Kauffman and Dubrovnik polynomials from BMW algebras

Fix distinct components $K_r,K_s$ and let their projected shadows be ${\gamma }_r,{\gamma }_s$. Smooth each of the finitely many self-crossings of ${\gamma }_r$ inside a small disk disjoint from ${\gamma }_s$; the disk exists because there are no triple points, and the smoothing leaves every mixed intersection with ${\gamma }_s$ unchanged. After finitely many such smoothings, ${\gamma }_r$ is a finite disjoint union of embedded circles. As ${\gamma }_s$ traverses any embedded circle, every transverse intersection changes whether it is inside or outside that circle, and the closed traversal returns to its initial region. Hence the original number of mixed intersections of $K_r,K_s$ is even. Each determinant crossing sign is $+1$ or $-1$, both congruent to $1$ modulo $2$, so the signed mixed-crossing sum $S_{rs}$ is even. Summing over $s\mathrm{\ne }r$ shows that the total mixed-crossing sum $S_r$ involving $K_r$ is even. Reversing $K_r$ leaves its self-crossing signs fixed by [F2] and negates its mixed crossing signs, so $w$ changes by $-2S_r\in 4\mathbb{Z}$. Thus $i^w$ is unchanged; $c$ and the unoriented value $D$ are unchanged as well. Therefore the formula defining $L$ is independent of reversing $K_r$, and repeating for every component proves orientation independence.

For a positive curl, $D$ gains a factor $a$ and $w$ increases by $1$; substituting $a_D=-iA$ and $z_D=iZ$ in [F5] shows that $L$ gains the factor $i{a}_D=A$. For a negative curl, $D$ gains $a^{-1}$ and $w$ decreases by $1$, so $L$ gains $i^{-1}{a}_D^{-1}=A^{-1}$. Adding a disjoint trivial circle increases $c$ by $1$ and multiplies $D$ by $d$, so the factor $(-1{)}^c$ makes $L$ acquire the factor $-d$. Set ${\delta }_L=-d$. Under $a_D=-iA$ and $z_D=iZ$, the coefficient relation $z_D(d-1)=a_D-a_D^{-1}$ becomes $Z(1-d)=A+A^{-1}$, hence $Z({\delta }_L+1)=A+A^{-1}$. This gives the Kauffman circle rule using the universal relation without inverting $Z$.

Consider a positive crossing and write $w=w(K_{+})$ and $c=c(K_{+})$. Switching it gives $w(K_{-})=w-2$ and leaves $c$ fixed. Each smoothing deletes that crossing, so before any orientation adjustment its writhe is $w-1$. The orientation-preserving smoothing $K_0$ has $c+1$ components at a self-crossing and $c-1$ at a mixed crossing. The other smoothing $K_{\mathrm{\infty }}$ has $c$ components at a self-crossing and $c-1$ at a mixed crossing. To orient $K_{\mathrm{\infty }}$ at a self-crossing, first use the orientation-preserving smoothing to split the original component into two closed components, then join their arcs by the other smoothing and reverse one of the two orientations. This changes signs at exactly the crossings between that component and the other split component or any remaining component; the pairwise sums are even by step 1.1, so $w(K_{\mathrm{\infty }})\equiv w-1(\mathrm{m}\mathrm{o}\mathrm{d}4)$. At a mixed crossing, reverse one whole incident component in the original diagram. Its writhe changes by minus twice a mixed-crossing sum, which is divisible by $4$ by step 1.1; this reversal changes the displayed crossing from positive to negative and makes $K_{\mathrm{\infty }}$ the orientation-preserving smoothing. Removing that negative crossing raises writhe by $1$, so $w(K_{\mathrm{\infty }})\equiv w+1(\mathrm{m}\mathrm{o}\mathrm{d}4)$. By [F5], the phase $-i^w(-1{)}^c$ multiplying the Dubrovnik value changes by the factor $i$ on $K_0$ and by $-i$ on $K_{\mathrm{\infty }}$. It also changes sign when the crossing is switched, since $w(K_{-})=w-2$. Therefore the Dubrovnik difference skein in [F1] becomes the Kauffman sum skein $L_{+}+L_{-}=Z(L_0+L_{\mathrm{\infty }})$ after substituting $a=-iA$ and $z=iZ$, equivalently $A=ia$ and $Z=-iz$. The calculation uses the positive member $K_{+}$ of the crossing pair; since the resulting sum-skein is symmetric in $K_{+}$ and $K_{-}$, it applies whether the input diagram displays the positive or negative member.

