Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 19 / 1 / Solution

Put $\delta =-A^2-A^{-2}$. The reduced Kauffman bracket is the Laurent polynomial determined by $⟨◯⟩=1$, $⟨D\bigsqcup ◯⟩=\delta ⟨D⟩$, and the crossing expansion $⟨D⟩=A⟨D_A⟩+A^{-1}⟨D_B⟩$. Fix the A smoothing so that a positive curl has multiplier $-A^3$; rotating the local crossing interchanges the smoothing descriptions. Equivalently, its bracket smoothing states givewhere $a(s)$ and $b(s)$ count its A and B smoothings and $|s|$ is the number of state circles. Resolving all crossings proves that the recursive rules define the same polynomial independently of the order of expansion.

Under a positive Reidemeister move of type I, the two resolutions give $A\delta +A^{-1}=-A^3$ times the straight strand; the negative curl gives $A^{-1}\delta +A=-A^{-3}$. Thus the bracket itself is not an unframed link invariant. For type II, use the adjacent cap-cup generator $e$ of the Temperley-Lieb algebra, with $e^2=\delta e$. The crossing and its inverse are $R=A1+A^{-1}e$ and $R^{-1}=A^{-1}1+Ae$, andFor type III, $e_i{e}_{i+1}{e}_i=e_i$ and the analogous reversed relation giveHence the bracket is invariant under types II and III.

For an oriented diagram, let $w(D)$ be its writhe of a link diagram, the sum of its signed crossings. Types II and III preserve writhe of a link diagram, while a positive or negative curl changes it by one or minus one. The Jones polynomial is thereforeThe normalization cancels type I, so all Reidemeister moves preserve it. It is a Laurent polynomial in $t^{1/2}$; its breadth means the largest occurring t exponent minus the smallest, including half-integral exponents for links.

The Jones polynomial skein relation follows directly from the same normalization. Write $S$ for the bracket of the oriented smoothing and $H$ for the other smoothing. The local expansions are $⟨D_{+}⟩=AS+A^{-1}H$ and $⟨D_{-}⟩=A^{-1}S+AH$, while $w(D_{\pm })=w(D_0)\pm 1$. ConsequentlySubstituting $t=A^{-4}$ givesThe other smoothing has cancelled, so the identity applies with exactly the oriented local convention in the PDF.