Skein modules for links in 3-manifolds

For the core $L:C\hookrightarrow X$ of a framed link, write its annular projection as $g:C\to \mathrm{int}(\mathcal{A})$. Since $L(C)$ is compact in the interior, its projection has positive distance from $\mathrm{\partial }\mathcal{A}$. Fix an auxiliary smooth embedding $x:C\hookrightarrow {\mathbb{R}}^3$. Perturb the two coordinates of $g$ by a finite-dimensional family of polynomials of degree at most five in $x$, multiplied by a smooth cutoff supported in $\mathrm{int}(\mathcal{A})$ and equal to one near $g(C)$. At any three distinct points $x_1,x_2,x_3$, the polynomials $p_i(x)={\prod }_{j\mathrm{\ne }i}\parallel x-x_j{\parallel }^2$ vanish to first order at the other two points and are nonzero at $x_i$; multiplying by affine functions therefore prescribes independently the value and first derivative of both projected coordinates at these points. Thus the family is a submersion on every one-, two-, and three-point first-jet space.

Apply [F4] to the derivative-zero stratum, the pair-equality diagonal, the nontransverse-pair stratum, and the triple-equality diagonal. The derivative-zero stratum has codimension two on the one-dimensional source, so it is avoided. Pair equality has codimension two on the two-dimensional distinct-pair source and is transverse, hence gives isolated double points; tangency imposes one additional independent equation, so its codimension is three and it is avoided. Triple equality has codimension four on the three-dimensional distinct-triple source, so it is avoided. Choose a good parameter $a$ in a ball about $0$ on which every core $(g_{ta}(x),L_3(x))$, $0\le t\le 1$, remains embedded in $X$; compactness, openness of embeddings in the $C^1$ topology, and positive boundary clearance give such a ball. The parameter path then gives a smooth core isotopy $h_t$.

Let $F$ denote the original union of framing annuli. To transport the given framing along $h_t$, the map $(x,t)\mapsto (h_t(x),t)$ has compact embedded trace in $X\times [0,1]$, with spatial projection compactly contained in $X$. In local product coordinates along this trace, extend the spatial core velocity across the normal coordinates, multiply by cutoffs, and combine the extensions with a finite partition of unity. This gives a time-dependent vector field $V_t$ on $X$, supported in a fixed compact subset of $X$ and agreeing with ${\dot{h}}_t$ along every $h_t(C)$. Its flow $H_t$ exists on $[0,1]$ and satisfies $H_t\circ h_0=h_t$ by uniqueness of solutions. Hence each original framing annulus is carried to an embedded annulus $H_t(F)$ in $X$. This gives an ambient isotopy of the framed link, so the framing remains transported and embedded throughout the perturbation.

The transverse double pairs are finite. Immersion gives, near each point of the compact source, a source arc on which the projection is injective; finitely many such arcs exclude double pairs near the diagonal in $C\times C$. Away from that diagonal, the double-pair set is closed in a compact set and is discrete by transversality, hence finite. This proves the annular regular-diagram assertion. For a link in $S^3$, choose orientations on its finitely many components, move it off $\mathrm{\infty }$, and apply [F1].

