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Knots

Introduction

There is a really good web site for knots, called “The KnotPlot Site”, created by Robert Scharein. In particular, under the heading "Knot tables" on the page http://www.cs.ubc.ca/nest/imager/contributions/scharein/knot-theory/knot-theory.html there is a beautiful display of 16 knots. Here they are:

The top row shows the trefoil knot, the figure-eight knot, two knots with five crossings, three knots with six crossings, and one with seven. The second row shows six more knots with seven crossings, and two with eight crossings. Altogether, there are 21 knots with eight crossings, and 49 with 9 crossings. In fact, if we just count prime knots, we have:

Crossing number
Number of prime knots
3
1
4
1
5
2
6
3
7
7
8
21
9
49
10
165
11
552
12
2 176
13
9 988
14
46 972
15
253 293
16
1 388 710

We will be following the excellent book The Knot Book by Colin Adams, in working through a sequence of ideas and exercises. On the way, we will discover what the crossing number is, what a prime knot is, and we will also begin to see how this extraordinary table was drawn up. I have used the same exercise numbers as the book, even when I have slightly modified the wording.

Projections of knots

Firstly, note that the pictures we have been looking at are not knots at all; they are projections of knots.

Exercise 1.7: Show that by changing the crossings from over to under or vice versa, any projection of a knot can be made into a projection of the unknot.

An alternating knot is one having an alternating projection: as you walk around the knot the crossings alternate between under and over.

Exercise 1.6a: Show that by changing the crossings from over to under or vice versa, any projection of a knot can be made into a projection of an alternating knot.

Exercise 1.6b: In a projection with n crossings, what is the maximum number of crossings that would have to be changed in order to make the knot alternating?

Reidemeister moves

A given knot might be projected in many different ways. How do we know when two projections are of the same knot? In 1926 the topologist Kurt Reidemeister proved that two projections of the same knot can be related by a sequence of moves, which we now call the Reidemeister moves.

Here is the first Reidemeister move, R1:

Here is the second Reidemeister move, R2:

Here is the third Reidemeister move, R3:

This is a very powerful theorem, but its limitation is that we cannot predict how many such moves will be needed in any particular case.

Exercise 1.11: Find a sequence of Reidemeister moves which untangle the unknot shown below.

Three-colourings

We know now how to show that two projections are of the same knot. Suppose, however, that we have two projections which we suspect are of different knots. How would we show that? There are many ways; perhaps the simplest is to use the idea of three-colourability. At each crossing of a projection there are three arcs: the over-crossing arc and the two parts of the under-crossing arc. Now imagine drawing a projection using three different coloured pencils. The projection is called three-colourable if you can draw it such that at every crossing the three arcs are always either the same colour or use all three colours. It is fairly easy to see that the types 1 and 2 Reidemeister moves preserve three-colourings.

Exercise 1.23: Show that the type 3 Reidemeister move preserves three-colourability.

Exercise: Show that the trefoil knot is distinct from the unknot.

Exercise 1.21: Which of the three six-crossing knots are three-colourable?

Joining two knots

Given two knots K and L we can join them together to make a new knot which we call K#L. The knots K and L are called the factors of K#L. Here is a picture of two trefoil knots joined together:


Of course, if U is the unknot then U#K is always equal to K, for any knot K. So if we think of # as a sort of multiplication, U behaves a little like the number 1. A prime knot is one which cannot be represented as the join of two knots (unless one of them is the unknot).

Exercise 1.25: Show that the join of any knot with a three-colourable knot gives a new three-colourable knot.

Unsolved question: Find a way to generalize three-colourability to show that if the join of two knots is the unknot, then one of the factors must have been the unknot.

This shows us that # differs from multiplication in that there are no inverses.
Exercise 3.15: Show that if two knots are both alternating then so is their join.

Unknotting number

We have seen that in any projection it is possible to change some of the crossings to make a projection of the unknot. Given a knot K, imagine studying all of its projections in order to find the one which needs the smallest number of crossing changes to unknot it. This number is called the unknotting number u(K). It can be very difficult to find, even for quite simple knots.

Unsolved question: Find a simple proof that a knot with unknotting number 1 is prime.

Unsolved question: Is it true that a knot with unknotting number 2 cannot made by joining three factor knots?

Exercise: Show that u(K#L) is less than or equal to u(K)+u(L).
It is an old conjecture that in fact u(K#L)=u(K)+u(L).

Crossing number

Given a knot K, imagine studying all of its projections in order to find the one which has the smallest number of crossings. This number is called the crossing number c(K).

Unsolved question: Show that c(K#L)=c(K)+c(L).

