Geometric Topology (Math 2141, Math 2151)

Problem Set 3 - Solutions

1. The following picture indicates the necessary change.

2. Start with the picture from Question 1 and increase the number of twists as in the picture below. By ensuring that the knot is alternating, the crossing number will tend to infinity.

3. In the knot from Question 1, all the crossings are left-turns, so the writhe is -7. In the knot

all crossings are left-turns, so the writhe is -5. In the knot the top two crossings are right-turns, and the remaining crossings are left-turns, so the writhe is -2. In the Borromean rings, we first have to choose orientations to calculate the writhe. However, independent of the orientation we get that the writhe is zero, as between two components, we always get a crossing with sign +1 and one crossing with sign −1.

4. Depending on the orientation all crossings between components are either left- or right-turns. Thus the linking number is either 2 or -2.

5. Since two components of the Borromean rings alone are always unlinked, we get that the linking number is 0.

6. Let L = L1 t · · · t Ln and D = D1 t · · · t Dn a diagram for L such that each Di is a diagram for Li. Let us write C to be the set of all crossings of D involving more than one component, and for i < j let Cij be the set of all crossings between Di and Dj. If γ ∈ C is a crossing, we write ε(γ) ∈ {±1} for the sign of the crossing. Then by definition X lk(L) = lk(Li t Lj) 1≤i

because every γ ∈ C sits in exactly one Cij and each Cij ⊂ C. But this is what we have to show.

7. (a) The following is a diagram for the trefoil knot with 2 bridges. (b) This can be shown by an induction on the number of crossings in D. If there is only one crossing in D, the statement is clear. Now assume that D has n crossings and the statement holds for diagrams with less than n crossings. Since we have only one bridge, a subdiagram looks like

The right endpoint has to be connected to one of the undercrossings, as we have a knot diagram. If it were connected to the first crossing from the right, we could immediately remove that crossing and the result would follow by induction. Note that we cannot connect it to the second crossing from the right, as an underpass from the first crossing could not be connected to the rest of the diagram without producing a new crossing. Therefore it has to be connected to the 2m + 1st crossing from the right for some m ≥ 1.

But among the first 2m crossings from the right there has to be a subdiagram of the form , as we have to connect the under- passes without producing a new crossing. But then we can remove two crossings from the diagram without changing the knot, so the statement follows by induction. (c) The figure-8 knot has bridge number 2:

−→ −→

8. (a) Represent K by a diagram which realizes the crossing number. As every bridge has to go over at least one crossing, there can only be as many bridges in this diagram as there are crossings. Therefore b(K) ≤ c(K). (b) If the diagram realizing the crossing number is not alternating, there has to be a bridge which goes over at least two crossings, as there are just as many overpasses as there are underpasses. In that case we have less bridges than crossings so b(K) < c(K). If we have an alternating diagram and a nontrivial knot, the diagram contains a part as in the picture below to the left. After the alteration, we have one bridge less and so b(K) < c(K).

9. The following is a picture of the reef knot with three bridges, obtained by composing two pictures of the trefoil knot with two bridges

If we do the same with the granny knot, we get four bridges, however, by rotating the right trefoil by 180o, we get the following picture. 10. We have * + * + hDi = A + A−1

= −A10 + A−1h left trefoil i = −A10 + A6 − A2 − A−6.