The same construction has the relative parameterized form needed for a whole isotopy. Write its cores as $h_t:C\to X$ and its supplied framing annuli as $E_t$. If $h_t^{\mathrm{\prime }}$ is sufficiently $C^1$-close to $h_t$ and equal to it on fixed endpoint collars, set $h_{r,t}=h_t+r(h_t^{\mathrm{\prime }}-h_t)$ for $0\le r,t\le 1$ in the ambient ${\mathbb{R}}^3$ coordinates. Compactness of $C\times [0,1]$ gives uniform immersion bounds, separation of distant source points, and clearance from $\mathrm{\partial }X$; consequently all $h_{r,t}$ are embeddings in $X$ for a sufficiently small perturbation. The trace $(x,r,t)\mapsto (h_{r,t}(x),r,t)$ is embedded. In finitely many spatial normal-coordinate charts along this trace extend ${\mathrm{\partial }}_r{h}_{r,t}$ constantly in the spatial normal coordinates, and use smooth cutoffs and a finite partition of unity to obtain a vector field $V_{r,t}(y)$, smooth jointly in all variables, with one fixed compact support in $X$ and value ${\mathrm{\partial }}_r{h}_{r,t}(x)$ at $y=h_{r,t}(x)$. The charts can be taken on slightly extended parameter intervals, so the construction is smooth at parameter faces as well. On the $t$-endpoint collars this velocity is zero; multiply the extension by a $t$-cutoff zero on smaller such collars and one outside the original collars. This preserves its prescribed core values and makes $V_{r,t}=0$ on the smaller collars. Integrate in the auxiliary variable $r$, with initial condition ${\mathrm{\Psi }}_{0,t}=\mathrm{i}\mathrm{d}$. Compact support gives existence for the full interval, and smooth parameter dependence and uniqueness give ${\mathrm{\Psi }}_{r,t}{h}_t=h_{r,t}$. Each ${\mathrm{\Psi }}_{r,t}$ is an ambient diffeomorphism; it is the identity for $t$ near either endpoint. Therefore $E_t^{\mathrm{\prime }}={\mathrm{\Psi }}_{1,t}(E_t)$ is a smooth family of embedded framing annuli over $h_t^{\mathrm{\prime }}$ agreeing exactly with the supplied framed family on endpoint collars. If $E_t=G_t(E_0)$ is the supplied ambient framed isotopy, then ${\mathrm{\Psi }}_{1,t}\circ G_t$ is itself an ambient isotopy realizing this transported family, with the same endpoint maps. For each $r$, the analogous family ${\mathrm{\Psi }}_{r,t}(E_t)$ is framed, so this is a deformation of the whole original framed isotopy relative its endpoints, not merely a lift of one perturbed slice. No full twist is inserted at an endpoint.

For a supplied framed isotopy in $X$, reparametrize it to be constant near its endpoints and choose regular endpoint diagrams using steps 1.1, 2.1, 3.1, and 3.2. Regard $X$ as an open subset of ${\mathbb{R}}^3$ by the annulus-product coordinates. Its compact core track has positive distance from the boundary of $X$. Apply [F3] with an arbitrarily $C^1$-small perturbation fixed on endpoint collars, small enough for both this clearance and step 4.1. That step lifts the actual entire perturbed core isotopy to a framed family with exactly the same framed endpoints. Every exceptional event and every regular interval remains in $X$; by [F3] its projected annular diagrams differ by a finite sequence of R1, R2, R3, and planar isotopies in disks of $\mathrm{int}(\mathcal{A})$. For a supplied framed isotopy in $S^3$, first arrange that its core track misses the point at infinity. Indeed its smooth compact two-dimensional parameter source has image of empty three-dimensional interior: in finitely many source and target charts a bounded derivative covers its image by $O(N^2)$ balls of radius $O(1\mathrm{/}N)$, of total three-volume $O(1\mathrm{/}N)$. Thus there is a point $q$ outside that core track arbitrarily close to infinity; no claim is made that $q$ misses the unshrunk annuli. Write the supplied annulus family as $e_t:C\times [-1,1]\to S^3$, with $e_t(x,0)=h_t(x)$. The compact core track has positive round-metric distance from $q$. Uniform continuity of $e$ on its compact parameter domain therefore gives one $\epsilon >0$ such that $e_t(C\times [-\epsilon ,\epsilon ])$ misses $q$ for every $t$. Replace each annulus representative by this thin core band. Its normal derivative is multiplied by the positive number $\epsilon$, so its specified normal-section homotopy class and endpoint framing integer are unchanged. If literal annulus representatives are retained, at each endpoint $i=0,1$ the family $e_i(x,((1-s)+s\epsilon )u)$, $0\le s\le 1$, is an actual isotopy of embedded annuli fixing the core. Extend its velocity in local spatial charts and use finite cutoffs as in step 3.1 to obtain an ambient realization; adjoining this shrink at the initial end and its inverse at the terminal end restores the original annuli. These endpoint paths fix the core diagram and change no framing integer. The middle thin-annulus family now has compact image disjoint from $q$. Choose a rotation of the round $S^3$, arbitrarily close to the identity, carrying $q$ to infinity, and a small path of rotations from the identity to it. The regular endpoint projections stay regular along this path when it is chosen small enough, by openness of regular diagrams for these compact endpoints. Rotate the entire supplied framed family, obtaining a core track compactly contained in ${\mathbb{R}}^3$, and join its endpoints to the original framed endpoints by these rotation paths. The endpoint paths have only planar isotopy of their regular decorated diagrams. Apply [F3] and the relative lift of step 4.1 to the middle ${\mathbb{R}}^3$ isotopy (with compact support in ${\mathbb{R}}^3$ instead of $X$); its flows extend to $S^3$ by fixing infinity. This gives the required finite sequence for the original framed endpoints, rather than using a core-only equivalence to infer framing preservation.

