On the Jones Polynomial

constructing link invariants via braids & von Neumann algebras

A Monica Queen Thesis

This expository essay is aimed at introducing the Jones polynomial.
We will see the encapsulation of the Jones polynomial, which will involve topics in functional analysis and geometrical topology; making this essay an interdisciplinary area of mathematics.
The presentation is based on a lot of different sources of material (check references), but we will mainly be giving an account on Jones' papers ([25],[26],[27],[28],[29]) and Kauffman's papers ([33],[34],[35]).
A background in undergraduate Linear Algebra, Linear Analysis, Abstract Algebra, Commutative Algebra, & Topology is essential. It would also be useful if the reader has a background in Knot Theory & Operator Algebras.

Ars longa, vita brevis.
This essay is dedicated to my grandmother Grace, who passed away during the start of writing this.

§ INTRODUCTION

The objective of this M.Sc. thesis, as are most, is to prove to the university of satisfactory scholastic work. However, the author accepts that such egotistical objectives will not keep the writer, not to mention the peruser, engaged. Accordingly, the genuine motivation behind this thesis is to introduce the writer's discoveries and thoughts insightfully and to impart them to other people. Subsequently, ideally, this will add to the assemblage of information regarding the matter.

Thus the aims of this essay are two-fold. The first is to give a solid introduction on knot theory and give the combinatorial method of the Jones polynomial (via Kauffman's bracket [33] and skein relation [15, 20, 26]). This gives us a good base on what the Jones polynomial is, and the importance of the polynomial invariant.
For instance, it was until the Jones polynomial was created that the trefoil knot was proven to be distinct from its reflection.
The second aim is to give a brief explanation on how Vaughan Jones originally came to the realisation that there was a link between the von Neumann algebras he was working on and the construction of a most interesting link invariant.
This is given by first realising that all links can be represented by closed braids; and by defining a trace on the presentation from the braid group $\Bcal_n$ to the quotient algebra of $\Bcal_n$ called the Temperley-Lieb algebras $\TL_n$.
This method is based on Jones' original construction in 1984 [25, 26].

Purely going on the notion of physical space, the ether, and atoms as vortex knots, the mathematical theory of knots was formulated by physicists Peter Tait, William Thomson, and James Maxwell in the late 1800s [40, 60].
While their motivation for studying knots and links (i.e., atoms being knotted in ether) was widely discredited in 1887 by Michelson and Morley [46], where they proved that the ether does not exist, the amount of contributions made by them was vast enough.
It was until after the inexistence of the ether was proven that mathematicians started studying knots. And in the time of writing, knot theory has major applications linked to chemistry and biology, such as knots in DNA and proteins. For a nice read of introductory knot theory; see Adams' book [2].

To put it simply, knot theoryknot theory, is the study of distinguishing links (disjoint union of closed smooth curves in three-dimensional space) up to ambient isotopy (intuitively, this means that we can move the link around without allowing it to pass by itself or cutting it). Distinguishing links is usually done using link invariants (algebraic objects that do not change under ambient isotopy). There are many link invariants such as the coloring invariant and the crossing number invariant - these are easy to visualise, however they are not good enough. The crossing number invariant, for instance, is difficult to determine since its definition states that it has to be over the minimal link diagram (so this is summed over the minimum diagram of all possible diagrams - i.e., infinity). The Jones polynomial, however, is the most interesting and one of the most powerful knot invariants that are linked to many areas of mathematics and science.
Vaughan Jones constructed his polynomial invariant in 1984 [26]. For that, he won the Fields medal in 1990 and gave his speech wearing his home country's (New Zealand) rugby jersey: a historical day any knot theorist can tell you about.

Jones, however, did not make the connection on his own.
In fact, while Jones was giving a lecture on von Neumann algebras in Geneva in 1982, Didier Hatt-Arnold, a graduate student, was the one who raised the similarities between the algebra he was working on and the braid group [16].
This then motivated Jones to explore possible representations of the braid group into his algebra he was working on. Finally leading him to the construction of the marvellous link invariant two years later [26].

The fascination of this link invariant, from a knot theory point of view, is that it could detect mirror links. Since 1928, the only known polynomial link invariant was the Alexander polynomial [6], and while profound in its own way, detecting a link's mirror was not one of its strengths. However, the main fascination with this link invariant is the fact that it came from a completely different field of mathematics. This lead to other link invariants that arose from areas such as category theory.

While, at first, it may have seemed as a surprising result from the subject of von Neumann algebras, it became apparent that this subject has many amazing facts that are linked with other subjects such as Topological Quantum Field Theory.
However, recalling that the mathematical theory of knots was formulated by physicsts, it should not be surprising when physics saves the day again.
As Jones said,

“God may or may not play dice, but She sure loves a von Neumann algebra.” [30]

Despite the ease of defining the Jones polynomial in Chapter <ref> via the Kauffman bracket and the skein relation, we should mention that it is generally known that the Jones polynomial of a link is $NP$-hard [24]. However, when calculating the Jones polynomial of a braid representation of a link, it can be computed in polynomial time [48]. We will not discuss computation in this paper, but we see in the final chapter that it is in fact a lot easier to calculate the Jones polynomial via its original construction than the more visual and diagrammatic ones. This is not to say that the Kauffman bracket is not useful, in fact the reason it is liked more so today than the original construction is because of its astounding correlation into category and representation theory; see Khovanov's paper [39].

In Chapter <ref>, we will start with a brief overview on knot theory which includes all the essential information we need for the remainder of the essay.

In Chapter <ref>, we will give two important definitions of the Jones polynomial: via the Kauffman bracket [33] and the Skein relation version, which John Conway conjured in the 1960s for the Alexander polynomial [15].

The next chapter (Chapter <ref>) is on braid theory. We start with the diagrammatic definition of braids and see how it correlates with the Artin braid group, which is the algebraic definition. We also see a correlation between the symmetric group and the braid group.

In Chapter <ref>, we relate the previous two chapters. We will associate to every knot a braid, and prove Alexander's theorem, which provides a pseudo-converse by showing that every braid closure is equivalent to a link.
We conclude with an overview of Markov's moves and Markov's theorem. This gives an equivalence relation that will help us in making a one-to-one correspondence with braids and knots. However, two braids do not have to be braid equivalent in order to be Markov equivalent.
Markov's theorem states that the braid closures are Reidemeister equivalent if and only if the braids are Markov equivalent.
Thus the one-to-one correspondence will quotient the braid group with the Markov equivalence class.

The second part of the thesis focuses on more of Jones' work: we begin by defining the necessary analytical preliminaries in Chapter <ref> and discuss the Temperley-Lieb algebras in algebraic and diagrammatic settings in Chapter <ref>.

With the foundations set, we now have the necessary background in order to conclude the essay (Chapter <ref>), i.e., constructing the Jones polynomial via the original method.

§ KNOTS & LINKS

We begin with a brief overview of knot theory. We discuss all preliminary definitions and results needed to define the Jones polynomial and motivate our chosen thesis in the coming chapters.

This chapter is based on the presentations given in [55], [3], [4], [13], [15], [16], [20], [26], [28], [29], [33], [34], [35], [37], [38], [41], [43], [44], and [49].

Most of our discussion is based on that in the author's exposition on knot theory <cit.>.

§.§ Defining a knot and link

A knotknot is a smooth embedding of a circle $\SS^1$ into $\RR^3$ or $\SS^3$.
$\SS^3$ is essentially $\RR^3$ with its one point compactification. We will mostly use $\RR^3$.
Here, a smooth embeddingembedding means a smooth map which is injective and whose differential is nowhere zero.
Pictorially, Kauffman & Lomonaco Jr. <cit.> drew the embedded circles in $\RR^3$ as the following.

(11,138) .. controls (11,124.19) and (22.19,113) .. (36,113) .. controls (49.81,113) and (61,124.19) .. (61,138) .. controls (61,151.81) and (49.81,163) .. (36,163) .. controls (22.19,163) and (11,151.81) .. (11,138) – cycle ;
(69,122) – (134.04,109.38) ;
[shift=(136,109), rotate = 529.02] [color=rgb, 255:red, 0; green, 0; blue, 0   ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
(67,151) – (140.12,177.32) ;
[shift=(142,178), rotate = 199.8] [color=rgb, 255:red, 0; green, 0; blue, 0   ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
(217.64,66.09) .. controls (231.68,65.52) and (243.53,76.44) .. (244.1,90.47) – (249.03,211.22) .. controls (249.03,211.22) and (249.03,211.22) .. (249.03,211.22) – (147.36,215.37) .. controls (133.32,215.95) and (121.48,205.03) .. (120.9,191) – (115.97,70.25) .. controls (115.97,70.25) and (115.97,70.25) .. (115.97,70.25) – cycle ;
(153,101) .. controls (227,51) and (217,116) .. (181,94) ;
(189,103) .. controls (135,148) and (138,52) .. (168,84) ;
(145,107) .. controls (117.05,126.47) and (166.16,147.54) .. (189.08,125.77) .. controls (212,104) and (263,62) .. (211,81) ;
(202,88) – (196,93) ;
(185,160) .. controls (245,147) and (256,179) .. (214,179) ;
(166,162) .. controls (118,170) and (223,225) .. (209,164) ;
(209,150) .. controls (207,134) and (188.53,144.78) .. (179,153) .. controls (169.47,161.22) and (174.09,179.7) .. (205,175) ;

knot definition with embedded circles in $\RR^3$

In the above figure, the two embeddings are knots. It is easy to see that they are not the same. However, we will need to define what knots being “the same” means before claiming such (see next section).

The definition of a knot is equivalent to the following:

* A knot is a subset of $\RR^3$ that consists of piecewise linear simple closed curves.

* A knot is a subset of $\RR^3$, which is diffeomorphic to $\SS^1$.

Here, piecewise linearpiecewise linear means that the curve can be written as a finite union of line segments such that the segments are disjoint from one another, except when consecutive, i.e.,

\[K=[a_0,a_1]\cup[a_1,a_2]\cup\cdots\cup[a_n,a_0]\]

Here, diffeomorphismdiffeomorphism means that the map and its inverse are both smooth. Homeomorphismhomeomorphism, on the other hand, is when the map and its inverse are both continuous.[So this means that all diffeomorphisms are homeomorphisms, but not the other way around; because, differentiability implies continuity, but continuity does not imply differentiability.]

We can define a knot via homeomorphism (i.e., a subset of $\RR^3$ which is homeomorphic to a circle).
However, this requires us to define tame knotstame knots which are knots that can be defined smoothly or piecewise-linearly; and wild knotswild knots which are knots that are not tame.

To avoid such complications, we will always assume smoothness.

The most basic knot is the unknot, written as $0_1$, which is drawn below.
We also have the trefoil $3_1$ and the trefoil's reflection $\overline{3_1}$. We call $3_1$ the left-handed trefoil, whereas we call its reflection $\overline{3_1}$, the right-handed trefoil (fact: the right-handed trefoil and its reflection are two distinct knots; see Proposition <ref>).

(0,0) circle (0.8);

$0_1$   $3_1$   $\overline{3_1}$

examples of knots

The following is not a knot, since the differential at $(*)$ is undefined. However, for the sake of completion, we mention that since this is a subset of $\RR^3$ and is homeomorphic to a circle, we can see that it is an example of a wild knot.

(26,92) .. controls (3,64) and (-5,37) .. (27,18) .. controls (59,-1) and (85,38) .. (72,52) ;
(385,62) .. controls (457,72) and (448,137) .. (417,149) .. controls (386,161) and (329,157) .. (219,157) .. controls (109,157) and (36,167) .. (23,147) .. controls (10,127) and (38.37,80.97) .. (65,61) ;
(45,102) .. controls (57.39,110.98) and (102,107) .. (116,95) .. controls (130,83) and (130,84) .. (138,47) ;
(57,24) .. controls (33,29) and (38,45) .. (46,51) .. controls (54,57) and (104,60) .. (130,58) ;
(132,17) .. controls (108.97,17.49) and (108,27) .. (114,33) .. controls (120,39) and (154,40) .. (197,33) ;
(75,17) .. controls (126,-3) and (147,6) .. (140,32) ;
(150,14) .. controls (178.67,4.85) and (209,20) .. (210,32) .. controls (211,44) and (188,55) .. (146,57) ;
(215,28) .. controls (261,25) and (265,44) .. (317,54) ;
[fill=rgb, 255:red, 0; green, 0; blue, 0   ,fill opacity=1 ] (397.5,63.25) .. controls (397.5,61.73) and (398.73,60.5) .. (400.25,60.5) .. controls (401.77,60.5) and (403,61.73) .. (403,63.25) .. controls (403,64.77) and (401.77,66) .. (400.25,66) .. controls (398.73,66) and (397.5,64.77) .. (397.5,63.25) – cycle ;
(336,39) node [anchor=north west,scale=2][inner sep=0.75pt]   [align=left] $\displaystyle \dotsc $;
(395,39) node [anchor=north west,scale=2][inner sep=0.75pt]   [align=left] $\displaystyle *$;

wild knot

Such knots can have pathological behaviour, hence why we mentioned that we will always assume smoothness throughout this paper.

A linklink of $m$ components is a smooth embedding of $n$ disjoint circles into $\RR^3$.
Thus, a knot is just a link with one component.

An unlink is a finite number of disjointed unknots. We usually write an $m$-unlink as $\KPA^m$ (where $m$ denotes the number of disjointed unknots). The figure to the right is an example of an unlink $\KPA^2$.

[draw opacity=0] (90.01,115.72) .. controls (84.07,117.84) and (77.67,119) .. (71,119) .. controls (40.07,119) and (15,94.15) .. (15,63.5) .. controls (15,32.85) and (40.07,8) .. (71,8) .. controls (101.93,8) and (127,32.85) .. (127,63.5) .. controls (127,79.02) and (120.57,93.06) .. (110.2,103.13) – (71,63.5) – cycle ;
(90.01,115.72) .. controls (84.07,117.84) and (77.67,119) .. (71,119) .. controls (40.07,119) and (15,94.15) .. (15,63.5) .. controls (15,32.85) and (40.07,8) .. (71,8) .. controls (101.93,8) and (127,32.85) .. (127,63.5) .. controls (127,79.02) and (120.57,93.06) .. (110.2,103.13) ;
[draw opacity=0] (109.49,11.85) .. controls (115.36,9.56) and (121.72,8.22) .. (128.39,8.02) .. controls (159.31,7.13) and (185.09,31.24) .. (185.98,61.88) .. controls (186.86,92.52) and (162.52,118.08) .. (131.61,118.98) .. controls (100.69,119.87) and (74.91,95.76) .. (74.02,65.12) .. controls (73.57,49.61) and (79.6,35.39) .. (89.66,25.02) – (130,63.5) – cycle ;
(109.49,11.85) .. controls (115.36,9.56) and (121.72,8.22) .. (128.39,8.02) .. controls (159.31,7.13) and (185.09,31.24) .. (185.98,61.88) .. controls (186.86,92.52) and (162.52,118.08) .. (131.61,118.98) .. controls (100.69,119.87) and (74.91,95.76) .. (74.02,65.12) .. controls (73.57,49.61) and (79.6,35.39) .. (89.66,25.02) ;

(0,0) circle(0.4);
(0.9,0) circle(0.4);

Hopf Link            Unlink of two components
examples of links

Note that it is not always the case that a theorem for knots works for links as well. However, for the purposes of this paper, we will use them interchangeably, unless stated otherwise.

§.§ Link equivalency

We now discuss link equivalency. Essentially, we need two links to be equivalent if we can go from one link to the other without forming any self-intersections or breaking the definition of a link. So, we begin by explaining why homotopy equivalency is too weak for links.

Any two knots are homotopy equivalenthomotopy, i.e., for knots $K_1\colon{S^1\to\RR^3}$ and $K_2\colon{S^1\to\RR^3}$, there exists a continuous function $H\colon\RR^1\times[0,1]\to\RR^3$ such that $H(x,0)=K_1(x)$ and $H(x,1)=K_2(x)$. So, under homotopy, all knots would be trivial by pulling on the ends until the “knotted” part disappears (also known as the gentleman's knot - see Figure <ref>).

