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

$$V_{D_{2}}=V_{D}\otimes A$$

(65)

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 $V_{D_{2}}:$

		$\displaystyle m_{a}:$	$\displaystyle V_{D_{2}}\longrightarrow V_{D}$		(66)
		$\displaystyle\Delta_{a}:$	$\displaystyle V_{D}\longrightarrow V_{D_{2}}$		(67)
		$\displaystyle\iota_{a}:$	$\displaystyle V_{D}\longrightarrow V_{D_{2}}$		(68)

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,

		$\displaystyle m_{a}:$	$\displaystyle V_{D_{2}}\longrightarrow V_{D}\{-1\}$		(69)
		$\displaystyle\Delta_{a}:$	$\displaystyle V_{D}\longrightarrow V_{D_{2}}\{-1\}$		(70)
		$\displaystyle\iota_{a}:$	$\displaystyle V_{D}\longrightarrow V_{D_{2}}\{1\}$		(71)

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

$$\jmath_{a}\stackrel{{\scriptstyle\mbox{\scriptsize def}}}{{=}}\Delta_{a}-\iota_{a}m_{a}\Delta_{a}:V_{D}\longrightarrow V_{D_{2}}$$

(72)

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

		$\displaystyle\jmath_{a}\mathbf{1}=$	$\displaystyle(\Delta_{a}-\iota_{a}m_{a}\Delta_{a})\mathbf{1}=X\otimes\mathbf{1}-\mathbf{1}\otimes X+cX\otimes X$
		$\displaystyle\jmath_{a}X=$	$\displaystyle(\Delta_{a}-\iota_{a}m_{a}\Delta_{a})X=X\otimes X$

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

$$m_{a}\jmath_{a}=m_{a}(\Delta_{a}-\iota_{a}m_{a}\Delta_{a})=0$$

(74)

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

	$\displaystyle V_{D_{1}}(\ast 0)$	$\displaystyle\cong$	$\displaystyle V_{D_{2}}$		(75)
	$\displaystyle V_{D_{1}}(\ast 1)$	$\displaystyle\cong$	$\displaystyle V_{D}\{-1\}$		(76)

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 $V_{D_{1}}$

$$\xi_{a}:V_{D_{1}}(\ast 0)\longrightarrow V_{D_{1}}(\ast 1)$$

(77)

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

$$\begin{CD}V_{D_{1}}(\ast 0)@>{\xi_{a}}>{}>V_{D_{1}}(\ast 1)\\ @V{}V{\cong}V@V{}V{\cong}V\\ V_{D_{2}}@>{m_{a}}>{}>V_{D}\{-1\}\end{CD}$$

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:

$$V_{D_{1}}=V^{\prime}\oplus V^{\prime\prime}$$

(78)

where

	$\displaystyle V^{\prime}(\ast 0)$	$\displaystyle=$	$\displaystyle\jmath_{a}(V_{D})$		(79)
	$\displaystyle V^{\prime}(\ast 1)$	$\displaystyle=$	$\displaystyle 0$		(80)
	$\displaystyle V^{\prime\prime}(\ast 0)$	$\displaystyle=$	$\displaystyle\iota_{a}(V_{D})$		(81)
	$\displaystyle V^{\prime\prime}(\ast 1)$	$\displaystyle=$	$\displaystyle V_{D_{1}}(\ast 1)$		(82)

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

$$V_{D_{1}}\otimes E_{\mathcal{I}^{\prime}}=(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})\oplus(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$$

(83)

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

$$\overline{C}(V_{D_{1}}\otimes E_{\mathcal{I}^{\prime}})=\overline{C}(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})\oplus\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$$

(84)

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

	$\displaystyle\overline{C}(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})$	$\displaystyle=$	$\displaystyle\overline{C}(V^{\prime}(\ast 0)\otimes E_{\mathcal{I}})$
		$\displaystyle=$	$\displaystyle\overline{C}(V_{D}\{1\}\otimes E_{\mathcal{I}})$
		$\displaystyle=$	$\displaystyle\overline{C}(V_{D}\otimes E_{\mathcal{I}})\{1\}$