A planar isotopy transports the blackboard strip continuously. In R2 and R3 the participating strips deform inside the move ball without a full rotation of the normal direction. In an A-type R1 move the blackboard strip gains one full twist, and in the opposite B-type move it loses one; this is the local curl convention in [F6]. Since the framed link isotopy transports the specified annulus without changing its homotopy class, the recorded integer correction changes by the negative of that blackboard turn. Thus the move sequence preserves the framed isotopy class, and all framing changes in the statement are exact.

In the positive curl from the Statement, the $A$-smoothing is $L$ with one zero-framed trivial circle and the $B$-smoothing is $L$. By [F1] and [F2], $[L^{+}]=A\delta [L]+A^{-1}[L]=(A(-A^2-A^{-2})+A^{-1})[L]=-A^3[L]\mathrm{.}$

In the negative curl, the $A$-smoothing is $L$ and the $B$-smoothing is $L$ with one zero-framed trivial circle. Hence $[L^{-}]=A[L]+A^{-1}\delta [L]=(A+A^{-1}(-A^2-A^{-2}))[L]=-A^{-3}[L]\mathrm{.}$ These are equalities in the framed-link quotient and use no ambient-isotopy identification that forgets the framing.

Define $\tilde{f}:{\mathcal{F}}_R(M)\to {\mathcal{F}}_R(N)$ on basis elements by $\tilde{f}([L])=[f(L)]$ and extend $R$-linearly. A diffeomorphism sends ambient-isotopic framed annuli to ambient-isotopic framed annuli and sends the framing annuli by its derivative, so this is well-defined on the free modules.

Fix framed-link representatives $P$ in $\mathrm{\Sigma }\times (0,1)$ and $L$ in the interior of $M$. To check independence of isotopy representatives, take any isotopy track of either framed link. Its framing annuli have compact trace in the interior. Place the trace of the product-factor link in a sufficiently thin outer collar layer disjoint from the trace of the $M$-factor link. The disjoint-union family is a smooth isotopy of framed annuli in $M$. In finitely many submanifold charts of its compact trace, extend the annulus velocity to ambient charts with cutoffs and combine these extensions with a partition of unity; the resulting compactly supported time-dependent vector field has a flow that realizes this isotopy by ambient diffeomorphisms. Thus the union class is independent of both representatives. For the quotient relations, rescale $P$ into a boundary layer disjoint from $L$. A change of $P$ by either defining relation becomes the identical local relation after the collar embedding by [F2]; a change of $L$ by either relation remains the identical relation with the disjoint collar link outside its ball. The same local check in two successive collar layers shows that stacking descends to the skein quotients. Hence collar gluing gives a bilinear action. Decreasing the collar width uses an orientation-preserving product rescaling isotopy supported in a thin boundary layer and disjoint from $L$, so the resulting skein class is unchanged.

