Scalar evaluation of framed Kauffman link diagrams

We first construct the original, sum-skein version over $F=\mathbb{Q}(b,t)$. Set $h=(b+b^{-1})\mathrm{/}t-1$. Its desired normalization is $L(◯)=1$, positive and negative curls have factors $b,b^{-1}$, an extra trivial circle has factor $h$, and its skein equation is $L(D)+L(S_{c}D)=t{C}_c(D)$, where $S_c$ switches crossing $c$ and $C_c(D)=L(E_{c}D)+L(e_{c}D)$ is the sum over its two smoothings. All constructions are by induction on the crossing number $N$. At a given $N$, first define single-component diagrams; then define diagrams with increasing numbers of components. The crossing-free value on $r\ge 1$ circles is $h^{r-1}$. Throughout the induction the lower-crossing functions satisfy the stated equations and invariance under moves whose diagrams have fewer than $N$ crossings.

Assuming the function known below $N$, switching crossing $c$ transports a proposed value $x$ to $T_c(x)=-x+t{C}_c(D)$. Repeating this switch is the identity because its two smoothings do not depend on the crossing sign. Two switches $c,j$ commute as transports: applying the lower-crossing skein relation at the remaining crossing in each smoothing gives $C_c(D)+C_c(S_{j}D)=t{\sum }_{u\in \{E_c,e_c\},v\in \{E_j,e_j\}}L(uvD)=C_j(D)+C_j(S_{c}D)$. Consequently $T_j{T}_c(x)=x-t{C}_c(D)+t{C}_j(S_{c}D)=x-t{C}_j(D)+t{C}_c(S_{j}D)=T_c{T}_j(x)$. Thus an expansion along a fixed set of switches is independent of their order. Explicitly, for an ordered list $c_1,\dots ,c_k$ and $D_j=S_{c_j}\cdots S_{c_1}D$, its value obtained from a prescribed terminal value $u(D_k)$ is $(-1{)}^{k}u(D_k)+t{\sum }_{j=1}^k(-1{)}^{j-1}{C}_{c_j}(D_{j-1})$. This identity follows successively from the two-term transport rule, so it uses only already-defined evaluations of diagrams with at most $N-1$ crossings.

For a directed point $p$ on a one-component diagram, switch exactly those crossings first encountered underneath when following the component from $p$. The terminal diagram $U_p$ is descending and is an unknot: lift its traversed part above the untraversed part and draw that part away successively until only framing curls remain. Give it terminal value $b^{w(U_p)}$, where $w$ is oriented writhe. Define $Q(D,p)$ by the switch expansion of step 2.1 with this terminal value, and define $L(D)=(Q(D,p)+Q(D,\bar{p}))\mathrm{/}2$, where $\bar{p}$ reverses the direction at the same point. Use this average also when $D$ is descending for one directed point; $b^{w(U_p)}$ is a terminal scalar, not a second prescription for $L(D)$. For a multi-component diagram having a component over all the others, prescribe $L(D)=hL(C)L(D-C)$, with $C$ considered alone. If no component is over all the others, for every component $C$ choose a point and either direction, switch its undercrossings with other components so that it lies over all the others, and use terminal value $hL(C)L(D-C)$. Define $L(D)$ as the average of these $2r$ switch expansions, where $r$ is the number of components. The terminal factors are already defined: single-component diagrams are defined first at level $N$, and deleting one component decreases component number if the remaining crossing number is still $N$. The expansion sums have fewer crossings.

The multi-component switch set for lifting a fixed component is independent of point and direction, so step 2.1 makes its expansion independent of these choices. For a knot, moving the point past one encounter with a crossing toggles exactly that crossing in the descending switch set. The two terminal descending diagrams differ by switching this crossing, which is the first or last encounter from the new point. Their terminal values satisfy the required sum-skein equation: if their writhes are $v+1,v-1$, the two smoothings have values $b^v$ and $h{b}^v$. To see this, the smoothing that separates the traversed initial portion produces two descending unknots with one over the other; the other is their connected sum with possibly one direction reversed. Their total self-writhe is $v$, since the mutual crossing-sign sum of the two separated portions is zero: one lies over the other, and their algebraic planar intersection number is zero. Both diagrams reduce by pulling these overlying arcs away, followed by removing curls, so their evaluations follow from lower-crossing invariance and curl factors. The equation is $b^{v+1}+b^{v-1}=t(1+h)b^v$. By commuting the toggled switch to the beginning of the expansion, this terminal equality makes $Q(D,p)$ unchanged. Moving the point around all arcs proves point independence. Averaging the two directions makes direction independence explicit. Agreement with the terminal scalar on descending diagrams is deferred until after regular-isotopy invariance; point independence alone does not identify the two directed expansions.

