Kauffman bracket state sum and the Jones polynomial

Since $C(D)$ is finite by [F1], the set of all functions $S:C(D)\to \{A,B\}$ is finite with exactly $2^{\mathrm{\mid }C(D)\mathrm{\mid }}$ elements. Each state replaces every crossing in a finite collection of disjoint crossing disks by one of the two specified pairings. The resulting finite 1-manifold has at least one circle because the input diagram has at least one component. Hence each term $A^{a(S)-b(S)}{\delta }^{\mathrm{\mid }S\mathrm{\mid }-1}$ is a defined Laurent polynomial.

The displayed sum is indexed by the set of states itself, so reordering a list of crossings merely gives a bijective reindexing of the same finite sum. By [F2], a planar isotopy carries crossings, local charts, states, and resulting circles bijectively to those of the isotoped diagram, preserving $a(S)$, $b(S)$, and $\mathrm{\mid }S\mathrm{\mid }$. Therefore the formula is well-defined on planar-isotopy classes and is choice-free.

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. For the opposite crossing, rotating the source chart by a positive quarter-turn exchanges the two endpoint pairings, so its A/B smoothings are respectively $e_i$ and $I$. By [F1] these expansions are $C_i=AI+A^{-1}{e}_i$ and $C_i^{-}=A^{-1}I+A{e}_i$. Using [F2] and [F3], $C_i{C}_i^{-}=(AI+A^{-1}{e}_i)(A^{-1}I+A{e}_i)=I+(A^2+A^{-2}+\delta )e_i=I\mathrm{.}$ The same multiplication in the other order gives $C_i^{-}{C}_i=I$, since both factors are polynomials in $e_i$. Thus the two opposite crossings in an R2 pair cancel in either order.

Put $x=e_i$ and $y=e_{i+1}$. Expanding and using [F3] gives $C_i{C}_{i+1}{C}_i=A^{3}I+A(2x+y)+A^{-1}(xy+yx+\delta x)+A^{-3}x,$ while interchanging $x$ and $y$ gives $C_{i+1}{C}_i{C}_{i+1}=A^{3}I+A(2y+x)+A^{-1}(yx+xy+\delta y)+A^{-3}y\mathrm{.}$ Their difference is $A^{-1}(A^2+\delta +A^{-2})(x-y)=0$. Hence the two sides of the braid-like R3 move have equal bracket operators.

Near the three-strand event, choose a linear longitudinal coordinate nonzero on each of the three tangent lines, shrink the event disk, and orient the unoriented local tracks in its increasing direction. All three tracks are then monotone; these temporary arrows do not affect the bracket. Label the tracks $1,2,3$ at the incoming end. The two R3 sides meet their pairs in orders $(12,13,23)$ and $(23,13,12)$, and a strict height order permits exactly the exponent triples $+++,---,++-,-++,+--,--+$. The triples $+-+$ and $-+-$ would require a cyclic height order. Put $a=C_i$ and $b=C_{i+1}$. Besides $aba=bab$ from step 1.2 and its inverse from step 1.1, the four needed identities are $ab{a}^{-1}=b^{-1}ab,a^{-1}ba=ba{b}^{-1},a{b}^{-1}{a}^{-1}=b^{-1}{a}^{-1}b,a^{-1}{b}^{-1}a=b{a}^{-1}{b}^{-1}\mathrm{.}$ The first follows by multiplying $aba=bab$ on the left by $b^{-1}$ and on the right by $a^{-1}$; the second follows by multiplying on the left by $a^{-1}$ and on the right by $b^{-1}$; the last two are their inverses. Thus each of the six height orders gives equality with exactly its actual over/under choices. For R2, choose a longitudinal coordinate along the common tangent at the bigon event, shrink the disk, and temporarily orient both tracks increasingly; the two local crossings have opposite types and cancel by step 1.1. Reversing any original strand arrow leaves all smoothing weights unchanged. Adjoining the unchanged exterior tangle and closing its boundary preserves these equalities: expand each crossing into its two smoothings, use bilinearity, and evaluate each resulting crossingless closed diagram with one factor $\delta$ per circle as in [F2]. This gives equal sums ${\sum }_S{A}^{a(S)-b(S)}{\delta }^{\mathrm{\mid }S\mathrm{\mid }}$ on the two sides. Every complete smoothing of the nonempty diagram has $\mathrm{\mid }S\mathrm{\mid }\ge 1$, and $\delta =-A^2-A^{-2}$ is nonzero in the integral domain $\mathbb{Z}[A^{\pm 1}]$; hence these sums are $\delta ⟨D⟩$ and $\delta ⟨D^{\mathrm{\prime }}⟩$. Cancelling $\delta$ proves $⟨D⟩=⟨D^{\mathrm{\prime }}⟩$, so every R2/R3 variant preserves the bracket.

