ScalingStacks

Recall that in Section 2.2 we defined algebra $A$ as a free module of rank $2$ over the ring $R=\mathbb{Z}[c],$ generated by $\mathbf{1}$ and $X,$ with the multiplication rules

$$\mathbf{1}\mathbf{1}=\mathbf{1},\hskip 14.45377pt\mathbf{1}X=X\mathbf{1}=X,\hskip 14.45377ptX^{2}=0.$$

(182)

Gradings of $\mathbf{1}$ and $X$ are equal to $1$ and $-1$, respectively, so that the multiplication in $A$ is a graded map of degree $-1.$

Denote by $A\mbox{-mod}_{0}$ the category with objects being graded $A$-modules and morphisms being grading-preserving homomorphisms of graded $A$-modules. Note that $A\mbox{-mod}_{0}$ is an abelian category. Denote by $\{n\}$ the automorphism of $A$-mod that shifts the grading down by $n.$ Let $A$-mod be the category of graded $A$-modules and graded maps. $A$-mod has the same objects as $A\mbox{-mod}_{0},$ but more morphisms.

Given a graded $A$-module $M,$ define the multiplication map

$$m_{M}:A\otimes_{R}M\longrightarrow M$$

(187)

by

$$m_{M}(\mathbf{1}\otimes t)=t,\hskip 14.45377ptm_{M}(X\otimes t)=Xt,\hskip 14.45377ptt\in M.$$

(188)

The multiplication map $m_{M}$ is a degree $-1$ map with the grading on $A\otimes_{R}M$ defined as the product grading of gradings of $A$ and $M$.

To a graded $A$-module $M$ associate a map

$$\Delta_{M}:M\longrightarrow A\otimes M$$

(189)

(recall that all tensor products are over $R=\mathbb{Z}[c]$) by

$$\Delta_{M}(t)=X\otimes t+\mathbf{1}\otimes Xt+cX\otimes Xt,\hskip 21.68121ptt\in M.$$

(190)

Then $\Delta_{M}$ equips $M$ with the structure of a cocommutative comodule over $A.$ Map $\Delta_{M}$ has degree $-1.$ The following relation between $\Delta_{M}$ and $m_{M}$ is straightforward to check

$$\Delta_{M}m_{M}=(Id_{A}\otimes m_{M})(\Delta\otimes Id_{M})=(m\otimes Id_{M})(Id_{A}\otimes\Delta_{M})$$

(191)

Denote by $\iota_{M}$ the map $M\longrightarrow A\otimes M$ given by $\iota_{M}(t)=\mathbf{1}\otimes t$ for $t\in M.$

Introduce an $A$-module structure on $A^{\otimes n}\otimes M$ by

$$a(x\otimes y)=x\otimes ay\mbox{ for }a\in A,x\in A^{\otimes n},y\in M,$$

(192)

where $ay$ means $m_{M}(a\otimes y).$ Then $m_{M},\Delta_{M}$ and $\iota_{M},$ after appropriate shifts by $\{1\}$ or $\{-1\},$ are maps of graded $A$-modules.

Proof: Let us check (194), for instance. We have a decomposition

$$A\otimes M=(\mathbf{1}\otimes M)\oplus(X\otimes M).$$

(196)

Denote by $p$ the projection $A\otimes M\to X\otimes M,$ orthogonal to $\mathbf{1}\otimes M.$ Since $\iota_{M}M=\mathbf{1}\otimes M,$ it suffices to check that

$$p(\Delta_{M}-\iota_{M}m_{M}\Delta_{M}):M\to X\otimes M$$

(197)

is an $A$-module isomorphism. This map is given by

$$t\longmapsto X\otimes(1+cX)t,\hskip 7.22743pt\mbox{ where }t\in M.$$

(198)

The inverse map is

$$X\otimes t\longmapsto(1-cX)t$$

(199)

Decompositions (193) and (195) can be verified analogously. $\square.$

Let $\mathcal{M}_{1}$ be the category whose objects are one-dimensional manifolds that are unions of a finite number of circles and one interval. An ordering of the ends of this interval is fixed. Morphisms between objects $\alpha$ and $\beta$ of $\mathcal{M}_{1}$ are oriented surfaces whose boundary is the union of $\alpha,\beta$ and 2 intervals that join corresponding ends of the intervals of $\alpha$ and $\beta.$ An example is depicted below.

This surface represents a morphism from an interval to a union of a circle and an interval. We require that a surface can be presented as a composition of disjoint unions of surfaces $S_{2}^{1},S_{1}^{2},S_{0}^{1},S_{1}^{0},S_{2}^{2},S_{1}^{1},$ defined in Section 2.3, and surfaces $T_{1},T_{2},$ depicted below

We compose morphisms in this category by concatenating surfaces.

