Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'd love to have a list of 'small' 2-knots, for some sense of small. It's not clear what one should filter by, but there are two obvious candidates

  • Write a movie presentation, and count the frames.
  • Project the 2-knot to R^3, and count the triple points.

Does anyone know if this has been attempted? Such a list could be quite useful.

share|improve this question
add comment

2 Answers 2

up vote 7 down vote accepted

There is a table in Yoshikowa's paper based upon the representation as graphs with certain markings of vertices. The markings indicate which directions are the A and B smoothings. The table is quite small.

Kamada's book on Surface braids has a nice table of 2-knots that don't have triple points. There are tons of these. So your candidate for small is sort of strange. Braid index might be better.

Quandle cocycles are the only known ways to get lower bounds for triple points, and there are not nearly enough 3-cocycles that have been calculated.

Last spring Dennis Roseman told me a neat way to generate "random" 2-knots. I don't think he has written anything down about it though.

So if you look at movies, you have to be careful about what you define as an event. In the CRS point of view each of Reid. I,II,III, birth, death, saddle, switchback, and psi (pitchfork) moves is an event, but exchanges of distant critical points are also events. Any such event leaves its trace on the chart.

There are other notions of simplicity. For example, there is a notion of thickness. Project into 3-space and subsequently onto the plane (that gives the chart). Take a spear perpendicular to the plane and see how many times it generically passes through the surface. Find the maximum of these, and then minimize over all projections.

Probably the real reason for making a census would be to discover new invariants. So it is a good problem in that sense.

share|improve this answer
    
A problem with the Yoshikawa technique is that having one of those decorated knotted 4-valent graphs is not enough to specify an embedding $S^2 \to S^4$ -- when you resolve the 4-valent graph either way you get a trivial link. One also has to know the spanning discs for those two trivial links. Change your spanning discs and (in general) you change the knot. –  Ryan Budney Aug 8 '10 at 23:55
    
As far as I know, the assumption is not that one adds a generic spanning disk to the trivial link $L_t$. The assumption is that one adds a trivial disk system to $L_t$, for the definition of a trivial disk system see section 8.5 of "Braids and Knot Theory in Dimension Four" by Kamada. Any two trivial disk systems bounding a given trivial link $L_t$ are ambient isotopic, Proposition 8.6 of Kamada. Thus, the 2-knot type is independent of the trivial disk system which closes the trivial link $L_t$, Kamada Proposition 9.11. –  Kelly Davis Aug 3 '11 at 10:30
add comment

Another complexity filter could be the number of 4-dimensional simplices in a triangulation of the complement.

2-knots complements have simple ideal triangulations. The proof is very similar to the proof for 1-knots, viewed through the lens of Morse theory on manifolds with boundary. ie: the Wirtinger presentation is viewed as the fundamental group computation coming from the cell decomposition of a Morse height function (in the stratified sense) on the knot complement.

It'd be nice if there was a SnapPea-like algorithm for triangulating 2-knot complements. As far as I know those kinds of details haven't been worked out by anyone. But it should be reasonably doable.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.