How to find the braids that when closed make the $6_1$ knot.

Question

I have the $6_1$ knot and my question is how can I easily find the braids that when closed make this knot, what's the easiest way in general for any knot.

Answer 1

What you want is Alexander's Theorem, which describes explicitly how to convert every knot (or link) into a closed braid. The process is pretty simple. Here is a nice description.

I'll say briefly what you have to do, and you can try it out yourself on a knot projection of the $6_1$ knot.

Draw a projection of the knot on the Euclidean plane. Perturb the knot so that its projection is in general position with respect to the rays based at the origin. The outcome of this perturbation is that the projection does not cross the origin, the projection is tangent to rays at some even number $2K$ of points, and these points subdivide the projection into $2K$ subsegments that travel alternately clockwise and counterclockwise around the origin. If $K=0$, i.e. if there are no points of tangency, then the knot goes only clockwise (or only counterclockwise) around the origin, and the closed braid is evident.

So the work comes when $K \ge 1$. That work consists of isotoping the knot to reduce $K$ inductively to zero. Roughly speaking what you want to do is to take each counterclockwise segment $S$ and isotope the knot so as to push $S$ across the origin, making it into a clockwise segment, reducing $K$ by $1$. This is tricky in general, because when you attempt to isotope $S$, other segments might be dragged along with it. What you do is to subdivide $S$ into small enough pieces each of which has at most one overcrossing or under crossing in the projection, and then you can drag each segment across the origin individually, one-at-a-time, and avoid dragging anything else.

Perhaps, if you start with a simple enough projection of the $6_1$ knot, it won't be very tricky.