*Lecture 3 (05.10.2015)*

To visualize knots/links, we will use certain *diagrams* instead of three-dimensional pictures. To produce a (two-dimensional) diagram from a (three-dimensional) link, we choose a (two-dimensional) plane in space and project the link onto the plane such that

- over each point of the diagram, there lie at most two points of the geometrical link. In particular, at each point in the diagram, at most two strands cross — these points are called the
*crossings*of the diagram, - at each crossing, it is indicated which strand passes over/under the other.

Using the export function of knotplot, you can check how links (curves in three-dimensional space) are converted to two-dimensional diagrams.

Here are some examples of diagrams:

And here are some examples of “not diagrams” (not to be confused with knot diagrams 😉 ) because these violate the rules we imposed above:

Clearly each geometrical link $L$ has many different diagrams, depending onto which plane we project. The diagrams of a topological link $\cal L$ are defined to be all the diagrams of all its representatives $L\in\cal L$ — that is, we can take any representative geometrical link, deform it by ambient isotopy as we like, and then project onto any plane.

Given a diagram $D$, we can not reconstruct the geometrical link $L$ it was projected from, because the relative height information in the direction perpendicular to the plane of projection is lost. However, since over/under crossings are indicated in $D$, we can unambiguously recover the type of $L$ from the diagram, i.e. the equivalence class $[L]$. That is, we can reconstruct a unique *topological* link from a diagram. Consequently, it makes sense to talk about *equivalence (ambient isotopy) of diagrams*: Two diagrams are called equivalent (ambient isotopic) if and only if they arise as projections from links that are equivalent (ambient isotopic). We will discuss this equivalence of diagrams further in the next lecture.

For now, let us introduce a basic quantity of a diagram: Its *crossing number.*

Definition 1.10:

- The
crossing number $|D|$ of a diagram$D$ is the number of all its crossings.- The
crossing number$C(L)$of a geometrical link$L$ is defined as $$C(L):=\min\{|D|\;:\;D \text{ is a diagram of some geometrical link } L’ \text{ with } L’\cong L\}.$$ Sometimes the crossing number is also called theorderof $L$.

There exist many diagrams that are equivalent but have different crossing number. In fact, given a diagram $D$, one can always find equivalent diagrams $D’$ such that $|D’|$ is larger (as much larger as we want) than $|D|$. To understand that mechanism, consider the following picture:

Here the dashed circle indicates that not the whole diagram of the link is shown, just the small part that is contained in the circle. In the depicted move, the part outside the circle remains the same. But the transformation in the circle clearly has the properties that a) it increases the crossing number by 1, and b) it leads to a diagram that is equivalent to the previous one.

Hence links do not have a “maximal crossing number”. But in view of the *minimum* taken in Definition 1.10. ii), crossing number is a well-defined quantity. It should be clear that crossing number is a link invariant. But to get used to the concepts, we give a formal proof of this fact.

Lemma 1.11:Crossing number is a link invariant.

*Proof: * Let $L_1,L_2$ be two geometrical links that are equivalent, $L_1\cong L_2$. We have to show that

$$c(L_1)=\{|D|\,:\,D \text{ is a diagram of some link } L’ \text{ with } L’\cong L_1\}$$

coincides with $$c(L_2)=\{|D|\,:\,D \text{ is a diagram of some link } L” \text{ with } L”\cong L_2\}$$

But a link $L’$ that is equivalent to $L_1$ is also equivalent to $L_2$, in fact $L’\cong L_1\cong L_2$ $\Rightarrow L’\cong L_2$. Similarly, a link $L”$ that is equivalent to $L_2$ is also equivalent to $L_1$. Hence the two sets of diagrams over which the minimum is taken to compute $c(L_1)$ and $c(L_2)$ are the same. In particular, $c(L_1)=c(L_2)$. This completes the proof.

Some remarks:

- The proof shows that crossing number is in fact a property of topological links. That is always the case with link invariants (or with general equivalence classes, for that matter). Suppose you have a function $F$ from the family of all geometrical links to some set $S$, and you want to define a “corresponding” function $\cal F$ from the set of all topological links to $S$. The obvious way to define $\cal F$ in terms of $F$ is as follows: Take a topological link, i.e. an equivalence class $[L]$, and define ${\cal F}([L]) := F(L)$. Does this make sense? Remember that on the left hand side, $[L]$ is a set of many (equivalent) geometrical links, whereas on the right hand side, only a single geometrical link, namely $L$, appears. So we have to show that if we pick some other geometrical link $L’\in [L]$, then our definition is consistent, i.e. it gives the same answer $F(L’)=F(L)$. Since $[L]$ consists precisely of all links that are equivalent to $L$, the definition of $\cal F$ will be consistent if and only if $F$ is a link invariant. That is, link invariants are nothing but well-defined functions from the family of all topological links to some set $S$.
- After the somewhat boring number of components, crossing number is another and much more interesting link invariant.

