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

This question is about the relation between the notions of boundary link and ribbon link.

For the definition of ribbon link see: ribbon links - counterexamples.

An n-component link $L=L_1\cup\dots\cup L_n$ is said to be a boundary link if there exists an orientable surface consisting of n disjoint components $S=S_1\cup\dots\cup S_n$ such that for each $i\leq n$ we have $\partial S_i=L_i$.

There exists boundary links which are not ribbon and ribbon links which are not boundary (see Rolfsen book chapter 5 pag. 140)

It can be shown that every pure ribbon link is a boundary link. (A ribbon link is said to be pure if every ribbon singularity involves only one disc.)

Now my problem is: when is a boundary link ribbon? Clearly every component of such a link must be a ribbon knot so my question is:

  • $\textbf{Is every boundary link with ribbon components a ribbon link?}$

Assume the answer to the previous question is yes. Then the following question makes sense:

  • $\textbf{Is it true that a link is purely ribbon iff it is a boundary link}$ $\textbf{ with ribbon components?}$
share|improve this question

1 Answer 1

up vote 5 down vote accepted

The answer to your first question is no. There are non-ribbon boundary links whose components are unknotted! Indeed the Bing double of a knot is a boundary link with unknotted components, but it has recently been proven by several authors that the Bing double of the figure-8 knot is not slice (hence not ribbon.) See for example this paper.

Edit: The Bing double of the trefoil is also not slice, which is a less subtle calculation using the signature.

share|improve this answer
    
Thanks for your answer. I didn't know anything about this "Bing double" construction. Interesting source of (counter)examples! –  Paolo Aceto Dec 18 '10 at 2:45

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.