Computing an Invariant for Knots via Braid Words? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:01:35Z http://mathoverflow.net/feeds/question/90621 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90621/computing-an-invariant-for-knots-via-braid-words Computing an Invariant for Knots via Braid Words? Aeryk 2012-03-08T19:45:11Z 2012-03-08T19:58:23Z <p>I've been reading up on Knot Theory (which is not my area of expertise) and am stuck in the following bit of logic:</p> <p>Statement 1: Every knot can be represented as a braid. </p> <p>Statement 2: There's a straightforward way of doing this.</p> <p>Statement 3: Dynnikov gives an algorithm for telling if two braid words are equivalent via computing the "Dynnikov Coordinates".</p> <p>So it seems to me that in practice, if you have two knots and want to determine if they are equivalent, you turn them both into braids and compare Dynnikov Coordinates. </p> <p>Is this right? If so, why isn't this considered the ultimate knot invariant? What obvious point am I missing?</p> http://mathoverflow.net/questions/90621/computing-an-invariant-for-knots-via-braid-words/90623#90623 Answer by Kevin Walker for Computing an Invariant for Knots via Braid Words? Kevin Walker 2012-03-08T19:58:23Z 2012-03-08T19:58:23Z <p>Two braids represent the same link iff they are equivalent under the two "Markov moves". One of the Markov moves is conjugation within a fixed braid group \$B_n\$, and I expect Dynnikov coordinates would be helpful there. The other Markov move is a stabilization from \$B_n\$ to \$B_{n+1}\$ (related to a Reidemeister type I move), and I'm guessing that's where the flaw in your argument is. See also Qiaochu Yuan's comment.</p>