Talk:Knot polynomial
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what with the twisting number???
As you define it, the polynomials are not invariants! Just look at the figure 8 knot, which is the unknot twisted once. You have to normalize by dividing the polynome by the Kaufman relation. Also, what of disjoint sum of links? mousomer
What does it mean to "drop any one row" ??? Which row should I drop??
- Whatever row you wish.
- The 'n' rows you get from 'n' crossings are linearily dependent.
- Any 'n-1' of the rows are linearily independent.
- So just drop any one row+column to get a regular matrix. mousomer
This article really could use some improvement. First of all, it never mentions that the important part is not that these invariants are polynomials in one variable but that each coefficient is an invariant of a very special type, called Vassiliev invariant, or finite type invariant. Almost all of them can be obtained from the theory of quantum groups; they are also related to topological quantum field theory. This is a very rapidly developing field, with a huge body of literature available (such as Bar-Natan's paper in Topology). I know this field; however, I do not feel I could rewrite this article. I would rather write a new article, on Vassiliev invariants, from scratch. --Kirillov 1 Jan 2001 kirillov at math sunysb edu
Hmmm... this page is a disaster. Perhaps we should just start over, with some separate articles... --ScottMorrison 04:46, 5 Apr 2005 (UTC)
I agree this page is a disaster. I would like to delete the "Reasoning" section entirely, since it's mostly wrong (e.g., in the motivation for considering knot polynomials) when it's not irrelevant (e.g., in mentioning knot energies). The introduction needs to be rewritten as well, to include the relation to quantum field theory. For the Alexander polynomial, the algebraic-topological definition should be given as the primary definition. Probably the specific polynomials should get their own page. Any objections to me doing this? --Dylan Thurston 8 May 2205