BMW basis by descending Kauffman tangles

Work first over ${\mathrm{\Lambda }}_0$ and write $s=-z_0$, so $s(\delta -1)=\lambda -{\lambda }^{-1}$, where $z_0$ denotes the skein variable called $z_0$ in the statement. The coefficient ring is the domain $A$ of [F3], under $a=\lambda ,s=-z_0,d=\delta$; the coefficient injection and its factorization arguments were proved in that supplier. The scalar evaluation of Scalar evaluation of framed Kauffman link diagrams, with $a=\lambda$, $s=-z_0$ and $d=\delta$, identifies the closed-diagram skein module with ${\mathrm{\Lambda }}_0$. Its positive crossing is the usual positive braid crossing, with identity smoothing first. A positive braid closed as a two-strand braid gives a positive self-curl with factor $\lambda$; if the same crossing is followed by a cup-cap, the cup forces opposite strand directions, giving a negative self-curl with factor ${\lambda }^{-1}$. Thus the helper's framing rules give exactly $G_i{E}_i={\lambda }^{-1}{E}_i$ and the displayed source convention.

Choose a direction, order and initial point on every closed component as well as the standard directions and order of the boundary strands. A crossing is incorrect if first encountered underneath. The skein relation replaces an incorrect crossing by its reversal and two diagrams with one fewer crossing. Induct first on the total crossing number and, for the unchanged-crossing term, on incorrect crossings of this fixed traversal. Reversal changes no component or traversal and removes exactly that incorrect crossing. Smoothing terms use the outer induction, with newly chosen traversals if needed. Hence the quotient is spanned by descending diagrams. A descending component is unknotted; successively lifting its traversed portion above its remaining portion and pulling that initial arc away leaves only framing curls. Earlier components lie above later ones, so closed components can be separated and removed. Removing curls multiplies by powers of $\lambda$, and removing circles by $\delta$. The surviving diagrams are descending with no closed components and no self-crossings.

For a fixed connector, draw simple arcs in separate height layers ordered by their initial endpoints and join them to the fixed boundary intervals. This gives a descending, loop-free, self-crossing-free representative $T_c$. Any two such representatives can be put in those same layers: each earlier strand is over every later strand, so changing its height introduces no intersection. Within its layer, a simple arc with specified endpoints in a disk deforms to any other such arc, while other arcs are in different layers. Its framing has no self-curl. These deformations give framed isotopy, equivalently regular isotopy of the diagrams. Thus the class $T_c$ is unique. Step 2.1 proves that these classes span.

Close all boundary strands outside the rectangle without new crossings and evaluate by step 1.1, multiplying by ${\delta }^{-n}$; call the resulting linear functional $ϵ$. Let $\ast$ reflect a diagram top to bottom. For any choice of diagrams $U_c$ with connector $c$, form $B=(ϵ(U_c{U}_d^{\ast }){)}_{c,d}$. Specialize $\lambda =1,z_0=0$ to $\mathbb{Z}[{\delta }^{\pm 1}]$. Switching crossings and deleting curls then have no effect, so the entry becomes ${\delta }^{k_c+k_d+\mathrm{\ell }(c,d)-n}$, where $k_c$ is the number of closed components of $U_c$ and $\mathrm{\ell }(c,d)$ counts the alternating cycles in the union of the two matching edge sets on the same $2n$ vertices. This is at most $n$, and equals $n$ precisely when every cycle has two vertices, that is, $c=d$. Factor the units ${\delta }^{k_c}$ from row $c$ and ${\delta }^{k_d}$ from column $d$. The remaining matrix has diagonal $1$ and off-diagonal terms of strictly negative $\delta$-degree. The determinant of this remaining matrix has constant term $1$, because every nonidentity permutation uses an off-diagonal entry. Therefore the specialized determinant is nonzero. Therefore $\det B\mathrm{\ne }0$ in the domain ${\mathrm{\Lambda }}_0$. A relation among the $U_c$, paired with every $U_d^{\ast }$, has zero coefficient vector over $\mathrm{Frac}({\mathrm{\Lambda }}_0)$ and hence over ${\mathrm{\Lambda }}_0$. This proves independence of any representatives with distinct connectors, in particular the $T_c$.