(66.74,72.24) .. controls (39.92,85.15) and (19.62,74.23) .. (14.55,57.34) .. controls (9.47,40.45) and (21.21,30.35) .. (37.02,44.92) ;
(29.04,67.77) .. controls (39.8,47.18) and (49.34,31.51) .. (67.47,33.5) .. controls (85.59,35.49) and (89.94,55.35) .. (78.34,63.3) ;
(45.72,50.88) .. controls (90.66,71.25) and (90.66,99.06) .. (80.51,109.99) .. controls (70.37,120.91) and (53.69,123.89) .. (38.47,117.93) .. controls (23.25,111.97) and (19.62,88.13) .. (25.42,79.19) ;
(172.89,32.81) .. controls (167.76,39.55) and (158.35,40.39) .. (152.36,34.49) .. controls (146.38,28.6) and (148.94,21.03) .. (154.93,22.71) ;
(154.93,29.44) .. controls (160.92,19.34) and (165.2,18.5) .. (170.33,18.5) .. controls (175.46,18.5) and (181.45,22.71) .. (178.03,26.92) ;
(163.48,24.81) .. controls (217.16,45.55) and (205.19,93.53) .. (196.05,109.29) .. controls (186.91,125.04) and (165.84,138.99) .. (141.88,125.52) .. controls (117.93,112.05) and (118.79,74.17) .. (148.09,36.18) ;
(281.65,16.41) .. controls (280.44,18.04) and (278.21,18.25) .. (276.8,16.82) .. controls (275.38,15.39) and (275.99,13.55) .. (277.4,13.96) ;
(277.4,15.59) .. controls (278.82,13.14) and (279.83,12.93) .. (281.05,12.93) .. controls (282.26,12.93) and (283.68,13.96) .. (282.87,14.98) ;
(280.44,14.06) .. controls (287.03,20.82) and (311.42,29.38) .. (324.21,39.22) .. controls (336.99,49.05) and (345.94,68.72) .. (342.1,98.22) .. controls (338.27,127.72) and (317.81,134.74) .. (307.58,137.55) .. controls (297.36,140.36) and (267.95,138.96) .. (255.17,127.72) .. controls (242.38,116.48) and (237.3,103.62) .. (237.27,79.96) .. controls (237.23,56.3) and (257.06,39.42) .. (275.16,18) ;
(380.42,75.85) .. controls (380.42,42.74) and (397.01,15.91) .. (417.47,15.91) .. controls (437.93,15.91) and (454.51,42.74) .. (454.51,75.85) .. controls (454.51,108.96) and (437.93,135.8) .. (417.47,135.8) .. controls (397.01,135.8) and (380.42,108.96) .. (380.42,75.85) – cycle ;
[fill=black]   (417.47,15.91) .. controls (417.47,15.13) and (417.85,14.51) .. (418.33,14.51) .. controls (418.81,14.51) and (419.19,15.13) .. (419.19,15.91) .. controls (419.19,16.68) and (418.81,17.3) .. (418.33,17.3) .. controls (417.85,17.3) and (417.47,16.68) .. (417.47,15.91) – cycle ;
(495.91,73.16) .. controls (495.91,40.22) and (512.49,13.52) .. (532.95,13.52) .. controls (553.41,13.52) and (570,40.22) .. (570,73.16) .. controls (570,106.1) and (553.41,132.8) .. (532.95,132.8) .. controls (512.49,132.8) and (495.91,106.1) .. (495.91,73.16) – cycle ;
(96.7,73) node [anchor=north west][inner sep=0.75pt]    $\rightarrow$;
(210,73) node [anchor=north west][inner sep=0.75pt]    $\rightarrow$;
(350.7,73) node [anchor=north west][inner sep=0.75pt]    $\rightarrow$;
(463.7,73) node [anchor=north west][inner sep=0.75pt]    $\rightarrow$;

the gentleman's knot

So, our definition of link equivalence needs to be stronger than that of homotopy.

Two maps $A,B\subseteq\RR^3$ are said to be ambient isotopicambient isotopy if there exists a continuous map $H\colon\RR^3\times[0,1]\to\RR^3$ such that the map $h_t\colon\RR^3\to\RR^3$ defined by $h_t(x)=H(x,t)$ is a diffeomorphism for all $t\in[0,1]$ with $h_0$ equal to the identity map, and $h_1(A)=B$. We call the map $H$, the ambient isotopy map between $A$ and $B$.

In Definition <ref>, ambient isotopy is essentially homotopy where it is a smooth family of diffeomorphisms.

Two links are equivalentequivalent!links if they are
ambient isotopic.
When two links $L_1$ and $L_2$ are equivalent, we write $L_1\sim{L_2}$.

Fisher [19] proved that any orientation-preservingdiffeomorphism $f$ of $\RR^n$ is smoothly ambient isotopic to the identity. This allows us to characterise ambient isotopies as orientation-preserving diffeomorphisms; in particular, links are equivalent if there exists an orientation $-$ preserving piecewise linear diffeomorphism $h\colon\RR^3\to\RR^3$ such that $h(L_1)=L_2$, where $L_1$ and $L_2$ are links.

For those with a background in differential geometry, a diffeomorphism between two oriented manifolds $M$ and $N$ is said to be orientation-preserving if the pullback of the orientation on $N$ is the same, up to equivalence, of the orientation on $M$ <cit.>.

Any two links which are equivalent have the same link typelink type: the equivalence class of links.
Similarly, we call the equivalence class of knots knot typeknot type.

Ambient isotopy is an equivalence relation. (So, link equivalence is an equivalence relation.)

Let $L_1$, $L_2$, and $L_3$ be any links.

For reflexivity, this is obvious.

For symmetry, we suppose $L_1\sim{L_2}$. Then, we have a function $h_t$ such that $h_0(L_1)=L_1$ and $h_1(L_1)=L_2$ where the map and its inverse are smooth. It follows that $h^{-1}_0(L_2)=L_2$ and $h^{-1}_1(L_2)=L_1$, which means $L_2\sim{L_1}$.

Finally, for transitivity, suppose $L_1\sim{L_2}$ and $L_2\sim{L_3}$. Then, we have two functions $h_t$ and $f_t$ such that $h_t$ is the same as $h_t$ from the symmetry above; and $f_0(L_2)=L_2$ and $f_1(L_2)=L_3$. So, it then follows that $f_t\circ{h_t}(x)=f_t(h_t(x))$ is also a diffeomorphism (map compositions are smooth if both maps are smooth), so then $f_0(h_0(L_1))=f_0(L_1)=L_1$ and $f_1(h_1(L_1))=f_1(L_2)=L_3$. Thus, $L_1\sim{L_3}$.

Thus we have the following definition of an unlink and unknot.

The unknotunknot $0_1$ is a knot that is equivalent to a circle.
So, the $m$-unlinkunlink $\KPA^m$ is a link that is equivalent to $m$ disjointed unknots.

§.§ Link diagrams and Reidemeister's theorem

In this section, we discuss elementary methods of projecting a link onto a two-dimensional plane. The reason for doing so is because it would not be reasonable for us to work in $\RR^3$, as it is harder to imagine three-dimensional objects in our head. Before projecting the link, we first need to define the most elementary method of finding an equivalent link. This elementary method is called the triangle move. Here is a precise definition.

Consider a link $L$ in $\RR^3$, we find a planar triangle in $\RR^3$ that intersects $L$ in exactly one of the edges of the triangle. We then delete that edge from $L$, and replace it by the other two edges of the triangle. The inverse process of this move is done by replacing the two edges of the triangle with one (literally the opposite). We call this the triangle-movetriangle-move and denote it by $\Delta$ and its inverse process by $\Delta^{-1}$, see the figure below.

[ultra thick] (0,0)–(2,0);
[dashed,thin] (2,0)–(1,2)–(0,0);

(4.5,2) node[scale=3.33] $\overset{\Delta}{\longmapsto}$;
(4.5,0) node[scale=3.33] $\underset{\Delta^{-1}}{\longmapsfrom}$;
[dashed,thin] (0,0)–(2,0);
[ultra thick] (2,0)–(1,2)–(0,0);

Two links are equivalent if and only if they differ by a finite sequence of the triangle-moves or the inverse of the moves. A detailed proof can be found in <cit.>.

It is easy to show that the $\Delta$-move induces an equivalence relation. (It is closely analogous to the proof from Proposition <ref>.)

Now we can define the link projection maplink projection $\rho\colon\RR^3\to\RR^2$. As explained in <cit.>, this is done by:

* small triangle-moves such that each line segment of the link projects to a line in $\RR^2$,

* that the projections of two such segments intersect in at most one point which for disjoint segments is not an end point, and

* that no point belongs to the projections of three segments.

Given such a situation, the image of the link in $\RR^2$ together with “over/under” information (this refers to the relative heights above $\RR^2$ of the two inverse images of a crossing, which is always indicated in diagrams by breaks in the under-passing segments) at the crossings is called a link diagram of the given link. So, the only types of crossings in a link diagram that are allowed are essentially shown in Figure <ref>. And the crossings that are not allowed are triple points, tangents, and vertical edges (shown in Figure <ref>).

[ultra thick,draw=white,double=black,double distance=0.5pt] (0,0)–(2,0);

[ultra thick,draw=white,double=black,double distance=0.5pt] (1,-1)–(1,1);

crossings that are allowed

(-1,-1) .. controls (0,0) .. (0,1) .. controls (0,0) .. (1,-1);

[ultra thick,draw=white,double=black,double distance=0.5pt] (1,-1)–(-1,1);
[ultra thick,draw=white,double=black,double distance=0.5pt] (-1,0)–(1,0);

(-1,0) .. controls (0,0.5) .. (1,0);

vertical edge; triple point; tangent

crossings that are not allowed

A link universelink universe is a union of closed piece-wise linear curves in the plane such that there are finitely many self-intersections and that all self-intersections are transverse (i.e., at each crossing, there are exactly four half-curves with that crossing as an end point, so that none of them are tangent to another).

A link diagramlink diagram is essentially a link universe with a choice, i.e., the points of self-intersection are equipped with height information with the lower strand being indicated with a break in the curve.
If the link diagram has one component (i.e., a projection of a knot), then we call it a knot diagramknot diagram.

A crossing in a link universe would then have two crossing choices in a link diagram.

(0,-1.4) node[scale=1.33] link universe;

[white,double=black,ultra thick,double distance=0.5pt] (1,-1)–(-1,1);

[white,double=black,ultra thick,double distance=0.5pt] (-1,-1)–(1,1);

(5.5,-1.4) node[scale=1.33] crossing choices;
(5.5,-1.9) node[scale=1.33] link diagrams;

If links $L_1$ and $L_2$ are equivalent, then they are related by a sequence of triangle moves, as previously mentioned. After moving all the vertices of all the triangles by a very small amount, it can be assumed that the projections of no three of the vertices lie on the line in $\RR^2$ and the projections of no three edges pass through a single point. So then each triangle projects to a triangle.

The following example shows that using triangle moves is dreary and time consuming. This motivates the usefulness of the upcoming theorem.

Consider the following piecewise linear link.

[double=black,white,ultra thick,double distance=0.5pt] (2,1)–(-1,0.5);

It is easy to see that this is equivalent to the unknot by a twist. However, with the triangle moves, we first find a planar triangle that intersects our link in one of the edges of the triangle. The intersected edge is thickened for clarity.

[double=black,white,ultra thick,double distance=1pt] (2,1)–(-1,0.5);
[dashed] (2,1)–(0.5,1.5)–(-1,0.5);

And we can now replace the thick edge with the two dashed edges on our triangle.

[double=white,white,ultra thick,double distance=0.5pt] (2,1)–(-1,0.5);
[dashed] (2,1)–(-1,0.5);
[thick] (2,1)–(0.5,1.5)–(-1,0.5);

Thus, we see that this is equivalent to the unknot.

Thankfully, there is an easier method of showing link equivalency with Reidemeister's theorem. The theorem says that two link diagrams are equivalent if and only if they are related by a finite sequence of three moves (see Figure <ref>). The threorem also says that every link has a link diagram.
This statement is quite fascinating as it simplifies the triangle moves.
Here is a precise statement of the theorem.

* Any link $L$ is equivalent to a link which has a link diagram.

* Two links $L_1,L_2$ with diagrams $D_1,D_2$ are equivalent if and only if their diagrams are related by a finite sequence of Reidemeister movesReidemeister moves and planar isotopiesplanar isotopies.

[black] (0,2) – (0,-2);
[white,ultra thick] (0,0) circle(2);
[black,dashed] (0,0) circle(2);

[black,scale=4] at (-3.5,0) $\bm{\sim}$;
(a) at (0.2,0) ;
(0,2) .. controls (0,1) and (a.2 north west) ..
(a.center) .. controls (a.16 south east) and (a.16 north east) ..
(a) .. controls (a.2 south west) and (0,-1) ..
[white,ultra thick] (0,0) circle(2);
[black,dashed] (0,0) circle(2);

Type I - $(\RI)$

[black] (100:2) – (260:2);
[black] ( 80:2) – (280:2);
[white,ultra thick] (0,0) circle(2);
[black,dashed] (0,0) circle(2);

[black,scale=4] at (-3.5,0) $\bm{\sim}$;
[black] (100:2) .. controls ( 0.6,0) .. (260:2);
[draw=white,double=black,ultra thick,,double distance=0.5pt] ( 80:2) .. controls (-0.6,0) .. (280:2);
[white,ultra thick] (0,0) circle(2);
[black,dashed] (0,0) circle(2);

Type II - $(\RII)$

[draw=white,double=black ,ultra thick,double distance=0.5pt] (  0:2) .. controls (270:0.7) .. (180:2);
[draw=white,double=black  ,ultra thick,double distance=0.5pt] ( 60:2) .. controls (150:0.7) .. (240:2);
[draw=white,double=black,ultra thick,double distance=0.5pt] (120:2) .. controls ( 30:0.7) .. (300:2);
[white,ultra thick] (0,0) circle(2);
[black,dashed] (0,0) circle(2);

[black,scale=4] at (-3.5,0) $\bm{\sim}$;
[draw=white,double=black ,ultra thick,double distance=0.5pt] (  0:2) .. controls ( 90:0.7) .. (180:2);
[draw=white,double=black  ,ultra thick,double distance=0.5pt] ( 60:2) .. controls (330:0.7) .. (240:2);
[draw=white,double=black,ultra thick,double distance=0.5pt] (120:2) .. controls (210:0.7) .. (300:2);
[white,ultra thick] (0,0) circle(2);
[black,dashed] (0,0) circle(2);

Type III - $(\RIII)$

Reidemeister moves

The figures above show regions of the diagram, when the move is made the rest of the diagram is left unchanged (hence why we put dashed lines: to remind ourselves that we should only think of the enclosed region for the moment).

We note that some texts include planar isotopy as a Type O - $(\RO)$ move. Intuitively, this means moving the knot around by stretching or compressing.

The diagrams are said to be regularly isotopicregularly isotopic if a Type I move is not used. So the theorem shows that there is a bijective correspondence between the set of equivalence classes, i.e.,
(piecewise linear links in $\RR^3$ with $\Delta$-move equivalence) $\leftrightarrow$ (piecewise linear link diagrams in $\RR^2$ with R-move equivalence).

We show that any triangle-move can be reduced to a sequence of Reidemeister moves and planar isotopies. There are two cases: first being a planar isotopy - a regular triangle-move (Figure <ref>); the second is when the strands of the link come in contact with our chosen triangle which gives the three Reidemeister moves as shown in Figure <ref>.

[thin,dashed] (0,0)–(2,0)–(1,2)–(0,0);
[draw=white,double=black,ultra thick,,double distance=1.2pt] (2,0)–(-0.5,2);
[ultra thick] (0,0)–(2,0);
[thin,dashed] (2,0)–(1,2);
(2.8,1.5) node[scale=2] $\overset{\Delta}{\longmapsto}$;
(2.8,0.5) node[scale=2] $\underset{\Delta^{-1}}{\longmapsfrom}$;
(3,-1) node[scale=2] $\RI$;

[ultra thick] (0,0)–(1,2);
[draw=white,double=black,ultra thick,,double distance=1.2pt] (2,0)–(-0.5,2);
[thin,dashed] (0,0)–(2,0);
[ultra thick] (1,2)–(2,0);

[thin,dashed] (0,0)–(2,0)–(1,2)–(0,0);
[ultra thick] (0,0)–(1,2);
[ultra thick,draw=white,double=black,double distance=1.2pt] (1.6,2.1)–(1.2,-0.3);
(3,1.5) node[scale=2] $\overset{\Delta}{\longmapsto}$;
(3,0.5) node[scale=2] $\underset{\Delta^{-1}}{\longmapsfrom}$;
(3,-1) node[scale=2] $\RII$;

[thin,dashed] (0,0)–(2,0)–(1,2)–(0,0);
[ultra thick] (1,2)–(2,0);
[ultra thick] (2,0)–(0,0);
[ultra thick,draw=white,double=black,double distance=1.2pt] (1.6,2.1)–(1.2,-0.3);

[thin,dashed] (0,0)–(2,0)–(1,2)–(0,0);
[ultra thick] (0,0)–(1,2);
[ultra thick,draw=white,double=black,double distance=1.2pt] (1.4,2.1)–(1,-0.3);
[ultra thick,draw=white,double=black,double distance=1.2pt] (2.1,2.1)–(0.4,-0.3);
(3,1.5) node[scale=2] $\overset{\Delta}{\longmapsto}$;
(3,0.5) node[scale=2] $\underset{\Delta^{-1}}{\longmapsfrom}$;
(3,-1) node[scale=2] $\RIII$;

[thin,dashed] (0,0)–(2,0)–(1,2)–(0,0);
[ultra thick] (1,2)–(2,0)–(0,0);
[ultra thick,draw=white,double=black,double distance=1.2pt] (1.4,2.1)–(1,-0.3);
[ultra thick,draw=white,double=black,double distance=1.2pt] (2.1,2.1)–(0.4,-0.3);

Triangle moves $\leftrightarrow$ Reidemeister moves

Our strategy for proving that a function is a link invariant (see <ref>), is to show that it preserves the Reidemeister moves.

§.§ Orientation

We now discuss orientations, and see how the figure eight knot is equivalent to its reflection.

An orientationorientation for a link is a choice of direction along each of the components. This is indicated on the corresponding link diagram by adding an arrow to each component. The components of an $n$-component link can be oriented in $2^n$ ways.

The signsign $\epsilon(C)=\pm1$ of an oriented crossing $C$ is given by the figure below.

[thin, ->] (0,0)–(2,2);
[ultra thick,draw=white,double=black,double distance=0.5pt] (0,2)–(2,0);
[ultra thin,->] (0,2)–(2,0);
[thin,->,dashed] (0.2,0) .. controls (1,0.8) .. (1.7,0);
(1,0.2) node[scale=1.7] $+$;
(1,-0.5) node[scale=2] positive crossing;

[thin,->] (0,2)–(2,0);
[ultra thick,draw=white,double=black,double distance=0.5pt] (0,0)–(2,2);
[ultra thin,->] (0,0)–(2,2);
[thin,->,dashed] (0.2,2) .. controls (1,1.2) .. (1.7,2);
(1,1.8) node[scale=1.7] $-$;
(1,-0.5) node[scale=2] negative crossing;

sign of crossing

Figure <ref> can be thought of as moving along the orientation of the under-strand until we reach a crossing where we need to “jump” to the over strand; if we go to the right at the “jump” then the sign is positive, and if we go to the left then it is negative.

The writhewrithe $\omega(D)$ of a link diagram $D$ is the sum of all the crossing signs, i.e. $\omega(D)=\sum_{C\textnormal{ a crossing in }D}\epsilon(C)$.

The following shows an example of finding the writhe of the left-handed trefoil, Hopf link, and the unlink (with a twist).