The resulting function satisfies the sum-skein equation at every crossing. For a knot choose the directed point so the selected crossing is first in the expansion of the diagram in which it is incorrect; the two expansions differ by that initial transport, giving the equation. For a link, compare the expansions used in step 3.1. If the crossing is a self-crossing of the component being lifted, apply the lower-complexity equation to its single-component terminal factor and to every smoothing term. If it belongs only to the remaining link, apply the equation to that terminal factor and to every smoothing term. If its two strands lie on different components and the lifted component is one of them, commute that crossing to the first switch position; the two terminal lift diagrams agree and the initial transport gives the equation. If the lifted component is neither one, the lower-crossing equation applies to all smoothing terms and to the remaining-link factor. For clarity, the consistency with an already overlying component has a separate two-layer comparison. Suppose $A$ is above the rest and the selected lift is $B$. Commute the switches of $B$ with components other than $A$ before its mutual switches with $A$. Smoothing one of these earlier switches leaves $A$ over the resulting lower-crossing diagram, so induction factors out $hL(A)$; their alternating sum is exactly the lift expansion for $D-A$. For the remaining comparison both $A$ and $B$ are above the remaining link $R$. If $R$ is nonempty, every smoothing of an $A,B$ crossing merges them into a single component above $R$; induction factors its value as $hL(R)$ times its isolated knot value. Both terminal orders have the identical value $h^{2}L(A)L(B)L(R)$. If $R$ is empty, there is no factor $hL(R)$ to extract, and both terminal orders instead have value $hL(A)L(B)$. It remains to compare the two-component lift expansion with $hL(A)L(B)$. Switch self-crossings in $A$ and $B$ until each is descending; the difference of these two quantities changes sign under each such switch, since every smoothing correction has fewer crossings and agrees by induction. For descending $A,B$ with $A$ above $B$, straighten the descending two-component diagram to two disjoint circles by nonincreasing moves and curl removals. This comparison respects each such move directly at the level of its switch expansion: for a cancelling crossing pair $u,v$, the correction is $t(C_u(D)-C_v(S_{u}D))=0$, since the two opening smoothings agree and the other two give the same signed curl; a move III changes the order of two switches and is exactly the transport-square identity of step 2.1, with its smoothed diagrams related by a cancelling pair. Terminal writhe and scalar products obey the same curl factors. Thus the difference is zero at the end and hence at the start. This proves the required two-component comparison, and then the preceding factorizations prove the direct-overlying prescription for the original link. Every averaging summand therefore obeys the same skein equation.

To verify the curl rule, place the point outside the curl and commute its switch, if any, to the end of the expansion. At lower crossing number the skein equation and descending values recover every directed expansion from $L$, by step 2.1 and induction. For either direction, if the curl of sign $\epsilon \in \{1,-1\}$ is first encountered over, its terminal factor is $b^{\epsilon }$ and every smoothing outside the curl retains it, giving $b^{\epsilon }L(D_0)$, where $D_0$ has the curl removed. If first encountered under, switch it; the switched curl has factor $b^{-\epsilon }$ by the preceding case. Its two smoothings are $D_0$ and $D_0\bigsqcup ◯$, so the transport gives $(-b^{-\epsilon }+t(1+h))L(D_0)=b^{\epsilon }L(D_0)$. Here the lower-crossing circle rule applies, and $t(1+h)=b+b^{-1}$. This proves the curl rule for both directed summands and hence their average; the component-lift products give the same calculation for links. An extra disjoint circle contributes $h$ in every terminal product and every smoothing term, so induction gives the circle rule as well. A descending arc can be pulled away without adding crossings by moves III and cancelling pairs by moves II, leaving curls. When used in step 4.1 these reductions concern diagrams with fewer than $N$ crossings, so they use only the induction hypothesis.

For move II, switching both crossings of a cancellable bigon leaves the evaluation unchanged. Indeed, expand at one crossing and then the other: the switched-diagram terms have positive combined sign, and the two smoothing sums cancel. The smoothings in which the bigon opens are the same diagram on either expansion; the other two smoothings each produce the same curl of the same sign on the same strand, so their values agree by step 6.1. Choose the basepoint and component lift so that the overpassing local arc is traversed or lifted first. Neither crossing is then switched in that local portion; all switches occur outside it, each smoothing term has fewer crossings, and the terminal descending or layered product loses the bigon without changing its writhe or factors. Induction gives invariance. If the chosen lift instead meets the underpassing arc, use the just-proved simultaneous switching identity to make it the overpassing arc; the same argument applies. This covers a bigon on one component and on two distinct components, and proves both directions of move II.

