Complexity of surfaces bounding knots in 4-ball and 3-sphere respectively - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:35:15Z http://mathoverflow.net/feeds/question/91569 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91569/complexity-of-surfaces-bounding-knots-in-4-ball-and-3-sphere-respectively Complexity of surfaces bounding knots in 4-ball and 3-sphere respectively Ryan Budney 2012-03-18T20:35:16Z 2012-03-19T17:19:38Z <p>I'm interested in a complexity question related to problems like the slice-ribbon problem. </p> <p>To be specific, if $K \subset S^3$ is a knot, it might be non-trivial yet still bound a smoothly-embedded compact 2-dimensional ball $B \subset D^4$, where we consider $S^3 = \partial D^4$, and the ball is required to be properly embedded, meaning $B \cap D^4 = \partial B = K$. This is called a (smoothly) <em>slice knot</em>. </p> <p>The slice-ribbon conjecture states that the only way this can happen is if you can find $B$ so that the distance function $d : D^4 \to \mathbb R$ given by $d(x) = |x|$ has no local maxima on $B$, i.e. it is Morse with only saddle points and local minima on $B$, and no other critical points. </p> <ul> <li>Q1: Do people have computations describing how many such saddle points you need for knots? In a ribbon diagram for the knot this is the number of components in the self-intersection set. I'm particularly interested in how this number would relate to a minimal-complexity planar knot diagram for the knot -- where you minimize crossing number or some other similar feature of a knot diagram. I want this complexity to be related to the number of tetrahedra you would need in a triangulation of $S^3$ so that the knot is a "normal curve" in the triangulation, meaning transverse to the triangulation and linear inside each tetrahedron. </li> </ul> <p>More generally, I'm interested in the computational problem of finding candidate knots that are smoothly slice yet not ribbon. In 3-manifold theory there is a "normalization" process for normal surfaces in a triangulated manifold. This involves some isotopy, and embedded surgery. There is a much more primitive type of "normalization" process that works in any triangulated manifold (of any dimension) with respect to any submanifold -- simply subdivide the ambient triangulation and perturb the submanifold to be transverse. After some number of subdivisions, the submanifold will have to appear to be linear in each top-dimensional simplex. This is the key idea in Whitehead's proof that smooth manifolds admit triangulations.</p> <ul> <li>Q2: Are there any good estimates for how many subdivisions it takes to "normalize" a submanifold in the above Whitehead sense? </li> </ul> <p>Q2 can be taken to be quite general. For example, for curves in surfaces, you can always normalize without any subdividing at all, all it takes is an isotopy. After minimizing the number of intersection points with the 1-skeleton of the triangulation, the only curve that needs more than a small isotopy (in the sense of the mapping space topology) would be the boundary of a disc embedded in the interior of a triangle. </p> <p>For example, is there an interesting topological invariant of a knot that gives good lower-bounds on the number of tetrahedra you need to triangulate $S^3$ to make it appear to be locally linear? Things like the maximum of the curvature aren't topological, and I consider crossing number in a planar diagram to be not computationally-friendly enough. </p> http://mathoverflow.net/questions/91569/complexity-of-surfaces-bounding-knots-in-4-ball-and-3-sphere-respectively/91600#91600 Answer by Paolo Aceto for Complexity of surfaces bounding knots in 4-ball and 3-sphere respectively Paolo Aceto 2012-03-19T09:25:11Z 2012-03-19T17:19:38Z <p>This is an answer to the first question.</p> <p>Let's indicate with $r(K)$ the ribbon number of a knot, i.e.the minimum number of ribbon singularities needed to realize a ribbon disc spanning $K$. We have $$r(K)\geq g(K)$$ This is shown by Fox here:http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm10_01_08.pdf</p> <p>Mizuma has shown that under certain conditions on the Alexander and Jones polynomial you can assume that $r(K)\geq 3$. This is Theorem 1.5 here:http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/1782ojm.pdf It is a very special situation and I don't think that much more is known in the general case.</p> <p>Maybe it is worth noting that given a band diagram for a ribbon disc one can add a fake ribbon singularity near each singularity and then eliminate both with a tubing operation. This produces a Seifert surface whose genus equals the number of the original ribbon singularities in the band diagram. The definition of a band diagram and a picture of this trick can be found here:http://etd.adm.unipi.it/theses/available/etd-07062011-061816/unrestricted/Polynomial_invariants_of_ribbon_links_and_symmetric_unions.pdf (pag. 27)</p>