Geometric Topology 7

The Kauffman Bracket

Recall that a regular isotopy between link diagrams $D,D'$ is defined by a sequence of moves $R_0,R_1,R_2,R_3$. Let $O$ denote a diagram of the unknot with no crossings (a circle), and let $OD$ denote the disjoint union of plane diagrams.

Proposition: The rules:

allow one to associate to each link diagram $D$ a Laurent polynomial $\boxed{D}$ in $A$ (so one can have positive and negative powers of $A$ and constants) that depends only on the regular isotopy class of $D$, so that $D \approx D' \implies \boxed{D} = \boxed{D'}$.

The usual notation for the Kauffman bracket is $\langle D \rangle$, but using boxes will make it easier to surround parts of the diagrams. In either case, it is customary not to include the variable, which is always denoted '$A$'.

Example: The bracket of the disjoint union of $n \geq 1$ circles equals $(-A^2 - A^{-2})^{n-1}$.

A Word About Splittings

Each boxed term in rule (iii) denotes the Kauffman bracket of an entire diagram with the strands joined inside the boxes as shown, and no changes outside the boxes. The rule relates the bracket of $3$ diagrams, in which a given crossing $X$ is eliminated ('split') in its two possible planar ways.

If we regard the overpass as the $x$-axis (in either direction), a positive splitting (whose resulting diagram has a coefficient $A^{+1}$) corresponds to opening up a channel between the odd quadrants. It is therefore achieved by placing a 'splitting marker' between $Q_1$ and $Q_3$, which can be imagined to be the line with Cartesian equation $y = +x$. Similarly a negative splitting corresponds to joining the even quadrants so as to insert the splitting marker $y = -x$.

Note that orientation plays no role in the definition of the bracket. In particular, the crossing $X$ in (iii) has no sign associated to it, and by rotating the page any crossing can be viewed as if it looks like $X$. However if the underlying diagram is chess-boarded then $X$ acquires a chess sign, and joining the black regions will result in a splitting whose sign equals that chess sign (positive $\iff$ $Q_1,Q_3$ are black).

Invariance Under $R_2$

If a diagram has more than one crossing, the order in which one applies (iii) is irrelevant. For example, one can apply (iii) to each of $2$ crossings of a diagram $D$ to decompose $\boxed{D}$ into the sum of $4$ terms, with coefficients:

Resolving the next $2$ crossings gives $4$ terms, but (ii) causes $3$ of these to cancel out. Hence, the double underpass can be removed without changing the bracket polynomial.

Invariance Under $R_3$

Apply (iii) to the diagonal crossing, then $R_2$ invariance to the diagram with coefficient $A$. This gives a symmetrical configuration, which establishes $R_3$ invariance.

Starting with a diagram with $c$ crossings, we can keep 'splitting' so as to end up with $2^c$ diagrams, each one $D_i$ (the result of a state) with no crossings. Thus $D_i \approx n_i O$ and $\boxed{D_i} = (-A^2 -A^{-2})^{n_i - 1}$ where each $n_i$ where $1 \leq i \leq 2^c$ is a positive.

The Effect of $R_1$

Let $D'$ be a link diagram incorporating a twist, and let $D$ be the diagram in which $R_1$ has been used to eliminate the twist (keeping the ends fixed). The latter is negative in the sense its writhe-sign is $-1$, chosen to conform with our diagram of (iii). We want to relate $\boxed{D'}$ to $\boxed{D}$.

If the crossing $X$ is split positively (so a splitting marker is placed horizontally), we get a single connected arc inside the box, and so $D$ altogether. If $X$ is split negatively (vertically splitting marker), we get an extra circle resulting in a diagram $D'' \sim OD$. So:

Now $w(D') = w(D) - 1$, so it follows that:

Conclusion: Given a diagram $D$ of a link $L$, the quantity $(-A)^{-3w(D)}\boxed{D}$ is invariant by moves $R_0,R_1,R_2,R_3$. It therefore depends only on the ambient isotopy class of $L$: it is (up to change of variable) the Jones polynomial of $L$.

Examples: Hopf Link and Trefoil Diagrams

Splitting the two crossings of a diagram of the Hopf link gives $2^2$ states and:

This allows us to determine the bracket of a diagram of the right-handed trefoil $R3_1$:

Exercises:

1. The bracket of the $3$-leaved diagram of $L3_1$ is $A^7 - A^3 - A^{-5}$.
2. If $mD$ is the diagram $D$ in which all crossings have been reversed then $\boxed{mD}$ is obtained from $\boxed{D}$ by replacing $A$ with $A^{-1}$.

