BMW descending matchings are independent by closure pairing

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 $[d]$.