Geometric Topology 8

Summing Over States

Fix a diagram $D$ of a link $L$ . One can compute the Kauffman bracket of $D$ by choosing one of its crossings to split using (iii), and then applying the same step to each of the two resulting diagrams. Proceeding in this way we finish with $2^c$ diagrams, each of which has no crossings left, is therefore a disjoint union of 'circles'. Because the changes only take place around each crossing, the order in which we choose the crossings is irrelevant, as in the proof about $R_2$ invariance.

A choice of splittings for every crossing is called a state. A splitting is positive if and only if the diagram that results is paired with $A$ . So a state can be regarded as a map:

s: \{1,2,3,...,c\} \rightarrow \{+1,-1\}

that assigns to each crossing the sign in which it is to be split. There are $2^c$ such states. One can indicate by $s(D)$ the diagram that results when $s$ is applied to $D$ , and by $|s|$ (pronounced 'mod $s$ ') the number of circles in $s(D)$ . Each state involves $p$ positive crossings and $n$ negative ones, with $p+n = c$ and $p-n = \sum_{i=1}^c s(i)$ .

The Big Sum

Repeated application of rule (iii) yields the "big sum": Proposition:

\boxed{D} = \sum_{2^c \text{ states}}A^{p(s) - n(s)}(-A^2 - A^{-2})^{|s|-1}.

There are two special states, namely $s_+$ with $s_+(i) = +1$ for all $i$ (so all splittings positive) and likewise $s_-$ with all splittings negative.

Exercise: In general, if $s,s'$ are states whose values (splittings) disagree at exactly one crossing then $|s'| = |s| \pm 1$ .

Alternating Links

Suppose that a link has an alternating diagram $D$ that is positively chess-boarded with $B/W$ black/white regions respectively. A positive splitting at every crossing will unite all the black regions and result in unknots around the white ones, so $|s_+| = W$ . Similarly $|s_-|=B$ .

Lemma: If $D$ is reduced then the highest power of $A$ in $\boxed{D}$ will arise from $s_+$ and Kauffman term:

A^c(-A^2 - A^{-2})^{W-1} = \pm A^{c+2W-2} \pm ...

Proof: Set $h = c+2W-2$ . Let $s'$ denote a state with exactly one negative splitting. We know that $|s'| = |s_+| \pm 1$ , so the highest power of $A$ in the Kauffman term of $s'$ will either equal $h$ or $h-4$ . We must prove it is the latter, otherwise there is a danger that terms in $A^h$ could cancel each other out. But we claim that the hypothesis 'reduced' forces $|s'| = |s|-1$ . For, having made a positive splitting, the resulting two arcs must form part of two distinct circles, otherwise the splitting would be an isthmus, with an arc joining one side to the other. If that splitting is changed to negative, the white regions in $Q_2,Q_4$ are united, two circles become one, and $|s'|=|s_+| -1$ . $\blacksquare$

Laurent Span

This is the difference between the highest and lowest powers in a Laurent polynomial, so we can talk about the $A$ -span in $\boxed{D}$ , or the $t$ -span in $V(L)$ .

Suppose that $D$ is both reduced and connected. Then it has $B+W = c+2$ regions by Euler's formula. We have seen that the highest power of $A$ in $\boxed{D}$ arises from $h=c+2W-2$ . Similarly the lowest power will be $\ell = -c - 2B + 2$ . So the $A$ -span is:

h - \ell = 2c + 2(W+B-2) = 2c + 2(c+2-2) = 4c.

To pass from the Kauffman bracket $\boxed{D}$ to the Jones polynomial $V(K)$ we multiply by $(-A)^{-3w(D)}$ (which does not affect the span), and replace $A$ by $t^{-1/4}$ . In conclusion:

Corollary: If a link $L$ has a connected reduced alternating diagram with $c$ crossings, $c$ equals the $t$ -span of $V(L)$ , and so in these circumstances $c$ depends only on the ambient isotopy class of $L$ .

Examples:

V(6_2) = t-1 + 2t^{-1} - 2t^{-2} + 2t^{-3} - 2t^{-4} + t^{-5} \\ V(\text{right Hopf}) = -t^{5/2} - t^{1/2} \ \text{but: } \ V(OO) = -t^{-1/2} - t^{1/2}

Historical Remarks

In 1987, Kauffman, Murasugi and Thistlethwaite improved the corollary around Laurent span by relating $V$ to spanning trees in a graph arising from the chess-boarding, so as to prove:

Theorem: If a link $L$ has a reduced alternating diagram $D$ with $c$ crossings then $c = cr(L)$ is its crossing number.

This was one of P.G. Tait's conjectures dating from 1898.

It is an open question as to whether any knot with $V(K) =1$ is necessarily trivial, i.e. ambient isotopic to an unknot. There do exist non-trivial links with $V(L) = 1$ . There are examples of distinct knots $K_1,K_2$ with $V(K_1) = V(K_2)$ , for example:

-t^4 + 2t^3 - 2t^2 + 2t + t^{-2} - 2t^{-3} + 2t^{-4} - 2t^{-5} + t^{-6}

is the Jones polynomial shared by the Conway and K-T knots, each with $cr = 11$ .

To cite an easier result, the Jones polynomial behaves well with respect to the connect sum of links, namely $V(K_1 \# K_2) = V(K_1)V(K_2)$ . In particular this confirms there are exactly three ambient isotopy classes formed by connecting two trefoil knots.