## Invariants

How can we tell if two tangles (or links, or knots) are different? That we cannot move the strings around as we are allowed and get from one tangle to another? We find invariants which can tell the difference. The best way to explain what an invariant is, is to give an example. The component number of a tangle is the number of strings in the tangle. Clearly if two tangles have different number of strings then they are not the same. For example the trefoil knot has component number 1 and the Hopf link has component number 2.

Trefoil Knot

An invariant is some mathematical object, like a number or a polynomial that we can associate to tangles (or links, or knots) that depends only on the tangle-type. For instance the component number of a tangle doesn’t change when the strings move about or are stretched. Therefore, it is an invariant.

The component number is rather a blunt invariant. What if we want to tell the difference between tangles with the same component number? Let’s define an invariant for links with component number 2. We will call it the linking number. The linking number is actually an invariant for “oriented” links with component number 2. Oriented means that each string in the tangle comes with a preferred direction. We indicate this in a diagram by drawing an arrow on each string.

Whenever two different strings cross we can use the right hand rule to assign a positive or negative value to the crossing. Put your thumb in the direction (according to the orientation) of the over-strand and your fingers in the direction of the under-strand. If your palm is facing up (away from the screen) then it is a positive crossing and if your palm is facing down (towards the screen) then it is a negative crossing.

Signs of Oriented Crossings

Now think of our link as having components (strings) called A and B. The linking number Lk(A,B) is the sum of the signs of the crossings in which A crosses over B. In order to see that the linking number is an invariant we need to analyze its behavior under Reidemeister moves.

Reidemeister 1

Consider the first Reidemeister move. The left part of the equation has a crossing, but it comes from only 1 component, so it contributes 0 to the linking number. The same applies to the right part of the equation. The middle part of the equation has no crossings and so it contributes 0 to the linking number. Thus the linking number is invariant under Reidemeister 1.

Reidemeister 2

Consider the second Reidemeister move. There are two cases, either the strands come from the same component or different components. In the first case, the left side of the equation contributes 0 to the linking number. In the second case, no matter which orientation there is on the strands, the two crossings have opposite signs and so contributes 0 to the linking number. In either case, the right side of the equation has no crossings and so contributes 0 to the linking number. Thus the linking number is invariant under Reidemeister 2.

Reidemeister 3

Consider Reidemeister 3. Notice that each pair of strands cross in the same way but in different places on each side of the equation. Thus, no matter which components the strands belong to, nor which orientation we give them, each side contributes the same to the linking number. Thus the linking number is invariant under Reidemeister 3.

Thus, the linking number Lk(A,B) is an invariant of 2 component oriented links. Even better, it’s symmetric Lk(A,B)=Lk(B,A). So we can calculate it by summing the signs of the crossings where B crosses over A.

We can easily calculate the linking number for the oriented Hopf link pictured above, Lk(A,B)=-1.

What happens if we try to calculate the self-linking number of a knot Lk(K,K). Unfortunately it is no longer invariant under Reidemeister 1, since the argument we had used to prove invariance required that we were calculating linking number Lk(A,B) between different components A and B. You can see that the arguments for Reidemeister 2 and 3 did not require that the components were different so that the self-linking number, which we shall call the writhe Wr(K)=Lk(K,K), is invariant under Reidemeister 2 and 3. Furthermore, it does not depend on the orientation, since switching the orientation will not change the sign of a crossing (the orientation switches on both strands, so the sign is preserved).

In order to remedy the problem of non-invariance of the writhe under Reidemeister 1, we introduce a new property of tangles, “framing”. If orientation can be thought of as arrows going parallel to the tangle, then framing can be thought of as arrows going perpendicular to the tangle. If we extend the tangle along these arrows we obtain a “ribbon”, that is a tangle whose components are 2-dimensional surfaces. Now the self-linking number makes sense, as the linking number of the two edges of the ribbon.

We can project framed tangles in such a way that the ribbon is flattened in the projection. Then we need only draw the ribbon using a string as before and we can extend that string perpendicularly in the plane of projection. The resulting framing is called the “blackboard framing”. Such diagrams represent equivalent tangles if and only if they are connected by a sequence of Reidemeister 2 & 3 moves and the framed Reidemeister 1 move.

Framed Reidemeister 1

Notice that no matter what orientation is chosen, both sides have negative crossings. Since the sign of the crossing cannot change, the writhe is invariant under framed Reidemeister 1. Thus, the writhe, Wr(K), is an invariant of framed knots.

We have introduced two interesting new invariants, the linking number Lk(A,B) and the writhe Wr(K) but in order to do so we had to add more structure to tangles, orientation and framing. That these structures are natural as well as closely related is hinted at by our study of invariants. The linking number is sensitive to orientation but not framing and the writhe is sensitive to framing but not orientation. We will have more to say about these features of tangles in the future.

### 3 Responses to “Invariants”

1. Brandon Williams Says:

Yo, I’m at laguardia waiting for my flight and thought I would check out your posts.

I think you need to flip a crossing in that last picture…

2. Eitan Says:

Fixed, thanks. Have a good flight.

3. professional landscape lighting company Says:

Great post. I’m going through some of these issues as well..