Category $\mathcal{M}_{1}$ is a module-category over $\mathcal{M},$ defined in Section 2.3. The bifunctor $\mathcal{M}\otimes\mathcal{M}_{1}\to\mathcal{M}_{1}$ is defined on objects and morphisms by taking disjoint unions.

Recall from the previous section that $A$-mod denotes the category of graded $A$-modules and graded homomorphisms. Given a graded $A$-module $M$, define a monoidal functor

$$F_{M}:\mathcal{M}_{1}\longrightarrow A\mbox{-mod}$$

by assigning the graded $A$-module $A^{\otimes n}\otimes M$ to a union of $n$ circles and one interval, maps $\Delta_{M}$ and $m_{M}$ (defined in the previous section) to the elementary surfaces $T_{1}$ and $T_{2}$:

$$F_{M}(T_{1})=\Delta_{M},\hskip 7.22743ptF_{M}(T_{2})=m_{M}.$$

(200)

To the other six elementary surfaces $S_{2}^{1},S_{1}^{2},S_{0}^{1},S_{1}^{0},S_{2}^{2},S_{1}^{1}$ (Section 2.3) associate the same maps as for the functor $F$:

$$\begin{array}[]{lll}F_{M}(S_{2}^{1})=m,&F_{M}(S_{1}^{2})=\Delta,&F_{M}(S_{0}^{1})=\iota,\\ F_{M}(S_{1}^{0})=\epsilon,&F_{M}(S_{2}^{2})=\mbox{Perm},&F_{M}(S_{1}^{1})=\mbox{Id}.\end{array}$$

(201)

A $(1,1)$-tangle is a proper smooth embedding $e:T\hookrightarrow\mathbb{R}^{2}\times[0,1]$ of a finite collection $T$ of circles and one interval $[0,1]$ into $\mathbb{R}^{2}\times[0,1]$ such that the boundary points of $[0,1]$ go to the corresponding boundary component of $\mathbb{R}^{2}\times[0,1]:$

$$e(0)\in\mathbb{R}^{2}\times\{0\},\hskip 14.45377pte(1)\in\mathbb{R}^{2}\times\{1\}.$$

(202)

Two $(1,1)$-tangles are called equivalent if they are isotopic via an isotopy that fixes the boundary. Oriented $(1,1)$-tangles are $(1,1)$-tangles with a chosen orientation of each component, orientation of $[0,1]$ always chosen in the direction from $0$ to $1.$

Define a marked oriented link (in $\mathbb{R}^{3}$) as an oriented link with a marked component. Obviously, there is a natural one-to-one correspondence between marked oriented links and oriented $(1,1)$-tangles: the closure of a $(1,1)$-tangle is a marked oriented link. Denote this map from oriented $(1,1)$-tangles to oriented marked links by $\mbox{cl}.$

A plane diagram $D$ of an oriented $(1,1)$-tangle $L$ is a generic projection of $L$ onto $\mathbb{R}\times[0,1].$

If $D$ is a plane diagram of an oriented $(1,1)$-tangle, define $x(D)$ and $y(D)$ in the same way as for plane diagrams of oriented links (see Section 2.4).

Fix a graded $A$-module $M.$ Let $n$ be the number of double points of $D$, so that $n=x(D)+y(D)$ and $\mathcal{I}$ the set of double points of $D.$ To $M$ and $D$ associate a commutative $\mathcal{I}$-cube $V_{D}^{M}$ over the category $A\mbox{-mod}_{0}$ of graded $A$-modules as follows.

For $\mathcal{L}\subset\mathcal{I}$ the $\mathcal{L}$-resolution $D(\mathcal{L})$ of $D$ consists of a disjoint union of circles and an interval. The functor $F_{M}$ (see Section 8.2) assigns a graded $A$-module to $D(\mathcal{L}).$ Define

$$V_{D}^{M}(\mathcal{L})=F_{M}(D(\mathcal{L}))\{-|\mathcal{L}|\}.$$

(203)

Maps between $V_{D}^{M}(\mathcal{L})$ for various subsets $\mathcal{L}$ are defined by the procedure completely analogous to the one described in Section 4.2. Due to shifts $\{-|\mathcal{L}|\},$ these maps of graded $A$-modules are grading-preserving, rather than just graded maps, so that $V_{D}^{M}$ is a commutative cube over $A\mbox{-mod}_{0}.$

