ScalingStacks

Let $D$ be a diagram with $n-1$ double points and let $D_{1}$ be a diagram constructed from $D$ by adding a right-twisted curl. Denote by $a$ the new crossing that appears in the curl. Let $\mathcal{I}$ be the set of crossings of $D$ and $\mathcal{I}^{\prime}$ the set of crossings of $D_{1}.$ We have a natural bijection of sets $\mathcal{I}\to\mathcal{I}^{\prime}\setminus\{a\}$ and use it to identify these two sets.

Crossing $a$ can be resolved in two ways. $0$-resolution gives a diagram, isotopic to $D$ and canonically identified with $D$. $1$-resolution produces a diagram, denoted $D_{2},$ which is a disjoint union of $D$ and a simple circle.

Note that diagrams $D$ and $D_{2}$ are the same as diagrams $D$ and $D_{2}$ from Section 5.1 and we will be using cube maps $m_{a},\Delta_{a},\iota_{a}$ defined in that section. Also define a map

where $\epsilon_{a}$ is associated to the surface

This surface has one critical point relative to the height function and it is a local maximum. The cube map $\epsilon_{a}$ changes the grading by $1$ and becomes grading preserving after an appropriate shift:

Let $\aleph$ be the map

$\aleph$ is graded of degree $1.$

Proof: It suffices to check this when $D$ is a simple circle. Then

The $R$-submodule of $A\otimes A$ generated by these two vectors complements $\Delta(A)$ and there is direct sum decomposition of $R$-modules

$\square$

Denote by $\wp$ the cube map

Note that $\wp$ is a graded map of degree $-1.$

Proof: Map $\wp\Delta_{a}:V_{D}\longrightarrow V_{D}$ is the zero map because

(the second equality uses that $\epsilon_{a}\Delta_{a}=\mbox{Id}.$)

The equality (92) is checked similarly:

The third equality in the computation above follows from the identities

The fifth equality is implied by $m_{a}\Delta_{a}m_{a}\Delta_{a}=0.$ This identity follows from the nilpotence property $m\Delta m\Delta=0$ of the structure maps $m$ and $\Delta$ of $A.$

Using the splitting (89) of $V_{D_{2}}$ and Lemma 1, we can decompose the $\mathcal{I}^{\prime}$-cube $V_{D_{1}}$ as a direct sum of two $\mathcal{I}^{\prime}$-cubes as follows:

where

Tensoring (95) with $E_{\mathcal{I}^{\prime}}$ we get a splitting of skew-commutative $\mathcal{I}^{\prime}$-cubes

This induces a splitting of complexes associated to these skew $\mathcal{I}^{\prime}$-cubes

Proof: The complex $\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$ is isomorphic to the cone of the identity map of the complex $\overline{C}(V_{D}\otimes E_{\mathcal{I}})[-1].$

$\square$

Proof: We have a chain of isomorphisms of complexes

The first isomorphism here follows from (96) and is obtained by fixing an isomorphism between skew-commutative $\mathcal{I}$-cubes $E_{\mathcal{I}^{\prime}}(\ast 1)$ and $E_{\mathcal{I}}.$ The second isomorphism comes from an isomorphism $V^{\prime}(\ast 1)=V_{D}\{-2\},$ induced by $\aleph.$

$\square$

Proof: We have

$\square$

Note that $x(D_{1})=x(D)+1$ and $y(D_{1})=y(D).$ By (45)

and

Therefore, complexes $C(D)$ and $C(D_{1})$ are quasiisomorphic. Q.E.D.