Let $K$ be a slice knot in $S^3 = \partial B^4$. Then $K$ bounds a smoothly properly embedded disk $D$ in $B^4$. Let $\nu(D)$ denotes the tubular neighborhood of $D$.

Or we consider ribbon disks by excluding index two critical points.

Do we know $4$-dimensional handle decompositions of the slice disk exterior $B^4 - \mathrm{int}{ \nu(D)}$ for any slice $K$? At least for any ribbon disk exterior?


1 Answer 1


We can produce Kirby diagrams for the complement of a ribbon surface $S$. Indeed, there is a procedure that is described in Gompf-Stipsicz "4-manifolds and Kirby calculus". I'll briefly explain that, for more details see pg 211-213 of that book.

First we perturb the projection $B^4 \simeq \mathbb{D}^3\times I \to I$ to a Morse function for the embedded ribbon surface $S$. One can prove that a critical point of index $k$ for $S$ corresponds to attaching an handle of index $k+1$ to the complement $B^4\setminus \nu S$ (see Prop. 6.2.1).

Consider a diagram for the ribbon knot together with the bands specifying the ribbon moves. When we do band surgery along these bands we end up with a bunch unknots. The disks they bound corresponds to the 0-handles of $S$, while the bands are the (2-dimensional) 1-handles of $S$. Now, in order to build the complement $B^4\setminus \nu S$ we start with a 0-handle and then attach a (4-dimensional) 1-handle for each disk above followed by a 0-framed 2-handle for each band. The attaching circle of the 2-handle has to follow the band (really it is a perturbation of the band such that it links the unknots (now appearing as dotted circles)).

I do not know what you can do for a generic slice surface since in principle you would have the extra complication of attaching 3-handles. In the book is mentioned that it is not known any example handle decomposition for a topological slice disk which is not ribbon.

  • $\begingroup$ Thus, the number of bands determine the number of $4$-dimensional $1$- and $2$-handles. For example, if we have stevedore knot or square knot, then the ribbon disk exterior has one $0$-handle and two $1$-handles and two $2$-handles, right? $\endgroup$
    – user160180
    Jan 7, 2021 at 11:47
  • $\begingroup$ You have to specify the bands first. The number of 4-dim 1-handles equals the number of disks/unknots that you get once you perform all the ribbon moves. The number of 2 handles is the same as the number of bands. $\endgroup$ Jan 7, 2021 at 11:55
  • $\begingroup$ For example If you perform a ribbon move to the Stevendore knot as shown in example 1.7 of maths.ed.ac.uk/~v1ranick/papers/sliceknots2.pdf then you end up with a 0 handle, 2 1-handles and a single 2-handle. $\endgroup$ Jan 7, 2021 at 11:59
  • $\begingroup$ I see. The same picture can be drawn for the square knot by adding a vertical band from the left to the right trefoil. The result is the unlink. $\endgroup$
    – user160180
    Jan 7, 2021 at 12:10
  • 1
    $\begingroup$ The 2-handle will link algebraically once a dotted circles but generally there will be more geometric intersections. So I don't know, is like trying to show that the manifold is not $\mathbb{S}^1\times \mathbb{D}^3$. $\endgroup$ Jan 7, 2021 at 12:51

You must log in to answer this question.