[rotate=  0] (a) at (0,-1) ;
[rotate=120] (b) at (0,-1) ;
[rotate=240] (c) at (0,-1) ;
(a) .. controls (a.4 north west) and (c.4 north east) .. (c.center);
(b) .. controls (b.4 north west) and (a.4 north east) .. (a.center);
(c) .. controls (c.4 north west) and (b.4 north east) .. (b.center);
(a.center) .. controls (a.16 south west) and (c.16 south east) .. (c);
(b.center) .. controls (b.16 south west) and (a.16 south east) .. (a);
(c.center) .. controls (c.16 south west) and (b.16 south east) .. (b);
[->] ( 90:2.035) – +(  0:0.01);

(0,-2.5) node[scale=1.8] $\omega(D_1)=-3$;
(c.east) node[anchor=south east] $-1$;
(a.west) node[anchor=north] $-1$;
(b.west) node[anchor=south west] $-1$;

(-0.25,0) +( 70: 0.5) arc( 70:410: 0.5);
( 0.25,0) +( 70:-0.5) arc( 70:410:-0.5);

[->] (-0.25, 0.50) – +(-0.01,0);
[->] ( 0.25, 0.50) – +( 0.01,0);
(0,-.85) node[scale=1.8/3] $\omega(D_2)=2$;
(0,0.55) node[scale=1/3] $+1$;
(0,-0.55) node[scale=1/3] $+1$;

(0,0) +(60: 0.8) arc(60:300: 0.8);
(2,0) +(60:-0.8) arc(60:300:-0.8);
(2,0) ++( 60:-0.8) – ( 60:0.8);
[draw=white,double=black,ultra thick,double distance=0.4pt] (2,0) ++(300:-0.8) – (300:0.8);
[->] (0,0.8) – +(-0.01,0);

(1,-2.5) node[scale=1.8] $\omega(D_3)=1$;
(1,0.5) node $+1$;

writhe example

The reversereverse of a link is done by reversing the orientations of all the components in its corresponding diagram. For a link $L$, we denote the reverse by $rL$.

It is easy to see that the writhe of the diagrams in Example <ref> remains the same when reversing the orientation.

We can generalise this remark. By looking at a single crossing, we can see that changing all orientations of all the components in a link does not change the sign. So the signs of the following are the same.

\[
\epsilon\left(\signpositive\right)
= +1 =
\epsilon\left(\reflectbox{\signnegative}\right)
\]

\[
\epsilon\left(\signnegative\right) = -1 =
\epsilon\left(\reflectbox{\signpositive}\right)
\]

Thus, we have $\omega(L)=\omega(rL)$. And so this means the writhe of an unoriented knot diagram is well defined.

The reflectionreflection of a link is done by replacing all over-strands with under-strands at each crossing and vice versa of the original link's link diagram. For link $L$, we say $\overline{L}$ is the reflection of $L$.

By the end of the next chapter, we will show that the trefoil and its reflection are in fact two distinct knots.
For now, however, we will show that both the figure eight knot and its reflection are equivalent.

The figure eight knot $4_1$ is equivalent to its reflection $\overline{4_1}$.

We show this using the Reidemeister moves.

(100.7,131) .. controls (132.22,133) and (134.84,33) .. (91.08,20) .. controls (47.31,7) and (54.31,47) .. (70.94,70) ;
(56.06,123) .. controls (37.68,92) and (117.34,85) .. (70.07,35) ;
(51.69,25) .. controls (24.55,9) and (-15.71,132) .. (82.32,132) ;
(85.82,86) .. controls (119.96,142) and (60.44,155) .. (56.94,137) ;

(90.7,125.8) .. controls (122.22,127.8) and (124.84,27.8) .. (81.08,14.8) .. controls (37.31,1.8) and (44.31,41.8) .. (60.94,64.8) ;
(63,75.9) .. controls (82,52.9) and (65,31.9) .. (55,25.9) ;
(41.69,18.8) .. controls (35.17,14.96) and (23.91,29.46) .. (30,47.9) .. controls (36.09,66.34) and (74,84.9) .. (68,95.9) .. controls (62,106.9) and (13.07,59.9) .. (10.04,71.9) .. controls (7,83.9) and (17,128.9) .. (72.32,125.8) ;
(71.57,75.35) .. controls (116.57,125.35) and (52.57,153.35) .. (44.5,130) ;
(57,82.9) – (52,89.9) ;
(47,96.8) .. controls (43.5,102) and (41.5,108) .. (42.5,116) ;

(88.7,124.8) .. controls (120.22,126.8) and (109.59,27.15) .. (79.08,13.8) .. controls (48.57,0.45) and (48.57,21.45) .. (46.57,27.45) ;
(61.57,72.45) .. controls (80.57,49.45) and (73.57,17.45) .. (57.57,16.45) ;
(42.57,17.45) .. controls (5.57,21.45) and (56.57,30.45) .. (57.57,41.45) .. controls (58.57,52.45) and (21.91,28.46) .. (28,46.9) .. controls (34.09,65.34) and (72,83.9) .. (66,94.9) .. controls (60,105.9) and (11.07,58.9) .. (8.04,70.9) .. controls (5,82.9) and (15,127.9) .. (70.32,124.8) ;
(72.57,68.72) .. controls (124.57,112.72) and (49.57,155) .. (42.5,129) ;
(55,81.9) – (50,88.9) ;
(45,95.8) .. controls (41.5,101) and (39.5,107) .. (40.5,115) ;
(45.57,34.45) – (44.57,40.45) ;
(44.57,46.45) .. controls (45.57,54.45) and (54.57,52.45) .. (63.57,59.45) ;

(90.2,124.8) .. controls (121.72,126.8) and (111.09,27.15) .. (80.58,13.8) .. controls (50.07,0.45) and (50.07,21.45) .. (48.07,27.45) ;
(73.95,37.32) .. controls (77.95,23.32) and (75.07,17.45) .. (59.07,16.45) ;
(44.07,17.45) .. controls (7.07,21.45) and (80.58,36.85) .. (86.95,58.32) .. controls (93.32,79.78) and (78.95,98.32) .. (64.95,99.32) .. controls (50.95,100.32) and (19.95,62.32) .. (9.54,70.9) .. controls (-0.88,79.48) and (16.5,127.9) .. (71.82,124.8) ;
(86.95,88.32) .. controls (108.95,120.32) and (51.07,155) .. (44,129) ;
(46.5,95.8) .. controls (43,101) and (41,107) .. (42,115) ;
(45.95,35.32) .. controls (43.95,46.32) and (56.07,52.45) .. (65.07,59.45) ;
(72.95,66.32) .. controls (78.95,70.32) and (76.95,75.32) .. (80.95,80.32) ;
(52.35,86.52) .. controls (55.35,80.52) and (69.95,73.32) .. (72.07,48.45) ;

(65.2,126.8) .. controls (96.72,128.8) and (86.09,29.15) .. (55.58,15.8) .. controls (25.07,2.45) and (25.07,23.45) .. (23.07,29.45) ;
(48.95,39.32) .. controls (52.95,25.32) and (50.07,19.45) .. (34.07,18.45) ;
(19.07,19.45) .. controls (-17.93,23.45) and (55.58,38.85) .. (61.95,60.32) .. controls (68.32,81.78) and (52.4,96.13) .. (39.95,101.32) .. controls (27.5,106.5) and (33.5,117.5) .. (52.5,122.5) ;
(61.95,90.32) .. controls (83.95,122.32) and (23.5,156.5) .. (19,131) ;
(27.35,88.52) .. controls (16.5,102.5) and (17.5,123.5) .. (19,131) ;
(20.95,37.32) .. controls (18.95,48.32) and (31.07,54.45) .. (40.07,61.45) ;
(47.95,68.32) .. controls (53.95,72.32) and (51.95,77.32) .. (55.95,82.32) ;
(27.35,88.52) .. controls (30.35,82.52) and (44.95,75.32) .. (47.07,50.45) ;

(78.2,132.8) .. controls (109.72,134.8) and (105.5,71.32) .. (102.5,61.32) .. controls (99.5,51.32) and (91.5,37.32) .. (76.5,25.32) ;
(64.5,47.72) .. controls (66.5,41.72) and (69.5,37.72) .. (65.5,34.72) ;
(23.5,15.32) .. controls (3.5,10.32) and (6.5,23.32) .. (15.5,26) .. controls (24.5,28.68) and (26.42,18.12) .. (35.46,11.92) .. controls (44.5,5.72) and (89.5,4.72) .. (83.5,13) .. controls (77.5,21.28) and (65.5,24) .. (56.5,37) .. controls (47.5,50) and (69.86,49.15) .. (74.95,66.32) .. controls (80.04,83.49) and (65.4,102.13) .. (52.95,107.32) .. controls (40.5,112.5) and (46.5,123.5) .. (65.5,128.5) ;
(74.95,96.32) .. controls (96.95,128.32) and (36.5,162.5) .. (32,137) ;
(40.35,94.52) .. controls (29.5,108.5) and (30.5,129.5) .. (32,137) ;
(69.5,19.32) .. controls (56.5,7.32) and (40.5,19.32) .. (40.5,35.32) .. controls (40.5,51.32) and (44.5,62.32) .. (53.07,67.45) ;
(60.95,74.32) .. controls (66.95,78.32) and (64.95,83.32) .. (68.95,88.32) ;
(40.35,94.52) .. controls (43.35,88.52) and (57.5,79.72) .. (62.5,56.72) ;
(58.5,27.72) .. controls (55.5,24.72) and (53.5,23.72) .. (49.5,22.72) ;
(33.5,18.32) – (39.5,20.32) ;

(55.7,139.8) .. controls (87.22,141.8) and (83,78.32) .. (80,68.32) .. controls (77,58.32) and (69,44.32) .. (54,32.32) ;
(42,54.72) .. controls (44,48.72) and (47,44.72) .. (43,41.72) ;
(19.9,24.78) .. controls (5.8,17.25) and (23.9,5.78) .. (35.9,4.78) .. controls (47.9,3.78) and (67,11.72) .. (61,20) .. controls (55,28.28) and (43,31) .. (34,44) .. controls (25,57) and (47.36,56.15) .. (52.45,73.32) .. controls (57.54,90.49) and (42.9,109.13) .. (30.45,114.32) .. controls (18,119.5) and (24,130.5) .. (43,135.5) ;
(52.45,103.32) .. controls (74.45,135.32) and (14,169.5) .. (9.5,144) ;
(17.85,101.52) .. controls (7,115.5) and (8,136.5) .. (9.5,144) ;
(47,26.32) .. controls (34,14.32) and (18,26.32) .. (18,42.32) .. controls (18,58.32) and (22,69.32) .. (30.57,74.45) ;
(38.45,81.32) .. controls (44.45,85.32) and (42.45,90.32) .. (46.45,95.32) ;
(17.85,101.52) .. controls (20.85,95.52) and (35,86.72) .. (40,63.72) ;
(36,34.72) .. controls (33,31.72) and (31,30.72) .. (27,29.72) ;

(62.7,143.8) .. controls (94.22,145.8) and (90,82.32) .. (87,72.32) .. controls (84,62.32) and (76,48.32) .. (61,36.32) ;
(26.9,28.78) .. controls (12.8,21.25) and (30.9,9.78) .. (42.9,8.78) .. controls (54.9,7.78) and (74.45,15.45) .. (63.45,24.45) .. controls (52.45,33.45) and (54.45,55.45) .. (66.45,70.45) .. controls (78.45,85.45) and (49.9,113.13) .. (37.45,118.32) .. controls (25,123.5) and (31,134.5) .. (50,139.5) ;
(59.45,107.32) .. controls (81.45,139.32) and (21,173.5) .. (16.5,148) ;
(24.85,105.52) .. controls (14,119.5) and (15,140.5) .. (16.5,148) ;
(54,30.32) .. controls (41,18.32) and (25,30.32) .. (25,46.32) .. controls (25,62.32) and (29,73.32) .. (37.57,78.45) ;
(45.45,85.32) .. controls (51.45,89.32) and (49.45,94.32) .. (53.45,99.32) ;
(24.85,105.52) .. controls (27.85,99.52) and (67.45,64.45) .. (33.45,35.45) ;

(69.37,143) .. controls (99.37,143) and (94,82.85) .. (91,72.85) .. controls (88,62.85) and (80,48.85) .. (65,36.85) ;
(30.9,29.32) .. controls (16.8,21.78) and (34.9,10.32) .. (46.9,9.32) .. controls (58.9,8.32) and (78.45,15.98) .. (67.45,24.98) .. controls (56.45,33.98) and (58.45,55.98) .. (70.45,70.98) .. controls (82.45,85.98) and (53.37,114.13) .. (30.37,123) ;
(63.45,107.85) .. controls (75.37,121) and (60.37,153) .. (50.37,155) .. controls (40.37,157) and (33.37,147) .. (28.37,138) .. controls (23.37,129) and (23.37,114) .. (28.85,106.05) ;
(18.37,127) .. controls (3.37,133) and (12.37,154) .. (29.37,149) ;
(58,30.85) .. controls (45,18.85) and (29,30.85) .. (29,46.85) .. controls (29,62.85) and (33,73.85) .. (41.57,78.98) ;
(49.45,85.85) .. controls (55.45,89.85) and (53.45,94.85) .. (57.45,99.85) ;
(28.85,106.05) .. controls (32.37,96) and (71.45,64.98) .. (37.45,35.98) ;
(40.37,146) .. controls (48.37,145) and (48.37,144) .. (56.37,143) ;

(52.13,142.82) .. controls (91.23,154.75) and (84.23,84.67) .. (80,71.65) .. controls (75.77,58.63) and (69,47.65) .. (54,35.65) ;
(17.9,28.12) .. controls (3.8,20.58) and (21.9,9.12) .. (33.9,8.12) .. controls (45.9,7.12) and (65.45,14.78) .. (54.45,23.78) .. controls (43.45,32.78) and (51.27,63.98) .. (57.45,69.78) .. controls (63.63,75.58) and (60.63,78.58) .. (56.63,86.58) ;
(51.23,131.02) .. controls (53.23,146.02) and (17.13,169) .. (19.13,146) ;
(36.63,75.42) .. controls (11.23,105.02) and (70.63,79.42) .. (72.63,95.42) .. controls (74.63,111.42) and (58.23,120.82) .. (49.23,125.82) .. controls (40.23,130.82) and (20.13,133) .. (19.13,146) ;
(45,29.65) .. controls (32,17.65) and (16,29.65) .. (16,45.65) .. controls (16,61.65) and (23.63,67.42) .. (31.63,76.42) ;
(36.63,75.42) .. controls (45.23,66.02) and (44.23,49.02) .. (24.45,34.78) ;
(55.23,94.62) .. controls (48.23,110.62) and (32.63,112.58) .. (36.73,127.02) ;
(39.73,134.02) .. controls (42.73,138.02) and (41.73,137.02) .. (43.73,139.02) ;
(35.43,81.78) .. controls (38.43,84.78) and (45.63,82.42) .. (45.43,87.78) ;
(46.23,94.62) – (47.63,101.42) ;
(48.23,108.02) .. controls (49.23,114.02) and (48.23,114.02) .. (49.23,120.02) ;

(60.23,92.65) .. controls (62.23,105.65) and (7.23,121) .. (14.23,157) .. controls (21.23,193) and (82.13,147.13) .. (86.67,133.81) .. controls (91.2,120.49) and (83.31,75.94) .. (81,68.83) .. controls (78.69,61.72) and (70,44.83) .. (55,32.83) ;
(21.9,27.3) .. controls (7.8,19.77) and (25.9,8.3) .. (37.9,7.3) .. controls (49.9,6.3) and (69.45,13.97) .. (58.45,22.97) .. controls (47.45,31.97) and (53.23,55.65) .. (58.23,61.65) .. controls (63.23,67.65) and (60.23,76.65) .. (60.63,85.77) ;
(55.23,130.2) .. controls (57.23,145.2) and (21.13,168.18) .. (23.13,145.18) ;
(40.63,74.6) .. controls (15.23,104.2) and (74.63,78.6) .. (76.63,94.6) .. controls (78.63,110.6) and (62.23,120) .. (53.23,125) .. controls (44.23,130) and (24.13,132.18) .. (23.13,145.18) ;
(49,28.83) .. controls (36,16.83) and (20,28.83) .. (20,44.83) .. controls (20,60.83) and (27.63,66.6) .. (35.63,75.6) ;
(40.63,74.6) .. controls (49.23,65.2) and (48.23,48.2) .. (28.45,33.97) ;
(39.43,80.97) .. controls (42.43,83.97) and (49.63,81.6) .. (49.43,86.97) ;
(50.23,93.8) – (51.63,100.6) ;
(52.23,107.2) .. controls (53.23,113.2) and (52.23,113.2) .. (53.23,119.2) ;

(59.23,93.65) .. controls (61.23,106.65) and (6.23,122) .. (13.23,158) .. controls (20.23,194) and (77.45,146.78) .. (85.67,134.81) .. controls (93.88,122.83) and (82.31,76.94) .. (80,69.83) .. controls (77.69,62.72) and (69,45.83) .. (54,33.83) ;
(20.9,28.3) .. controls (6.8,20.77) and (24.9,9.3) .. (36.9,8.3) .. controls (48.9,7.3) and (68.45,14.97) .. (57.45,23.97) .. controls (46.45,32.97) and (52.23,56.65) .. (57.23,62.65) .. controls (62.23,68.65) and (59.23,77.65) .. (59.63,86.77) ;
(54.23,131.2) .. controls (56.23,146.2) and (20.13,169.18) .. (22.13,146.18) ;
(27.7,35.83) .. controls (26.7,105.83) and (74.7,84.83) .. (75.63,95.6) .. controls (76.57,106.37) and (61.23,121) .. (52.23,126) .. controls (43.23,131) and (23.13,133.18) .. (22.13,146.18) ;
(48,27.83) .. controls (35,15.83) and (24.3,26.83) .. (19,43.83) .. controls (13.7,60.83) and (17.26,66.05) .. (20.91,74.75) .. controls (24.56,83.44) and (31.7,89.83) .. (48.7,97.83) ;
(51.23,108.2) .. controls (52.23,114.2) and (51.23,114.2) .. (52.23,120.2) ;