For move III, label the three local arcs by their height order. First give them the height order in which they are first encountered in a chosen traversal or component-lift expansion. There are then no local switches, and corresponding switches outside the disk produce identical terminal values and lower-crossing smoothing terms related by the same move, giving invariance by induction. The six possible height orders are connected by exchanging adjacent heights. Such an exchange switches one corresponding crossing on both sides of the move. Apply the sum-skein equation at those crossings: each sum of the old and switched evaluations is $t$ times a pair of smoothing evaluations. The two pairs agree. One smoothing cancels the two remaining crossings by move II or identifies them by planar isotopy; the other slides a cup past them. Here is an endpoint description of the cup slide, independent of any BMW presentation. In the following diagram words the factors are stacked from top to bottom in their written left-to-right order. In a three-strand rectangle let $e_j$ denote the actual crossing-free cup and cap joining adjacent positions $j,j+1$, and $x_j$ the actual crossing at those positions. The diagrams $e_1{x}_2{x}_1$ and $x_2{x}_1{e}_2$ both reduce to the diagram with upper cup $(1,2)$, lower cap $(2^{\mathrm{\prime }},3^{\mathrm{\prime }})$, and an untwisted strand from $3$ to $1^{\mathrm{\prime }}$. Slide the arc attached to the cup across the two crossing disks; the two successive over/under encounters then bound a bigon and cancel by move II. Likewise $x_1{x}_2{e}_1$ and $e_2{x}_1{x}_2$ reduce to upper cup $(2,3)$, lower cap $(1^{\mathrm{\prime }},2^{\mathrm{\prime }})$, and the strand from $1$ to $3^{\mathrm{\prime }}$. Reflecting the crossing data gives the same cancellations for $x_1^{-1},x_2^{-1}$. These are identities of the smoothed diagrams by planar deformation and move II, concerning at most $N-1$ crossings, rather than assumed identities in an abstract algebra. Thus the equality of old evaluations is equivalent to equality of switched evaluations. The adjacent-height exchanges reach every type III configuration, proving move III invariance for every traversal, component choice and outside diagram. Consequently a descending knot $U$ now has $L(U)=b^{w(U)}$: pull its descending arc away by nonincreasing moves II and III until only curls remain, then use step 6.1. This uses the just-proved invariance at $N$, not a direct prescription in step 3.1. Finally expand the established skein equation along the switches for any directed point $p$. The terminal value is now $L(U_p)=b^{w(U_p)}$, so step 2.1 gives $Q(D,p)=L(D)$, for both directions. Thus the average agrees with every directed expansion, including on descending knots. This closes the simultaneous crossing-number induction over $F$.

Extend the field by $i^2=-1$, replace $b$ by $ia$ and $t$ by $-is$, and for a nonempty unoriented diagram $D$ temporarily choose orientations and put $D^{\mathrm{♭}}(D)=-i^{-w(D)}(-1{)}^{c(D)}L(D;ia,-is)$, where $c(D)$ is its number of components. This does not depend on orientations. Reversing one whole component changes writhe by minus twice its mutual crossing-sign sum with all the other components, hence by a multiple of four by [F2]. The phase is unchanged. The same formula is unchanged by moves II and III because writhe and component number are unchanged by these moves and $L$ is invariant. A positive curl multiplies the phase by $i^{-1}$ and $L$ by $ia$, giving factor $a$; a negative curl gives $i(ia{)}^{-1}=a^{-1}$. Its value on one trivial circle is $1$.