By [F1], $D$ is unchanged under planar isotopy and Reidemeister II and III. The determinant writhe is unchanged under II and III by [F2, F4], so the formula in [F5] gives the corresponding regular-isotopy rules for $L$. The positive and negative curl factors are those computed in step 1.2.

For a positive type-I move, $A^{-(w+1)}AL=A^{-w}L;$ for a negative type-I move, $A^{-(w-1)}{A}^{-1}L=A^{-w}L\mathrm{.}$ Moves II and III leave $w$ unchanged, hence also preserve $F$. Therefore $F$ is unchanged by every listed move, while the conversion identity is the defining relation in [F5].

The relation defining $T$ is $q-q^{-1}=z$, hence $q^2-zq-1=0$ and $q^{-1}=q-z$. Therefore $T\cong S[q]\mathrm{/}(q^2-zq-1)$. Division by this monic quadratic gives a unique remainder $u+vq$ with $u,v\in S$, so $T$ is free over $S$ on $1,q$ and the constant inclusion is injective.

With $\rho =a$, $\delta =d$ and $z_B=z_D=z$, the BMW coefficient relation becomes exactly $z(d-1)=a-a^{-1}$, the defining relation of $S$. Thus every coefficient assignment is compatible. No skein variable is inverted.

In the fixed chart the downward positive Artin crossing has over-strand from top right to bottom left. Its determinant sign is positive, and its orientation-preserving smoothing is the through pairing $D_0$; the other smoothing is the cup-cap $D_{\mathrm{\infty }}$. Consequently the BMW relation maps to $G_i-G_i^{-1}=z_B(1-E_i)=z_D(D_0-D_{\mathrm{\infty }})$, as required. The bracket's names $A$ and $B$ remain those of [F3]; in this chart the Dubrovnik names $D_0,D_{\mathrm{\infty }}$ are respectively $A,B$.

Composing the positive crossing with a cup-cap reverses the direction of exactly one incident branch when the resulting self-crossing is oriented along its strand. The determinant sign is therefore negative and its curl coefficient is $a^{-1}={\rho }^{-1}$, agreeing with $g_i{e}_i=e_i{g}_i={\rho }^{-1}{e}_i$. The inverse crossing gives $a=\rho$. In $E_i{G}_{i\pm 1}{E}_i$ the surviving self-curl has the opposite sign and coefficient $a=\rho$, agreeing with the adjacent untwisting relation. A double cup-cap creates one trivial circle, so its coefficient is $d=\delta$. These checks distinguish a curl after bending from a positive braid closure, whose strands retain their downward orientations and whose curl coefficient is $a$.

The inverse of the positive crossing is its negative crossing: stacking them creates a type-II bigon, which cancels by regular isotopy. For $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$, the supports of $g_i$ and $g_j$ are disjoint, so they commute; the supports of $g_i$ and $e_j$ are disjoint, so those commute as well; and two distant cup-cap tangles commute by planar isotopy. The adjacent braid relation is the type-III move on three monotone strands, and the two possible orders of stacking the cup-cap tangles in $e_i{e}_{i\pm 1}{e}_i=e_i$ reduce by planar isotopy to the same single cup and cap. These checks preserve the ordered boundary endpoints.

The BMW skein relation maps to $G_i-G_i^{-1}=z_B(1-E_i)=z_D(D_0-D_{\mathrm{\infty }})$: here $z_B=z_D$, the identity is the through smoothing $D_0$, and $E_i$ is the cup-cap smoothing $D_{\mathrm{\infty }}$. In the fixed southwest-to-northeast over-strand chart, write the projected oriented tangent coordinates as $(t,x)$ with $t$ increasing downward. The over and under tangents of the downward crossing can be taken as $(1,-1)$ and $(1,1)$; their determinant is $2>0$ by [F4], and its oriented smoothing is through. The bracket names in [F3] are $A,B$ for these two pairings, respectively.