The tangle quotient over any ${\mathrm{\Lambda }}_0$-algebra $S$ is the base change of its universal quotient. Indeed it is a cokernel of the map from the free module on all defining regular-isotopy, skein, curl and loop relations to the free module on all diagrams, and [F4] says tensor product preserves this cokernel. Thus $K{T}_n(S)=K{T}_n({\mathrm{\Lambda }}_0){\otimes }_{{\mathrm{\Lambda }}_0}S$ and the classes $T_c$ form an $S$-basis. The same base-change argument applies to the generators-and-relations presentation of BMW. It suffices to prove the universal BMW-to-tangle map injective. In the rest of the proof write $W_j$ for the universal BMW algebra, with $g_i-g_i^{-1}=s(1-e_i)$, $g_i{e}_i={\lambda }^{-1}{e}_i$, $e_i{g}_{i\pm 1}{e}_i=\lambda e_i$, and write $\varphi (g_i)=G_i$, $\varphi (e_i)=E_i$. The tangle generators obey these relations: inverse crossings cancel by II, three crossings give III, disjoint disks commute, a cup and cap satisfy the snake move, and the remaining cup slides and curls give the mixed and untwisting relations. This defines $\varphi$.

We give the required algebraic reductions before using tangle independence. For adjacent $i,j$, the defining relations and snake relation give $g_j{e}_i{g}_j=g_i^{-1}{e}_j{g}_i^{-1}$: first $g_j{e}_i=g_i^{-1}{e}_j{e}_i$, and $e_j{e}_i{g}_j=e_j{e}_i{e}_j{g}_i^{-1}=e_j{g}_i^{-1}$. They also give $g_j{e}_i{e}_j=g_i^{-1}{e}_j$ and its word reversal $e_j{e}_i{g}_j=e_j{g}_i^{-1}$, by multiplying $g_i{g}_j{e}_i=e_j{e}_i$ on the right by $e_j$ and using $e_j{e}_i{e}_j=e_j$. These are identities in the presentation, not consequences pulled back from diagrams. Finally $g_i^2=1+s{g}_i-s{\lambda }^{-1}{e}_i$, and $g_i^{-1}=g_i-s+s{e}_i$.

We prove $W_{m+1}=W_m\{1,g_m,e_m\}{W}_m$, with braces meaning linear span. Induct on $m$; it is immediate at $m=1$ from the quadratic reduction in step 6.1. To prove closure at the next rank, use the lower-rank expression for the factor between two occurrences of $g_m$ or $e_m$. Its outer $W_{m-1}$ factors commute with these two generators, so only $\chi \eta \psi$ need reduction, with $\chi ,\psi \in \{g_m,e_m\}$ and $\eta \in \{1,g_{m-1},e_{m-1}\}$. For $\eta =1$, quadratic, loop and untwisting relations suffice. For $\eta =g_{m-1}$ the four reductions are braid, $g_m{g}_{m-1}{e}_m=e_{m-1}{e}_m$, $e_m{g}_{m-1}{g}_m=e_m{e}_{m-1}$, and $e_m{g}_{m-1}{e}_m=\lambda e_m$. For $\eta =e_{m-1}$ they are $g_m{e}_{m-1}{g}_m=g_{m-1}^{-1}{e}_m{g}_{m-1}^{-1}$, $g_m{e}_{m-1}{e}_m=g_{m-1}^{-1}{e}_m$, its word reversal, and $e_m{e}_{m-1}{e}_m=e_m$. Every result contains at most one rank-$m$ generator between $W_m$ factors. This proves the asserted span.

Let $\alpha$ reverse words and let $S$ shift all indices by one; reflection $r_{2k}$ sends an index $i$ to $2k-i$. These preserve the relations. Moving distant generators and applying braid gives $g_i^{\pm 1}(g_m\cdots g_1)=(g_m\cdots g_1)g_{i+1}^{\pm 1}$ for $i<m$, and the analogous cup slide $e_i(g_m\cdots g_1)=(g_m\cdots g_1)e_{i+1}$. The latter uses $e_i{g}_{i+1}{g}_i=e_i{e}_{i+1}=g_{i+1}{g}_i{e}_{i+1}$ at the only adjacent positions. Reversing words gives the corresponding identities for $g_1\cdots g_m$. Another useful identity is $g_i{g}_{i+1}{e}_i{e}_{i+2}=g_{i+2}{g}_{i+1}{e}_i{e}_{i+2}$: both equal $e_{i+1}{e}_i{e}_{i+2}$ by the defining mixed relations and distant commutation.

