Now that I got a mutant-discriminating invariant... - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:46:50Z http://mathoverflow.net/feeds/question/67048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67048/now-that-i-got-a-mutant-discriminating-invariant Now that I got a mutant-discriminating invariant... Hauke Reddmann 2011-06-06T15:50:30Z 2011-07-04T20:22:13Z <p>...what can I do with the darn thing?</p> <p>Background: I read that still no Vassiliev Invariant with mutant-discriminating power is known (correct me if this is outdated). Now, my research lead to a whole load of mutant-discriminating invariants but I preferably work with <em>unoriented</em> knots. (Since my S matrices for unoriented and oriented knots look rather related, I could also provide with the oriented analogue - I just never checked whether that is also mutant-discriminating.) Anyway. Here is the S matrix, it's a rather simple 4D one:<br> S={{-c^(-1), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, c*(-c^(-2) + c^2), 0, 0, 0, 0, 0, 0, 0, 0, -c, 0, 0, 0},<br> {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, c^(-3), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 0, c^3, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},<br> {0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, c^3, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, c^(-3), 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0},<br> {0, 0, 0, -c, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0},<br> {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -c^(-1)}};<br> cap=cup={{0, 0, 0, (-I)*c}, {0, 0, 1, 0}, {0, 1, 0, 0}, {I/c, 0, 0, 0}}; writhe normalization=c^3, c is a free parameter. </p> <p>Now question 1: For example, if I have the skein relation for the Alexander polynomial, I can do a lot of tangle calculus with that until I arrive at a skein relation where I can reverse an arrow and the equation still holds, thus effectively landing in my familiar unoriented knot area again. Could this also work for some Vassiliev invariant of sufficient high order (or a linear combination of several VI)? The snag is of course VI are defined via Gauss diagrams, not skein relations, and I have no idea whether "reversing one arrow" does make sense at all in that context.</p> <p>Question 2: How do I check <em>practically</em> if a certain S matrix (say, for oriented knots - I have tons of them either) is also a VI? OK, it has to vanish for n doublepoints but how do I actually <em>compute</em> that (I can only prove that S is <em>not</em> a VI by computing the value of a random n-doublepoint knot - if the result is nonzero, fine, but if not, I must try another...and another...Does a result exist that a degree n VI is proven as such if you just check some f(n) "independent" knots?)</p> <p>(Soft) Question 3 of course is my first sentence. Is having a mutant-discriminating invariant of good <em>practical</em> use for knot taxonomists? After all, there is more than one way to distinguish two knots...</p> http://mathoverflow.net/questions/67048/now-that-i-got-a-mutant-discriminating-invariant/67075#67075 Answer by Noah Snyder for Now that I got a mutant-discriminating invariant... Noah Snyder 2011-06-06T18:46:45Z 2011-06-06T18:46:45Z <p>First remember that knot polynomials don't give VI without modifying them somewhat. You first do the change of variables q=e^h, then you expand your invariant as a power series, and finally the coefficients of this power series are VI. The key reason that this works is that the R-matrix and its inverse agree modulo h. I don't know your notation and examples well enough to see whether a similar argument would work here (after changing variables in a similar way). </p>