When the positive crossing is followed by a cup-cap, the two branches belong to the same bent strand. Relative to their downward braid orientations exactly one branch reverses, so its determinant self-crossing sign is negative. Thus $G_i{E}_i=E_i{G}_i=a^{-1}{E}_i={\rho }^{-1}{E}_i$; the negative crossing gives the positive self-curl and coefficient $a=\rho$. For $E_i{G}_{i+1}{E}_i$, follow the surviving strand from the top position $i+2$: it traverses the crossing from position $i+2$ to $i+1$, turns along the lower cup to position $i$, returns up the uncrossed strand at position $i$ to the upper cap, and traverses the other crossing branch from $i+1$ to $i+2$. Both encounters are downward in the crossing disk, so the self-crossing has positive determinant sign and gives $a{E}_i=\rho E_i$. For $E_i{G}_{i-1}{E}_i$, start at the top position $i-1$: go down the crossing's left branch, around the lower cup and up position $i+1$, then around the upper cap and down the crossing's right branch. Again both crossing branches are traversed downward, giving the same positive curl. Stacking $E_i$ twice creates a disjoint trivial circle and leaves $E_i$, so $E_i^2=d{E}_i=\delta E_i$. The self-curl sign after a cup is distinct from the positive self-curl formed by closing a positive two-strand braid.

For the remaining adjacent relations, use three consecutive positions $1,2,3$. In $G_1{G}_2{E}_1$, the upper cup joins top endpoints $2,3$, the lower cap joins bottom endpoints $1,2$, and the remaining strand joins top $1$ to bottom $3$. That strand passes under both arms of the upper cup. Slide the cup toward the top: its two crossings with that strand form a type-II pair and cancel, leaving the planar tangle $E_2{E}_1$. Hence $G_1{G}_2{E}_1=E_2{E}_1$. To reverse stacking, reflect top to bottom AND interchange over and under at every crossing. This is a half-turn in three-space about a horizontal axis in the projection plane; it preserves regular isotopy, fixes $G_j$ and $E_j$, and reverses products. It therefore gives $E_1{G}_2{G}_1=E_1{E}_2$. Likewise, left-to-right reflection together with over/under interchange fixes crossing chirality and sends $G_1,G_2$ to $G_2,G_1$ and $E_1,E_2$ to $E_2,E_1$. Applying it to the two identities gives $G_2{G}_1{E}_2=E_1{E}_2$ and $E_2{G}_1{G}_2=E_2{E}_1$. Translating the three-position rectangle gives both signs of the adjacent index in [F1]. Neither operation is a lone planar reflection, which would invert the crossings. Together with steps 1.1, 1.2 and 2.1, these verify every defining relation for $n\ge 2$.

For $n=0$ and $n=1$, the BMW algebra is the scalar algebra $T$ by [F1], with no crossing or cup-cap generators. Sending a scalar to the same scalar times the empty tangle or the straight tangle is unital and $T$-linear, so it defines the stated boundary homomorphism.

If a tangle diagram has no crossings, every closed component is a disjoint trivial circle and the remaining arcs form a crossingless connector; removing each circle by [F2] leaves a scalar multiple of its standard matching tangle. This is the base case for induction on the lexicographically ordered pair (crossing count, bad-crossing count).

For a diagram with crossings, choose a finite traversal order of its components and orient each arc from its smaller boundary label to its larger one; for closed components use any finite encoded orientation and basepoint. A crossing is bad when its first encountered branch is under. If there is a bad crossing, the skein relation expresses the diagram as its switched diagram plus two smoothing terms. Switching preserves the crossing count and lowers the bad-crossing count by one; each smoothing lowers the crossing count by one. Thus every term is lower in the stated lexicographic order, and the induction hypothesis expresses each as a linear combination of standard descending representatives. The choices here are finite choices made from the encoded diagram, not an appeal to an infinite choice principle.