The preceding span gives $e_m{W}_m{e}_m=W_{m-1}{e}_m$. Its middle factors reduce to $e_m^2=\delta e_m$, $e_m{g}_{m-1}{e}_m=\lambda e_m$, and $e_m{e}_{m-1}{e}_m=e_m$; reverse inclusion uses invertibility of $\delta$. It also gives $W_{m+1}{e}_m=W_m{e}_m$: for left stability under $g_m$, expand $W_m$ over $W_{m-1}$ and use $g_m{e}_m={\lambda }^{-1}{e}_m$, $g_m{g}_{m-1}{e}_m=e_{m-1}{e}_m$, and $g_m{e}_{m-1}{e}_m=g_{m-1}^{-1}{e}_m$. Stability under $e_m$ follows from the first identity, and $g_m^{-1}$ follows from the skein expression. Word reversal gives all right-handed identities.

Put $f_1=e_1$ and $f_k=(g_1\cdots g_{2k-2})f_{k-1}{e}_{2k-1}(g_{2k-2}\cdots g_1)$. Define $C_j=e_j{e}_{j+2}\cdots e_{2k-j}$ for $1\le j\le k$. Then $f_k=C_k{C}_{k-1}\cdots C_2{C}_1{C}_2\cdots C_k$. To verify this formula inductively, substitute the formula for $f_{k-1}$ into the recursion. Move $e_{2k-1}$ into the central odd-index product and move the other $e$-columns through the two $g$-strings using step 7.2. The remaining middle identity is $(g_1\cdots g_{2j})(e_1{e}_3\cdots e_{2j-1})=(e_2{e}_4\cdots e_{2j})(e_1{e}_3\cdots e_{2j-1})$: group its left side into $(g_1{g}_2{e}_1)(g_3{g}_4{e}_3)\cdots (g_{2j-1}{g}_{2j}{e}_{2j-1})$, replace each by $e_{2i}{e}_{2i-1}$, and commute the even terms left. Apply also its word reversal. The resulting columns are exactly the displayed diamond expression. Hence $\alpha (f_k)=r_{2k}(f_k)=f_k$, and its diagram $F_k=\varphi (f_k)$ is the crossing-free nested pairing $i\leftrightarrow 2k+1-i$ at top and bottom, by successively applying these cup slides.

The positive permutation braid $g_{\pi }$ is well defined for $\pi \in S_n$: use a reduced adjacent-transposition expression. Any two strands in its positive braid cross at most once, and conversely a positive braid with that property has crossing count equal to the inversion number of its permutation. In our positive left-over-right convention, increasing top-vertex order makes them descending: a pair that crosses once has its initially left strand still on the left just before that crossing, so that strand crosses over. Their uniqueness as braid words can also be checked by a monotone sorting procedure. Bring the strand with the rightmost target endpoint to its canonical rightmost route, sliding its crossings past crossings of other strands; a disjoint slide is distant commutation, and a slide past a crossing sharing one strand is the three-strand braid move. Since two strands cross at most once, the partners to be passed are exactly the inversion partners prescribed by the permutation, so this removes the same list of crossings in any diagram. Delete that strand and repeat. Thus both braid words reduce by distant and braid relations to the same recursively sorted wiring, giving equality in $W_n$, without pulling an arbitrary tangle isotopy back through $\varphi$. If $\pi$ is an $(a,b)$ shuffle, its first $a$ and last $b$ images increase. Every permutation factors uniquely as a shuffle followed by a block permutation, with lengths adding, by sorting its two blocks; hence its positive lift factors likewise. Finally if ${\pi }^{-1}(i+1)={\pi }^{-1}(i)+1=j+1$, then $g_i{g}_{\pi }=g_{\pi }{g}_j$ and $e_i{g}_{\pi }=g_{\pi }{e}_j$. For the cup identity, induct on the length of $\pi$. A leftmost letter distant from $i$ commutes with $e_i$; the other possibilities are $g_{i+1}{g}_i$ or $g_{i-1}{g}_i$, and the cup-slide equality in step 7.2 moves the cup past that pair. Removing it decreases length by two, completing the induction.