Proof: We have

	$\displaystyle\overline{C}(D_{1})$	$\displaystyle=$	$\displaystyle\overline{C}(V_{D_{1}}\otimes E_{\mathcal{I}^{\prime}})$
		$\displaystyle=$	$\displaystyle\overline{C}(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})\oplus\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$
		$\displaystyle=$	$\displaystyle\overline{C}(V_{D}\otimes E_{\mathcal{I}})\{1\}\oplus\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$
		$\displaystyle=$	$\displaystyle\overline{C}(D)\{1\}\oplus(\mbox{Acyclic complex})$

$\square$

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

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

(85)

and

	$\displaystyle C(D_{1})$	$\displaystyle=$	$\displaystyle\overline{C}(D_{1})[x(D_{1})]\{2x(D_{1})-y(D_{1})\}$
		$\displaystyle=$	$\displaystyle\overline{C}(D_{1})[x(D)]\{2x(D)-y(D)-1\}.$

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

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

$$\epsilon_{a}:V_{D_{2}}\longrightarrow V_{D}$$

(86)

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:

$$\epsilon_{a}:V_{D_{2}}\longrightarrow V_{D}\{1\}$$

(87)

Let $\aleph$ be the map

$$\aleph=\iota_{a}-c\iota_{a}m_{a}\Delta_{a}:V_{D}\longrightarrow V_{D_{2}}$$

(88)

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

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

		$\displaystyle\aleph(\mathbf{1})$	$\displaystyle=\mathbf{1}\otimes\mathbf{1}-2c\mathbf{1}\otimes X$
		$\displaystyle\aleph(X)$	$\displaystyle=\mathbf{1}\otimes X$

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

$$A\otimes A=R\cdot\aleph(\mathbf{1})\oplus R\cdot\aleph(X)\oplus\Delta(A)$$

$\square$

Denote by $\wp$ the cube map

$$\wp=m_{a}-m_{a}\Delta_{a}\epsilon_{a}:V_{D_{2}}\longrightarrow V_{D}.$$

(90)

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

$$\wp\Delta_{a}=m_{a}\Delta_{a}-m_{a}\Delta_{a}\epsilon_{a}\Delta_{a}=m_{a}\Delta_{a}-m_{a}\Delta_{a}=0$$

(93)

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

The equality (92) is checked similarly:

	$\displaystyle\wp\aleph$	$\displaystyle=$	$\displaystyle(m_{a}-m_{a}\Delta_{a}\epsilon_{a})(\iota_{a}-c\iota_{a}m_{a}\Delta_{a})$
		$\displaystyle=$	$\displaystyle m_{a}\iota_{a}-cm_{a}\iota_{a}m_{a}\Delta_{a}-m_{a}\Delta_{a}\epsilon_{a}\iota_{a}+cm_{a}\Delta_{a}\epsilon_{a}\iota_{a}m_{a}\Delta_{a}$
		$\displaystyle=$	$\displaystyle{{\mathrm{Id}}}-cm_{a}\Delta_{a}+cm_{a}\Delta_{a}-c^{2}m_{a}\Delta_{a}m_{a}\Delta_{a}$
		$\displaystyle=$	$\displaystyle{{\mathrm{Id}}}-c^{2}m_{a}\Delta_{a}m_{a}\Delta_{a}$
		$\displaystyle=$	$\displaystyle{{\mathrm{Id}}}$

The third equality in the computation above follows from the identities

$$m_{a}\iota_{a}={{\mathrm{Id}}},\hskip 21.68121pt\epsilon_{a}\iota_{a}=-c.$$

(94)

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:

$$V_{D_{1}}=V^{\prime}\oplus V^{\prime\prime}$$

(95)

where

	$\displaystyle V^{\prime}(\ast 0)$	$\displaystyle=$	$\displaystyle 0$		(96)
	$\displaystyle V^{\prime}(\ast 1)$	$\displaystyle=$	$\displaystyle\aleph(V_{D})\{-1\}\subset V_{D_{2}}\{-1\}=V_{D_{1}}(\ast 1)$		(97)
	$\displaystyle V^{\prime\prime}(\ast 0)$	$\displaystyle=$	$\displaystyle V_{D}=V_{D_{1}}(\ast 0)$		(98)
	$\displaystyle V^{\prime\prime}(\ast 1)$	$\displaystyle=$	$\displaystyle\Delta_{a}(V_{D})\{-1\}\subset V_{D_{2}}\{-1\}=V_{D_{1}}(\ast 1)$		(99)