The image of an oriented crossing ball is an oriented ball. Its transported chart preserves the over/under crossing and both endpoint pairings, so $\tilde{f}([L_{\times }]-A[L_A]-A^{-1}[L_B])$ is the corresponding defining relation in $N$. A zero-framed trivial circle bounding a disk in the ball maps to another zero-framed trivial circle bounding the image disk. Thus $\tilde{f}({\mathcal{R}}_A(M))\subseteq {\mathcal{R}}_A(N)$ and $\tilde{f}$ descends to $f_{\ast }$. Applying the same construction to $f^{-1}$ gives the inverse; applying it to a composite gives the composite map on every link generator.

Place two algebra links in successive subcollars and the $M$-link deeper inside. Both $(P\ast Q)\cdot L$ and $P\cdot (Q\cdot L)$ are represented by the same ordered union, up to an orientation-preserving product rescaling, and the empty link acts as the identity. Thus the action is associative and unital. If an orientation-preserving diffeomorphism carries one specified collar productwise to another, it sends each stacked or glued link union to the corresponding union, so its module maps intertwine the actions.

The state sum in [F1] is invariant under planar isotopy because that isotopy bijects crossings, states, and state circles. By [F3], diagrams of the same framed link are connected by a finite move sequence. For RII and RIII, [F2] leaves the raw bracket unchanged and the framing correction is unchanged. For an A-type R1 kink, the local state contribution is $A\delta +A^{-1}=-A^3$; for the opposite B-type kink it is $A+A^{-1}\delta =-A^{-3}$. For the A-type kink [F3] decreases the correction by one and for the B-type kink it increases the correction by one, so the product $(-A^3{)}^m⟨D⟩$ is unchanged. Hence $E(L)$ is independent of diagram and framing correction.

At a crossing, partition the finite state set by whether that crossing is smoothed A or B. This gives $⟨D_{\times }⟩=A⟨D_A⟩+A^{-1}⟨D_B⟩$. In a framed skein triple the exterior framed strips agree; place the same twist marks on those common arcs and none on the blackboard-framed local tangles. Their total correction $m$ is the same for the crossing and both smoothings. Multiplying the state identity by $\delta (-A^3{)}^m$ proves that $E$ sends the crossing relation in [F4] to zero.

If $L$ is nonempty, adjoining a disjoint zero-framed trivial circle adds one circle to every state and multiplies the bracket by $\delta$. Thus $E(L\bigsqcup O)=\delta E(L)$. For the empty link, its one-circle diagram has bracket $1$, so the defining assignment gives $E(O)=\delta E(\mathrm{\varnothing })$. The evaluation therefore annihilates every trivial-circle relation in [F4]. Its $R$-linear extension to the free module factors through the quotient ${\mathcal{K}}_A(S^3)$ and sends the empty class to $1$.

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 $\mathrm{\Psi }$.

We prove by induction on the number of crossings of a regular diagram for a blackboard-framed link that its class is a scalar multiple of $[\mathrm{\varnothing }]$. If the diagram has a crossing, the crossing relation in [F1] expresses its class as the sum of the two smoothing classes with coefficients $A$ and $A^{-1}$. The skein convention gives each smoothing its blackboard framing, and each has one fewer crossing, so the induction applies. A crossingless diagram is a finite disjoint union of planar circles. Choose an innermost circle; its planar disk has no other component in its interior and is disjoint from the remaining link. Its blackboard framing is the zero framing, so [F1] removes it with factor $\delta$. Remove this component and repeat; the crossingless link reduces to a scalar multiple of the empty link. For a general framed link, [F2] supplies a diagram and [F3], applied finitely many times to its framing-correction integer, reduces its class to a scalar multiple of the corresponding blackboard-framed diagram. Hence every class lies in $R[\mathrm{\varnothing }]$.

