# Computing an Invariant for Knots via Braid Words?

I've been reading up on Knot Theory (which is not my area of expertise) and am stuck in the following bit of logic:

Statement 1: Every knot can be represented as a braid.

Statement 2: There's a straightforward way of doing this.

Statement 3: Dynnikov gives an algorithm for telling if two braid words are equivalent via computing the "Dynnikov Coordinates".

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.

Is this right? If so, why isn't this considered the ultimate knot invariant? What obvious point am I missing?

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If you have two knots and two braids representing them, knowing that they're the same braid is a sufficient condition for knowing that the two knots are the same, but it's not necessary. The necessary condition was written down by Markov (en.wikipedia.org/wiki/Braid_theory#Closed_braids). –  Qiaochu Yuan Mar 8 '12 at 19:53

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.