If there is no bad crossing, the diagram is totally descending. Earlier components pass over later ones, so separate their interiors into successive height layers, keeping the boundary endpoints fixed. Within one layer, the first-encounter-over condition lets the already traversed part of a component be pulled above its untraversed part: it is an unknotted arc relative to its endpoints, or an unknotted closed curve. Straighten each arc to a simple planar arc in its layer and each closed curve to a circle by ambient isotopy. This simplification may use type-I moves on open arcs as well as on closed curves; it is not asserted to be regular isotopy. In a finite generic diagram movie, planar isotopy and moves II and III preserve the skein class, while each deletion of a positive or negative curl contributes $a$ or $a^{-1}$ by [F2] (an insertion contributes the inverse factor). Thus the original diagram equals $a^r$ times the resulting layered diagram for some integer $r$. Separate and delete its $c$ trivial closed circles by the free-circle rule, leaving $a^r{d}^c$ times a loop-free, self-crossing-free descending diagram with the same open connector.

The base case, induction step, and descending terminal case together show by well-founded induction that every diagram reduces to a linear combination of standard representatives. For each connector choose simple arcs in the rectangle and put their interiors in height layers ordered by their initial endpoints; a generic projection gives a loop-free, self-crossing-free descending representative. Any two such representatives with that connector can be put in the same ordered layers. In a disk, two simple arcs with the same endpoints are isotopic relative to those endpoints; apply this within each layer, leaving the other layers fixed and retaining the boundary collars. These isotopies keep each arc simple and its blackboard framing untwisted, so their generic projections use only planar isotopy and moves II and III, with no curl factor. Thus the loop-free, self-crossing-free representatives are unique up to regular isotopy. There are $(2n-1)!!$ perfect matchings on $2n$ endpoints, obtained by pairing the first endpoint in $2n-1$ ways and repeating; for $n=0$ the empty matching is unique and for $n=1$ the straight strand is unique.

Define $f:R\to T$ by $q\mapsto q$, $r\mapsto a$, $\mathrm{\Delta }\mapsto d$. Its defining relation maps to zero by [F3], and its unit generators map to units. Conversely define $h:T\to R$ by $a\mapsto r$, $d\mapsto \mathrm{\Delta }$, $q\mapsto q$, and $z\mapsto q-q^{-1}$. The relation of $S$ maps to the relation of $R$, and $q-q^{-1}-z$ maps to zero. On each displayed ring generator, both composites are the identity. Thus $f,h$ are inverse ring isomorphisms, without inverting $z$ or requiring a flatness argument.

Transport the source presentation along $f$. Send each $g_i^{\pm 1}$ and $e_i$ to the identically named generator over $T$. The braid, commutation and cup-slide relations are literally identical, while the remaining coefficients become $q-q^{-1}$, $a^{-1}$, $a$, and $d$, exactly as in [F2] and [F3]. The inverse coefficient map $h$ gives an inverse map of presented algebras, because it fixes the generator names and reverses each coefficient substitution. This verifies the source comparison without assuming that either tangle realization is injective.

Transport the basis in [F1] through this algebra isomorphism. If $b_m$ denotes the transported basis element with matching index $m$, every source element has one expression ${\sum }_m{t}_m{b}_m$ with $t_m\in T$, and these expressions are unique. This is the finite source normal form supplied by the already proved local universal theorem, whose ideal/shuffle proof concerns the abstract algebra rather than a dimension inference from a tangle quotient. A perfect matching pairs the first endpoint in $2n-1$ ways and then matches the remaining endpoints; hence there are $N_n=(2n-1)!!$ indices, and one empty index at $n=0$. The scalar definitions give the same rank-one conclusion at $n=1$.

