What would the slice-ribbon conjecture imply for 4-dimensional topology?
I've heard people speak of the slice-ribbon conjecture as an approach to the 4-dimensional smooth Poincare conjecture, and to the classification of homology 3-spheres which bound homology 4-balls. But I've never understood what they were talking about.

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    $\begingroup$ For those of us out of the loop, what is this conjecture? $\endgroup$ Nov 28, 2009 at 21:30
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    $\begingroup$ A smooth knot in $S^3$ is slice if it bounds a smoothly embedded $D^2$ in $D^4$ -- the $D^2$ in $D^4$ is called the slice disc. A ribbon knot is a slice knot where the $D^2$ in $D^4$ has a special position -- the function that gives you the distance from the origin in $D^4$ is a morse function on the slice disc, and the associated cobordism starts at the knot, has 1-handle attachments then is followed by 2-handle attachments. ie: there is no mixed-order handle attachments. $\endgroup$ Nov 28, 2009 at 21:45
  • $\begingroup$ The slice-ribbon conjecture is that every slice knot must be ribbon: garden.irmacs.sfu.ca/?q=op/slice_ribbon_problem_0 The basis for the conjecture seems to be just that every known slice knot happens to also be ribbon ("experimental evidence"), and I do not know a conceptual reason to believe it to be true. $\endgroup$ Nov 29, 2009 at 1:21

3 Answers 3


I think of the ribbon-slice conjecture as a wish that would simplify certain 4D questions. Let me explain this in 3 examples.

  1. Given an embedded "ribbon disk" in 4-space (where the Morse function has no local maxima) one can push it up into 3-space and obtain an immersed disk (whose boundary is still the given knot) where the singularities are mild: these are the so called "ribbon singularities", arcs of double points such that on one of the sheets, the arc lies in the interior. (Picture...) One would actually call this immersed disk in 3-space a "ribbon" (that is allowed to cut through itself). It contains the information about the embedded disk in 4-space by pushing one sheet of each ribbon singularity into the forth dimension. There is a fairly obvious algorithm how to create all such ribbons in 3-space, starting from the unlink and adding bands. No such simple 3D-picture exists for arbitrary slice disks and one may wish that any slice knot is ribbon.
  2. A knot K is slice if and only if there is a ribbon knot R such that the connected sum K # R is ribbon. One may wish that one didn't have to stabilize.
  3. Consider the monoid M of oriented knots under connected sum. If -K is the reversed mirror image of K then K # (-K) is ribbon. So it's very tempting to try to turn M into a group (where -K would become the inverse of K) by identifying two knots K' and K if K' # (-K) is ribbon. But the wish doesn't come true: this is not an equivalence relation and if we force it to be one then by 2 we end up with the knot concordance group (where two knots K' and K are identified if K' # (-K) is slice).

It is amazing that there are no proposed counter examples to this conjecture, not even for links.


I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

2nd edit: As far as I know, the slice-ribbon conjecture has no major consequences. As I describe above, it's more of an ''outer-marker'' type of conjecture. It's a measure of how well we understand knotting of 2-dimensional things in 4-dimensional things.

3rd edit: Here is a type of mild consequence that was pointed out to me recently. In my arXiv preprint on embeddings of 3-manifolds in $S^4$ there's Construction 2.9 which creates embeddings of certain 3-manifolds $M$ in homotopy 4-spheres. The first step is to find a contractible $4$-manifold $W$ that bounds the 3-manifold $M$, then you double $W$ to get a homotopy $S^4$. If the link used in the construction is a ribbon link, the contractible manifold $W$ admits a handle decomposition with one 0-handle, $n$ 1-handles and $n$ 2-handles (for some $n$) and no higher dimensional handles. So the homotopy $S^4$ constructed that contains $M$ is diffeomorphic to $S^4$ provided the corresponding presentation of $\pi_1 W$ is trivializable by Andrews-Curtis moves (handle slides for the handle presentation). This argument will appear in the next draft of the paper, which should appear before January.

  • $\begingroup$ Wasn't there some construction of a counterexample to the 4-dimensional Poincare conjecture from a counterexample to slice-ribbon? (I've heard somebody mention something like this, but did not understand) $\endgroup$ Dec 1, 2009 at 13:23
  • $\begingroup$ FYI, in Kirby's problem list Cameron Gordon decomposes the slice-ribbon problem into two problems -- see Problem 4.22 on the list. The intermediate notion is "homotopically ribbon". I'm unaware of this construction you're referring to. $\endgroup$ Dec 1, 2009 at 18:01

I have a question concerning Peter Teichner's answer. Aren't there candidate counterexamples, for instance in the following paper in `Topology and its Applications':

Some well-disguised ribbon knots

Robert E. Gompf, 1 and Katura Miyazakib, , 2

Abstract For certain knots J in S1 × D2, the dual knot J* in S1 × D2 is defined. Let J(O) be the satellite knot of the unknot O with pattern J, and K be the satellite of J(O) with pattern J*. The knot K then bounds a smooth disk in a 4-ball, but is not obviously a ribbon knot. We show that K is, in fact, ribbon. We also show that the connected sum J(O) # J*(O) is a nonribbon knot for which all known algebraic obstructions to sliceness vanish.

  • $\begingroup$ Thanks, I didn't know about that paper. Yes, those are proposed counter-examples. I did a little Googling around to see if anyone has shown those knots to be slice but I haven't found anything. $\endgroup$ Aug 4, 2010 at 0:41

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