Low Dimensional Topology

December 2, 2010

Slice-Ribbon Conjecture in danger!

Filed under: Uncategorized — dmoskovich @ 12:41 pm

A few days ago, I found gold in my inbox.
It was in a mass-mailing by Geometry & Topology Publications, announcing seven new papers. It was one of these, Fibered knots and potential counterexamples to the Property 2R and Slice-Ribbon Conjectures by Bob Gompf, Martin Scharlemann and Abigail Thompson, which really blew me away.
The paper is a gripping read. As I see things, its breakthrough result is the discovery of a family of slice knots which are unlikely to be ribbon, bringing my confidence in the veracity of the Slice-Ribbon Conjecture down from around 60% (what can I say? I was an optimist!) to around 5%. After more than 30 years, the Slice-Ribbon Conjecture is encircled and in imminent fear of annihilation.

A knot is slice if it bounds a disc in the 4-ball. It’s ribbon if it bounds a self-intersecting disc in S^3 with only ribbon singularities. It’s pretty easy to see that every ribbon knot must be slice (add one dimension and resolve the ribbon singularities using it), but there’s no a-priori reason to suspect that every slice-knot would be ribbon… except that experimentally, every slice knot we know of happens to also be ribbon. We know loads of slice knots, which don’t have much in common except the fact that they’re slice… and they’re all ribbon. All slice-knots which are 2-bridge are ribbon by a result of Lisca, and by a result of Greene and Jabucka, so are 3-stranded pretzel knots. And so, there’s strong experimental support for Fox’s:

Slice-Ribbon Conjecture: Every slice knot is ribbon.

The Slice-Ribbon Conjecture is a wish, which would simplify various 4-dimensional questions. In Israel one would call such a thing a pink dream. I wish it were true.
This paper (which I shall call GST for Gompf-Scharlemann-Thompson, not for Goods and Services Tax) tells us to wake up from our dream. It proposes a family of potential counterexamples to the Slice-Ribbon Conjecture, coming from proposed counterexamples to the Generalized Property R Conjecture.
Recall that David Gabai proved:

Property R: If 0-framed surgery on a knot K\subset S^3 yields $S^1\times S^2$ then K is the unknot.

Generalized Proporty R is the generalization of this statement to links:

Generalized Property R Conjecture: If surgery on an n-component link L gives rise to a connect sum of n copies of S^1\times S^2, then L becomes the unlink after handle-slides.

Sadly, GST tells us that this conjecture has about as much chance of being true as the Andrews-Curtis Conjecture, which looks about as plausible as P=NP. More concretely, a square knot interleaved with the connect-sum of an (n,n+1) torus knot with its mirror (call this 2-component link L_{n,1}) is a likely counterexample to the Generalized Property R Conjecture, which they show is implied by the fact that a balanced presentation of its link group is a long-known likely counterexample to the Andrews-Curtis Conjecture.
If a link were to satisfy the Generalized Property R Conjecture, then all of its band-sums (including knots) would be ribbon, but the hypothesis implies only that all such band-sums must be slice, by an (easy) result of Jonathan Hillman, quoted as Theorem 2.3 of GST. No issue arises with smooth structure in these examples, and so L_{n,1} have no reason at all to be ribbon, although they are certainly slice. In fact, I’d be really surprised if they would turn out to be ribbon. And it wouldn’t contradict the experimental evidence, because the simplest of these knots (for n\geq 3) has 53 crossings (Figure 2 of GST), and so is far beyond the range of any experiments conducted up until this point in time.
There are a zillion ribbon-obstructions with which we could now bombard L_{n,1} for n=3,4,5, and if one of these works then the Slice-Ribbon Conjecture is toast. In fact, this is exactly what I think will happen. Somebody will write a computer programme to evaluate a ribbon obstruction on one of these (it looks horrible to do by hand!), and they’ll be no reason to suspect it will vanish.
Finally, something which doesn’t interest me personally as much (although it relates to the smooth 4-dimensional Poincare Conjecture): GST suggests a weaker Generalized Property R conjecture in which we’re allowed to stabilize by introducing canceling Hopf pairs. It wouldn’t be surprising if that were true, because 4-manifold topology behaves much more nicely after stabilization.
But what an exciting paper! If Slice-Ribbon survives this particular onslaught, I’ll eat tip my hat!

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6 Comments »

  1. Could you give some references to the zillions of ribbon obstructions?

    Comment by Ian Agol — December 3, 2010 @ 1:41 am | Reply

    • :D
      Thanks for the question! Of course, being a “coloured knots” person, I was thinking of Cappell and Shaneson’s formula for the Rohlin invariant of the irregular dihedral cover. The formula can be extended to any irregular covering space associated to a finite metabelian group (use intersection homology instead of homology in the construction, because the cobordism they use has singularities when the group is more general) (at least if the commutator subgroup is a cyclic group to some power, of order coprime to the order of the abelianization). I started doing this but never carried it through- anyway, there’s no reason for an obstruction associated to a dihedral cover not to be enough. It’s a ribbon obstruction, because it contains terms which depend on the choice of characteristic knot (which is information about the colouring), which will contribute zero if the knot is ribbon, by the way the cobordism is constructed (the detailed construction appeared a decade later, in Linking Numbers in Branched Covers).
      Given a Seifert matrix for the knot, and another one for the characteristic knot, you can calculate the expression. You can find the characteristic knot because its homology class is the Alexander dual to the cohomology class represented by the colouring under the universal coefficient theorem. So I expect it can all be programmed into a computer without too much trouble (easier said than done, like everything in this world).
      So there you have a zillion ribbon obstructions. One ribbon obstruction for each representation of the knot group onto a dihedral group whose commutator subgroup is cyclic of order p, where p ranges over the first zillion odd primes.
      Maybe there are many other ribbon obstructions known in the literature- but these are the ones I was thinking of.

      Comment by dmoskovich — December 3, 2010 @ 7:47 am | Reply

      • Looks like Friedl has some ribbon obstructions based on metabelian reps.:

        http://front.math.ucdavis.edu/0305.5402

        Comment by Ian Agol — December 3, 2010 @ 4:04 pm

      • Here’s a completely different ribbon obstruction which seems to have been unjustly ignored (mathscinet lists no citations):
        Theorem H of V. Turaev, Multiplace generalizations of the Seifert form of a classical knot, Math. USSR, Sb. 44(3) (1983), 335–361.

        Comment by dmoskovich — December 22, 2010 @ 11:28 am

    • Yeah… invariants association to metabelian reps should crack it if these knots are indeed not ribbon (and if they are indeed counterexamples to Generalized Property R, I see no reason why they would be). And if not, Cappell-Shaneson should extend also to polycyclic groups, just by iterating their construction, so one can also look deeper into the derived series of the knot group.

      Comment by dmoskovich — December 4, 2010 @ 6:17 pm | Reply

  2. A knot $K$ is called homotopy ribbon if it bounds a locally flat disk $D$ such that the map $\pi_1(S^3\setminus K)\to \pi_1(D^4\setminus D)$ is surjective.
    If a knot is ribbon, then it is also homotopy ribbon. There are various obstructions to being homotopy ribbon, e.g. the ones Ian mentioned, but presumably such invariants are not helpful in this case.

    Comment by Stefan Friedl — December 7, 2010 @ 5:29 pm | Reply


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