For $i<k$ one has $g_i^{\pm 1}{f}_k=g_{2k-i}^{\pm 1}{f}_k$ and $e_i{f}_k=e_{2k-i}{f}_k$. For the $g$ identity use induction on $k$. At $k=2$, $g_1{g}_2{e}_1{e}_3=g_3{g}_2{e}_1{e}_3$ by step 7.2. For $2\le i<k$, commute $g_i$ through the recursion to obtain $(g_1\cdots g_{2k-2})g_{i-1}{f}_{k-1}{e}_{2k-1}(g_{2k-2}\cdots g_1)$, replace $g_{i-1}$ by $g_{2k-i-1}$ using induction and commute back, obtaining $g_{2k-i}{f}_k$. For $i=1,k>2$, substitute the recursion once more for $f_{k-1}$; the braid relations rewrite the two initial strings as $(g_2\cdots g_{2k-3})(g_1\cdots g_{2k-2})$. The adjacent middle block $g_{2k-3}{g}_{2k-2}{e}_{2k-3}{e}_{2k-1}$ can be replaced by $g_{2k-1}{g}_{2k-2}{e}_{2k-3}{e}_{2k-1}$ using step 7.2. Move $g_{2k-1}$ left and the two $e$'s right, recovering $g_{2k-1}{f}_k$. Multiplying the resulting identity by the commuting inverses gives its inverse version. For the $e$ identity and $2\le i<k$ the same recursion applies with the cup-slide identity. The remaining $i=1$ case is an explicit column cancellation: at $k=2$ it is $e_1{e}_2{e}_1{e}_3{e}_2=e_1{e}_3{e}_2=e_3{f}_2$. At $k\ge 3$, commute $e_1$ past $C_k\cdots C_3$; then $e_1{C}_2{C}_1=(e_4{e}_6\cdots e_{2k-2})C_1$, by $e_1{e}_2{e}_1=e_1$. Moreover $C_3(e_4{e}_6\cdots e_{2k-2})C_1=C_1$: commute matching odd generators together and cancel successively $e_3{e}_4{e}_3$, then $e_5{e}_6{e}_5$, through the final pair. Thus $e_1{f}_k=C_k\cdots C_4{C}_1{C}_2\cdots C_k$, an expression fixed by index reflection. Reflecting gives $e_{2k-1}{f}_k=e_1{f}_k$. Word reversal and shifting give all right and shifted versions.

The map $\varphi$ is surjective. A descending basis diagram with $k$ upper cups and $k$ lower caps can be factored into a braid, then $(E_1{E}_3\cdots E_{2k-1})\otimes B$, then another braid, with $B$ on the $n-2k$ through strands. Move the boundary ends of each cup or cap to consecutive positions using braids, then straighten each strand; the separate layers ensure these moves introduce no extra framing curls. A braid is a product of elementary positive or negative adjacent crossings, by reading its crossings in height order. This factorization puts every $T_c$ in the image of $\varphi$.

It follows that $W_{2k}{f}_k=W_k{f}_k$ and $W_{2k+1}S(f_k)=W_{k+1}S(f_k)$, and likewise on the right. To prove the first equality, show $W_k{f}_k$ stable under the remaining generators. For indices $i>k$ they commute with $W_k$ and step 9.1 replaces their action on $f_k$ by index $2k-i<k$. For index $k$ write $f_k=e_{k}r$ using its first diamond column, and apply $W_{k+1}{e}_k=W_k{e}_k$ from step 8.1. The shifted diamond starts with $e_{k+1}$, and its reflection index is $2k+2-i$, giving the same argument at rank $2k+1$.

For $r=n-2k\ge 0$, let $I_n^{(r)}$ be the two-sided ideal generated by $e_1{e}_3\cdots e_{2k-1}$; set $I_n^{(-1)}=I_n^{(-2)}=0$ as required by parity. They give the increasing parity filtration ending at $I_n^{(n)}=W_n$. The recursion for $f_k$ expresses it as an invertible braid on each side of $e_1{e}_3\cdots e_{2k-1}$, so it generates the same ideal. If $n>2k$, its shift generates the same ideal, since shifting these cups is conjugation by $g_{n-1}\cdots g_1$ using step 7.2. Therefore step 10.1 gives $I_{2k}^{(0)}=W_k{f}_k{W}_k$ and $I_{2k+1}^{(1)}=W_{k+1}S(f_k)W_{k+1}$. The image of $I_n^{(r)}$ consists of tangles factoring through $r$ intermediate strands. Descending reduction separately in the two factors expresses them using representatives whose connectors have at most $r$ through pairs.

