ScalingStacks

Let $U$ be a closed disk in the plane $\mathbb{R}^{2}$ and $\dot{U}$ the interior of $U$ so that $U=\partial U\cup\dot{U}.$ Let $T^{\prime}$ be a tangle in $(\mathbb{R}^{2}\setminus\dot{U})\times[0,1]$ with $m$ points (where $m$ is even) on the boundary $\partial U\times[0,1]$ and $T$ a generic projection of $T^{\prime}$ on $\mathbb{R}^{2}\setminus\dot{U}.$ The intersection of $T$ with $\partial U$ consists of $m$ points. Denote them by $p_{1},\dots,p_{m}$ (see an example on the diagram below, $\partial U$ is shown by a dashed circle).

Let $\mathcal{I}$ be the set of double points of $T.$ Pick two systems $Q_{0}$ and $Q_{1}$ of $\frac{m}{2}$ simple disjoints arcs in $U$ with ends in points $p_{1},\dots p_{m}$:

Then $Q_{0}\cup T$ and $Q_{1}\cup T$ (here and further on we denote them by $P_{0}$ and $P_{1}$ respectively) can be considered as two plane diagrams of links in $\mathbb{R}^{3}:$

To $P_{0}$ and $P_{1}$ there are associated $\mathcal{I}$-cubes $V_{P_{0}}$ and $V_{P_{1}}.$

Let $S$ be a compact oriented surface in $U\times[0,1]$ such that the boundary of $S$ is the union of $Q_{0}\times\{0\},Q_{1}\times\{1\}$ and $(p_{1}\cup\dots\cup p_{m})\times[0,1].$ To $S$ we associate an $\mathcal{I}$-cube map

as follows. For each $\mathcal{L}\subset\mathcal{I}$ we must construct a map

and check the commutativity of diagrams (29).

To $\mathcal{L}$ there is associated a resolution $T(\mathcal{L})$ of double points of $T.$ Thus $T(\mathcal{L})$ is a collection of simple closed curves and arcs in $\mathbb{R}^{2}\setminus\mbox{\.{U}}$ with ends in $p_{1},\dots,p_{m}.$ Then (by (40)

where $F$ is the functor described in Section 2.3 ( $T(\mathcal{L})\cup Q_{0}$ and $T(\mathcal{L})\cup Q_{1}$ are collections of simple closed curves on the plane, so that we can apply functor $F$ to them).

Let $S^{\prime}$ be a surface in $\mathbb{R}^{2}\times[0,1]$ which is $S$ inside $U\times[0,1]$ and $T(\mathcal{L})\times[0,1]$ outside $U\times[0,1].$ Map $F(S^{\prime})$

is a graded map of $R$-modules of degree $\chi(S^{\prime})=\chi(S)-\frac{m}{2}.$ Define $\psi_{S,\mathcal{L}}$ as this map, shifted by $|\mathcal{L}|$:

The commutativity condition (29) is immediate. We sum up our result as

Everything in this section extends to the case when the diagrams $Q_{0}$ and $Q_{1}$ are allowed to have simple closed circles in addition to $\frac{m}{2}$ simple disjoint acts joining points $p_{1},\dots p_{m}.$ For instance, $Q_{0}$ may look like

In this more general case to each compact oriented surface $S$ in $U\times[0,1]$ such that the boundary of $S$ is the union of $Q_{0}\times\{0\},Q_{1}\times\{1\}$ and $(p_{1}\cup\dots\cup p_{m})\times[0,1],$ in exactly the same fashion as before, we associate an $\mathcal{I}$-cube map

This map is a graded map of cubes over $R{\mbox{-mod}_{0}}$ of degree equal to the Euler characteristic of $S$ minus $\frac{m}{2}.$

Tensoring the map $\psi_{S}$ with the identity map of the skew-commutative $n$-cube $E_{\mathcal{I}}$ and passing to associated complexes, we obtain a map of complexes of graded $R$-modules

In general this map is not a morphism in the category $\mbox{Kom}(R{\mbox{-mod}_{0}})$ of complexes of graded $R$-modules and grading-preserving homomorphism, as it shifts the grading by $\chi(S)-\frac{m}{2},$ but $\psi^{\prime}_{S}$ becomes a morphism in $\mbox{Kom}(R{\mbox{-mod}_{0}})$ when the grading of $\overline{C}(P_{0})$ or $\overline{C}(P_{1})$ is appropriately shifted.