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

$$V_{D_{1}}\otimes E_{\mathcal{I}^{\prime}}=(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})\oplus(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$$

(100)

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

$$\overline{C}(V_{D_{1}}\otimes E_{\mathcal{I}^{\prime}})=\overline{C}(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})\oplus\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$$

(101)

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

	$\displaystyle\overline{C}(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})$	$\displaystyle=$	$\displaystyle\overline{C}(V^{\prime}(\ast 1)\otimes E_{\mathcal{I}})[-1]$
		$\displaystyle=$	$\displaystyle\overline{C}(V_{D}\{-2\}\otimes E_{\mathcal{I}})[-1]$
		$\displaystyle=$	$\displaystyle\overline{C}(V_{D}\otimes E_{\mathcal{I}})[-1]\{-2\}$
		$\displaystyle=$	$\displaystyle\overline{C}(D)[-1]\{-2\}$

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

	$\displaystyle\overline{C}(D_{1})$	$\displaystyle=$	$\displaystyle\overline{C}(V_{D_{1}}\otimes E_{\mathcal{I}^{\prime}})$
		$\displaystyle=$	$\displaystyle\overline{C}(V^{\prime}\otimes E_{\mathcal{I}^{\prime}})\oplus\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$
		$\displaystyle=$	$\displaystyle\overline{C}(D)[-1]\{-2\}\oplus\overline{C}(V^{\prime\prime}\otimes E_{\mathcal{I}^{\prime}})$
		$\displaystyle=$	$\displaystyle\overline{C}(D)[-1]\{-2\}\oplus(\mbox{Acyclic complex})$

$\square$

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

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

(102)

and

	$\displaystyle C(D_{1})$	$\displaystyle=$	$\displaystyle\overline{C}(D_{1})[x(D_{1})]\{2x(D_{1})-y(D_{1})\}$
		$\displaystyle=$	$\displaystyle\overline{C}(D_{1})[x(D)+1]\{2x(D)-y(D)+2\}$

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

Let $D$ and $D_{1}$ be two diagrams that differ as depicted below

In this section we will construct a quasi-isomorphism of complexes $C(D)$ and $C(D_{1}).$

We assume that $D$ has $n-2$ double points. Consequently, $D_{1}$ has $n$ double points. Let $\mathcal{I}^{\prime}$ be the set of double points of $D_{1},$ let $\mathcal{I}$ be $\mathcal{I}^{\prime}\setminus\{a,b\}$ where $a$ and $b$ are double points of $D_{1}$ depicted above. We identify $\mathcal{I}$ with the double points set of $D.$

Denote by $d$ the differential of the complex $\overline{C}(D_{1}).$ Consider diagrams $D_{1}(\ast 00),D_{1}(\ast 01),D_{1}(\ast 10),D_{1}(\ast 11)$ obtained by resolving double points $a$ and $b$ of $D_{1}:$

(e.g., $D_{1}(\ast 01)$ is constructed from $D_{1}$ by taking $0$-resolution of $a$ and $1$-resolution of $b,$ etc.) Each of these four diagrams has $\mathcal{I}$ as the set of its double points.

To a diagram $D_{1}(\ast uv)$ where $u,v\in\{0,1\}$ there is associated the complex $\overline{C}(D_{1}(\ast uv))$ of graded $R$-modules. Denote by $d_{uv}$ the differential in this complex:

$$d_{ab}:\overline{C}(D_{1}(\ast uv))\longrightarrow\overline{C}(D_{1}(\ast uv)).$$

(103)

We denote by $d_{uv}^{(i)}$ the differential in shifted complexes

$$d_{uv}^{(i)}:\overline{C}(D_{1}(\ast uv))[i]\{i\}\longrightarrow\overline{C}(D_{1}(\ast uv))[i]\{i\}\mbox{ for }i\in\mathbb{Z}.$$

(104)