Define the ring homomorphism $\sigma :S\to \mathbb{Z}[x^{\pm 1}]$ by $a\mapsto 1$, $z\mapsto 0$, and $d\mapsto x$; the defining relation maps to $0=0$, so the map is well defined. For any closed diagram in an entry of $M_n$, the skein rule in [F1] becomes $P_{+}=P_{-}$ under $\sigma$. Switch finitely many crossings until the closed diagram is totally descending; these switches do not change its specialized value. Pulling the first traversed strands above the later ones unties every component, and the specialized curl factor is $1$. The result is a disjoint union of $\mathrm{\ell }(m,m^{\mathrm{\prime }})$ trivial circles, where $\mathrm{\ell }(m,m^{\mathrm{\prime }})$ is the number of connected components in the union of the two endpoint matchings. Hence $\sigma (M_n[m,m^{\mathrm{\prime }}])=x^{\mathrm{\ell }(m,m^{\mathrm{\prime }})}$. The closure is a closed diagram at every $n$, including the empty diagram for $n=0$.

If $m=m^{\mathrm{\prime }}$, their union consists of $n$ doubled edges, so $\mathrm{\ell }(m,m)=n$. If $m\mathrm{\ne }{m}^{\mathrm{\prime }}$, each connected component of their union is an alternating even cycle, and at least one has length at least four; therefore $\mathrm{\ell }(m,m^{\mathrm{\prime }})\le n-1$. Let $N=(2n-1)!!$ be the number of matchings, with $N=1$ for $n=0$. In the determinant expansion of $\sigma (M_n)$, the identity permutation contributes $x^{nN}$. Every nonidentity permutation has at least one off-diagonal entry of degree at most $n-1$, while every other entry has degree at most $n$, so its term has degree at most $nN-1$. The identity term is therefore the unique leading term with coefficient $1$, and $\det (\sigma (M_n))\mathrm{\ne }0$. Thus $\det M_n\mathrm{\ne }0$ in $S$. For $n=0$ the matrix is $[1]$, so the same conclusion holds.

Since $S$ is a domain by [F1], multiplication by $\det M_n\mathrm{\ne }0$ is injective on $S$. By [F2], every $t\in T$ has a unique form $u+vq$ with $u,v\in S$. If $(\det M_n)t=0$, uniqueness gives $(\det M_n)u=(\det M_n)v=0$ in $S$, hence $u=v=0$; so $\det M_n$ is a non-zero-divisor in $T$.

Suppose ${\sum }_m{t}_m{T}_m=0$ in the tangle skein algebra over $T$. Close this relation against each reflected standard matching and apply the scalar evaluation; closure respects the skein, curl, and circle relations, so the resulting equations are $M_n^{\mathsf{T}}t=0$. Multiplication by the adjugate gives $(\det M_n)t=0$. Step 3.1 shows each coordinate $t_m$ is zero. Therefore the standard tangles are linearly independent over $T$. The same argument includes $n=0$ and $n=1$, whose matrices are respectively $[1]$ and $[d]$.

By [F1], the tangle class ${\mathrm{\Phi }}_n(x)$ depends only on the BMW algebra element $x$, so closing it and applying $P$ defines a $T$-linear scalar. For $x,y\in BM{W}_n$, the two closed tangles obtained by closing the stacked products ${\mathrm{\Phi }}_n(x){\mathrm{\Phi }}_n(y)$ and ${\mathrm{\Phi }}_n(y){\mathrm{\Phi }}_n(x)$ differ only by moving the chosen cut between the factors once around the closed tangle. This is an isotopy of the closed diagram and preserves its framing; therefore $P(\widehat{{\mathrm{\Phi }}_n(xy)})=P(\widehat{{\mathrm{\Phi }}_n(yx)})$, proving cyclicity after multiplication by $d^{-n}$.

Under tower inclusion, the new straight strand closes to a disjoint trivial circle. By [F2], $P(\widehat{{\mathrm{\Phi }}_{n+1}({\iota }_n(x))})=dP(\widehat{{\mathrm{\Phi }}_n(x)})$. Thus $d^{-(n+1)}dP(\widehat{{\mathrm{\Phi }}_n(x)})={\tau }_n(x)$, which proves tower compatibility.

