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

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.

True. In a connected sum of knots, a bracket smoothing state from each summand contributes a state of the joined diagram. Joining the two distinguished state circles reduces their total count by one, so the factors ${\delta }^{s-1}$ multiply. The smoothing weights also multiply and the writhe of a link diagram adds. Thus the Jones polynomial of a connected sum is

False. The zero polynomial is already a counterexample. The Jones polynomial evaluation at one of a nonempty link with $\ell$ components isTo prove this, specialize the Jones polynomial skein relation at $t=1$: switching a crossing preserves the value. Switch crossings until the diagram is an unlink without changing its component count. The Kauffman bracket loop rule, equivalently the Jones polynomial of a split union, gives the displayed evaluation. Hence arbitrary Laurent polynomials cannot occur.

False as written for arbitrary links. The two-component unlink has a diagram with no crossings butThe correct connected-diagram statement does hold, in particular for every knot diagram. If $s_A,s_B$ count the all-$A$ and all-$B$ bracket smoothing state circles, changing $j$ smoothings increases the circle count by at most $j$. The maximum and minimum $A$ exponents therefore lie between $n+2s_A-2$ and $-n-2s_B+2$. This gives the Kauffman bracket breadth bound ${\mathrm{br}}_A⟨D⟩\le 2n+2s_A+2s_B-4$. For a connected projection, the Turaev surface has Euler characteristic $s_A+s_B-n\le 2$, so the bound is $4n$. Writhe of a link diagram normalization changes only a monomial, and $t=A^{-4}$ divides the breadth by four, giving $\mathrm{br}{V}_L\le n$.

For a diagram with $k$ projection components, sum the same Euler characteristic inequality over the components. Then $s_A+s_B\le n+2k$ and the Jones polynomial breadth bound for a disconnected diagram is $\mathrm{br}{V}_L\le n+k-1$. The extra $k-1$ is indispensable, as the unlink shows.

True with the stated Conway normalization. Fix $t^{1/2}=i$ at $t=-1$. The Jones polynomial skein relation becomesThe Conway polynomial satisfies ${\nabla }_{L_{+}}(z)-{\nabla }_{L_{-}}(z)=z{\nabla }_{L_0}(z)$. Put $W_L=(-1{)}^{\ell (L)-1}{\nabla }_L(2i)$. An oriented smoothing changes the number of components by one, so these quantities satisfy precisely $W_{L_{+}}-W_{L_{-}}=-2i{W}_{L_0}$. Both $W$ and $V(-1)$ are one on the unknot and zero on every multi-component unlink: for $V$, the split-union factor vanishes at this substitution.

These skein relations and unlink values determine the invariants uniquely. Switching a first undesired crossing gives a diagram closer to a descending diagram, while the smoothing term has fewer crossings; induction first on crossing count and then on undesired crossings reduces everything to unlinks. Thus $W_L=V_L(-1)$. For a knot there is one component and the Conway-normalized Alexander polynomial givesThis is the signed Jones evaluation at minus one. Replacing ${\Delta }_K$ by an arbitrary Laurent unit multiple would destroy the signed conclusion.