The commutative $\mathcal{I}$-cube $V_{D_{1}}$ can be viewed as a commutative square of $\mathcal{I}^{\prime}$-cubes

$$\begin{CD}V_{D_{1}(\ast 00)}@>{\phi_{2}}>{}>V_{D_{1}(\ast 01)}\{-1\}\\ @V{\phi_{1}}V{}V@V{}V{\phi_{3}}V\\ V_{D_{1}(\ast 10)}\{-1\}@>{\phi_{4}}>{}>V_{D_{1}(\ast 11)}\{-2\}\end{CD}$$

where $\phi_{i},1\leq i\leq 4$ denote the corresponding cube maps. Recall that these cube maps are associated to certain elementary surfaces (see Sections 4.2, 4.3) that have one saddle point relative to the height function and no other critical points. For example, $\phi_{1}$ is associated to the surface

The maps $\phi_{i}$ induce maps $\psi_{i}$ between complexes:

	$\displaystyle\psi_{1}$	$\displaystyle:$	$\displaystyle\overline{C}(D_{1}(\ast 00))\longrightarrow\overline{C}(D_{1}(\ast 10))\{-1\}$
	$\displaystyle\psi_{2}$	$\displaystyle:$	$\displaystyle\overline{C}(D_{1}(\ast 00))\longrightarrow\overline{C}(D_{1}(\ast 01))\{-1\}$
	$\displaystyle\psi_{3}$	$\displaystyle:$	$\displaystyle\overline{C}(D_{1}(\ast 01))[-1]\{-1\}\longrightarrow\overline{C}(D_{1}(\ast 11))[-1]\{-2\}$
	$\displaystyle\psi_{4}$	$\displaystyle:$	$\displaystyle\overline{C}(D_{1}(\ast 10))[-1]\{-1\}\longrightarrow\overline{C}(D_{1}(\ast 11))[-1]\{-2\}$

We can decompose $\overline{C}(D_{1}),$ considered as a $\mathbb{Z}\oplus\mathbb{Z}$-graded $R$-module (see the end of Section 3.1), into the following direct sum of $\mathbb{Z}\oplus\mathbb{Z}$-graded $R$-modules.

	$\displaystyle\overline{C}(D_{1})$	$\displaystyle=$	$\displaystyle\overline{C}(D_{1}(\ast 00))\oplus\overline{C}(D_{1}(\ast 01))[-1]\{-1\}$
		$\displaystyle\oplus$	$\displaystyle\overline{C}(D_{1}(\ast 10))[-1]\{-1\}\oplus\overline{C}(D_{1}(\ast 11))[-2]\{-2\}$

Let’s say a few words about this decomposition: $\overline{C}(D_{1})$ is the direct sum of $R$-modules $V_{D_{1}}(\mathcal{L})$ which sit in the vertices of the $\mathcal{I}^{\prime}$-cube $V_{D_{1}}.$ Since we presented this cube as a commutative square of $\mathcal{I}$-cubes $V_{D_{1}(\ast uv)}\{-u-v\}$ for $u,v\in\{0,1\},$ the above decomposition results. Well, almost. Indeed, when we pass from $\mathcal{J}$-cubes to complexes we tensor with the fixed skew cube $E_{\mathcal{J}}.$ To define the left hand side of the above formula we tensor $V_{D_{1}}$ with the skew $\mathcal{I}^{\prime}$-cube $E_{\mathcal{I}^{\prime}},$ while for the right hand side similar tensor products are formed with the skew $\mathcal{I}$-cube $E_{\mathcal{I}}.$ Therefore, we must say how we identify $R$-modules which sit in the vertices of $E_{\mathcal{I}^{\prime}}$ with $R$-modules sitting in the vertices of $E_{\mathcal{I}}.$ For $D_{1}(\ast 00):$ we map $E_{\mathcal{I}}(\mathcal{L})$ where $\mathcal{L}\subset\mathcal{I}$ to $E_{\mathcal{I}^{\prime}}(\mathcal{L})$ by sending $z\in o(\mathcal{L})$ to $z\in o(\mathcal{L}).$ For $D_{1}(\ast 10)$: map $E_{\mathcal{I}}(\mathcal{L})$ where $\mathcal{L}\subset\mathcal{I}$ to $E_{\mathcal{I}^{\prime}}(\mathcal{L}a)$ by sending $z\in o(\mathcal{L})$ to $za\in o(\mathcal{L}a).$ Similarly for $D_{1}(\ast 01).$ For $D_{1}(\ast 11)$: map $E_{\mathcal{I}}(\mathcal{L})$ to $E_{\mathcal{I}^{\prime}}(\mathcal{L}ab)$ by sending $z\in o(\mathcal{L})$ to $zab\in o(\mathcal{L}ab).$ This is not a canonical choice, since we could have sent $z$ to $zba$ and would have gotten minus the original map. So, to define the latter map, we implicitly fix an ordering of $a$ and $b.$