We verify the difference-skein signs rather than importing the Dubrovnik conversion. If the chosen global orientations make the crossing in $L_{+}$ negative, interchange $L_{+},L_{-}$ and also $L_0,L_{\mathrm{\infty }}$; proving the equation after both interchanges is equivalent to the original equation. We may therefore take this crossing positive and write $w=w(L_{+})$ and $c=c(L_{+})$. Switching gives writhe $w-2$ and the same component number, so $D^{\mathrm{♭}}(L_{+})-D^{\mathrm{♭}}(L_{-})=-i^{-w}(-1{)}^c(L(L_{+})+L(L_{-}))$. The sum-skein equation gives the common factor $(-i^{-w}(-1{)}^c)(-is)$ times the sum of the two smoothing evaluations. For the oriented smoothing $L_0$, writhe is $w-1$ and component parity reverses; its phase is therefore exactly $(-i^{-w}(-1{)}^c)(-i)$. The other smoothing has the opposite phase. If the original crossing is between distinct components, orient it after first reversing one entire incident component: that changes writhe by a multiple of four, changes the removed crossing to negative, and merging gives writhe congruent to $w+1$ modulo four and reverses component parity. If the crossing is a self-crossing, the oriented smoothing splits its component into two; to orient the other smoothing reverse one of these portions. Their mutual crossing-sign sum is even by [F2], so this reversal changes the remaining writhe by a multiple of four. The other smoothing then has writhe congruent to $w-1$ and retains component parity. In both cases its phase is the negative of the phase for $L_0$. Substitution proves $D^{\mathrm{♭}}(L_{+})-D^{\mathrm{♭}}(L_{-})=s(D^{\mathrm{♭}}(L_0)-D^{\mathrm{♭}}(L_{\mathrm{\infty }}))$.

Under the substitution, the original extra-circle factor becomes $h=((ia)+(ia{)}^{-1})\mathrm{/}(-is)-1=-(a-a^{-1})\mathrm{/}s-1$. The extra factor $(-1)$ from component number changes this to $1+(a-a^{-1})\mathrm{/}s$. Let $d$ denote this element and put $P(D)=d{D}^{\mathrm{♭}}(D)$ for nonempty $D$, and $P(\mathrm{\varnothing })=1$. It has the desired curl, skein and loop rules, including the first circle. We justify every factorization fact needed for the coefficient injection. A polynomial ring $B[x]$ over a UFD $B$ is a UFD: factor the content using factorization in $B$, factor its primitive part over $\mathrm{Frac}(B)[x]$ by [F5], and clear denominators and contents of the factors. Their product is primitive by [F4], so comparison of contents makes the remaining scalar a unit of $B$. Irreducibility of the primitive factors is [F4]; comparing contents and the fraction-field factorizations proves uniqueness. Starting from $\mathbb{Z}$ by [F3] and iterating proves $\mathbb{Z}[a,d,s]$ is a UFD. Inverting $a$ and $d$ preserves this property: factor each numerator, discard factors associated to $a$ or $d$ since they become units, and compare two such factorizations by clearing powers of $a,d$ and using uniqueness in the polynomial ring. A remaining prime stays prime because divisibility after clearing those powers is ordinary divisibility by a prime different from $a,d$. Thus $B=\mathbb{Z}[a^{\pm 1},d^{\pm 1}]$ and $B[s]$ are UFDs. The element $d-1$ is prime in $B$, since substituting $d=1$ has kernel $(d-1)$ and domain quotient $\mathbb{Z}[a^{\pm 1}]$. It does not divide $a-a^{-1}$, whose value at $d=1$ is nonzero. Hence $f=s(d-1)-(a-a^{-1})$ is primitive over $B$, and it is irreducible over $\mathrm{Frac}(B)$ because it is linear. By [F4] it is irreducible and therefore prime in $B[s]$. Its quotient $A$ is a domain. Its localization at the nonzero element $s$ is injective: $s^{k}u=0$ in a domain implies $u=0$. Eliminating $d$ there gives a subring of $\mathbb{Q}(i)(a,s)$ with $d=1+(a-a^{-1})\mathrm{/}s$, and thus the claimed injection. No choice principle is needed in these finite factorizations.

In fact every value $P(D)$ belongs to $A$. Induct on crossing number, and at fixed crossing number on the incorrect crossings of a chosen traversal: the difference-skein rule changes an incorrect crossing to a correct one plus two diagrams with fewer crossings, with coefficients $\pm s$. A fully descending diagram is a disjoint union of unknots with framing curls, so its value is a product of powers of $a^{\pm 1}$ and $d$. The induction therefore uses only coefficients in $A$. The same reduction proves uniqueness of a function satisfying the stated rules. Extend $P$ linearly to the free $A$-module on diagrams. Its kernel contains the defining skein, curl and loop relations and regular-isotopy differences, so it descends to their quotient; its value on the empty diagram is $1$. The descending reduction shows the empty diagram spans this quotient, and the descended function is a left inverse to the scalar map $A\to K{T}_0(A)$. This map is consequently injective as well as surjective, proving the freeness assertion.