Annular state expansion for the Kauffman bracket skein module

Let $B_i$ and $B_o$ be the inner and outer boundary circles of the planar annulus. For any embedded state circle $C$, apply [F5] to $C$ and $B_i$. Their complement has a disk component on the $B_i$ side, a disk component on the $C$ side, and an annulus component. Since $B_o$ lies outside the disk bounded by $B_i$, it lies either in the annulus component or in the disk component on the $C$ side. In the first case the disk component on the $C$ side lies inside the disk bounded by $B_o$ and outside the disk bounded by $B_i$, so $C$ bounds a disk in $\mathcal{A}$. In the second case $C$ separates $B_i$ from $B_o$, and the annulus component between $C$ and $B_i$ lies inside the disk bounded by $B_o$ and outside the disk bounded by $B_i$. The two disk sides of $C$ then contain $B_i$ and $B_o$, respectively, so neither is contained in $\mathcal{A}$; hence $C$ is essential exactly in this second case. Thus an essential $C$ has $B_i$ and $B_o$ on opposite disk sides. For two disjoint essential state circles, apply [F5] to the pair. Each disk side containing $B_i$ contains the disk bounded by $B_i$, while $B_o$ lies on the opposite side of both; therefore the curves are nested and the complementary annulus between them lies outside the disk bounded by $B_i$ and inside the disk bounded by $B_o$, hence in $\mathcal{A}$. Ordering the finitely many essential circles by disk inclusion shows every consecutive pair bounds such an annular band, so they are parallel. Thus $u(S)$ and $e(S)$ are unambiguous finite counts, and each state contributes a Laurent-polynomial multiple of $z^{e(S)}$.

The state set is finite. At a chosen crossing, partition it into states using the A-smoothing and states using the B-smoothing. In a framed skein triple the exterior annular strips agree; place the same twist marks on those common arcs and none on the blackboard-framed local crossing or smoothings. Their total correction $m$ is common to all three, so the partition gives $\mathrm{\Psi }(D_{\times },m)=A\mathrm{\Psi }(D_A,m)+A^{-1}\mathrm{\Psi }(D_B,m)$. Adding a zero-framed trivial circle increases $u(S)$ by one in every state, so it multiplies $\mathrm{\Psi }$ by $\delta$. Hence the state expansion annihilates both relation generators in [F2].

Annular planar isotopy preserves which state circles bound disks. Under R2 and R3, the local tangle identities in [F4] remain equal after attaching the same exterior annular tangle; evaluating each resulting contractible loop by $\delta$ and each essential circle by $z$ is an $R$-linear operation, so the state expansions agree. For an A-type R1 kink the local state factor is $A\delta +A^{-1}=-A^3$; for the opposite B-type kink it is $A+A^{-1}\delta =-A^{-3}$. By [F3] the correction decreases by one for the A-type kink and increases by one for the B-type kink, so the factor $(-A^3{)}^m$ cancels the raw state-sum change. The value is therefore independent of diagram and framing correction.

By [F3], every framed-link generator has a regular annular diagram and framing correction. Steps 2.1 and 3.1 show that the state expansion is independent of that choice and vanishes on both generators of the defining submodule [F2]. Its $R$-linear extension therefore factors through the quotient and gives the stated map $\mathrm{\Psi }$.