(57.23,94.65) .. controls (59.23,107.65) and (4.23,123) .. (11.23,159) .. controls (18.23,195) and (75.45,147.78) .. (83.67,135.81) .. controls (91.88,123.83) and (80.31,77.94) .. (78,70.83) .. controls (75.69,63.72) and (67,46.83) .. (52,34.83) ;
(18.9,29.3) .. controls (8.72,14.65) and (22.9,10.3) .. (34.9,9.3) .. controls (46.9,8.3) and (77.18,4.28) .. (55.45,24.97) .. controls (33.72,45.65) and (74.72,67.65) .. (59.72,84.65) ;
(26.72,39.65) .. controls (51.72,65.65) and (30.72,77.65) .. (56.55,90.94) .. controls (82.37,104.23) and (78.72,113.65) .. (72.72,122.65) .. controls (66.72,131.65) and (36.88,145) .. (31.88,128) ;
(46,28.83) .. controls (33,16.83) and (20.72,33.65) .. (13.72,50.65) .. controls (6.72,67.65) and (18.88,92) .. (25.88,112) ;

(42.78,84.68) .. controls (40.23,71.78) and (94.52,54.09) .. (85.99,18.43) .. controls (77.46,-17.24) and (22.31,32.38) .. (14.62,44.69) .. controls (6.92,57.01) and (20.44,102.36) .. (23.05,109.37) .. controls (25.67,116.37) and (35.07,132.87) .. (50.57,144.22) ;
(83.88,148.33) .. controls (94.68,162.54) and (80.69,167.49) .. (68.74,169) .. controls (56.8,170.51) and (26.71,175.82) .. (47.54,154.23) .. controls (68.37,132.63) and (26.47,112.41) .. (40.73,94.78) ;
(75.62,138.33) .. controls (49.53,113.42) and (70,100.53) .. (43.63,88.36) .. controls (17.26,76.18) and (20.51,66.62) .. (26.12,57.37) .. controls (31.73,48.12) and (60.96,33.51) .. (66.68,50.28) ;
(56.82,149.96) .. controls (70.32,161.39) and (81.87,144.07) .. (88.14,126.78) .. controls (94.41,109.5) and (81.21,85.69) .. (73.36,66.01) ;

(35.68,121) .. controls (-0.32,123) and (-3.32,23) .. (46.68,10) .. controls (96.68,-3) and (88.68,37) .. (69.68,60) ;
(86.68,113) .. controls (107.68,82) and (16.68,75) .. (70.68,25) ;
(91.68,15) .. controls (122.68,-1) and (168.68,122) .. (56.68,122) ;
(52.68,76) .. controls (13.68,132) and (81.68,145) .. (85.68,127) ;

There exists some knots that have both its reverse and mirror to be distinct to its original knot. One such example is the knot $K=9_{32}$. In fact, $K,rK,\overline{K},\overline{rK}$ are all distinct oriented knots (<cit.>). However, the reverse of a knot is often equivalent to its original knot.

The reflection of a link diagram negates the writhe of the original link diagram.

We look at a single crossing and see what happens to the sign when the crossing is reflected,

\[\epsilon\left(\signpositive\right)=-\epsilon\left(\reflectbox{\signpositive}\right).\]

The writhe of a link preserves under regular isotopy, but not under ambient isotopy.

A function that is preserved under regular isotopy but not under ambient isotopy, means that we need to show the writhe does not change under $\RII$ and $\RIII$ moves, but does change under $\RI$ moves.

\begin{align*}
\RI: \qquad &\omega\left(\reflectbox{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.6cm,scale=0.5]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}}\!\right) =
%   \rimove
\omega\left(\knotsinmath{\begin{scope}[scale=0.2]
\draw[black,->] (0,2) -- (0,-2);
\end{scope}}\right) + 1\\
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}\right) =
\omega\left(\knotsinmath{\begin{scope}[scale=0.2]
\draw[black,->] (0,2) -- (0,-2);
\end{scope}}\right) - 1
\end{align*}

For $\RII$, given any orientation, we can see

\begin{align*}
\omega\left(\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.4cm,scale=0.5]
\draw    (14,48) .. controls (53.6,18.3) and (74.58,18.23) .. (112.84,47.12) ;
\draw [shift={(114,48)}, rotate = 217.35] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (39,41.23) .. controls (52,53.23) and (69,51.23) .. (83,37.23) ;
\draw    (20,20.23) -- (29,29.23) ;
\draw    (94,24.23) -- (106.77,7.81) ;
\draw [shift={(108,6.23)}, rotate = 487.87] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\end{scope}}\right) = \omega\left(\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.4cm,scale=0.5]
\draw    (14,48) .. controls (53.6,18.3) and (74.58,18.23) .. (112.84,47.12) ;
\draw [shift={(114,48)}, rotate = 217.35] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (39,41.23) .. controls (52,53.23) and (69,51.23) .. (83,37.23) ;
\draw    (29,29.23) -- (16.32,14.74) ;
\draw [shift={(15,13.23)}, rotate = 408.81] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (94,24.23) -- (102,12.23) ;
\end{scope}}\right) &= \omega\left(\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.4cm,scale=0.5]
\draw    (107,44.23) .. controls (76.31,17.5) and (48.56,15.28) .. (15.02,50.16) ;
\draw [shift={(14,51.23)}, rotate = 313.36] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (39,41.23) .. controls (52,53.23) and (69,51.23) .. (83,37.23) ;
\draw    (20,20.23) -- (29,29.23) ;
\draw    (94,24.23) -- (106.77,7.81) ;
\draw [shift={(108,6.23)}, rotate = 487.87] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\end{scope}}\right) = \omega\left(\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.4cm,scale=0.5]
\draw    (107,44.23) .. controls (76.31,17.5) and (48.56,15.28) .. (15.02,50.16) ;
\draw [shift={(14,51.23)}, rotate = 313.36] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (39,41.23) .. controls (52,53.23) and (69,51.23) .. (83,37.23) ;
\draw    (17.26,14.79) -- (29,29.23) ;
\draw [shift={(16,13.23)}, rotate = 50.91] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (94,24.23) -- (102,12.23) ;
\end{scope}}\right)\\
&= \omega\left(\resolveriimove\right) = 0 \text{, for the given orientation.}
\end{align*}

And it should be clear that it is also true for its reflection, i.e., $\omega\left(\riimove\right)=0=\omega\left(\resolveriimove\right)$, given any orientation.
Now $\RIII$ is analogous to what we did for $\RII$. So the details are left for the reader, i.e., given any orientation we have,

\[
\omega\left(\knotsinmath{\begin{scope}[scale=0.2]
\draw[draw=white,double=black ,ultra thick,double distance=0.5pt] (  0:2) .. controls (270:0.7) .. (180:2);
\draw[draw=white,double=black  ,ultra thick,double distance=0.5pt] ( 60:2) .. controls (150:0.7) .. (240:2);
\draw[draw=white,double=black,ultra thick,double distance=0.5pt] (120:2) .. controls ( 30:0.7) .. (300:2);
\end{scope}
}\right) = 0 = \omega\left(\knotsinmath{
\begin{scope}[scale=0.2]
\draw[draw=white,double=black ,ultra thick,double distance=0.5pt] (  0:2) .. controls ( 90:0.7) .. (180:2);
\draw[draw=white,double=black  ,ultra thick,double distance=0.5pt] ( 60:2) .. controls (330:0.7) .. (240:2);
\draw[draw=white,double=black,ultra thick,double distance=0.5pt] (120:2) .. controls (210:0.7) .. (300:2);
\end{scope}
\]

§.§ Invariants

By only considering the Reidemeister moves, it is difficult to distinguish links. This is because there is no method that can tell us whether we have not made the right number of Reidemeister moves yet. So the best way to distinguish links is by using invariants, which are algebraic ways of characterising these topological objects.

An invariantinvariant $f$ is a function

\[f\colon\{\text{equivalence classes of topological objects}\}\to{S}\]

for some set $S$, with the property $f(A)=f(B)$ whenever $A$ and $B$ are equivalent.

So a link invariant is just an invariant function $f\colon\{\text{link type}\}\to{S}$ for some set $S$.

Thus, to prove that two links are in fact distinct (the main problem of knot theory), we need to show that they produce different outputs in a link invariant.

Invariant functions can only tell the objects apart - they can not show equivalency.

The crossing numbercrossing number is a link invariant, such that it is defined as the minimal number of crossings needed for a diagram of the link $L$, where the minimum is taken over the infinity of possible diagrams of a link. We write $c(L)$ for the link $L$.

For low crossing numbers, we can find the knots by tabulating all knot diagrams; see the following result. In general, however, it is difficult to know which knots have the same knot type. This is solved by associating a knot to an invariant.

There are no knots with crossing number one or two.

We first draw the link universe with one component and one crossing and then show all the possible choices. The strategy here is to draw a link universe with the number of crossings we need without having to create more than one component.

(a) at (0,0) ;
(a.center) .. controls (a.8 north west) and (a.8 south west) .. (a.center);
(a.center) .. controls (a.8 north east) and (a.8 south east) .. (a.center);
(0,-0.5) node[scale=0.7] link universe;

(a) at (0,0) ;
(a.center) .. controls (a.8 north west) and (a.8 south west) .. (a);
(a) .. controls (a.8 north east) and (a.8 south east) .. (a.center);
(0,-0.5) node[scale=0.7,rotate=4] link diagram;
(0,-0.8) node[scale=0.7,rotate=4] unknot;

(a) at (0,0) ;
(a) .. controls (a.8 north west) and (a.8 south west) .. (a.center);
(a.center) .. controls (a.8 north east) and (a.8 south east) .. (a);
(0,-0.5) node[scale=0.7,rotate=-4] link diagram;
(0,-0.8) node[scale=0.7,rotate=-4] unknot;

We clearly see that the first and second choices are unknots (via $\RI$). Therefore, there are no knots with crossing number one.
We leave the details for crossing number two to the reader, as it follows the same argument.

The unlinking numberunlinking number is another link invariant. It is defined to be the minimum number of crossing changes (from “over” to “under” or vice versa) needed to change link $L$ to the unlink, where the minimum is taken over all possible sets of crossing changes in all possible diagrams of $L$. We write this as $u(L)$.

So, like the crossing number invariant, the unlinking number is very hard to calculate.

We define the complexitycomplexity of a link $\kappa(L)$ to be the minimal ordered pair of integers $(c(L),u(L))$, where $c(L)$ is the crossing number of link $L$ and $u(L)$ is the unlinking number of $L$.
We order complexities lexocographically, i.e., $(a,b)<(c,d)$ if and only if $(a<c)$ or ($a=c$ and $b<d$).

The most basic complexity, is that of the unknot, i.e., $\kappa\left(\KPA\right)=(0,0)$.

For the following trefoil,

[rotate=  0]
(a) at (0,-1) ;

(b) at (0,-1) ;

(c) at (0,-1) ;

(a.center) .. controls (a.4 north west) and (c.4 north east) .. (c);
(b.center) .. controls (b.4 north west) and (a.4 north east) .. (a);
(c.center) .. controls (c.4 north west) and (b.4 north east) .. (b);
(a) .. controls (a.16 south west) and (c.16 south east) .. (c.center);
(b) .. controls (b.16 south west) and (a.16 south east) .. (a.center);
(c) .. controls (c.16 south west) and (b.16 south east) .. (b.center);

We can see that it has complexity $(3,1)$, by changing one of its crossings as shown below,

[rotate=  0] (a) at (0,-1) ;
[rotate=120] (b) at (0,-1) ;
[rotate=240] (c) at (0,-1) ;
(a) .. controls (a.4 north west) and (c.4 north east) .. (c);
(b.center) .. controls (b.4 north west) and (a.4 north east) .. (a.center);
(c.center) .. controls (c.4 north west) and (b.4 north east) .. (b);
(a.center) .. controls (a.16 south west) and (c.16 south east) .. (c.center);
(b) .. controls (b.16 south west) and (a.16 south east) .. (a);
(c) .. controls (c.16 south west) and (b.16 south east) .. (b.center);

For the figure-eight knot $4_1$, it is not hard to see that it has complexity $(4,1)$ by also changing one of its crossings.

So, $\kappa(0_1)<\kappa(3_1)<\kappa(4_1)$.

Every link $L$ with a corresponding diagram $D$ has an unlinking number.

We begin by choosing a starting point on all of the components on the link diagram. Following a clockwise orientation, we follow each component of the link diagram from the starting point.
When a crossing appears, if it is an overpass, we change it to an underpass, otherwise we keep it as it is.
Once we are done, we have an unlink. This shows us that the unlinking number $u(L)\leq{c(L)}$.

There are many other link invariants, but in this paper we will mainly be focusing on the Jones polynomial and the Kauffman bracket.

§ THE KAUFFMAN BRACKET CONSTRUCTION

We will now define the Jones polynomial. This will be based on the Kauffman bracket [33, 34, 35].
We will then show that this admits a skein relation [15, 20, 26]. We will not go deep into the theory of skein relations. The interested reader is directed to [15] and [20].

Our discussion is based on that in Kauffman's paper [33], Jones' paper [26], Jones' book [28], Conway's paper [15], and HOMFLY-PT's paper [20, 53].

We note that HOMFLY-PT is an acronym for the eight people who wrote the two papers, namely, Hoste, Ocneanu, Millett, Freyd, Lickorish, and Yetter. The PT are the other two who wrote a different paper on the same subject: Pyztycki and Traczyk.

Notation <cit.>.
Formally, when we look at an enclosed region of a link such that there are four distinct points (NW, NE, SE, SW), then we call the enclosed region a tangletangle.

[dashed] (0,0) circle (0.5);
(0,0) node $L$;
(0.7,0.7) node[anchor=south west] NE;
(-0.7,0.7) node[anchor=south east] NW;
(0.7,-0.7) node[anchor=north west] SE;
(-0.7,-0.7) node[anchor=north east] SW;

We call it an elementary $2$-tangletangle!elementary $2$-tangle if the tangle consists of exactly one crossing or none. Such as the following.

[dashed] (0,0) circle(1.5);
[-] (-1,-1) to [bend left=40] (1,-1);
[-] (1,1) to [bend left=40] (-1,1);

(0,-2) node[scale=1.5] $\tangle{\infty}$;
[dashed] (0,0) circle(1.5);
[-] (-1,-1)–(1,1);
[ultra thick, draw=white,double=black,double distance=0.5pt] (1,-1)–(-1,1);
[-] (-0.9,0.9)–(-1,1);
(0,-2) node[scale=1.5] $\tangle{-1}$;

[dashed] (0,0) circle(1.5);
[-] (1,-1)–(-1,1);
[ultra thick,draw=white,double=black,double distance=0.5pt] (-1,-1)–(1,1);
[-] (-0.9,-0.9)–(1,1);
(0,-2) node[scale=1.5] $\tangle{1}$;

[dashed] (0,0) circle(1.5);
[-] (-1,-1) to [bend right=40] (-1,1);
[-] (1,-1) to [bend left=40] (1,1);
(0,-2) node[scale=1.5] $\tangle{0}$;

elementary $2$-tangles

These elementary $2$-tangles are usually recognised as the skein relation of a link (see <ref>).
For more details on tangles, the interested reader is directed to Conway's paper [15].

A Laurent polynomialLaurent polynomial is an expression of the form $p=\sum_{k=-N}^Na_kA^k$ for some natural number $N$ and some list of coefficients $a_{-N},\dotsc,a_N\in\ZZ$.

The aim of the Kauffman bracket is to define an easier, more combinatorial approach to the Jones polynomial.

§.§ `states model' and the Kauffman bracket polynomial

Our first definition of the Jones polynomial will require us to define the Kauffman bracket. We will begin by defining a state of a link diagram (as was presented in Kauffman's paper [34]).

Given a non-oriented link diagram, we notice that each crossing is divided into four regions. By convention, we label the regions of each crossing $A$ and $B$, by starting from the over-strand in a counterclockwise fashion. Both possible crossings are labelled below.

[ultra thick,draw=white,double=black,double distance=0.5pt] (-1,-1)–(1,1);
(0,0) node[scale=1.25,anchor=north] $A$;
(0,0) node[scale=1.25,anchor=west] $B$;
(0,0) node[scale=1.25,anchor=south] $A$;
(0,0) node[scale=1.25,anchor=east] $B$;
[->] (0.5,0.5) to [bend right=30] (0,1);
[->] (-0.5,-0.5) to [bend right=30] (0,-1);

[ultra thick,draw=white,double=black,double distance=0.5pt] (-1,1)–(1,-1);
(0,0) node[scale=1.25,anchor=west] $A$;
(0,0) node[scale=1.25,anchor=south] $B$;
(0,0) node[scale=1.25,anchor=east] $A$;
(0,0) node[scale=1.25,anchor=north] $B$;
[->] (-0.5,0.5) to [bend right=30] (-1,0);
[->] (0.5,-0.5) to [bend right=30] (1,0);

Given a non-oriented link diagram, a splitting markersplitting marker is defined as a marker that essentially smooths the crossing. We colour the splitting marker in red in the figure below, and show how we can get two different split diagrams for any crossing.

[draw=red] (-0.5,0)–(0.5,0);
[->] (1,0)–(2,0);
(2,1) .. controls (3,0) .. (4,1);
(2,-1) .. controls (3,0) .. (4,-1);

[draw=red] (0,0.5)–(0,-0.5);
[->] (1,0)–(2,0);
(2,1) .. controls (3,0) .. (2,-1);
(4,1) .. controls (3,0) .. (4,-1);

splitting marker

A statestate of a link $L$ is a choice of the splitting marker for each crossing in $L$.

So, by convention, we define an $A$-smoothing and a $B$-smoothing as the following.