Note that the above decomposition is not a direct sum of complexes, as the differential $d_{uv}^{(-u-v)}$ of $\overline{C}(D_{1}(\ast uv))$ differs from $d$ restricted to $\overline{C}(D_{1}(\ast uv))[-u-v]\{-u-v\}\subset\overline{C}(D_{1}),$ except when $u=v=1.$ Exactly, we have

$$\begin{array}[]{lllll}dx&=&d_{00}x+[-1]\psi_{1}x+[-1]\psi_{2}x&\mbox{ for }&x\in\overline{C}(D_{1}(\ast 00))\\ dx&=&-d_{01}^{(-1)}x-[-1]\psi_{3}x&\mbox{ for }&x\in\overline{C}(D_{1}(\ast 01))[-1]\{-1\}\\ dx&=&-d_{10}^{(-1)}x+[-1]\psi_{4}x&\mbox{ for }&x\in\overline{C}(D_{1}(\ast 10))[-1]\{-1\}\\ dx&=&d_{11}^{(-2)}x&\mbox{ for }&x\in\overline{C}(D_{1}(\ast 11))[-2]\{-2\}\end{array}$$

Some explanation: applying $\psi_{1}$ to $x\in\overline{C}(D_{1}(\ast 00))$ we get an element of $\overline{C}(D_{1}(\ast 10))\{-1\},$ so that we shift $\psi_{1}x$ by $[-1]$ to land it in $\overline{C}(D_{1}(\ast 10))[-1]\{-1\}\subset\overline{C}(D_{1}),$ etc. Various signs in the above formulas come from our previous four identifications of the skew cube $E_{\mathcal{I}}$ with codimension $2$ faces of $E_{\mathcal{I}^{\prime}}.$

Let $\alpha$ be the map of complexes

$$\alpha:\overline{C}(D_{1}(\ast 01))[-1]\{-1\}\longrightarrow\overline{C}(D_{1}(\ast 10))[-1]\{-1\}$$

(105)

associated to the surface

Considered as a map of $\mathbb{Z}\oplus\mathbb{Z}$-graded $R$-modules, $\alpha$ is grading-preserving.

Let $\beta$ be the map of complexes

$$\beta:\overline{C}(D_{1}(\ast 11))[-2]\{-2\}\longrightarrow\overline{C}(D_{1}(\ast 10))[-1]\{-1\}$$

(106)

associated to the surface

Note that $\beta$ is a graded map of degree $(-1,0).$

Let $X_{1},X_{2},X_{3}$ be $R$-submodules of $\overline{C}(D_{1})$ given by

	$\displaystyle X_{1}$	$\displaystyle=$	$\displaystyle\{z+\alpha(z)|z\in\overline{C}(D_{1}(\ast 01))[-1]\{-1\}\}$		(107)
	$\displaystyle X_{2}$	$\displaystyle=$	$\displaystyle\{z+dw|z,w\in\overline{C}(D_{1}(\ast 00))\}$		(108)
	$\displaystyle X_{3}$	$\displaystyle=$	$\displaystyle\{z+\beta(w)|z,w\in\overline{C}(D_{1}(\ast 11))[-2]\{-2\}\}$		(109)

Proof: Let us first check that $X_{1},X_{2}$ and $X_{3}$ are direct sums of their graded components. For $X_{2}$ it follows from the fact that $\overline{C}(D_{1}(\ast 00))$ is a direct sum of its graded components and $d$ is graded of degree $(1,0).$ Submodule $X_{3}$ is graded because $\overline{C}(D_{1}(\ast 11))[-2]\{-2\}$ is a direct sum of its graded components and $\beta$ is a graded map. Finally, $X_{1}$ is graded since $\alpha$ is grading-preserving.