# Bridge Presentation of a Knot

I'm looking for a way or a sort of hands-on technique to draw a bridge presentation of a Knot (or Link). Any leads or help will be really helpful. (At least for the case of 2 bridge knots)

The most common definition of bridge presentation is the minimum number of local maxima over all diagrams of a knot. Similarly, there will the same number of local minima. With this, we can think of all the maxima being at the same height and all the minima being at the same height, below the maxima.

Here is a diagram with 3 bridges. What is missing in the box will be what is called a braid. Braids only descend from top to bottom. Otherwise, there would be another bridge.

The problem here is that you can put in a braid so that the knot you get doesn't need all three of these bridges, so the bridge number can actually be lower than 3.

For two bridge knots specifically, you only use two semicircles on top and bottom. But it gets even easier for 2 bridge knots. (Again, be careful, this is just two bridge knots that have the following property.) In the braid for a 2-bridge knot, you can assume that one strand is a vertical line. Then the rest of the braid only has three strands, which alternate between twisting regions. See the image below.

For more on why this is possible, look up rational knots, which are the same as 2-bridge knots, and you can see more here and here.

Since you asked, here is an old hand drawn example of what you asked for. The knot that we start with is $7_7$, which is the rational knot $8/21$. That means we put that slope lines on a $I\times I$ square and $-8/21$ on the back. But for graph paper, we made it slope 1 on a 8 by 21 rectangle. So the first tangle in the upper left, we get the knot by attaching the two left hand side ends of the tangle and then the same for the right. Now, we have two corresponding things happen between each picture. We modify the diagram and we modify the fraction.

To go to the next image on the top, we see that we can pull red section back to where they meet one third of the way down the image. Then we can repeat this for all the rest of the arcs to get until just the arcs meeting the two top corners are left. Similarly, we can just repeat this whole process for the middle one third, to arrive at the second image.

For the fraction, we use partial fraction decomposition to see that $$\frac{8}{21} = \frac{1}{\frac{21}{8}} = \frac{1}{2+\frac{5}{8}}$$ The 2 represents the two whole squares we isolated that only have 1 crossing each.

Now, we repeat the whole process but with $\frac{5}{8}$ and all the rest until we get only ones in the numerators and the tangle decomposes to last image in the bottom right and the partial fraction decomposition (which got cut off) should be

$$\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{1+1/2}}}}$$

To go from the last image to the square tower (like figure 2 in Hatcher and Thurston's paper) you use a planar isotopy which is shown in Figure 8 of Kauffman and Lambropoulou's paper.

• that helped. Can you give me a reference which describes the techniques (or some more examples) of how to get from Fig.1 to Fig.2 in diagram in Hatcher's paper that you suggested? Commented Oct 11, 2017 at 15:36
• that was so helpful. Makes a lot of sense now. Thanks a ton. Commented Oct 13, 2017 at 3:28
• @SubhankarD. Glad to help. Commented Oct 13, 2017 at 3:35

These may be a good references: "Bridge presentation of virtual knots" January, 2008.