[ultra thick,draw=white,double=black] (-1,1)–(1,-1);
(0,0) node[scale=1.75,anchor=west] $A$;
(0,0) node[scale=1.75,anchor=south] $B$;
(0,0) node[scale=1.75,anchor=east] $A$;
(0,0) node[scale=1.75,anchor=north] $B$;
[->] (0,-1)–(-2,-3);
[->] (0,-1)–(2,-3);
(0,1) .. controls (1,0) .. (2,1);
(0,-1) .. controls (1,0) .. (2,-1);
[draw=red] (1.5,0)–(0.5,0);
(1,-2) node[scale=2] $A$-smoothing;

(0,1) .. controls (1,0) .. (0,-1);
(2,1) .. controls (1,0) .. (2,-1);
[draw=red] (1,0.5)–(1,-0.5);
(1,-2) node[scale=2] $B$-smoothing;

[ultra thick,draw=white,double=black] (-1,-1)–(1,1);
(0,0) node[scale=2,anchor=north] $A$;
(0,0) node[scale=2,anchor=west] $B$;
(0,0) node[scale=2,anchor=south] $A$;
(0,0) node[scale=2,anchor=east] $B$;
[->] (0,-1)–(-2,-3);
[->] (0,-1)–(2,-3);

(0,1) .. controls (1,0) .. (0,-1);
(2,1) .. controls (1,0) .. (2,-1);
[draw=red] (1,0.5)–(1,-0.5);
(1,-2) node[scale=2] $A$-smoothing;

(0,1) .. controls (1,0) .. (2,1);
(0,-1) .. controls (1,0) .. (2,-1);
[draw=red] (1.5,0)–(0.5,0);
(1,-2) node[scale=2] $B$-smoothing;

So for an $n$ crossing diagram $L$ we have a state $s$ is a bijective map $s\mapsto\{A,B\}^n$. This means we have $2^n$ states for an $n$ crossing diagram.

When we smooth out a crossing on a diagram $D$, we call the $A$-smoothed diagram $D_{\{A\}}$, and we call the $B$-smoothed diagram $D_{\{B\}}$. The subscripted set represents the smoothings of the respected crossings on $D$. For example, if we have two crossings $c_1,c_2$ on a diagram $D$, then the four states would be $D_{\{A,A\}}$, $D_{\{A,B\}}$, $D_{\{B,A\}}$, and $D_{\{B,B\}}$.

Since these are tangles, we can instead define the splitting of a single crossing. We immediately get the other crossings, by rotating by $\pi/2$.
For instance, if we define $\tangle{-1}\mapsto\{\tangle{\infty}_A,\tangle{0}_B\}$, then by rotating all the tangles and keeping $A$ and $B$ in place, we get $\tangle{1}\mapsto\{\tangle{0}_A,\tangle{\infty}_B\}$.

Consider the trefoil knot diagram $D$ drawn below. To find all the states of $D$, we need to resolve all three crossings $c_1$, $c_2$, and $c_3$ to find all possible $2^3$ states.

[every path/.style=,scale=0.02,yscale=-1]
(81,52) .. controls (94,158) and (-17,89) .. (23,59) ;
(85,100) .. controls (145,132) and (115,14) .. (53,45) ;
(80,25) .. controls (85,5) and (53,11) .. (41,27) .. controls (29,43) and (36,70) .. (66,87) ;
(77,110) node[scale=48,yscale=-1] $c1$;
(18,40) node[scale=48,yscale=-1] $c3$;
(94,25) node[scale=48,yscale=-1] $c2$;

So, the first two possible diagrams are obtained by considering crossing $c_1$ and finding the $A$-smoothing and $B$-smoothing.

(101,52.5) .. controls (121,125.5) and (186,11.5) .. (75,47.5) ;
(53,57.5) .. controls (20,66.5) and (48.03,100.44) .. (65.02,109.47) .. controls (82,118.5) and (96,108.5) .. (89.21,90.43) .. controls (82.42,72.36) and (73.63,63.29) .. (63.32,42.39) .. controls (53,21.5) and (95,-4.5) .. (99,31.5) ;
(219,71.5) .. controls (191,78.5) and (234,117.5) .. (261,107.5) .. controls (288,97.5) and (304,34.5) .. (242,58.5) ;
(268,63.5) .. controls (271,86.5) and (232,96.5) .. (230.32,53.39) .. controls (228.63,10.29) and (262,6.5) .. (266,42.5) ;

$D_{\{A\}}$                $D_{\{B\}}$

The next possible diagrams are obtained by resolving the second crossing $c_2$ of $D_{\{A\}}$ and $D_{\{B\}}$.

(53,57.5) .. controls (20,66.5) and (48.03,100.44) .. (65.02,109.47) .. controls (82,118.5) and (96,108.5) .. (89.21,90.43) .. controls (82.42,72.36) and (75.63,67.29) .. (63.32,42.39) .. controls (51,17.5) and (78.67,2.2) .. (89.83,14.85) .. controls (101,27.5) and (95,37.5) .. (80,45.5) ;
(341,58.5) .. controls (304,66.5) and (354,110.5) .. (381,100.5) .. controls (408,90.5) and (426,19.5) .. (386,57.5) ;
(386,57.5) .. controls (366,82.5) and (349,79.5) .. (350.32,46.39) .. controls (351.63,13.29) and (427,29.5) .. (362,51.5) ;
(107.5,50.75) .. controls (107.5,36.53) and (119.03,25) .. (133.25,25) .. controls (147.47,25) and (159,36.53) .. (159,50.75) .. controls (159,64.97) and (147.47,76.5) .. (133.25,76.5) .. controls (119.03,76.5) and (107.5,64.97) .. (107.5,50.75) – cycle ;
(226,61.5) .. controls (193,70.5) and (221.03,104.44) .. (238.02,113.47) .. controls (255,122.5) and (269,112.5) .. (262.21,94.43) .. controls (255.42,76.36) and (248.63,71.29) .. (236.32,46.39) .. controls (224,21.5) and (251.67,6.2) .. (262.83,18.85) .. controls (274,31.5) and (268,41.5) .. (253,49.5) ;
(464,72.5) .. controls (442,74.5) and (484,114.5) .. (506,108.5) .. controls (528,102.5) and (542,37.5) .. (520,27.5) ;
(487,59.5) .. controls (527,45.5) and (510,68.5) .. (501.85,77.01) .. controls (493.7,85.52) and (485,81.5) .. (475.32,54.39) .. controls (465.63,27.29) and (493,16.5) .. (520,27.5) ;

$D_{\{A,A\}}$            $D_{\{A,B\}}$          $D_{\{B,A\}}$          $D_{\{B,B\}}$

And so the $8$ states are thus obtained by resolving the last crossing $c_3$ of $D_{\{A,A\}}$, $D_{\{A,B\}}$, $D_{\{B,A\}}$, and $D_{\{B,B\}}$.

(84.5,51) .. controls (84.5,39.4) and (93.9,30) .. (105.5,30) .. controls (117.1,30) and (126.5,39.4) .. (126.5,51) .. controls (126.5,62.6) and (117.1,72) .. (105.5,72) .. controls (93.9,72) and (84.5,62.6) .. (84.5,51) – cycle ;
(50,97) .. controls (50,84.3) and (60.3,74) .. (73,74) .. controls (85.7,74) and (96,84.3) .. (96,97) .. controls (96,109.7) and (85.7,120) .. (73,120) .. controls (60.3,120) and (50,109.7) .. (50,97) – cycle ;
(30.5,47.75) .. controls (30.5,35.19) and (40.69,25) .. (53.25,25) .. controls (65.81,25) and (76,35.19) .. (76,47.75) .. controls (76,60.31) and (65.81,70.5) .. (53.25,70.5) .. controls (40.69,70.5) and (30.5,60.31) .. (30.5,47.75) – cycle ;
(218.5,72) .. controls (210.5,89) and (235.5,114) .. (206.5,111) .. controls (177.5,108) and (202.5,87) .. (200.5,67) .. controls (198.5,47) and (191.17,33.7) .. (211.83,30.85) .. controls (232.5,28) and (226.5,56) .. (218.5,72) – cycle ;
(230.5,76) .. controls (230.5,64.4) and (239.9,55) .. (251.5,55) .. controls (263.1,55) and (272.5,64.4) .. (272.5,76) .. controls (272.5,87.6) and (263.1,97) .. (251.5,97) .. controls (239.9,97) and (230.5,87.6) .. (230.5,76) – cycle ;
(335.5,49) .. controls (335.5,37.4) and (344.9,28) .. (356.5,28) .. controls (368.1,28) and (377.5,37.4) .. (377.5,49) .. controls (377.5,60.6) and (368.1,70) .. (356.5,70) .. controls (344.9,70) and (335.5,60.6) .. (335.5,49) – cycle ;
(315,96) .. controls (315,83.3) and (325.3,73) .. (338,73) .. controls (350.7,73) and (361,83.3) .. (361,96) .. controls (361,108.7) and (350.7,119) .. (338,119) .. controls (325.3,119) and (315,108.7) .. (315,96) – cycle ;
(472.5,77) .. controls (464.5,94) and (489.5,119) .. (460.5,116) .. controls (431.5,113) and (456.5,92) .. (454.5,72) .. controls (452.5,52) and (445.17,38.7) .. (465.83,35.85) .. controls (486.5,33) and (480.5,61) .. (472.5,77) – cycle ;

$D_{\{A,A,A\}}$            $D_{\{A,A,B\}}$          $D_{\{A,B,A\}}$      $D_{\{A,B,B\}}$

(242.68,57.31) .. controls (219.64,64.53) and (236.2,104.04) .. (259.85,104.52) .. controls (283.5,105) and (291.5,49) .. (272.5,26.47) ;
(242.68,57.31) .. controls (278.82,43.82) and (235.91,38.52) .. (235,48.15) .. controls (234.1,57.79) and (221.17,57.89) .. (212.42,31.77) .. controls (203.67,5.66) and (248.5,3) .. (272.5,26.47) ;
(12,91) .. controls (12,78.3) and (22.3,68) .. (35,68) .. controls (47.7,68) and (58,78.3) .. (58,91) .. controls (58,103.7) and (47.7,114) .. (35,114) .. controls (22.3,114) and (12,103.7) .. (12,91) – cycle ;
(21.5,36.75) .. controls (21.5,24.19) and (31.69,14) .. (44.25,14) .. controls (56.81,14) and (67,24.19) .. (67,36.75) .. controls (67,49.31) and (56.81,59.5) .. (44.25,59.5) .. controls (31.69,59.5) and (21.5,49.31) .. (21.5,36.75) – cycle ;
(153.5,62) .. controls (145.5,79) and (170.5,104) .. (141.5,101) .. controls (112.5,98) and (137.5,77) .. (135.5,57) .. controls (133.5,37) and (126.17,23.7) .. (146.83,20.85) .. controls (167.5,18) and (161.5,46) .. (153.5,62) – cycle ;
(362.5,63.5) .. controls (362.5,50.8) and (372.8,40.5) .. (385.5,40.5) .. controls (398.2,40.5) and (408.5,50.8) .. (408.5,63.5) .. controls (408.5,76.2) and (398.2,86.5) .. (385.5,86.5) .. controls (372.8,86.5) and (362.5,76.2) .. (362.5,63.5) – cycle ;
(338,63.5) .. controls (338,37.27) and (359.27,16) .. (385.5,16) .. controls (411.73,16) and (433,37.27) .. (433,63.5) .. controls (433,89.73) and (411.73,111) .. (385.5,111) .. controls (359.27,111) and (338,89.73) .. (338,63.5) – cycle ;

$D_{\{B,A,A\}}$        $D_{\{B,A,B\}}$        $D_{\{B,B,A\}}$          $D_{\{B,B,B\}}$

The Kauffman bracketKauffman bracket is a function,

\[\langle{D}\rangle\colon\{\text{unoriented link diagrams D}\}\to\ZZ[A,B,d],\]

that is an invariant under $\RII$ and $\RIII$, and is uniquely defined by the following properties,

* $\langle{\bigcirc}\rangle=1$,

* $\langle{D\cup\bigcirc}\rangle=d\langle{D}\rangle$,

* $\left\langle\KPB\right\rangle=A\left\langle\KPC\right\rangle+B\left\langle\KPD\right\rangle$,

* $\left\langle\KPE\right\rangle=A\left\langle\KPD\right\rangle+B\left\langle\KPC\right\rangle$

The bracket is well defined with states and splitting markers. After labelling the regions, when we split the diagram, we multiply the bracket of the split diagram with the region the splitting marker is coinciding with.

\[
\left\langle\begin{tikzpicture}[every path/.style={},scale=0.5,baseline=-\dimexpr\fontdimen22\textfont2\relax]
\clip (-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle;
\draw (-1,-1)--(1,1);
\draw[ultra thick,draw=white,double=black] (-1,1)--(1,-1);
\draw (0,0) node[scale=1.4,anchor=west] {$A$};
\draw (0,0) node[scale=1.4,anchor=south] {$B$};
\draw (0,0) node[scale=1.4,anchor=east] {$A$};
\draw (0,0) node[scale=1.4,anchor=north] {$B$};
\draw[thick,red,dashed] (-0.8,0)--(0.8,0);
\draw[blue,thick,dashed] (0,-0.8)--(0,0.8);
\end{tikzpicture}\right\rangle
\textcolor{red}{A}\left\langle\KPC\right\rangle+\textcolor{blue}{B}\left\langle\KPD\right\rangle
\]

In tangle notation, (iii) would be $\left\langle\tangle{-1}\right\rangle=A\left\langle\tangle{\infty}_A\right\rangle+B\left\langle\tangle{0}_B\right\rangle$ and by rotating, we get (iv). Thus, defining both is not necessary, but is convenient.

The Kauffman bracket is well-defined, i.e., we have the Kauffman bracket defined by its state-sumKauffman bracket!state-sum

\[\langle{D}\rangle=\sum_{\text{states }s\text{ of }D}A^{|a(s)|}B^{|b(s)|}d^{|s|-1},\]

where $a(s)$ is the number of $A$-smoothings in the state $s$, $b(s)$ is the number of $B$-smoothings in the state $s$, and $|s|$ is the number of components in the state $s$.

We show the uniqueness of the polynomial bracket by induction on the number of crossings in the link diagram.

For a diagram with no crossings, we clearly have $\left\langle\KPA\right\rangle=1$, since the only state would be the unknot itself.
Now if the link diagram $D$ is empty, then we need $\left\langle{D\cup\bigcirc}\right\rangle=1$, which is clearly satisfied.

If the link diagram $D$ is not empty, we see that any state $s$ of $D$ with $\bigcirc$ is the same as the state of $D\cup\bigcirc$.
We note
$\left|a\left(s\cup\KPA\right)\right|=|a(s)|$ and $\left|b\left(s\cup\KPA\right)\right|=|b(s)|$,

since they have the same crossings and so the smoothings will be the same (except for the extra circle which will not affect $a(s)$ or $b(s)$). So

\begin{align*}
\left\langle{D\cup\KPA}\right\rangle
\sum_{s}{A^{|a(s)|}B^{|b(s)|}d^{|s|+1-1}}\\
&= \sum_{s}{A^{|a(s)|}B^{|b(s)|}d^{|s|-1}d},
%  \left\langle{D\cup\bigcirc}\right\rangle
%  &=
%  \sum_{\text{states }s\text{ of }D}{A^{\alpha(s\cup\bigcirc)}(-A^2-A^{-2})^{|\text{components of state }s|+1-1}}\\
%  &= \sum_{\text{states }s\text{ of }D}{A^{\alpha(s)}(-A^2-A^{-2})^{|\text{components of state }s|-1}(-A^2-A^{-2})},
\end{align*}

which is clearly $d\left\langle{D}\right\rangle$. Thus for an $m$-unlink (no crossings), we have,

\[\left\langle{\KPA^m}\right\rangle=d^{m-1}.\]

We now suppose that the Kauffman bracket is in fact unique for all link diagrams with less than $c(D)$ crossings (inductive hypothesis).
Thus, we want to show that for any arbitrary link diagram $D$, we have $\left\langle{D}\right\rangle=A\left\langle{D_A}\right\rangle+B\left\langle{D_B}\right\rangle$. And by the inductive hypothesis,

\begin{align*}
A\langle{D_A}\rangle+B\langle{D_B}\rangle &= A\sum_{s\in{D_A}}A^{|a(s)|}B^{|b(s)|}d^{|s|-1}+B\sum_{s\in{D_B}}A^{|b(S)|}B^{|b(s)|}d^{|s|-1}\\
&= \sum_{s\in{D_A}}A^{|a(s)|+1}B^{|b(s)|}d^{|s|-1}+\sum_{s\in{D_B}}A^{|a(s)|}B^{|b(s)|+1}d^{|s|-1}
\end{align*}

And since $\sum_{s\in{D_A}}A^{|a(s)|+1}B^{|b(s)|}d^{|s|-1}$ is the same as $\tangle{1}\mapsto\{\tangle{0}_A,\tangle{\infty}_B\}$ and, similarly, for the other sum we have it being $\tangle{-1}\mapsto\{\tangle{\infty}_A,\tangle{0}_B\}$. Adding the states of $\tangle{1}$ with the states of $\tangle{-1}$ gives us the states of $D$. Thus, we have

\[= \sum_{s\in{D}}A^{|a(s)|}B^{|b(s)|}d^{|s|-1} = \langle{D}\rangle\]

And we are done.

The Kauffman bracket is invariant under regular isotopy if we have $B=A^{-1}$ and $d=-A^{2}-A^{-2}$.

