ScalingStacks

Let $D$ be a plane diagram with $n-1$ double points and let $D_{1}$ be a diagram constructed from $D$ by adding a left-twisted curl. Denote by $\mathcal{I}^{\prime}$ the set of double points of $D_{1},$ by $a$ the double point in the curl and by $\mathcal{I}$ the set of double points of $D.$ There is a natural bijection of sets $\mathcal{I}\to\mathcal{I}^{\prime}\setminus\{a\},$ coming from identifying a double point of $D$ with the corresponding double point of $D_{1}.$ We will use this bijection to identify the two sets $\mathcal{I}$ and $\mathcal{I}^{\prime}\setminus\{a\}.$

The crossing $a$ of $D_{1}$ can be resolved in two ways. The 0-resolution of $a$ is a diagram $D_{2}$ which is a disjoint union of $D$ and a circle. The 1-resolution is a diagram isotopic to $D$ and we will identify this diagram with $D.$

In this section we will define a quasi-isomorphism of the complexes $C(D)$ and $C(D_{1}).$ This quasi-isomorphism arises from a splitting of the $\mathcal{I}^{\prime}$-cube $V_{D_{1}}$ as a direct sum of two cubes, $V_{D_{1}}=V^{\prime}\oplus V^{\prime\prime}.$ This splitting will induce a decomposition of the complex $C(D_{1})$ into a direct sum of an acyclic complex and a complex isomorphic to $C(D).$

Recall that $V_{D},V_{D_{1}}$ and $V_{D_{2}}$ are the cubes associated with the diagrams $D,D_{1}$ and $D_{2}$ respectively. $V_{D_{1}}$ has index set $\mathcal{I}^{\prime},$ while $V_{D}$ and $V_{D_{2}}$ are $\mathcal{I}$-cubes.

From the decomposition of $D_{2}$ as a union of $D$ and a simple circle we get a canonical isomorphism of cubes

where $V_{D}\otimes A$ is the $\mathcal{I}$-cube obtained from $V_{D}$ by tensoring graded $R$-modules $V_{D}(\mathcal{L}),\mathcal{L}\subset\mathcal{I}$ with $A$ and tensoring the structure maps $\xi^{V_{D}}_{a}(\mathcal{L})$ with the identity map of $A$.

Let $U\subset\mathbb{R}^{2}$ be a small neighborhood of $a$ that contains the curl:

The picture above depicts how the diagram $D_{1}$ looks inside $U.$ The boundary of $U$ is shown by a dashed circular line. Intersections of $U$ with diagrams $D$ and $D_{2}$ are depicted below

Outside of $U$ diagrams $D,D_{1}$ and $D_{2}$ coincide. It is explained in Section 4.3 how surfaces in $U\times[0,1]$, satisfying certain conditions, give rise to cube maps. Using this construction we now define three cube maps between cubes $V_{D}$ and $V_{D_{2}}:$

The map $m_{a}$ is associated to the following surface:

Here and further on we depict surfaces embedded in $U\times[0,1]$ by a sequence of their cross-sections $U\times\{t\},t\in[0,1],$ the leftmost one being the intersection of the surface with $U\times\{0\}$, the rightmost being the intersection with $U\times\{1\}.$ For such a surface $S\in U\times[0,1]$ we will call the projection $S\to[0,1]$ the height function of $S$. These surfaces will have only nondegenerate critical points relative to the height function. We depict enough sections of $S$ to make it obvious what surface we are considering, sometimes adding extra information, i.e., that the above surface has one saddle point and no other critical points relative to the height function.

The intersections $S\cap U\times\{0\},S\cap U\times\{1\}$ of the surface $S$ depicted above with the boundary disks $U\times\{0\},U\times\{1\}$ are isomorphic to the intersections $D_{2}\cap(U\times\{0\})$, respectively $D\cap(U\times\{1\}).$ Thus, $S$ defines a map $m_{a}$ from the cube $V_{D_{2}}$ to $V_{D}.$

The cube map $\Delta_{a}$ is associated to the surface

This surface has one saddle point and no other critical points relative to the height function.

$\iota_{a}$ is associated to

The only critical point of the height function is a local minimum.

The cube maps $m_{a},\Delta_{a},\iota_{a}$ are graded maps and change the grading by $-1,-1,1$ respectively. So let’s keep in mind that $m_{a},\Delta_{a},\iota_{a}$ become grading-preserving if we appropriately shift gradings of our cubes, for example,

are grading-preserving maps of cubes over $R{\mbox{-mod}_{0}}.$

The composition $m_{a}\iota_{a}$ is equal to the identity map from $V_{D}$ to itself. Denote by $\jmath_{a}$ the map

The map $\jmath_{a}$ is a graded map of degree $-1.$

Proof: It is enough to consider the case when $D$ is a single circle. Then $\mathcal{I}^{\prime}=\{a\},\mathcal{I}=\emptyset,V_{D}=A$ and $\iota_{a}(V_{D})=\mathbf{1}\otimes A.$ But

and, thus, $A\otimes A$ is a direct sum of $\mathbf{1}\otimes A$ and the $R$-submodule spanned by $\jmath_{a}\mathbf{1}$ and $\jmath_{a}X.$

$\square$

Note that

because $m_{a}\iota_{a}=Id.$

The $\mathcal{I}^{\prime}$-cube $V_{D_{1}}$ contains $V_{D}$ and $V_{D_{2}}$ as subcubes of codimension $1.$ Namely, we have canonical isomorphisms

Recall from Section 3.2 that $V_{D_{1}}(\ast 0)$ denotes the $\mathcal{I}^{\prime}\setminus\{a\}$-cube (i.e. $\mathcal{I}$-cube) with $V_{D_{1}}(\ast 0)(\mathcal{L})=V_{D_{1}}(\mathcal{L})$ for $\mathcal{L}\subset\mathcal{I},$ etc.

Under these isomorphisms the structure map $\xi_{a}^{V_{D_{1}}}$ (denoted below by $\xi_{a}$) for the $\mathcal{I}^{\prime}$-cube $V_{D_{1}}$

is equal to the map $m_{a}$ of $\mathcal{I}$-cubes, i.e., the following diagram is commutative

Using the splitting (73) of $V_{D_{2}}$ we can decompose the $\mathcal{I}^{\prime}$-cube $V_{D_{1}}$ as a direct sum of two $\mathcal{I}^{\prime}$-cubes as follows:

where

Some explanation: in the formula (79) $\jmath_{a}(V_{D})$ is a subcube of $V_{D_{2}}$ and, due to (75), $\jmath_{a}(V_{D})$ sits inside $V_{D_{1}}$ as a subcube of codimension 1. Equation (80) means that $V^{\prime}(\ast 1)(\mathcal{L})=0$ for all $\mathcal{L}\subset\mathcal{I}^{\prime}.$ Thus, $V^{\prime}(\mathcal{L})=\jmath_{a}(V_{D}(\mathcal{L}))\subset V_{D_{1}}(\mathcal{L})$ for $\mathcal{L}\subset\mathcal{I}^{\prime},$ if $\mathcal{L}$ does not contain $a.$ If $\mathcal{L}$ contains $a,$ $V^{\prime}(\mathcal{L})=0.$

Tensoring (78) 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-commutative $\mathcal{I}^{\prime}$-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 $\overline{C}(V_{D}\otimes E_{\mathcal{I}})[-1]\{-1\}.$

$\square$

Proof: We have a chain of isomorphisms of complexes

Proof: We have

$\square$

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

and

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