The V.I. definition goes doublepoint=overpassunderpass (or was it the other way around? If it's 50:50, I score 0 always :). Would it lead anywhere to define doublepoint=overpass+underpass? (Even if it's equivalent. As I wouldn't have to guess then :) I ask because I hate arrows on knots.
Stoimenow tabulated values for the 3degree V.I., but only for knots. Has someone this data also for links? (Until 6 crossings already would be useful.)
Is this logic correct? Suppose you have a V.I. vanishing on all knots with >=2 doublepoints. Consider a clasp consisting of 2 doublepoints. From the definition then S^2+S^(2)2*S^0 = 0, where S is the usual braid generator resp. S matrix, and thus you can deduct a skein equation from the degree of a V.I. (and the higher the degree, the more of them).

$\begingroup$ retagged: general topology > quantum topology. General topology means pointset topology: en.wikipedia.org/wiki/General_topology $\endgroup$– Daniel MoskovichMay 31 '11 at 3:22
1) No. You can always mirrorreflect your definition across the plane that defines over/under.
2) I think so. Take a look in the Chmutov, Duzhin, Mostovoy survey on the arXiv. On page 92 they have the first ten nontrivial Vassiliev invariants, computed on sufficientlymany knots. If what you're looking for isn't near there, they likely have a reference for it.
3) Type one Vassiliev invariants are trivial for knots, and for links I believe linking number is the only nontrivial one. I'm not following your argument but I'm also not seeing where it's going. Perhaps this is what you're trying to prove?

$\begingroup$ THX. 1) So Vassiliev invariants and undirected knots don't go together? 2) is then superfluous for my goal (but still, that link is a goldmine, many many THX!). Considering 3), the type number wasn't relevant. Could as well have used a higher order for the example: 0=S^33S+3S^1S^3 or 0=S^44S^2+6S^04S^2+S^4  I just wanted to know whether you can handle the definition as if it were an ordinary skein equation. $\endgroup$ Jun 1 '11 at 14:09