The Jones Polynomial

We have already defined this implicitly for the diagram $D$ of an oriented link $L$, by combining the writhe and Kauffman bracket, so the result is unchanged by all $R$-moves.

The Jones polynomial $V(L)$ (sometimes denoted $V_L(t)$) is in fact a Laurent polynomial in the variable $t^{\frac{1}{2}}$ that is substituted in place fo $A^-2. It was introduced by Vaughan Jones in 1984, but it was quickly understood that it can be derived from Kauffman's bracket.

Definition: If $D$ is a diagram for $L$ then $V(L) = (-A)^{-3w(D)}\boxed{D}$ with $A=t^{-\frac{1}{4}}$.

$V(L)$ is an invariant of the isotopy class of $L$ as an oriented link, but orientation is irrelevant if $L$ is a knot.

We will see that if $L$ has an odd number of components then $V(t)$ is actually a Laurent polynomial in $t$. Otherwise it will be $t^{\frac{1}{2}}$ times such a Laurent polynomial. It has integer coefficients.

Exercise: The Jones polynomial of the mirror image of a link is found by subsituting $t^{-1}$ in place of $t$.

Examples of Knot Polynomials

Unlike the bracket, $V$ is a true knot invariant, so its value can be computed using any valid diagram, which is very convenient. Previous formulae imply that $V$ of any unknot equals $1$, and:

in accordance with the previous exercise.

Exercise: An easier-to-memorize formula reflects the fact that $4_1$ is achiral:

Results to come:

1. If $L$ is a link with $\ell \geq 1$ components then $V_L(1) = (-2)^{\ell -1}$, so a knot $K$ has $V_K(1) = 1$. This is easily proved using the skein relation to come.
2. If a knot $K$ has a (connected, reduced) alternating diagram with $c$ crossings then the difference between the highest and lowest powers of $t$ (the 'Laurent span') in $V(K)$ equals $c$.

Examples of Link Polynomials

We can show that:

• $V(\text{right Hopf link}) = -t^{1/2} - t^{5/2}$
• $V(\text{Solomon link}) = -t^{9/2} - t^{-5/2} + t^{-3/2} - t^{-1/2}$
• $V(\text{Whitehead link}) = -t^{-3/2} + t^{-1/2} - 2t^{1/2} + t^{3/2} - 2t^{5/2} + t^{7/2}$
• $V(\text{Borromean rings}) = -t^{-3} + 3t^{-2} - 2t^{-1} + 4 - 2t + 3t^2 - t^3$

The Skein Relation

Surround a crossing of an oriented link diagram $D$ by a box and orient the page so that arrows enter on the left and leave on the right. There are exactly $3$ ways to arrange the strands in the box with at most one crossing:

1. Top strand entering from the left is the overpass (and bottom strand entering from the left is the underpass) ($L_+$)
2. Top strand entering from the left is the underpass (and bottom strand entering from the left is the overpass) ($L_{-}$)
3. Top strand entering from the left and bottom strand entering from the left can be pulled apart and do not share a crossing ($L_0$)

In particular there is only one way to split an oriented crossing. There is a well-known relation between the Jones polynomials of the links one can form by rearranging the strands in such a box:

Proposition: Set $z = t^{1/2}-t^{-1/2}$. Then:

A Coherent Proof

Exercise: If $L_{\pm}$ are knots then $L_0$ is a link with two components. This helps explain the appearance of half-integral powers of $t$ as coefficients of $V(L_0)$.

The skein relation is proved by splitting the two crossings using Kauffman's rule (iii).

Let $w_0 = w(L_0)$ denote the writhe of the diagram representing $L_0$. The writhe sign of the featured crossing in $L_{\pm}$ is $\pm 1$, so the writhe of the entire diagram representing $L_{\pm}$ equals $w_0 \pm 1$. We know that:

Therefore:

Subtracting:

We have used the change of variable $t = A^{-4}$ freely. $\blacksquare$

Example: The Jones Polynomial of $5_1$

In order to compute the Jones polynomial of the knot $5_1$, we first note it has writhe $+5$, and its crossings are of the type seen in $L_+$. Set $z = t^{1/2} - t^{-t/2}$ again. If we focus on the top crossing of the knot and apply $R_2$ to obtain the version of the diagram with the top crossing flipped, we get the equation:

where $S$ is the right-handed Solomon link, with the right-handed orientation ($w = +4$).

Example: The Jones Polynomial of the Solomon Link

We can express $V(S)$ in terms of the polynomials:

of the right-handed trefoil knot and Hopf link. Focusing on the top-left crossing we get:

Exercise: Why is this $t^6$ times the previous polynomial for the Solomon link?

Putting it all together, $V(5_1) = -t^7 + t^6 - t^5 + t^4 + t^2$.