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

2$\begingroup$ For those of us out of the loop, what is this conjecture? $\endgroup$ – Theo JohnsonFreyd Nov 28 '09 at 21:30

2$\begingroup$ 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 1handle attachments then is followed by 2handle attachments. ie: there is no mixedorder handle attachments. $\endgroup$ – Ryan Budney Nov 28 '09 at 21:45

$\begingroup$ The sliceribbon 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. $\endgroup$ – Daniel Moskovich Nov 29 '09 at 1:21
I think of the ribbonslice conjecture as a wish that would simplify certain 4D questions. Let me explain this in 3 examples.
 Given an embedded "ribbon disk" in 4space (where the Morse function has no local maxima) one can push it up into 3space 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 3space a "ribbon" (that is allowed to cut through itself). It contains the information about the embedded disk in 4space 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 3space, starting from the unlink and adding bands. No such simple 3Dpicture exists for arbitrary slice disks and one may wish that any slice knot is ribbon.
 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.
 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 4manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3spheres bound homology 4balls 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 connectsums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2bridge knots in the concordance group of knots in $S^3$.
EDIT: Not exactly addressing your question, I think of the sliceribbon conjecture as a primitive 4dimensional 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 1handle and 2handle attachments, in that order). You can mess up a ribbon disc by taking connectsums with 2knots. So modulo connect sums with 2knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the sliceribbon problem.
2nd edit: As far as I know, the sliceribbon conjecture has no major consequences. As I describe above, it's more of an ''outermarker'' type of conjecture. It's a measure of how well we understand knotting of 2dimensional things in 4dimensional things.
3rd edit: Here is a type of mild consequence that was pointed out to me recently. In my arXiv preprint on embeddings of 3manifolds in $S^4$ there's Construction 2.9 which creates embeddings of certain 3manifolds $M$ in homotopy 4spheres. The first step is to find a contractible $4$manifold $W$ that bounds the 3manifold $M$, 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 0handle, $n$ 1handles and $n$ 2handles (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 AndrewsCurtis 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 4dimensional Poincare conjecture from a counterexample to sliceribbon? (I've heard somebody mention something like this, but did not understand) $\endgroup$ – Daniel Moskovich Dec 1 '09 at 13:23

$\begingroup$ FYI, in Kirby's problem list Cameron Gordon decomposes the sliceribbon 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. $\endgroup$ – Ryan Budney Dec 1 '09 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 welldisguised 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 4ball, 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 counterexamples. I did a little Googling around to see if anyone has shown those knots to be slice but I haven't found anything. $\endgroup$ – Ryan Budney Aug 4 '10 at 0:41