Suppose $\varphi$ is injective at ranks below $n$. If $n=2k$, choose preimages $u_c$ of the tangle basis at rank $k$. The elements $u_c{f}_k{u}_d$ span $I_n^{(0)}$ by step 11.1, and their images have distinct connectors: bending a $k$-connector along the nested rainbow gives a bijection with pairings of the $2k$ upper vertices, independently for the lower vertices. Thus their images are independent by step 4.1 and $\varphi$ is injective on $I_n^{(0)}$. For $n=2k+1$, use a basis at rank $k+1$ on either side of $S(f_k)$. Bending its $k$ paired positions leaves one free endpoint, so a $(k+1)$-connector corresponds bijectively to a pairing of $2k+1$ boundary vertices together with one free vertex: append that free vertex as the extra $(2k+2)$nd vertex to give the inverse bijection. Consequently the resulting upper and lower partial pairings give all distinct rank-one connectors. Again step 4.1 proves injectivity on $I_n^{(1)}$. For $n=0,1$, both algebras are the scalar ring, giving the induction bases.

For $r=n-2k>0$ let $V_n^{(r)}$ be the span of $g_{\pi }(w\otimes g_{\tau })\alpha (g_{\sigma })$, where $w\in I_{2k}^{(0)}$, $\pi ,\sigma$ are $(2k,r)$ shuffles and $\tau \in S_r$. When $k=0$, interpret $I_0^{(0)}={\mathrm{\Lambda }}_0$. Under the induction hypothesis, use the rank-zero basis of step 12.1 for $w$. Its images, together with the two shuffles and the through-strand permutation, have distinct connectors: the shuffles specify the $2k$ upper and lower cup positions, $w$ specifies their pairings, and $\tau$ specifies the remaining bijection. Thus $\varphi$ is injective on $V_n^{(r)}$. Combining these representatives with descending representatives for connectors having at most $r-2$ through pairs, step 4.1 gives $\varphi (V_n^{(r)})\cap \varphi (I_n^{(r-2)})=0$. Hence injectivity on $I_n^{(r-2)}$ implies injectivity on $I_n^{(r-2)}+V_n^{(r)}$.

For $1\le k<n\mathrm{/}2$, the induction hypothesis identifies $I_{2k}^{(0)}$ with its rank-zero tangle image. It is spanned by elements $(g_m\cdots g_{2k-2})e_{2k-1}{w}^{\mathrm{\prime }}$ with $1\le m\le 2k-1$ and $w^{\mathrm{\prime }}\in I_{2k}^{(0)}$, interpreting the product as empty at $m=2k-1$. To see this, choose descending representatives with the strand at vertex $2k$ above all other strands. If its partner is $m$, the positive string $G_m\cdots G_{2k-2}$ brings that partner to position $2k-1$, giving the indicated cup followed by a rank-zero diagram. The adjacent-partner case uses ${\delta }^{-1}{E}_{2k-1}$ to duplicate that cup without changing its value. The representative can have any ordering of its strands: step 2.1 still gives spanning with that chosen order. Lift this factorization through the known lower-rank isomorphism. The case $k=0$ needs no such factorization: its cup block is empty, so the mixed-block case below never arises.

We prove this sum equals $I_n^{(r)}$. At $r=1$, $W_1=V_1^{(1)}={\mathrm{\Lambda }}_0$ and the cup ideal is zero; hence $I_{2k}^{(0)}\otimes W_1\subset V_n^{(1)}$ directly, with $I_n^{(-1)}=0$. For $j\ge 2$, $W_j$ modulo its cup ideal is spanned by its positive permutation lifts: multiplication by $g_i$ increases length, giving $g_{s_i\pi }$, or decreases length and the quadratic identity of step 6.1 expresses the product using $g_{s_i\pi }$, $g_{\pi }$ and a cup-ideal term. Thus for $r\ge 2$, $W_r=V_r^{(r)}+W_r{e}_1{W}_r$. Tensoring its cup ideal with $I_{2k}^{(0)}$ lands in $I_n^{(r-2)}$, because its generator is $e_1{e}_3\cdots e_{2k+1}$. It follows that $I_{2k}^{(0)}\otimes W_r\subset I_n^{(r-2)}+V_n^{(r)}$. By length-additive block/shuffle factorization of positive permutation braids, the same containment holds after multiplying on either side by arbitrary positive permutation lifts: their block factors are absorbed in the two-sided ideal $I_{2k}^{(0)}$ and the algebra $W_r$.

