most recent 30 from http://mathoverflow.net 2013-05-19T23:49:36Z http://mathoverflow.net/feeds/question/66461 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66461/some-vassiliev-invariant-questions Hauke Reddmann 2011-05-30T15:07:31Z 2011-05-31T03:21:39Z

<ol> <li><p>The V.I. definition goes doublepoint=overpass-underpass (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.</p></li> <li><p>Stoimenow tabulated values for the 3-degree V.I., but only for knots. Has someone this data also for links? (Until 6 crossings already would be useful.)</p></li> <li><p>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).</p></li> </ol>

http://mathoverflow.net/questions/66461/some-vassiliev-invariant-questions/66479#66479 Ryan Budney 2011-05-30T18:36:06Z 2011-05-30T19:00:12Z

<p>1) No. You can always mirror-reflect your definition across the plane that defines over/under. </p> <p>2) I think so. Take a look in the <a href="http://front.math.ucdavis.edu/1103.5628" rel="nofollow">Chmutov, Duzhin, Mostovoy survey on the arXiv.</a> On page 92 they have the first ten non-trivial Vassiliev invariants, computed on sufficiently-many knots. If what you're looking for isn't near there, they likely have a reference for it. </p> <p>3) Type one Vassiliev invariants are trivial for knots, and for links I believe linking number is the only non-trivial 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?</p>