A link in $S^3$ is said to be slice if it bounds a collection of flat disks into the $4$-ball. Here "flat," means that there is a (locally) trivial normal bundle. This condition can be strengthened to "smoothly slice," meaning that the link bounds a smooth collection of disks. Even more restrictive is the condition of "ribbon," which means that the link bounds a collection of disks with only local maxima, and no local minima, as one moves into the $4$-ball. Thus we have an increasingly subtle series of questions. Is the link topologically slice? Is it smoothly slice? Is it ribbon? As far as I know, there are no algorithms for all three of these questions. My question is whether this is really true. For example, is there an algorithm to detect whether a link is ribbon? One interesting and venerable open problem in the field is whether all slice links are ribbon. This paper gives a sequence of examples of links which are smoothly slice but not obviously ribbon. If there were an algorithm for detecting "ribbonness," then one could apply it to these examples, and possibly disprove the slice=ribbon conjecture. It seems to me that detecting ribbon disks should not be that hard, perhaps using ideas akin to normal surface theory. You would just be looking for disks in the link complement with ribbon singularities. Has anyone thought about this?

Edit: As Ryan Budney points out, if a slice disk exists, it can be found algorithmically by iteratively subdividing a triangulation, but in the absence of upper bounds on how many times you need to do this, the algorithm can't return a negative answer that a link is not ribbon or slice.

Also, I meant to mention in my original post that the question of whether the Whitehead double of the Borromean rings is topologically slice is related to whether surgery works in 4 dimensions. Given how hard that question is, it would seem highly unlikely that there is an algorithm for the topological case!


3 Answers 3


This is addressing the last part of your question. There are a couple of issues with trying to implement normal surface theory for immersed surfaces. If an immersed surface is $\pi_1$-injective, then one may homotope it to be normal (meaning that the components of preimages of each tetrahedron is a normal disk - triangle or quadrilateral). In fact, one needs only the weaker condition that the surface is simple loop injective - every homotopically non-trivial curve in the surface is $\pi_1$-injective in the 3-manifold.

The first issue is that a ribbon disk might not be simple loop injective. You may always modify an immersed surface by surgeries along simple loops which are homotopically trivial in the 3-manifold to obtain a simple-loop injective surface. However, starting with an immersed ribbon disk (thought of as a planar surface in the knot complement), there is no guarantee that it is simple loop injective, and such surgery modifications may mess up the ribbon singularities. I thought about this a bit before, but I didn't see how to deal with this issue.

The second issue is that even if one had an immersed normal representative for the ribbon disk (with respect to a triangulation of the knot complement), how does one find such a ribbon disk algorithmically? The usual method for embedded normal surfaces is to analyze the normal surface solution space with compatible quadrilaterals, and notice that the normal coordinates uniquely determine the surface. One then uses linear programming to reduce this to a finite collection of possible surfaces of bounded complexity. For immersed surfaces, this strategy fails, and people haven't found a suitable generalization.


In the special case that your knot in $S^3$ is a fibered knot, there is an algorithm to determine if it is ribbon. It comes from a sequence of papers, due to Darren Long and Andrew Casson. The last paper in the sequence is:

  • Algorithmic compression of surface automorphisms. Invent. Math 81 295--303 (1985).

As far as I know we're still some ways from implementing this algorithm but it would be interesting to see someone try.


There is in principle an algorithm to determine if a link is smoothly slice, in that it will terminate on a smooth slice link but on a non smoothly slice link it will run forever.

The algorithm: put your link in $S^3$, triangulate $S^3$ so that the link is transverse and normal in the triangulation (normal meaning appearing linear in each tetrahedron). Extend this triangulation of $S^3$ to a triangulation of $D^4$. Do a search for "normal" 2-manifolds in this triangulated $4$-manifold that bound the link, such that every component is discs. Normal meaning "looks linear and transverse to the skeleton in each 4-dimensional simplex". This is a linear programming problem but of course, if the link is slice it's slice discs may not appear in this triangulation. So you subdivide the triangulation of $D^4$, barycentrically. After enough iterations of this, any slice discs for your link have to appear, by a general position argument / linearization argument.

This could be turned into a semi-useful algorithm if there were useful upper bounds on the number of subdivisions you have to do.

I suppose you could make a similar algorithm for determining if a link is ribbon -- but staying entirely in the realm of triangulations of $S^3$. I believe ribbon discs require subdivision to appear as solutions of the normal surface equations -- i.e. they don't "normalize" so it can't be a straightforward application of normal surface theory. There likely has to be some kind of acceptance that the triangulation will have to get more complicated. Off the top of my head I don't have an example but if such examples haven't been worked out, they should be readily found using Regina.

  • $\begingroup$ Like you say, this algorithm will find a smooth slice disk if one exists, but will never tell you that one doesn't exist. $\endgroup$
    – Jim Conant
    Commented Feb 19, 2011 at 12:23

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