Jones polynomials

The first polynomial invariant of knots was discovered by James Alexander in 1923. After that, knot theory was relatively quiet until Vaughan Jones discovered a completely new polynomial invariant in 1985; this was the catalyst for a burst of ideas and results. One of the key new ideas is that of a skein relation, which gives us a recursive method of defining and calculating a knot polynomial.

Given a knot K the Jones polynomial V(K) is calculated from a projection of K. It is called an invariant because if two projections have different polynomials then they must be projections of different knots. We calculate it by using the following two axioms.

Axiom 1

\begin{displaymath}V({\rm unknot}) = 1 \end{displaymath}

Axiom 2 Whenever three oriented knots $ K_+ , K_- , and K_0 $ are the same except near one crossing where they differ as shown below, then

\begin{displaymath} t^{-1}V(K_+) - tV(K_-) + (t^{-{1/2}} - t^{{1/2}})V(K_0) = 0 \end{displaymath}

This is called the skein relation.

We will show how to calculate $ V(T_-) $ where $ T_- $ is the trefoil, and the minus refers to the crossing indicated by the arrow in the picture, which also shows the orientation of the knot. We also show $ T_+ $ and $ T_0 $ :

After Reidemeister moves 2 and 1, $ T_+ $ becomes the unknot, and so $ V(T_+) = 1. $ $ T_0 $ is not a knot at all: it is a link. Luckily, the Jones polynomial works for these too.

Let us rename T0 as L -, where this minus sign refers to the crossing indicated. Again we also show L + and L0:

After Reidemeister move 2, L + becomes the disjoint union (denoted ) of two unknots. Here we use the following theorem, which is quite easy to prove from the two axioms.

Theorem 1

$ V(L $ unknot) = $(-t^{-1/2} - t^{1/2})V(L) $

We simply take L to be the unknot, and obtain $ V(L_{+}) = -t^{-1/2} - t^{1/2}.$ After Reidemeister move 1, L0 becomes the unknot, so

\begin{displaymath}V(L_{0}) = 1. \end{displaymath}

Hence the skein relation

\begin{displaymath}t^{-1}V(L_{+}) - tV(L_{-}) + (t^{-1/2} - t^{1/2})V(L_{0}) = 0 \end{displaymath}

becomes

\begin{displaymath}t^{-1}(-t^{-1/2} - t^{1/2}) - tV(L_{-}) + t^{-1/2} - t^{1/2} = 0. \end{displaymath}

Therefore

\begin{displaymath}V(L_{-}) = (1/t) (-t^{-3/2}- t^{-1/2} + t^{-1/2} - t^{-1/2}) = -t^{-5/2} - t^{-1/2} \end{displaymath}

Now we put this result into the following skein relation:

\begin{displaymath}t^{-1}V(T_{+}) - tV(T_{-}) + (t^{-1/2} - t^{1/2})V(T_{0}) = 0 \end{displaymath}

to obtain

\begin{displaymath}t^{-1} - tV(T_{-}) + (t^{-1/2} - t^{1/2})(-t^{-5/2} - t^{-1/2}) = 0 \end{displaymath}

Therefore

\begin{displaymath}V(T_{-}) = (1/t) (t^{-1} - t^{-3} - t^{-1} + t^{-2} + 1)= -t^{-4} + t^{-3} + t^{-1} \end{displaymath}

We can similarly calculate the Jones polynomials of all the knots in our table. The first few are:

Knot
Jones polynomial
31
 $ -t^{-4} + t^{-3} + t^{-1} $
41
 $ t^{-2} - t^{-1} + 1 - t + t^{2} $
51
 $ -t^{-7} + t^{-6} - t^{-5} + t^{-4} + t^{-2} $
52
 $ -t^{-6} + t^{-5} - t^{-4} + 2t^{-3} - t^{-2} + t^{-1} $
61
 $ t^{-4} - t^{-3} + t^{-2} - 2t^{-1} + 2 - t + t^{2} $
62
 $ t^{-5} - 2t^{-4} + 2t^{-3} - 2t^{-2} + 2t^{-1} - 1 + t $
63
 $ -t^{-3} + 2t^{-2} - 2t^{-1} + 3 - 2t + 2t^{2} - t^{3} $

The Jones polynomial behaves very neatly when two knots are joined: V(K#L)=V(K)V(L).

Exercise: Check some of these Jones polynomials for yourself.

If two projections have the same polynomial, does that mean that they are projections of the same knot? Unfortunately, the answer is no. In fact, we know very little about just what the Jones polynomial measures, as the following question reveals.

Unsolved question: Is there a knot other than the unknot which has Jones polynomial 1?

Dr Stephen Huggett
Department of Mathematics and Statistics
University of Plymouth
Plymouth
Devon
PL4 8AA

www.tech.plym.ac.uk/maths/staff/shuggett/home.html
s.huggett@plymouth.ac.uk