For $\RII$,

\begin{align*}
\left\langle\riimove\right\rangle
\end{align*}

We need $(A^2+B^2+ABd)\left\langle\rotatebox{85}{\!\!\!\!\!\resolveriimove}\right\rangle+AB\left\langle\resolveriimove\right\rangle=\left\langle\resolveriimove\right\rangle$, so $AB=1$ and $A^2+B^2+ABd=0$.
And we are done.
As for $\RIII$, we have

\begin{align*}
\left\langle\knotsinmath{\begin{scope}[scale=0.15]
\draw[draw=white,double=black ,ultra thick,double distance=0.5pt] (  0:2) .. controls (270:0.7) .. (180:2);
\draw[draw=white,double=black  ,ultra thick,double distance=0.5pt] ( 60:2) .. controls (150:0.7) .. (240:2);
\draw[draw=white,double=black,ultra thick,double distance=0.5pt] (120:2) .. controls ( 30:0.7) .. (300:2);
\end{scope}}\right\rangle
&= A\left\langle\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=0.3,yshift=-0.9cm]
\draw    (32,59.98) .. controls (49,23.98) and (95,23.98) .. (117,60.98) ;
\draw    (49,52.15) .. controls (63,59.15) and (75,60.15) .. (93,52.15) ;
\draw    (51,10.15) .. controls (63,22.15) and (80,23.15) .. (96,11.15) ;
\draw    (16,35.15) .. controls (30,42.15) and (16,35.15) .. (32,43) ;
\draw    (110,42.15) .. controls (120,34.15) and (114,39.15) .. (125,30.15) ;
\end{scope}}\right\rangle + B\left\langle\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=0.3,yshift=-0.9cm]
\draw    (96,10.65) .. controls (84,17.65) and (80.99,24.66) .. (89,32.66) .. controls (97,40.65) and (100,44.32) .. (118,64.65) ;
\draw    (49,52.15) .. controls (63,59.15) and (75,60.15) .. (93,52.15) ;
\draw    (56,9.65) .. controls (65,17.65) and (65,21.65) .. (55,32.65) .. controls (45,43.65) and (39,49.65) .. (25,64.65) ;
\draw    (16,35.15) .. controls (30,42.15) and (16,35.15) .. (32,43) ;
\draw    (110,42.15) .. controls (120,34.15) and (114,39.15) .. (125,30.15) ;
\end{scope}}\right\rangle
\end{align*}

By using the fact that the bracket is invariant under $\RII$, we find,

\begin{align*}
&= A\left\langle\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=0.3,yshift=-0.9cm]
\draw    (21.72,26.85) .. controls (42.43,49.7) and (97.9,36.72) .. (118.45,24.93) ;
\draw    (18.45,55.93) .. controls (30.45,51.93) and (54.45,47.93) .. (76.45,48.93) .. controls (98.45,49.93) and (129.45,52.93) .. (137.45,55.93) ;
\draw    (51,10.15) .. controls (63,22.15) and (80,23.15) .. (96,11.15) ;
\end{scope}}\right\rangle + B\left\langle\rotatebox{180}{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=0.3,yshift=-0.3cm]
\draw    (96,10.65) .. controls (84,17.65) and (80.99,24.66) .. (89,32.66) .. controls (97,40.65) and (100,44.32) .. (118,64.65) ;
\draw    (49,52.15) .. controls (63,59.15) and (75,60.15) .. (93,52.15) ;
\draw    (56,9.65) .. controls (65,17.65) and (65,21.65) .. (55,32.65) .. controls (45,43.65) and (39,49.65) .. (25,64.65) ;
\draw    (16,35.15) .. controls (30,42.15) and (16,35.15) .. (32,43) ;
\draw    (110,42.15) .. controls (120,34.15) and (114,39.15) .. (125,30.15) ;
\end{scope}}}\right\rangle\\
&= A\left\langle\rotatebox{180}{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=0.3,yshift=-0.3cm]
\draw    (32,59.98) .. controls (49,23.98) and (95,23.98) .. (117,60.98) ;
\draw    (49,52.15) .. controls (63,59.15) and (75,60.15) .. (93,52.15) ;
\draw    (51,10.15) .. controls (63,22.15) and (80,23.15) .. (96,11.15) ;
\draw    (16,35.15) .. controls (30,42.15) and (16,35.15) .. (32,43) ;
\draw    (110,42.15) .. controls (120,34.15) and (114,39.15) .. (125,30.15) ;
\end{scope}}}\right\rangle + B\left\langle\rotatebox{180}{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=0.3,yshift=-0.3cm]
\draw    (96,10.65) .. controls (84,17.65) and (80.99,24.66) .. (89,32.66) .. controls (97,40.65) and (100,44.32) .. (118,64.65) ;
\draw    (49,52.15) .. controls (63,59.15) and (75,60.15) .. (93,52.15) ;
\draw    (56,9.65) .. controls (65,17.65) and (65,21.65) .. (55,32.65) .. controls (45,43.65) and (39,49.65) .. (25,64.65) ;
\draw    (16,35.15) .. controls (30,42.15) and (16,35.15) .. (32,43) ;
\draw    (110,42.15) .. controls (120,34.15) and (114,39.15) .. (125,30.15) ;
\end{scope}}}\right\rangle\\
&= \left\langle\knotsinmath{\begin{scope}[scale=0.15]
\draw[draw=white,double=black ,ultra thick,double distance=0.5pt] (  0:2) .. controls ( 90:0.7) .. (180:2);
\draw[draw=white,double=black  ,ultra thick,double distance=0.5pt] ( 60:2) .. controls (330:0.7) .. (240:2);
\draw[draw=white,double=black,ultra thick,double distance=0.5pt] (120:2) .. controls (210:0.7) .. (300:2);
\end{scope}}\right\rangle
\end{align*}

Thus, the bracket is invariant under regular isotopy.

And by the uniqueness of the Kauffman bracket, we then have the following state sum, where $B$ and $d$ are replaced with $A^{-1}$ and $(-A^{2}-A^{-2})$, respectively, we have

\[\langle{D}\rangle=\sum_{\text{states }s\text{ of }D}A^{\alpha(s)}(-A^2-A^{-2})^{|\text{components of state }s|-1}\]

where $\alpha(s)=|a(s)|+|b(s)|$.

Consider the left-handed trefoil (the trefoil with the negative writhe). We will calculate its bracket polynomial using its inductive and state-sum definition. We begin with the inductive definition.

\begin{align*}
\left\langle\trefoil\right\rangle
\right)\\
A\langle\bigcirc\bigcirc\rangle + A^{-1}\langle\bigcirc\rangle\\
A^{-2}\left(A\langle\bigcirc\rangle+A^{-1}\langle\bigcirc\bigcirc\rangle\right) + A\langle\bigcirc\bigcirc\rangle + A^{-1}\langle\bigcirc\rangle\\
(-A^4-1)(A\langle\bigcirc\bigcirc\rangle+A^{-1}\langle\bigcirc\rangle) + 2A\langle\bigcirc\bigcirc\rangle + 3A^{-1}\\
(-A^5+A^{-3}+A)(-A^2-A^{-2})\langle\bigcirc\rangle - A^3 + 2A^{-1}\\
&= -A^{-5} - A^3 + A^7
\end{align*}

We now use the state-sum definition.

2cstate   1c$\alpha(s)$   1c$|\text{components of state}|$

(84.5,51) .. controls (84.5,39.4) and (93.9,30) .. (105.5,30) .. controls (117.1,30) and (126.5,39.4) .. (126.5,51) .. controls (126.5,62.6) and (117.1,72) .. (105.5,72) .. controls (93.9,72) and (84.5,62.6) .. (84.5,51) – cycle ;
(50,97) .. controls (50,84.3) and (60.3,74) .. (73,74) .. controls (85.7,74) and (96,84.3) .. (96,97) .. controls (96,109.7) and (85.7,120) .. (73,120) .. controls (60.3,120) and (50,109.7) .. (50,97) – cycle ;
(30.5,47.75) .. controls (30.5,35.19) and (40.69,25) .. (53.25,25) .. controls (65.81,25) and (76,35.19) .. (76,47.75) .. controls (76,60.31) and (65.81,70.5) .. (53.25,70.5) .. controls (40.69,70.5) and (30.5,60.31) .. (30.5,47.75) – cycle ;

$3$   $3$

(36.5,53) .. controls (28.5,70) and (53.5,95) .. (24.5,92) .. controls (-4.5,89) and (20.5,68) .. (18.5,48) .. controls (16.5,28) and (9.17,14.7) .. (29.83,11.85) .. controls (50.5,9) and (44.5,37) .. (36.5,53) – cycle ;
(48.5,57) .. controls (48.5,45.4) and (57.9,36) .. (69.5,36) .. controls (81.1,36) and (90.5,45.4) .. (90.5,57) .. controls (90.5,68.6) and (81.1,78) .. (69.5,78) .. controls (57.9,78) and (48.5,68.6) .. (48.5,57) – cycle ;

$1$   $2$

(38.5,27) .. controls (38.5,15.4) and (47.9,6) .. (59.5,6) .. controls (71.1,6) and (80.5,15.4) .. (80.5,27) .. controls (80.5,38.6) and (71.1,48) .. (59.5,48) .. controls (47.9,48) and (38.5,38.6) .. (38.5,27) – cycle ;
(18,74) .. controls (18,61.3) and (28.3,51) .. (41,51) .. controls (53.7,51) and (64,61.3) .. (64,74) .. controls (64,86.7) and (53.7,97) .. (41,97) .. controls (28.3,97) and (18,86.7) .. (18,74) – cycle ;

$1$   $2$

(41.5,55) .. controls (33.5,72) and (58.5,97) .. (29.5,94) .. controls (0.5,91) and (25.5,70) .. (23.5,50) .. controls (21.5,30) and (14.17,16.7) .. (34.83,13.85) .. controls (55.5,11) and (49.5,39) .. (41.5,55) – cycle ;

$-1$   $1$

(7,85) .. controls (7,72.3) and (17.3,62) .. (30,62) .. controls (42.7,62) and (53,72.3) .. (53,85) .. controls (53,97.7) and (42.7,108) .. (30,108) .. controls (17.3,108) and (7,97.7) .. (7,85) – cycle ;
(16.5,30.75) .. controls (16.5,18.19) and (26.69,8) .. (39.25,8) .. controls (51.81,8) and (62,18.19) .. (62,30.75) .. controls (62,43.31) and (51.81,53.5) .. (39.25,53.5) .. controls (26.69,53.5) and (16.5,43.31) .. (16.5,30.75) – cycle ;

$1$   $2$

(38.5,57) .. controls (30.5,74) and (55.5,99) .. (26.5,96) .. controls (-2.5,93) and (22.5,72) .. (20.5,52) .. controls (18.5,32) and (11.17,18.7) .. (31.83,15.85) .. controls (52.5,13) and (46.5,41) .. (38.5,57) – cycle ;

$-1$   $1$

(45.68,54.79) .. controls (22.64,62.01) and (39.2,101.52) .. (62.85,102) .. controls (86.5,102.48) and (94.5,46.48) .. (75.5,23.95) ;
(45.68,54.79) .. controls (81.82,41.3) and (38.91,36) .. (38,45.63) .. controls (37.1,55.27) and (24.17,55.37) .. (15.42,29.25) .. controls (6.67,3.14) and (51.5,0.48) .. (75.5,23.95) ;

$-1$   $1$

(52.5,54.5) .. controls (52.5,41.8) and (62.8,31.5) .. (75.5,31.5) .. controls (88.2,31.5) and (98.5,41.8) .. (98.5,54.5) .. controls (98.5,67.2) and (88.2,77.5) .. (75.5,77.5) .. controls (62.8,77.5) and (52.5,67.2) .. (52.5,54.5) – cycle ;
(28,54.5) .. controls (28,28.27) and (49.27,7) .. (75.5,7) .. controls (101.73,7) and (123,28.27) .. (123,54.5) .. controls (123,80.73) and (101.73,102) .. (75.5,102) .. controls (49.27,102) and (28,80.73) .. (28,54.5) – cycle ;

$-3$   $2$

Thus, the polynomial becomes

\begin{align*}
\left\langle{D}\right\rangle &= \sum_{\text{states }s\text{ of }D}A^{\alpha(s)}(-A^2-A^{-2})^{|\text{components of state }s|-1}\\
&= A^3d^2 + Ad + Ad + A^{-1} + Ad + A^{-1} + A^{-1} + A^{-3}d\\
&= d(A^3d+3A+A^{-3}) + 3A^{-1}\\
&= (-A^2-A^{-2})(-A^5-A+3A+A^{-3}) + 3A^{-1}\\
&= A^7+A^3-2A^3-2A^{-1}-A^{-1}-A^{-5}+3A^{-1} = A^7-A^3-A^{-5}
\end{align*}

Let us see what happens to the Kauffman bracket when we apply the Reidemeister moves.

The Kauffman bracket is not invariant under $\RI$.

\begin{align*}
\left\langle\rimove\right\rangle
\end{align*}

\begin{align*}
\left\langle\reflectbox{\rimove}\!\right\rangle
\end{align*}

So, the Kauffman bracket is not invariant under $\RI$.

We can use the facts we proved in Lemma <ref> for a faster calculation. For instance, let us see the trefoil again.

\[
\left\langle\trefoil\right\rangle
\]

we now apply two $\RI$ moves and multiply $(-A^3)^2=A^6$. So, we would get

\[
= A^7+A^{-1}\left\langle\hopflink\!\!\right\rangle
A^7 + A^{-1}\left(A\left\langle\reflectbox{\negativetwist}\!\!\right\rangle
\]

we then apply an $\RI$ move again to both and find,

\[
= A^7+A^{-1}(A(-A^3)+A^{-1}(-A^{-3}))=A^7-A^3-A^{-5}
\]

Hence we have simplified the calculation from Example <ref> by threefold.

§.§ Kauffman's bracket polynomial definition of the Jones polynomial

We have already seen from Lemma <ref> that the bracket is not invariant under $\RI$. We now discuss the method of `normalising' the bracket with respect to its writhe in order to make it invariant under ambient isotopy. Here is a precise definition.

The Jones polynomialJones polynomial!via Kauffman bracket is the function

\[V(L)\colon\{\text{oriented links L with corresponding diagrams D}\}\to\ZZ[t^{-1/2},t^{1/2}],\]

given by $V(L)=((-A)^{-3\omega(D)}\langle{|D|}\rangle)_{t^{1/2}=A^{-2}}$, where $|D|$ is the link diagram $D$ without orientation.

The Jones polynomial is a link invariant.

We check what happens to the Jones polynomial under the three Reidemeister moves (ambient isotopy).
We already know that the bracket polynomial is invariant under regular isotopy from Proposition <ref>, and we also know from Proposition <ref> that the writhe is invariant under regular isotopy.
Thus, we only need to show that the Jones polynomial is invariant under $\RI$.

Now remember from Lemma <ref>, if we apply an $\RI$ move to the diagram, then we need to multiply by a factor of $-A^{\pm3}$ depending on its writhe.

We now consider the positive twist, i.e.,

\[ \omega\left(\reflectbox{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.35cm,scale=0.3]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}}\!\right)=+1 \qquad\text{ and }\qquad \left\langle\reflectbox{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.35cm,scale=0.3]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}}\right\rangle=-A^3\left\langle\alonecurve\right\rangle \]

\begin{align*}
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}}\!\right) &= (-A)^{-3\omega\left(\reflectbox{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.35cm,scale=0.3]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}}\!\right)}\left\langle\reflectbox{\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.35cm,scale=0.3]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}}\!\right\rangle\\
&= (-A)^{-3\left(\omega\left(\alonecurve\right)+1\right)}\left(-A^{3}\right)\left\langle\alonecurve\right\rangle\\
&= (-A)^{-3\omega\left(\alonecurve\right)}\left\langle\alonecurve\right\rangle = V\left(\alonecurve\right)
\end{align*}

And similarly,

\begin{align*}
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}\right) &= (-A)^{-3\omega\left(\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.35cm,scale=0.3]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw [shift={(73,54.23)}, rotate = 203.75] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}\right)}\left\langle\knotsinmath{\begin{scope}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,yshift=-0.35cm,scale=0.3]
\draw    (36,52.23) .. controls (29,84.23) and (4,67.23) .. (15,47.23) .. controls (25.78,27.63) and (47.12,42.61) .. (71.5,53.57) ;
\draw    (47,11.23) -- (42,30.23) ;
\end{scope}}\right\rangle\\
&= (-A)^{-3\omega\left(\alonecurve\right)+3}(-A^{-3})\left\langle\alonecurve\right\rangle\\
&= (-A)^{-3\omega\left(\alonecurve\right)}\left\langle\alonecurve\right\rangle = V\left(\alonecurve\right)
\end{align*}

Thus, it is also invariant under $\RI$, making it a link invariant under ambient isotopy.

Let us calculate the Jones polynomial for the trefoil from our examples above,

\[V(\overline{3_1})=-A^{-3(-3)}(-A^{-5}-A^3+A^7)=A^4+A^{12}-A^{16}=-t^{-4}+t^{-3}+t^{-1}\]

There exists two distinct knots with crossing number three.

This also follows an argument analogous to the one in Proposition <ref>. We begin by drawing all possible link universes with three crossings such that

* there are no twists (since an $\RI$ can undo it thus leaving it with a lower crossing number)

* if there is a loop, it has to be connected to the whole of the knot, since we are trying to draw a link universe of one component.

So, we have the following link universe.

[rotate=  0] (a) at (0,-1) ;
[rotate=120] (b) at (0,-1) ;
[rotate=240] (c) at (0,-1) ;
(a.center) .. controls (a.4 north west) and (c.4 north east) .. (c.center);
(b.center) .. controls (b.4 north west) and (a.4 north east) .. (a.center);
(c.center) .. controls (c.4 north west) and (b.4 north east) .. (b.center);
(a.center) .. controls (a.16 south west) and (c.16 south east) .. (c.center);
(b.center) .. controls (b.16 south west) and (a.16 south east) .. (a.center);
(c.center) .. controls (c.16 south west) and (b.16 south east) .. (b.center);

By tabulating all $2^3$ knot diagrams, we find that there are only in fact two that are not the unknot.

(0,-2.5) node[scale=2,xscale=-1] $3_1$;

(0,-2.5) node[scale=2] $\overline{3_1}$;

So, we only need to show that $V(3_1)\neq{V(\overline{3_1})}$.