Check left stability of $J=I_n^{(r-2)}+V_n^{(r)}$ under $g_i,e_i$. Write a spanning element $x=g_{\pi }(w\otimes g_{\tau })\alpha (g_{\sigma })$. If ${\pi }^{-1}(i)>{\pi }^{-1}(i+1)$, remove the leftmost $g_i$ from $g_{\pi }$ to obtain a shorter shuffle ${\pi }_1$; then $e_i{g}_{\pi }={\lambda }^{-1}{e}_i{g}_{{\pi }_1}$ and $g_i{g}_{\pi }=g_{{\pi }_1}+s{g}_{\pi }-s{\lambda }^{-1}{e}_i{g}_{{\pi }_1}$. It therefore suffices to treat increasing preimages. Multiplication by $g_i$ then gives a positive permutation lift, handled by step 14.1. For $e_i$, if both preimages lie in one block, they are consecutive because $\pi$ is a shuffle. Move $e_i$ through $g_{\pi }$ using step 8.3; it acts inside $I_{2k}^{(0)}\otimes W_r$, so step 14.1 applies.

The remaining case is $u={\pi }^{-1}(i)\le 2k<{\pi }^{-1}(i+1)=v$. Let $\eta \in S_{2k}\times S_r$ move position $2k$ to $u$ and position $2k+1$ to $v$, preserving the order of the other positions within each block: explicitly $\eta (j)=j$ for $j<u$, $j+1$ for $u\le j<2k$, $\eta (2k)=u$, $\eta (2k+1)=v$, $\eta (j)=j-1$ for $2k+1<j\le v$, and $\eta (j)=j$ for $j>v$. Shuffle factorization gives $g_{\pi \eta }=g_{\pi }{g}_{\eta }$, and step 8.3 gives $e_i{g}_{\pi }=g_{\pi \eta }{e}_{2k}{g}_{\eta }^{-1}$. Absorb $g_{\eta }^{-1}$ into the block ideal and $W_r$. We must therefore show $g_{\pi \eta }{e}_{2k}(I_{2k}^{(0)}\otimes W_r)\alpha (g_{\sigma })\subset J$.

Put $H=g_{\pi \eta }(g_m\cdots g_{2k-2})$. The images $\pi \eta (j)$ increase for $j<2k$, so this is a positive permutation lift. Its final three relevant images are $\pi \eta (m),i,i+1$. Moving $e_{2k}$ past the $g$-string and applying the defining mixed relations gives $g_{\pi \eta }{e}_{2k}(g_m\cdots g_{2k-2})e_{2k-1}=H{g}_{2k-1}{g}_{2k}{e}_{2k-1}=H{g}_{2k-1}^{-1}{g}_{2k}^{-1}{e}_{2k-1}$. For the second equality, both products of the last two $g$'s times $e_{2k-1}$ equal $e_{2k}{e}_{2k-1}$; for adjacent $i,j$, expanding both inverses with step 6.1, using $g_i{e}_j{e}_i=g_j^{-1}{e}_i$, and reducing cups gives $(g_i^{-1}{g}_j^{-1}-g_i{g}_j)e_i=(s(\lambda -{\lambda }^{-1})-s^2(\delta -1))e_i=0$ by the coefficient relation. This supplies the inverse identity without assuming invariance of the presentation under crossing reversal. If $\pi \eta (m)<i$, the positive-sign string is a positive permutation braid; if $\pi \eta (m)>i+1$, the negative-sign string removes exactly those two crossings from $H$ and is a positive permutation braid. No other case exists since the images are distinct. Thus the entire expression is $g_{\theta }{e}_{2k-1}$ for a positive permutation lift. Step 14.1 now puts each term of step 13.2 in $J$. This proves the mixed-block case, and hence left stability. Word reversal preserves $J$, so it is two-sided.

The ideal $J$ lies in $I_n^{(r)}$ and contains its generator $e_1{e}_3\cdots e_{2k-1}$, which belongs to the block ideal in $V_n^{(r)}$. Its two-sided stability therefore proves $I_n^{(r)}=I_n^{(r-2)}+V_n^{(r)}$. Induct on $n$, beginning at $n=0,1$, and within a fixed $n$ on the through-strand parity $r$. Step 12.1 starts the inner induction at $r=0$ or $1$; step 13.1 and the equality just proved advance it by two. At $r=n$ it proves injectivity on $W_n$. Together with step 9.2 this proves the universal isomorphism. Finally base-change along $\lambda \mapsto \rho ,z_0\mapsto -(q-q^{-1}),\delta \mapsto \delta$ using step 5.1. The tangle basis tensors to the asserted BMW basis, and its cardinality is $(2n-1)!!$ by [F2], including the empty matching at $n=0$.