212142.pdf .

If for some strange reason you mean isotopies which aren't necessarily proper then they are always isotopic, as you can just shrink along a radial vector field until the derivative approximates the embedding (so every embedding is isotopic to a flat disk.