We have already calculated $V(\overline{3_1})$ form Example <ref>. And for the sake of not repeating onesself, we will soon see from Example <ref> that we have $V(3_1)=A^{-4}+A^{-12}-A^{-16}$. Thus, $V(3_1)\neq{V(\overline{3_1})}$, and so $3_1\not\sim\overline{3_1}$.

So, now we know that the trefoil and its reflection are not equivalent knots. And so we have the following result.

The Jones polynomial of a mirror link $\overline{L}$ of link $L$ has the polynomial $V(\overline{L})=(V(L))_{t=t^{-1}}$.

We know that the writhe of the reflection of the link $L$ is the negation of the writhe of $L$.
As for the Kauffman bracket, it is not hard to see that $A=B$ when looking at the reflection, since, $\left\langle\KPB\right\rangle=A\left\langle\KPD\right\rangle+B\left\langle\KPC\right\rangle$, whereas $\left\langle\KPE\right\rangle=A\left\langle\KPC\right\rangle+B\left\langle\KPD\right\rangle$.
Thus, $\langle\overline{D}\rangle=(\langle{D}\rangle)_{A=A^{-1}}$.
And so, $V(\overline{D})=(-A^{3\omega(D)}(\langle{D}\rangle)_{A=A^{-1}}=V(D)_{A=A^{-1}}=V(D)_{t=t^{-1}}$

§.§ The skein relation definition of the Jones polynomial

In this section, we will begin by exploring the Jones polynomial defined by its skein relation and we will also see the Kauffman bracket polynomial and how its normalisation (with respect to the writhe) is equivalent to the Jones polynomial.

The skein relation was shown by John Conway in the 1960's [15]. He defined the Alexander polynomial via the skein relation (now sometimes known as the Alexander-Conway polynomial). So what is a skein relation?

The skein relationskein relation is defined as the relation between three link diagrams which we call $L_+$, $L_-$ and $L_0$. These diagrams are the same, except in the neighbourhood of a single crossing. See the below figure.

[dashed] (0,0) circle(1.5);
[->] (1,-1)–(-1,1);
[ultra thick, draw=white,double=black,double distance=0.5pt] (-1,-1)–(1,1);
[->] (0.9,0.9)–(1,1);
(0,-2) node[scale=1.5] $L_+$;

[dashed] (0,0) circle(1.5);
[->] (-1,-1)–(1,1);
[ultra thick, draw=white,double=black,double distance=0.5pt] (1,-1)–(-1,1);
[->] (-0.9,0.9)–(-1,1);
(0,-2) node[scale=1.5] $L_-$;

[dashed] (0,0) circle(1.5);
[->] (-1,-1) to [bend right=40] (-1,1);
[->] (1,-1) to [bend left=40] (1,1);
(0,-2) node[scale=1.5] $L_0$;

skein relation

In other words, they are the three elementary $2$-tangles $\tangle{1}$, $\tangle{-1}$ and $\tangle{0}$.

To define an invariant function that admits a skein relation, we have to notice that the skein relation is related to the unknotting number, in that when we have a skein relation, the complexity of the other two link diagrams are strictly less than that of the original link diagram.

We now define an invariant function $P$ that admits a skein relation, and is unique. We will not go too deep into this theory. The interested reader is directed to [20] and [41].

It is said that a link invariant $P\colon\{\text{Oriented links in }\RR^3\}\to{A}$, where $A$ is a commutative ring, admits a skein relation invariantskein relation invariant if

* $P(0_1)=P(\KPA)=1$ and

* there exists three invertible elements $a_+$, $a_-$ and $a_0$ in $A$, such that whenever we have three links $L_+$, $L_-$ and $L_0$ that are skein related, then $a_+P(L_+)+a_-P(L_-)+a_0P(L_0)=0$.

The following theorem says that if the skein relation invariant polynomial is well defined (i.e., it is a link invariant), then it is uniquely determined by $a_+$, $a_-$ and $a_0$.

If the skein relation invariant polynomial $P$ defined above is in fact an invariant, then it is uniquely determined by $a_+$, $a_-$ and $a_0$.

The proof of the uniqueness of the skein relation invariant is closely analogous to the uniqueness of the Jones polynomial via skein theory, Proposition <ref>, and so is left as an exercise for the reader. The existence proof of this polynomial can be found in [41].

The skein relation invariant is homogeneous in $a_+$, $a_-$, and $a_0$, i.e., all of the terms have a total degree of zero of the variables.

The Jones polynomial invariant $V(L)$ admits a skein relation such that it is a function $V\colon\{\text{Oriented links in }\RR^3\}\to\ZZ[t^{-1/2},t^{1/2}]$ such that $V(0_1)=V(\KPA)=1$, and whenever three oriented links $L_+,L_-,L_0$ are skein related, then

\[t^{-1}V(L_+)-tV(L_-)+(t^{-1/2}-t^{1/2})V(L_0)=0.\]

And it is uniquely determined by those two conditions.

We first show that the Jones polynomial admits a skein relation, then we will show that it can be uniquely determined by the skein relation conditions.
The first property $V(\KPA)=1$ is trivial.
For the second property we need to remove $D_\infty$ terms found in the Kauffman bracket. So consider the diagrams below.

[dashed] (0,0) circle(1.5);
[-] (-1,-1) to [bend left=40] (1,-1);
[-] (1,1) to [bend left=40] (-1,1);

(0,-2) node[scale=1.5] $D_\infty$;
[dashed] (0,0) circle(1.5);
[-] (-1,-1)–(1,1);
[ultra thick, draw=white,double=black,double distance=0.5pt] (1,-1)–(-1,1);
[-] (-0.9,0.9)–(-1,1);
(0,-2) node[scale=1.5] $D_-$;

[dashed] (0,0) circle(1.5);
[-] (1,-1)–(-1,1);
[ultra thick,draw=white,double=black,double distance=0.5pt] (-1,-1)–(1,1);
[-] (-0.9,-0.9)–(1,1);
(0,-2) node[scale=1.5] $D_+$;

[dashed] (0,0) circle(1.5);
[-] (-1,-1) to [bend right=40] (-1,1);
[-] (1,-1) to [bend left=40] (1,1);
(0,-2) node[scale=1.5] $D_0$;

So we have $\left\langle{D_-}\right\rangle=\left\langle{\KPB}\right\rangle=A\left\langle{D_\infty}\right\rangle+A^{-1}\left\langle{D_0}\right\rangle$ and $\left\langle{D_+}\right\rangle=\left\langle\KPE\right\rangle=A\left\langle{D_0}\right\rangle+A^{-1}\left\langle{D_\infty}\right\rangle$. And so noting the substitution $t^{1/2}=A^{-2}$, we have

\[
\]

Now $\omega(D_0)+1=\omega(D_+)$ and $\omega(D_-)=\omega(D_0)-1$. So the above formula becomes

\[
\]

Now we substitute the relations we wrote above $\left\langle{D_-}\right\rangle$ and $\left\langle{D_+}\right\rangle$, so

\[
\]

So we are done.

We now show the uniqueness by induction on the number of crossings $n$ on an arbitrary link $L$.

We note that we already know $V(\KPA)=1$. And by considering any arbitrary link $L$, we will now see that for an $m$ component unlink $V(\KPA^m)=(t^{1/2}+t^{-1/2})^{m-1}$.

(16.17,17) – (52.98,17) – (52.98,40.93) – (16.17,40.93) – cycle ;
(53.47,22.28) .. controls (64.33,22.38) and (64.33,36.38) .. (53.47,36.28) ;
(71.52,28.48) .. controls (71.52,22.96) and (75.99,18.48) .. (81.52,18.48) .. controls (87.04,18.48) and (91.52,22.96) .. (91.52,28.48) .. controls (91.52,34.01) and (87.04,38.48) .. (81.52,38.48) .. controls (75.99,38.48) and (71.52,34.01) .. (71.52,28.48) – cycle ;
(123.17,17) – (159.98,17) – (159.98,40.93) – (123.17,40.93) – cycle ;
(177.02,30.47) .. controls (194.02,44.47) and (194.53,9.8) .. (182.53,20.8) .. controls (170.53,31.8) and (170.53,36.8) .. (160.47,36.28) ;
(172.53,26.8) .. controls (169.02,24.47) and (166.53,23) .. (159.53,23) ;
(223.17,17) – (259.98,17) – (259.98,40.93) – (223.17,40.93) – cycle ;
(278.02,27.78) .. controls (292.23,13.95) and (301.23,38.95) .. (288.23,35.95) .. controls (275.23,32.95) and (271,21.91) .. (260.23,22.95) ;
(273.53,31.44) .. controls (270.02,33.76) and (267.53,35.23) .. (260.53,35.23) ;
(164,32.83) .. controls (167.65,32.82) and (170.61,32.09) .. (172.88,30.63) .. controls (171.29,32.82) and (170.4,35.73) .. (170.18,39.37) ;
(262.45,32.13) .. controls (265.92,33.23) and (268.96,33.44) .. (271.57,32.74) .. controls (269.39,34.34) and (267.65,36.84) .. (266.34,40.24) ;
(59.59,32.06) .. controls (60.44,29.71) and (60.73,27.51) .. (60.45,25.48) .. controls (61.29,27.35) and (62.69,29.06) .. (64.66,30.62) ;
(67.01,29) .. controls (69.47,27.83) and (71.19,26.41) .. (72.17,24.74) .. controls (71.93,26.66) and (72.45,28.83) .. (73.71,31.25) ;
(29.17,19) node [anchor=north west][inner sep=0.75pt][yscale=-1]   [align=left] $\displaystyle L$;
(136.17,19) node [anchor=north west][inner sep=0.75pt]   [align=left][yscale=-1] $\displaystyle L$;
(236.17,19) node [anchor=north west][inner sep=0.75pt]   [align=left][yscale=-1] $\displaystyle L$;

$L_0=L\cup\KPA$        $L_+\sim{L}$        $L_-\sim{L}$

By the skein relation definition, we have $t^{-1}V(L)-tV(L)+(t^{-1/2}-t^{1/2})V(L\cup\KPA)=0$, and thus

\[V\left(L\cup\KPA\right)=\frac{t-t^{-1}}{t^{-1/2}-t^{1/2}}V(L)=-(t^{1/2}+t^{-1/2})V(L).\]

So by induction on $m$, we find $V(L\cup\KPA^m)=(-t^{1/2}-t^{-1/2})^mV(L)$.
Now we know that the link $L_0$ will have one less crossing than $L_+$ and $L_-$. So, without loss of generality, if our oriented link $L$ is $L_+$, then we have complexity $(c(L_+),u(L_+))$.
Since $c(L_0)<c(L_+)$, we have $\kappa(L_0)<\kappa(L_+)$. Now $c(L_-)=c(L_+)$, but $u(L_-)=c(L_+)-1$. So we have the complexities of the two links $L_-$ and $L_0$, strictly less than the complexity of $L_+$.
Now, by induction on the complexity, we can see that the Jones polynomial is in fact unique.

It can be seen that by substituting $a_+$ for $t^{-1}$, $a_-$ for $-t$, and $a_0$ for $t^{-1/2}-t^{1/2}$, the skein relation invariant $P$ becomes the Jones polynomial $V$.

We have shown in the above proof what the Jones polynomial of the disjoint union of a link and the $m$-unlink is, so we shall write it as the following corollary.

The disjoint union of a link $L$ and the $m$-unlink is given by $V\left(L\cup\KPA^m\right)=(-t^{1/2}-t^{-1/2})^mV(L)$.

We then have the following theorem, which we will prove at the end of the essay using nicer results with braid representations.

The two disjoint links $L_1$ and $L_2$ has the following Jones polynomial,

\[V(L_1\cup{L_2})=(-t^{1/2}-t^{-1/2})V(L_1)V(L_2)\]

Let us calculate the Jones polynomial of the figure eight knot $4_1$ drawn below via skein theory. We let the knot $4_1$ be the knot $L_+$ where we have circled our chosen neighbourhood.

(41,121) .. controls (5,123) and (2,23) .. (52,10) .. controls (102,-3) and (94,37) .. (75,60) ;
(92,113) .. controls (113,82) and (22,75) .. (76,25) ;
(97,15) .. controls (128,-1) and (174,122) .. (62,122) ;
(58,76) .. controls (19,132) and (87,145) .. (91,127) ;
(8.04,76.36) .. controls (11.54,72.43) and (13.82,68.37) .. (14.86,64.2) .. controls (15.05,68.5) and (16.46,72.93) .. (19.09,77.49) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (67,22.5) .. controls (67,11.73) and (75.73,3) .. (86.5,3) .. controls (97.27,3) and (106,11.73) .. (106,22.5) .. controls (106,33.27) and (97.27,42) .. (86.5,42) .. controls (75.73,42) and (67,33.27) .. (67,22.5) – cycle ;

(42,122) .. controls (-15,83) and (53,-29) .. (76,14) ;
(92,113) .. controls (113,82) and (7,53) .. (97,15) ;
(97,15) .. controls (128,-1) and (174,122) .. (62,122) ;
(58,76) .. controls (19,132) and (87,145) .. (91,127) ;
(13.95,75.26) .. controls (17.51,71.39) and (19.85,67.37) .. (20.96,63.21) .. controls (21.08,67.51) and (22.42,71.97) .. (24.98,76.56) ;
(77,56) .. controls (87,44) and (93,43) .. (87,31) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (59,21.5) .. controls (59,10.73) and (67.73,2) .. (78.5,2) .. controls (89.27,2) and (98,10.73) .. (98,21.5) .. controls (98,32.27) and (89.27,41) .. (78.5,41) .. controls (67.73,41) and (59,32.27) .. (59,21.5) – cycle ;

(41,121) .. controls (5,123) and (27,12.4) .. (52,10) .. controls (77,7.6) and (68,22.6) .. (104,8.6) ;
(91,113) .. controls (110.13,84.76) and (44.94,68.16) .. (58.97,40.88) .. controls (73,13.6) and (108,39.6) .. (79,64.6) ;
(104,8.6) .. controls (137,-11.4) and (174,122) .. (62,122) ;
(63,82.6) .. controls (20,125.6) and (87,146) .. (91,128) ;
(17.74,70.02) .. controls (21.45,66.29) and (23.93,62.35) .. (25.2,58.24) .. controls (25.16,62.55) and (26.33,67.05) .. (28.71,71.74) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (61,22.5) .. controls (61,11.73) and (69.73,3) .. (80.5,3) .. controls (91.27,3) and (100,11.73) .. (100,22.5) .. controls (100,33.27) and (91.27,42) .. (80.5,42) .. controls (69.73,42) and (61,33.27) .. (61,22.5) – cycle ;
(64.85,70.95) .. controls (64.86,67.63) and (64.17,64.8) .. (62.8,62.48) .. controls (64.87,64.29) and (67.62,65.6) .. (71.07,66.4) ;

$L_+$                        $L_-$                    $L_0$

Note that the knot $L_-$ is Reidemeiser equivalent to the unknot $0_1$, so $V(L_-)=1$. Whereas $L_0$ is Reidemeister equivalent to the Hopf link with the given orientations.
So, by the definition of the Jones polynomial, we have

\begin{align*}
t^{-1}V(4_1)-t+(t^{-1/2}-t^{1/2})V(L_0)&=0 \tag{*}\label{equation:Jonespolynomialof4_1}
\end{align*}

Now, we need to find what $V(L_0)$ is. So, we let $L_0$ be the Hopf link $H_-$ (since they are Reidemeister equivalent).

(41,121) .. controls (5,123) and (27,12.4) .. (52,10) .. controls (77,7.6) and (68,22.6) .. (104,8.6) ;
(104,8.6) .. controls (137,-11.4) and (174,122) .. (62,122) ;
(17.74,70.02) .. controls (21.45,66.29) and (23.93,62.35) .. (25.2,58.24) .. controls (25.16,62.55) and (26.33,67.05) .. (28.71,71.74) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (29,115.95) .. controls (29,103.16) and (39.36,92.8) .. (52.15,92.8) .. controls (64.94,92.8) and (75.3,103.16) .. (75.3,115.95) .. controls (75.3,128.74) and (64.94,139.1) .. (52.15,139.1) .. controls (39.36,139.1) and (29,128.74) .. (29,115.95) – cycle ;
(60.95,99.62) .. controls (57.97,101.67) and (55.87,104.02) .. (54.65,106.68) .. controls (54.99,103.71) and (54.46,100.45) .. (53.05,96.88) ;
(76.5,130) .. controls (72.5,151) and (46.78,147.2) .. (53.64,111.1) .. controls (60.5,75) and (90.5,73) .. (79.5,112) ;
(198,65.25) .. controls (198,37.5) and (220.5,15) .. (248.25,15) .. controls (276,15) and (298.5,37.5) .. (298.5,65.25) .. controls (298.5,93) and (276,115.5) .. (248.25,115.5) .. controls (220.5,115.5) and (198,93) .. (198,65.25) – cycle ;
[draw opacity=0] (290.52,19.35) .. controls (295.91,17.34) and (301.73,16.31) .. (307.79,16.47) .. controls (334.3,17.16) and (355.2,40.35) .. (354.47,68.25) .. controls (353.74,96.16) and (331.66,118.22) .. (305.14,117.52) .. controls (297.87,117.33) and (291.01,115.45) .. (284.91,112.24) – (306.46,67) – cycle ;
(290.52,19.35) .. controls (295.91,17.34) and (301.73,16.31) .. (307.79,16.47) .. controls (334.3,17.16) and (355.2,40.35) .. (354.47,68.25) .. controls (353.74,96.16) and (331.66,118.22) .. (305.14,117.52) .. controls (297.87,117.33) and (291.01,115.45) .. (284.91,112.24) ;
(260.5,92) .. controls (233.5,77) and (229.5,35) .. (264.5,29) ;
(192.31,70.71) .. controls (195.56,67.08) and (197.58,63.4) .. (198.35,59.67) .. controls (198.82,63.46) and (200.52,67.29) .. (203.45,71.18) ;
(243.72,52.13) .. controls (241.21,55.28) and (239.79,58.34) .. (239.45,61.34) .. controls (238.7,58.42) and (236.86,55.58) .. (233.93,52.82) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (249,99.85) .. controls (249,87.06) and (259.36,76.7) .. (272.15,76.7) .. controls (284.94,76.7) and (295.3,87.06) .. (295.3,99.85) .. controls (295.3,112.64) and (284.94,123) .. (272.15,123) .. controls (259.36,123) and (249,112.64) .. (249,99.85) – cycle ;
(458.5,75.8) .. controls (432.5,92.8) and (416.5,81.8) .. (418.26,72.06) .. controls (420.01,62.32) and (446.5,33.8) .. (462.35,36.45) .. controls (478.19,39.11) and (464.33,89.99) .. (485.47,92.12) .. controls (506.6,94.25) and (540.5,26.8) .. (486.5,57.8) ;
(417.89,59.79) .. controls (422.38,59.34) and (425.98,58.07) .. (428.69,55.99) .. controls (426.88,58.88) and (425.96,62.59) .. (425.95,67.11) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (449.2,77.25) .. controls (449.2,67.01) and (457.51,58.7) .. (467.75,58.7) .. controls (477.99,58.7) and (486.3,67.01) .. (486.3,77.25) .. controls (486.3,87.49) and (477.99,95.8) .. (467.75,95.8) .. controls (457.51,95.8) and (449.2,87.49) .. (449.2,77.25) – cycle ;

$H_-$                    $H_+$                    $H_0$

We note that the knot $L_0\sim_R0_1$. So, from the definition of the Jones polynomial we get,

\begin{equation}
t^{-1}V(H_+)-tV(H_-)+(t^{-1/2}-t^{1/2})V(H_0)=0 \tag{**}\label{equation:Hopflinkskein}
\end{equation}

and so we need to calculate the Jones polynomial of the unlink: $V(L_+)$. We let $L_+$ be $M_0$ since they are R-equivalent.