Closing ${\iota }_n(x)g_n$ turns the added positive crossing into a positive curl on the last closure arc of $\widehat{{\mathrm{\Phi }}_n(x)}$; closing ${\iota }_n(x)g_n^{-1}$ gives a negative curl on the same arc. By [F2] and [F3], the respective scalar factors are $a$ and $a^{-1}$. Normalization by $d^{-(n+1)}$ gives $a{d}^{-1}{\tau }_n(x)$ and $a^{-1}{d}^{-1}{\tau }_n(x)$.

In the closure of ${\iota }_n(x)e_n$, the added cup and cap and the two added closure arcs form a detour joining the top and bottom ends of the last strand of ${\mathrm{\Phi }}_n(x)$. Following that component through the cup, cap, and closure arcs gives exactly the same connection as the untwisted closure arc for ${\mathrm{\Phi }}_n(x)$; the other components and crossings are unchanged. Pulling this detour into the closing collar is a planar isotopy, so the unnormalized evaluations agree. The normalization changes from $d^{-n}$ to $d^{-(n+1)}$, giving $d^{-1}{\tau }_n(x)$.

By [F2] and [F3], the tangle skein algebra has a basis indexed by the perfect matchings of the $2n$ endpoints. There are $(2n-1)!!$ such matchings: choose the partner of the first endpoint in $2n-1$ ways and apply the same count to the remaining $2n-2$ endpoints, with one empty matching at $n=0$. The unique matching at $n=1$ is the straight strand. Hence the tangle algebra has the stated rank and boundary bases.

Fix a standard matching connector and let $k$ be the number of top-top pairs. The number of bottom-bottom pairs is also $k$, because the remaining endpoints must be paired top-to-bottom. If $k=0$, the connector is a permutation diagram, so its standard descending representative is a braid diagram; reading its finitely many crossing levels expresses it as a word in the crossing images $G_i^{\pm 1}={\mathrm{\Phi }}_n(g_i^{\pm 1})$. If $k>0$, choose a permutation braid $B_1$ that moves the $k$ top pairs to adjacent positions $(1,2),(3,4),\dots ,(2k-1,2k)$, and a permutation braid $B_2$ that routes the corresponding bottom pairs to those positions. The remaining through-strands form a braid $B$ on $n-2k$ strands. With the shifted copy of $B$ placed beside the cup-cap block, the tangle has factorization $B_1[(E_1{E}_3\cdots E_{2k-1})\otimes B]B_2\mathrm{.}$ Each $B_i$ and $B$ is a finite word in the $G_j^{\pm 1}$, each $E_j={\mathrm{\Phi }}_n(e_j)$ is a cup-cap generator, and the shifted factors use indices among $1,\dots ,n-1$. The permutation braid pieces are obtained by reading the existing upper and lower layers of $T_m$; moving the horizontal pairs to adjacent positions is a boundary-fixing braid isotopy, so it preserves the crossing data and total descending order. The factorization therefore has the same connector and is regularly isotopic to the standard representative. Uniqueness for a fixed connector from the spanning proof identifies its class with that representative. Thus every basis tangle is in the image of ${\mathrm{\Phi }}_n$, proving surjectivity by [F2].

Let $N$ be the number of matchings and choose the $N$ source basis elements supplied by [F5]. They define a surjection $u:T^N\to BM{W}_n(T)$, and step 2.1 gives the surjection ${\mathrm{\Phi }}_n:BM{W}_n(T)\to {\mathcal{K}\mathcal{T}}_n(T)$. Identify the target with $T^N$ using step 1.1. The composite $A={\mathrm{\Phi }}_{n}u:T^N\to T^N$ is a surjective square matrix over the commutative ring $T$. Choose a preimage of each of its finitely many standard basis vectors to obtain $B$ with $AB=I$. Then $\det (A)\det (B)=1$, so the adjugate formula makes $A$ invertible. If $x\in \ker {\mathrm{\Phi }}_n$, write $x=u(v)$; then $Av=0$, hence $v=0$ and $x=0$. Thus ${\mathrm{\Phi }}_n$ is injective as well as surjective. Its inverse transports the standard matching basis to the abstract BMW algebra, proving all the stated freeness and rank assertions, including $n=0,1$. No implication from target freeness alone is used: the finite source spanning bound [F5] is essential.