Example: For a diagram $D,$ depicted below,

resolutions of $D$ are

so that

	$\displaystyle V_{D}^{M}(\emptyset)$	$\displaystyle=$	$\displaystyle A\otimes M$
	$\displaystyle V_{D}^{M}(\mathcal{I})$	$\displaystyle=$	$\displaystyle M\{-1\}$

and the structure map $V_{D}^{M}(\emptyset)\longrightarrow V_{D}^{M}(\mathcal{I})$ is the multiplication map $m_{M}:A\otimes M\to M\{-1\}.$

Next we transform the commutative $\mathcal{I}$-cube $V_{D}^{M}$ into a skew commutative $\mathcal{I}$-cube by putting minus signs in front of some structure maps of $V_{D}^{M},$ or, equivalently, by tensoring it with $E_{\mathcal{I}}.$ Denote by $\overline{C}_{M}(D)$ the complex $\overline{C}(V_{D}^{M}\otimes E_{\mathcal{I}})$ of graded $A$-modules. Define

$$C_{M}(D)=\overline{C}_{M}(D)[x(D)]\{y(D)-2x(D)\}$$

(204)

Denote the $i$-th cohomology group of the complex $C_{M}(D)$ by $H^{i}(D,M).$ These cohomology groups are graded $A$-modules. Denote the $j$-th graded component of $H^{i}(D,M)$ by $H^{i,j}(D,M)$

Our proof of Theorem 1 immediately generalizes without essential modifications to a proof of Theorem 3. Proposition 38 is used to establish direct sum decompositions of $C_{M}(D)$, analogous to decompositions of $C(D)$, for suitable $D,$ given by Propositions 11, 14, 18.1, 21.1.

$\square$

Cohomology groups $H^{i}(D),$ defined in Section 4.2, is a special case of groups $H^{i}(D,M),$ as the next proposition explains.

$\square$

Given a finitely-generated graded $A$-module $M,$ define the graded Euler characteristic $\widehat{\chi}(M)$ by

$$\widehat{\chi}(M)=\sum_{j\in\mathbb{Z}}{\mathrm{dim}}_{\mathbb{Q}}(M_{j}\otimes_{\mathbb{Z}}\mathbb{Q})$$

(206)

$\square$

Given two graded $A$-modules $M,N$ and a grading-preserving homomorphism $f:M\to N$, it induces a map of commutative cubes $V_{D}^{M}\to V_{D}^{N},$ which, in turn, induces a map of complexes $C_{M}(D)\to C_{N}(D)$ and a map of cohomology groups $H^{i}(D,M)\to H^{i}(D,N).$ So, in fact, each diagram $D$ of an oriented long link defines functors $H^{i}_{D}$ from the category $A\mbox{-mod}_{0}$ of graded $A$-modules to itself, $H^{i}_{D}(M)=H^{i}(D,M).$ If two diagrams $D_{1},D_{2}$ are related by a Reidemeister move, constructions of Section 5 extend to functor isomorphism $H^{i}_{D_{1}}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}H^{i}_{D_{2}}.$ Let us frame this observation into a proposition.

Oriented long links with one component correspond one-to-one to oriented knots in $\mathbb{R}^{3}.$ Thus, Proposition 41 gives invariants of oriented knots in $\mathbb{R}^{3}.$ Moreover, if a diagram $D$ represents an oriented knot and $D^{\prime}$ is the diagram obtained from $D$ by reversing the orientation of the underlying curve, there is a natural in $M$ isomorphism $H^{i}(D,M)=H^{i}(D^{\prime},M).$ Consequently, for knots, isomorphism classes of functors $H^{i}_{D}$ do not depend on the orientation, and $H^{i}_{D}$ provide “functor-valued” invariants of non-oriented knots. Of course, these invariants depend on how the ambient 3-space is oriented.

Let $D$ be the 3-crossing diagram of the left-hand trefoil (knot $T_{2,3}$ in the notations of Section 6.2). The functors $H^{i}_{D}$ are written below

	$\displaystyle H^{-3}_{D}(M)$	$\displaystyle=$	$\displaystyle{\mathrm{ker}}\hskip 3.61371pt2X(M)\{8\},$
	$\displaystyle H^{-2}_{D}(M)$	$\displaystyle=$	$\displaystyle(M/(2XM))\{6\},$
	$\displaystyle H^{0}_{D}(M)$	$\displaystyle=$	$\displaystyle M\{2\},$
	$\displaystyle H^{i}_{D}(M)$	$\displaystyle=$	$\displaystyle 0\mbox{ for all other values of }i,$

where ${\mathrm{ker}}\hskip 3.61371pt2X(M)=\{t\in M|2Xt=0\}.$