An n-component link is said to be ribbon if it bounds a ribbon surface consisting of n discs. (a ribbon surface is an immersed surface with only ribbon singularities, see http://en.wikipedia.org/wiki/Ribbon_knot).

Let $L=L_1\cup,...,\cup L_n$ be an n-component link (n>1). The following conditions are necessary for L to be ribbon:

• for each pair $(L_i,L_j)$ we have $lk(L_i,L_j)=0$ (lk=linking number)
• each $L_i$ is itself a ribbon knot
• Let V(L) be the Jones Polynomial of L and $V(O^n)$ be the Jones polynomial of the trivial n-component link then $V(O^n)$ is a factor of V(L) i.e. $V(O^n)|V(L)$ (Eisermann arxiv.org/abs/0802.2287)

My questions are:

• Is there any example of a link which is not ribbon satisfying the previous conditions?

• Are there any other necessary conditions?

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The Borromean rings consists of unknots and pairwise linking numbers are zero. But I'm pretty sure the Borromean rings aren't ribbon nor are they slice -- basically by a fundamental group computation, that one strand represents a commutator in the complement of the other two. – Ryan Budney Oct 15 '10 at 14:51
I forgot to mention another less obvious condition. Let V(L) be the Jones Polynomial of L and V(O^n) be the Jones polynomial of the trivial n-component link if L is ribbon then V(O^n) is a factor of V(L) i.e. V(O^n)|V(L) (Eisermann arxiv.org/abs/0802.2287) – Paolo Aceto Oct 15 '10 at 15:20
Presumably there are restrictions on the Alexander module for the universal abelian cover, comparable to the restriction on slice knots that the Alexander polynomial has to have the form $f(t)f(t^{-1})$ for some polynomial $f(t)$. Hasn't something like this been worked out somewhere? – Ryan Budney Oct 15 '10 at 17:03