If $c=\mathrm{\mid }C(D)\mathrm{\mid }$, every state is a function from a $c$-element set to a two-element set, so the state set has exactly $2^c$ elements. This is finite, including the case $c=0$, where there is exactly one empty assignment. The algorithm processes every such function once and has no other loop.

For each visited state, the prescribed smoothing produces a finite crossingless union of circles. By [F2], $\mathrm{\mid }S\mathrm{\mid }-1$ is a nonnegative integer, so $A^{a(S)-b(S)}{\delta }^{\mathrm{\mid }S\mathrm{\mid }-1}$ is defined in $\mathbb{Z}[A^{\pm 1}]$. After any initial segment of the visits, the accumulator is the sum of exactly the terms already visited, by induction on the number of updates. After all $2^c$ visits it is the displayed state sum, which is returned. Therefore execution terminates with exactly the asserted output.

Label the crossing ports northwest $1$, northeast $2$, southeast $3$, and southwest $4$, in the chart of [F1]. The over-strand joins $4$ to $2$ and the under-strand joins $1$ to $3$. For the left exterior cap $(1,4)$ the remaining ports are $2,3$. Orient its through strand from $2$ to $3$: it traverses the over-strand northeast to southwest, then the cap, then the under-strand northwest to southeast. The tangent vectors may be taken as $(-1,-1)$ and $(1,-1)$, whose determinant is $2>0$. This is a positive kink. The A-pairing $(1,4)(2,3)$ closes one additional circle against the cap; the B-pairing $(1,2)(3,4)$ gives just the through arc. Hence its bracket factor is $A\delta +A^{-1}=-A^3$. Reversing the component negates both tangent vectors and preserves the determinant.

For the top exterior cap $(1,2)$ the remaining ports are $4,3$. Orient its through strand from $4$ to $3$: it traverses the over-strand southwest to northeast, then the cap, then the under-strand northwest to southeast. The tangent vectors $(1,1)$ and $(1,-1)$ have determinant $-2<0$, so this is a negative kink. Here the A-pairing gives only the through arc, while the B-pairing closes one additional circle. Its factor is $A+A^{-1}\delta =-A^{-3}$. The right and bottom cap cases are the respective $180$-degree rotations; orientation-preserving chart changes and reversal of the component preserve the calculation. These are all oriented type-I variants. The calculation is inside the move disk and permits every exterior connectivity and any additional components. Deletion uses the reciprocal Laurent-unit factor.

The positive kink changes writhe from $w$ to $w+1$, and the negative kink changes it to $w-1$. Their respective normalized values are $(-A^3{)}^{-(w+1)}(-A^3)⟨D⟩=(-A^3{)}^{-w}⟨D⟩,$ and $(-A^3{)}^{-(w-1)}(-A^{-3})⟨D⟩=(-A^3{)}^{-w}⟨D⟩\mathrm{.}$ Thus both type-I additions and their inverse deletions preserve $f_D$.

A type-II move creates or deletes two opposite signs, and a type-III move preserves all three signs, so writhe is unchanged. Their raw bracket is unchanged by [F4]. Planar isotopy transports the determinant crossing signs, smoothing states and state-circle counts bijectively. Therefore it preserves both factors of $f_D$, and the normalized value is invariant under every listed move.