(17.06,75.55) .. controls (17.06,52.99) and (34.56,34.7) .. (56.14,34.7) .. controls (77.72,34.7) and (95.22,52.99) .. (95.22,75.55) .. controls (95.22,98.11) and (77.72,116.4) .. (56.14,116.4) .. controls (34.56,116.4) and (17.06,98.11) .. (17.06,75.55) – cycle ;
(107.04,74.65) .. controls (107.04,52.09) and (124.54,33.8) .. (146.12,33.8) .. controls (167.7,33.8) and (185.2,52.09) .. (185.2,74.65) .. controls (185.2,97.21) and (167.7,115.5) .. (146.12,115.5) .. controls (124.54,115.5) and (107.04,97.21) .. (107.04,74.65) – cycle ;
(12.5,78.36) .. controls (15.03,75.65) and (16.64,72.86) .. (17.34,70) .. controls (17.56,72.94) and (18.71,75.98) .. (20.77,79.09) ;
(112.22,65.77) .. controls (109.53,68.12) and (107.77,70.59) .. (106.96,73.19) .. controls (106.83,70.46) and (105.76,67.59) .. (103.74,64.59) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (79.62,72.37) .. controls (79.62,60.91) and (88.5,51.63) .. (99.46,51.63) .. controls (110.42,51.63) and (119.3,60.91) .. (119.3,72.37) .. controls (119.3,83.82) and (110.42,93.11) .. (99.46,93.11) .. controls (88.5,93.11) and (79.62,83.82) .. (79.62,72.37) – cycle ;
(426.5,81.8) .. controls (413.5,96.8) and (366.5,96.8) .. (391.5,57.8) .. controls (416.5,18.8) and (421.51,75.14) .. (474.5,92.97) .. controls (527.5,110.8) and (498.5,10.8) .. (448.5,58.8) ;
(435.4,58.27) .. controls (436.13,64.73) and (437.91,69.88) .. (440.76,73.72) .. controls (436.85,71.19) and (431.88,69.96) .. (425.84,70.04) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (417.36,83.07) .. controls (410.11,67.75) and (416.14,49.69) .. (430.85,42.72) .. controls (445.55,35.76) and (463.35,42.52) .. (470.61,57.84) .. controls (477.87,73.16) and (471.83,91.22) .. (457.12,98.19) .. controls (442.42,105.15) and (424.62,98.38) .. (417.36,83.07) – cycle ;
(271.5,59.11) .. controls (258.5,44.11) and (211.5,44.11) .. (236.5,83.11) .. controls (261.5,122.11) and (266.51,65.77) .. (319.5,47.94) .. controls (372.5,30.11) and (343.5,130.11) .. (293.5,82.11) ;
(282.87,60.3) .. controls (282.21,66.77) and (280.48,71.93) .. (277.67,75.8) .. controls (281.55,73.23) and (286.51,71.96) .. (292.55,71.97) ;
[dash pattern=on 1.69pt off 2.76pt][line width=1.5]  (262.36,58.84) .. controls (255.11,74.16) and (261.14,92.22) .. (275.85,99.19) .. controls (290.55,106.15) and (308.35,99.38) .. (315.61,84.07) .. controls (322.87,68.75) and (316.83,50.69) .. (302.12,43.72) .. controls (287.42,36.76) and (269.62,43.52) .. (262.36,58.84) – cycle ;

$M_0$                $M_+$                $M_-$

Note that the knots $M_+$ and $M_-$ are R-equivalent to $0_1$, so we have $V(M_+)=V(M_-)=1$. So then using the skein theory definition of the Jones polynomial, we have $t^{-1}-t+(t^{-1/2}-t^{1/2})V(M_0)=0$. And so

\begin{align*}
V(M_0) &= \frac{t-t^{-1}}{t^{-1/2}-t^{1/2}}\\
&= \frac{(t^{-1/2}-t^{1/2})(-t^{1/2}-t^{-1/2})}{(t^{-1/2}-t^{1/2})}=-t^{1/2}-t^{-1/2}
\end{align*}

Now noting that we have $H_+\sim_RM_0$, we consider Equation (<ref>),

\begin{align*}
\qquad\qquad&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!t^{-1}(-t^{1/2}-t^{-1/2})-tV(H_-)+t^{-1/2}-t^{1/2}=0\\
V(H_-) &= t^{-1}(-t^{-3/2}-t^{1/2})\\
&= -t^{-5/2}-t^{-1/2}
\end{align*}

Now we can calculate the Jones polynomial of our figure eight knot. We consider Equation (<ref>),

\begin{align*}
V(4_1) &= t(t-(t^{-1/2}-t^{1/2})V(H_-))\\
&= t^2-(t^{1/2}-t{3/2})(-t^{-5/2}-t^{-1/2})\\
&= t^2+t^{-2}+1-t^{-1}-t
\end{align*}

The Alexander polynomialThe Alexander polynomial [6] also admits a skein relation [15]. From the skein relation invariant polynomial, we can get the Alexander polynomial by substituting $a_+=1$, $a_-=-1$ and $a_0=t^{-1/2}-t^{1/2}$ [20]. That is,

\[\nabla(L_+)-\nabla(L_-)+(t^{-1/2}-t^{1/2})\nabla(L_0)=0.\]

While the skein relation invariant polynomial determines both Alexander's and Jones' polynomial, there are no other relations between them, for instance

* $V(4_1)=V(11_{388})$, but $\nabla(4_1)\neq\nabla(11_{388})$.

* $V(11_{388})=V(\overline{11_{388}})$ and $\nabla(11_{388})=\nabla(\overline{11_{388}})$, but $P(11_{388})\neq{P(\overline{11_{388}})}$.

* $\nabla(11_{471})=1=\nabla(0_1)$, but $V(11_{471})\neq1$.

On another note, we notice that for any knot $K$, $V(K)_{t=-1}=\nabla(K)_{t=-1}$. This is the determinant of $K$.

The Jones polynomial is as strong as the 3-colouring invariant polynomial is weak.

The following conjecture is still open. Khovanov homology [39], however, does detect the unknot.

Does there exist a non-trivial knot $K$ such that $V(K)=1$?

From the author's exposition on knot theory [55], it was shown that the Alexander polynomial can be defined using a method known as “colouring”, and thus we have the following conjecture.

Can we define the Jones polynomial via colouring?

While these two definitions were amongst the last to be created with respect to its history, they are amongst the easiest since it is more of a combinatorial calculation, requiring little to no knowledge of more results.

The next chapters will introduce some results and notation we need, to define the original definition of the Jones polynomial. It is in the author's opinion that the original definition of the Jones polynomial (presented in the final chapter), is by far the easiest in terms of both comprehension and calculation.

§ THE BRAID GROUP B

Thus far, we only know of topological methods for studying links, which motivates the next two chapters where we will learn algebraic methods for studying links. This is because braids can be defined both algebraically and geometrically. So, the aim of the next two chapters will be to discuss the connection between braids and links (Alexander's and Markov's theorem), in hopes of constructing a link invariant. In fact, the original construction of the Jones polynomial [26] (see Chapter <ref>) utilises these algebraic methods.

We begin with introducing the notion of braids, which happen to be closely related to links with the exception that it cannot be tangled with itself (and is not closed).
Braid equivalence, similarly to link equivalence, is defined with ambient isotopy. Thus, allowing us to define invariant functions on braids.

The aim of this chapter is to form the braid group from a geometrical perspective onto an algebraic presentation. This is done by defining a multiplication (braid concatenation) on the set of braids $\Bcal_n$, thus giving us the group structure we need.
By defining a homomorphism from the braid group onto the symmetric group, we find that the kernel is an interesting group called the pure braid group.

This chapter is based on the presentations given in [3], [4], [10], [12], [13], [16], [17], [18], [27], [28], [29], [32], [37], [38], [43], [44], [49], [58], and [65].

§.§ Braids

Informally, braids consist of a top and a bottom bar, with strings tied to each bar, such that the strings always go downwards. There exists a permutation from the $n$ points on the top bar to the $n$ points on the bottom bar. Recall that a permutationpermutation of a set $S$ is a bijective map $f\colon{S\to{S}}$. Here is a precise definition.

An $n$-braidbraid is a set of $n$ injective (non-intersecting) smooth paths (called strands) in $\RR^2\times[0,1]$ connecting the points on the top bar with the points on the bottom bar (i.e., there exists a permutation from the $n$ points on the top bar to the $n$ points on the bottom bar).
Such that the projection of each of the paths onto the $\RR^2$ plane (shadow - over/under strands) is a diffeomorphism (i.e., no self-tangles: the strings never have a horizontal tangent vector).
An example of a 3-braid can be seen below in Figure <ref>.

[strands=3,braid start=(0,0)]
σ_1 σ_2 σ_1

Example of a 3-braid

Note that the end points of the braids are fixed. And a strand is always going down. So equivalency is essentially ambient isotopy while fixing the end points.

Similarly to links, two braids are equivalentequivalent!braids if they are ambient isotopic.

Similarly to links and knots, we call the equivalence class of braids, braid typebraid type.

Braid equivalence in the above definition is an equivalence relation.

It is obviously reflexive. It is symmetric since if braid $\alpha$ can be transformed into braid $\beta$, then by looking over the `moves' of the transformation, we can just go backwards. Finally, it is transitive since if braid $\alpha$ can be transformed into braid $\beta$ and then to braid $\gamma$, by looking at the `moves' of the two transformations, we can concatenate them and form one so that we can get a transformation from $\alpha$ to $\gamma$.

We can, just as with links, perform elementary moves, the $\Delta$-moves (Definition <ref>), on braids to show that two braids are equivalent. So by fixing the end points of the braids, we can move the strings however we would like, ensuring that we still satisfy the requirements of a braid and without breaking the strands.

The $\RII$ move on a braid via the triangle move:

[braid start=(0,0),strands=2] σ_1^-1σ_1

(1.5,-0.5) node $\sim$;

[thin] (2.25,0)–(2.25,-1);
[ultra thick,white,double=black,double distance=0.5] (2,0)–(2.5,-0.5)–(2,-1);
[thin,dashed] (2,0)–(2,-1);

(3,-0.5) node $\sim$;

[thin] (3.5,0)–(3.5,-1);
[thin] (4,0)–(4,-1);

It is easy to see the $\RIII$ move on a braid via the triangle move, and is left as an exercise for the reader.

The bars on the braids are usually drawn to emphasise that the diagram represents a braid. So this is not always necessary.

Now that we know we can perform the $\RII$ and $\RIII$ moves on a braid, it is typical to wonder if we can perform an $\RI$ move. Obviously we cannot since by definition, the strand cannot self-tangle. However, in the next chapter, we will see that when relating braids to knots (via closure), there exists a fascinating move that mimics the $\RI$ move when closing the braid.

Suppose we have an $n$-braid $\alpha$, with $A_1$ connected to $A^\prime_{1^\prime}$, $A_2$ connected to $A^\prime_{2^\prime}$, $\ldots$, $A_n$ connected to $A^\prime_{n^\prime}$. Then, we can assign to $\alpha$ a permutation,

\[\begin{pmatrix}1&2&\cdots&n\\1^\prime&2^\prime&\cdots&n^\prime\end{pmatrix}.\]

This is called the braid permutationbraid permutation.

The trivial $n$-braid, which we get by connecting $A_1$ to $A^\prime_1$, $A_2$ to $A^\prime_2$, $\ldots$, $A_n$ to $A^\prime_n$ corresponds to the identity permutation, $\begin{pmatrix}1&2&\cdots&n\\1&2&\cdots&n\end{pmatrix}$.

[thin] (0.5,0)–(4.5,0);
[thin] (1,0)–(1,-3);
[thin] (1.5,0)–(1.5,-3);
[thin] (2,0)–(2,-3);
[thin] (3,-1.5) node[scale=1.3] $\bm{\ldots}$;
[thin] (4,0)–(4,-3);
[thin] (0.5,-3)–(4.5,-3);

the trivial n-braid

The braid permutation for Figure <ref> is

\[\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}=(1\;3).\]

The braid permutation is invariant under ambient isotopy on braids.

If two braids are equivalent, their braid permutations are equal (this is clear since by definition, in order to transform one braid into another, we have to keep the end points fixed). Thus, this is a braid invariant.

§.§ Representations of groups

This section will give a brief on representations of groups.

A representationrepresentation of a finite group $G$ on an $n$-dimensional $\FF$-vector space $V$ is a group homomorphism $\phi\colon{G\to\text{GL}_n(V)}$.

Given a group $G$ that acts on set $X=\{x_1,x_2,\ldots,x_n\}$. We call the vector space $\FF(X)$ a permutation representation and consists of linear combinations.

So, recall that the group algebra $\CC[S_n]$ is generated by $a_i=(i\;i+1)$ for $i\leq{n-1}$. And has the following relations.

* $a_i^2=1$

* $a_ia_j=a_ja_i$ if $|i-j|>1$

* $a_ia_{i+1}a_i=a_{i+1}a_ia_{i+1}$

This is called the presentationpresentation of $\CC[S_n]$.

§.§ The braid group

In this section we will give the set of braids a group structure by defining a multiplication. This is useful because we can use algebraic instead of geometrical structures.

We let $\Bcal_n$ be the set of the equivalence classes of all $n$-braids.

Braid multiplicationbraid multiplication of $n$-braids is defined by concatenation of the braids by putting one braid on top of the other, i.e., $\Bcal_n\times\Bcal_n\to\Bcal_n$. By convention, $\alpha\beta$ means $\alpha$ on top of $\beta$.

(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\alpha$;

(-0.25,0.5) node[scale=1.33] $\cdot$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;

(-0.5,-0.25) node[scale=1.33] $=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0,-0.5) rectangle (1,-1.5);
(0.53,-1.75) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\alpha$;
(0.5,-1) node[scale=1.33] $\beta$;

It is typical to ask if (how) we can multiply an $n$-braid with an $m$-braid, for $n\neq{m}$. In fact we will need to do this in the next chapter when relating a braid to a knot.
It turns out that there exists a natural inclusion from $\Bcal_n$ to $\Bcal_{n+1}$ by adding a vertical string to the right of the braid. So, we have the inclusions,

\[\Bcal_1\subset\Bcal_2\subset\cdots\]

This is quite similar to the Temperley-Lieb algebra planar diagram relations in Chapter <ref>.
For the sake of completeness and less ambiguity, we shall define precise definitions we can use.

The braid inclusion maplinear operations on $\Bcal_n$!braid inclusion map $\incl$ $\incl_n\colon\Bcal_n\to\Bcal_{n+1}$ is defined by,

\[\incl_n\left(\knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}}\right) =\,\, \knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}\]

braid inclusion

So, we can multiply an $n$-braid with an $m$-braid for $n<m$, means that we have to use the inclusion map on the $n$-braid recursively until $n=m$. We have the following example.

Let $\alpha\in\Bcal_n$ and $\varrho\in\Bcal_{n+3}$, so to multiply them we need to perform the braid inclusion map on $\alpha$ thrice, i.e.,

\[\incl_n(\alpha)\in\Bcal_{n+1}\tto\incl_{n+1}(\incl_{n}(\alpha))\in\Bcal_{n+2}\tto\incl_{n+2}(\incl_{n+1}(\incl_{n}(\alpha)))\in\Bcal_{n+3}\]

visually we have,

\begin{align*}
\knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}} \cdot \knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (2,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (1.25,1.5)--(1.25,1);
\draw (1.25,0)--(1.25,-0.5);
\draw (1.5,0)--(1.5,-0.5);
\draw (1.5,1.5)--(1.5,1);
\draw (1.75,0)--(1.75,-0.5);
\draw (1.75,1.5)--(1.75,1);
\draw (0.85,1.25) node {$\cdots$};
\draw (0.85,-0.25) node {$\cdots$};
\draw (1,0.5) node[scale=1.33] {$\varrho$};
\end{scope}} &= \incl_{n+2}\circ\incl_{n+1}\circ\incl_{n}\left(\knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};