For example, the unknot has crossing number 0, because it has the diagram

which has zero crossings. If a knot has crossing number zero, it means that is is equivalent to the unknot.

Are there knots $K$ with crossing number $c(K)=1$? It was demonstrated in the lecture that this is not the case – any diagram with precisely one crossing is equivalent to a diagram with zero crossings, i.e. to the unknot. (One considers the picture of one crossing with four open ends and checks that all possibilities to connect the four ends result in either a diagram with exactly one crossing, which is however seen to be equivalent to a diagram with zero crossings, or a (lin) diagram with more than one crossing).

The same question for $c=2$ is left as an exercise:

Exercise 1.12:(Problem 1) Show that there exists no topological knot with crossing number two.

The trefoil has crossing number *at most three*: In its standard diagram

there are three crossings. Does there exist a diagram of the trefoil that has less than three crossings? According to what we saw above for crossing number 1 and 2, that would mean the crossing number of the trefoil would be zero, i.e. it would be equivalent to the unknot. Later we will find out that this is not the case, i.e. the crossing number of the trefoil is really three.

This discussion shows that although the crossing number is a link invariant, it is often very hard to determine for a a given link.

Sometimes it will be useful to give a diagram an **orientation**. In the case of a knot (= 1-link) this means that we give the curve a direction, indicated by an arrow on the diagram, which it keeps throughout the whole diagram. Hence there are two possible choices of orientation for each knot.

- In the case of an $n$-link, we give a direction to each of its $n$ components – in this case, there exist $2^n$ different orientations.
- For oriented links $L_1,L_2$, we have a natural notion of equivalence by requiring the existence of an ambient isotopy matching $L_1$ and $L_2$, including their orientations.
- In a oriented link diagram, we can distinguish between two different kinds of crossings, which gives rise to the definition of two numbers (writhe and linking number):

Definition 1.13:

- A crossing $c$ in an oriented link diagram is called
positive(or:a crossing of positive type), if the (oriented) overstrand has to be rotated counter-clockwise to match the (oriented) understrand. Otherwise the crossing is callednegative, ora crossing of negative type.Thesign of a crossingis defined as $$\varepsilon(c):=\begin{cases} +1 &\text{if } c \text{ is positive} \\ -1& \text{if } c \text{ is negative}. \end{cases}$$- The
linking numberof two components $K_1,K_2$ of an oriented link diagram is defined as $$\ell(K_1,K_2):=\frac{1}{2}\sum_c\varepsilon(c),$$ where the sum runs over all crossings $c$ in which $K_1$ passes over $K_2$, or vice versa. That is, crossings of $K_1$ with itself and crossings of $K_2$ with itself are not included in the sum.- The
writhe(ortwist number) of an oriented link diagram $L$ is defined as $$w(L):=\sum_c\varepsilon(c),$$ where the sum runs over all crossings of $L$.

There exist different ways for remembering how to determine the sign of a crossing. The one with the rule how the overstrand has to be rotated to match the understrand was used as the definition of the sign here, but we give two more (equivalent) characterisations:

First, one can take the point of view that the crossing in question is looked at from a direction such that the overstrand passes from the left to right (i.e., rotate the whole diagram until this is the case). If in this configuration, the understrand passes from bottom to top, the sign is positive, and negative otherwise.

Second, one can approach the crossing along the overstrand (“on the bridge”), in the direction given by its orientation. If the understrand passes below from the right to left, the sign is positive, and otherwise it is negative.

Note that writhe is *not* a link (or knot-) invariant. For example, below you see two equivalent diagrams (both corresponding to the knot type “unknot”)

The left diagram has writhe zero (no crossings), but the right diagram has writhe $-1$. Since both are equivalent, writhe cannot be an invariant of oriented knots.

Linking number, on the other hand, *is* an invariant (of oriented 2-links), as we will show next time. Regarding the definition of linking number, here is another question/exercise: Show that linking number is always an integer. This requires a proof because of the factor $\frac{1}{2}$ in the definition. If you need hints on how to do this, you can consider the book by Livingston, Exercise 2.6, page 37.