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. Thus its two smoothings are $I$ with weight $A$ and $e_i$ with weight $A^{-1}$. For the opposite crossing the pairings exchange. Hence the local operators are $C_i=AI+A^{-1}{e}_i,C_i^{-}=A^{-1}I+A{e}_i\mathrm{.}$ The relations [F5] and [F6] show these operators satisfy the Artin relations and are mutual inverses; distant generators commute by [F5].

Define ${\rho }_n({\sigma }_i)=C_i$ and $S_i:=-A^{-1}{C}_i=-I-q{e}_i$. With $t=q^2$ and $\delta =-(q+q^{-1})$, direct multiplication gives $S_i^2=I+(2q+q^2\delta )e_i=I+(q-q^3)e_i=(t-1)S_i+t\mathrm{.}$ The braid and distant-commutation relations for $S_i$ follow from [F5] and step 1.1. Under the coefficient map $\psi :\mathrm{\Lambda }\to R_A$ given by $v\mapsto t$ and $z\mapsto z_0$, the assignment $T_i\mapsto S_i$ therefore defines a compatible homomorphism ${\rho }_H$ from the specialized Hecke tower to the Temperley--Lieb diagram algebras. For a braid word, this map sends ${\pi }_n(\beta )$ to the product of the corresponding $S_i^{\pm 1}$.

The element $A^{-4}+1$ is a unit in $R_A$ and $\delta =-A^2(A^{-4}+1)$ is therefore a unit. Set ${\tau }_n={\delta }^{-n}{\mathrm{Tr}}_n$. By [F7], ${\tau }_n(1)=1$, ${\tau }_{n+1}(x\otimes 1)={\tau }_n(x)$, and ${\tau }_n$ is cyclic. Since $S_n=-I-q{e}_n$, the last closure rule of [F7] gives ${\tau }_{n+1}((x\otimes 1)S_n)=-{\tau }_n(x)-q{\delta }^{-1}{\tau }_n(x)=-\frac{1}{q^2+1}{\tau }_n(x)=z_0{\tau }_n(x)\mathrm{.}$ Thus $\tau \circ {\rho }_H$ satisfies all specialized Markov-trace rules, where ${\rho }_H(T_i)=S_i$.

The decomposition in [F10] is a direct sum of modules spanned by disjoint subsets of the Hecke standard basis. Under $\psi$, that basis remains a basis, so for $H_{\psi }(n):=H(n){\otimes }_{\mathrm{\Lambda },\psi }{R}_A$ the specialized tower still has $H_{\psi }(n+1)=H_{\psi }(n)\oplus H_{\psi }(n)T_n{H}_{\psi }(n)$. Any specialized Markov trace is therefore determined by its restriction to $H_{\psi }(n)$: restriction determines the first summand, while cyclicity and the Markov rule force ${\mathrm{tr}}_{n+1}(x{T}_{n}y)=z_0{\mathrm{tr}}_n(yx)$ on every product in the second summand. Starting at $H(1)$, induction proves uniqueness, without dividing by any specialized parameter. The scalar extension of the Ocneanu trace exists by [F8], so the trace in step 3.1 equals that specialized trace.

Expand each crossing of $\beta$ by $C_i$ or $C_i^{-}$ and close the resulting Temperley–Lieb diagrams. The unnormalized trace counts each closed state circle with a factor $\delta$, whereas the bracket state sum counts it with ${\delta }^{\mathrm{\mid }S\mathrm{\mid }-1}$; hence $⟨\widehat{\beta }⟩={\delta }^{-1}{\mathrm{Tr}}_n({\rho }_n(\beta ))={\delta }^{n-1}{\tau }_n({\rho }_n(\beta ))\mathrm{.}$ Since $C_i=-A{S}_i$ and $C_i^{-}=(-A{)}^{-1}{S}_i^{-1}$, multiplying along an Artin word gives ${\rho }_n(\beta )=(-A{)}^{e(\beta )}{\rho }_H({\pi }_n(\beta ))$. Step 4.1 now gives the displayed trace identity.