Define $\varphi :R\to {\mathcal{K}}_A(S^3)$ by $\varphi (r)=r[\mathrm{\varnothing }]$. Step 1.1 makes $\varphi$ surjective. If $\varphi (r)=0$, $R$-linearity and $E([\mathrm{\varnothing }])=1$ give $r=E(\varphi (r))=E(0)=0$. Thus $\varphi$ is injective as well as surjective, so the empty class is a free $R$-module generator.

The trivial-circle relation in [F1], with $L=\mathrm{\varnothing }$, gives $[O]=\delta [\mathrm{\varnothing }]$, as claimed.

Define $J$ by the stated $R$-linear assignment. On $z^n$ take the crossingless annular diagram of $n$ parallel essential circles with blackboard framing. It has $m=0$, $a=b=u=0$, and $e=n$, so [F3] gives $\mathrm{\Psi }(J(z^n))=z^n$. By linearity, $\mathrm{\Psi }\circ J$ is the identity on $R[z]$, and hence $J$ is injective.

Let $L$ be any framed link in the solid torus. By [F2], choose a regular annular diagram $D$ and write $m_j$ for the framing correction on its $j$th component, with $m={\sum }_j{m}_j$. Apply [F4] componentwise: each positive twist contributes $-A^3$, each negative twist contributes $-A^{-3}=(-A^3{)}^{-1}$, so $[L]=(-A^3{)}^m[D]$, where $D$ carries its blackboard framing. Induct on the number of crossings. At one crossing, [F1] replaces $D$ by its two smoothing diagrams with coefficients $A$ and $A^{-1}$, and each smoothing has one fewer crossing. Applying the induction hypothesis to each term yields $[D]={\sum }_S{A}^{a(S)-b(S)}[D_S],$ where each $D_S$ is a crossingless annular state with blackboard framing. Every embedded state circle in the annulus either bounds a disk or is essential and cuts the annulus into two annuli, so it is isotopic to a core circle. For finitely many disjoint essential circles, cutting along them gives annular regions between consecutive circles; thus they are parallel copies of the core. Among the contractible circles choose one whose disk contains no other state circle; its disk has a small thickening disjoint from the remaining link, so the trivial-circle relation in [F1] removes it with factor $\delta$. Repeating removes all contractible circles. Therefore $[L]=(-A^3{)}^m{\sum }_S{A}^{a(S)-b(S)}{\delta }^{u(S)}[z^{e(S)}]=J(\mathrm{\Psi }([L])),$ where the last equality is [F3]. Thus $J\circ \mathrm{\Psi }$ is the identity on every link generator, and by $R$-linearity on the whole quotient. Together with step 1.1 this proves $J$ and $\mathrm{\Psi }$ are inverse isomorphisms, so the displayed classes are an $R$-basis.

The quotient isomorphism and collar-action naturality are exactly [F1]. Under a diffeomorphism, the framing annulus is carried by the derivative along with its core, so this map preserves the full link framing rather than only the unframed core. A single full twist is changed according to [F2].

Apply [F2] to the zero-framed trivial circle $O$ in $S^3$. By [F3], $[O]=\delta [\mathrm{\varnothing }]$, so its positive twist represents $-A^3\delta [\mathrm{\varnothing }]$. The difference from $[O]$ is $(-A^3-1)\delta [\mathrm{\varnothing }]$. In $R$, $\delta =-(A^2+A^{-2})\mathrm{\ne }0$ and $-A^3-1\mathrm{\ne }0$; the product is nonzero because the product of two nonzero Laurent polynomials over the integral domain $\mathbb{Z}$ is nonzero (its highest-exponent coefficient is the product of their nonzero highest-exponent coefficients). Since $[\mathrm{\varnothing }]$ is a basis element by [F3], the two classes are unequal.

For the solid-torus core, [F2] gives the twisted class $-A^3[z]$. Its difference from $[z]$ is $(-A^3-1)[z]$, which is nonzero because $-A^3-1\mathrm{\ne }0$ in $R$ and $[z]$ is one of the free basis elements in [F4]. This proves the second displayed framing change is nontrivial.