The closure of the identity in $B_1$ has one circle; the state sum gives ${\delta }^{1-1}=1$, and its trace expression gives ${\delta }^0(-A{)}^0{\mathrm{tr}}_1(1)=1$. For positive stabilization, the Markov trace factor is $z_0=-1\mathrm{/}(q^2+1)$, so the bracket formula changes by $(-A)\delta z_0=-A^3\mathrm{.}$ For negative stabilization, [F9] and [F8] give trace factor $z_{-}:=t^{-1}(z_0+1-t)=-\frac{q^2}{q^2+1};$ the bracket formula changes by $\delta (-A{)}^{-1}{z}_{-}=-A^{-3}\mathrm{.}$ This proves the unknot value and both stabilization normalizations.

By [F1], each planar isotopy and each oriented Reidemeister move in a finite sequence preserves the normalized bracket value. Repeated equality along a finite sequence therefore preserves $f_D(A)$.

By [A1] and [F2], the two given diagrams are joined by such a finite sequence. Applying step 1.1 to that sequence gives equal normalized bracket values. By [F3], every nonempty oriented link has a regular diagram. Hence $f_D$ defines a value on every nonempty oriented link type.

A closed braid diagram has one crossing for each letter in a finite Artin word and no crossings at its closure arcs. Each positive generator contributes $+1$ to writhe by [F2]. In the reflected inverse half twist, the over- and under-strands exchange while their oriented projected tangent vectors retain their respective directions; exchanging the two vectors negates their determinant, so each inverse generator contributes $-1$. Summing over the word and using the homomorphism in [F3] and the finite word representation in [F8] gives $\mathrm{wr}(\widehat{\beta })=e(\beta )\mathrm{.}$

Put $q=A^{-2}$ and $t=q^2$. Then $\delta =-A^2-A^{-2}=-(q+q^{-1})=-(t+1)\mathrm{/}q$. By [F5] and step 1.1, $\begin{array}{rl}{\displaystyle f_{\widehat{\beta }}(A)}& {\displaystyle =(-A^3{)}^{-e(\beta )}{\delta }^{n-1}(-A{)}^{e(\beta )}{\mathrm{tr}}_n({\pi }_n(\beta ))}\\ {\displaystyle }& {\displaystyle =q^{e(\beta )}{\delta }^{n-1}{\mathrm{tr}}_n({\pi }_n(\beta ))}\\ {\displaystyle }& {\displaystyle =q^{e(\beta )}{(-\frac{q^2+1}{q})}^{n-1}{\mathrm{tr}}_n({\pi }_n(\beta ))\mathrm{.}}\end{array}$ This is exactly the library trace normalization with $s=q$ and $t=s^2$ from [F6]. Thus the two constructions agree on every braid closure.

Put $R_A=\mathbb{Z}[A^{\pm 1},(A^{-4}+1{)}^{-1}]$, as in [F5]. The target coefficient ring in [F6] is $T=\mathbb{Z}[s^{\pm 1},(s^2+1{)}^{-1}]$. The substitution $s\mapsto A^{-2}$, $t=s^2\mapsto A^{-4}$ defines a ring map $T\to R_A$. It is injective because distinct Laurent powers $s^k$ map to distinct Laurent monomials $A^{-2k}$, and the localization denominator maps to the unit $A^{-4}+1$. Hence the equality from step 2.1 is an equality in $T$, with $t=s^2$, rather than only after embedding the two values in a larger ring.

By [A1] and [F1], full AC supplies the countable-choice hypothesis of [F4]. By [F7], choose a braid closure for the given nonempty link $K$. The calculation in step 2.1 and its ring identification in step 3.1 prove equality on that closure; the invariance in [F4] and [F6] makes both values independent of the chosen regular diagram and braid representative. Therefore $V_K(t)=f_K(t^{-1\mathrm{/}4})$ for every nonempty oriented link.