ScalingStacks

We are given two diagrams with $n$ double points each, $D_{1}$ and $D_{2}$, that differ as depicted below.

In this section we will construct a quasi-isomorphism of complexes $C(D_{1})$ and $C(D_{2}).$

Let $\mathcal{I}^{\prime}$ be the set of double points of $D_{1}.$ We have $\mathcal{I}^{\prime}=\mathcal{I}\sqcup\{p_{1},q_{1},r_{1}\}$ where $\mathcal{I}$ are all double points not shown on the above picture. In particular, we can identify $\mathcal{I}\sqcup\{p_{2},q_{2},r_{2}\}$ with the set of double points of $D_{2}.$

For starters, consider the diagrams $D_{1}(\ast 0),D_{1}(\ast 1),D_{2}(\ast 0),D_{2}(\ast 1),$ obtained by resolving double points $r_{1}$ of $D_{1}$ and $r_{2}$ of $D_{2}$:

Note that diagrams $D_{1}(\ast 1)$ and $D_{2}(\ast 1)$ are isomorphic and that diagrams $D_{1}(\ast 0)$ and $D_{2}(\ast 0)$ represent isotopic links.

We decompose $\overline{C}(D_{1})$ and $\overline{C}(D_{2})$ into following direct sums:

These are direct sum decompositions of $\mathbb{Z}\oplus\mathbb{Z}$-graded $R$-modules, not complexes. The diagrams $D_{i}(\ast uv0)$ for $i=1,2$ and $u,v\in\{0,1\}$ are depicted below

For all $i,u,v$ as above, we identify the set of double points of $D_{i}(\ast uv0)$ with $\mathcal{I}.$ To fix the direct decomposition (122) we need identifications between the skew cube $E_{\mathcal{I}}$ and codimension $3$ facets of $E_{\mathcal{I}^{\prime}}.$ From the discussion in the previous section it should be clear how these identifications are chosen. For instance, for $D_{1}(\ast 110),$ we map $E_{\mathcal{I}}$ to a codimension $3$ facet of $E_{\mathcal{I}^{\prime}}$ via maps $E_{\mathcal{I}}(\mathcal{L})\to E_{\mathcal{I}^{\prime}}(\mathcal{L}p_{1}q_{1})$ given by $o(\mathcal{L})\ni z\longmapsto zp_{1}q_{1}\in o(\mathcal{L}\sqcup\{p_{1},q_{1}\}).$

Let $\tau_{1}$ be the map of complexes

associated to the surface

Relative to the height function this surface has two critical points, one of which is a saddle point and the other – a local minimum. Considered as a map of $\mathbb{Z}\oplus\mathbb{Z}$-graded $R$-modules, $\tau_{1}$ is grading-preserving.

Let $\delta_{1}$ be the map of complexes

associated to the surface

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

where $d_{1}$ denotes the differential of $\overline{C}(D_{1}).$ Warning: These $X_{1},X_{2},X_{3}$ have no relation to the complexes $X_{1},X_{2},X_{3}$ considered in Section 5.3.

Propositions 19-21 below can be proved in the same fashion as Propositions 17 and 18 of the previous section. For this reason and to keep this paper from being too lengthy the proofs are omitted.

Let $\tau_{2}$ be the map of complexes

associated to the surface

Considered as a map of $\mathbb{Z}\oplus\mathbb{Z}$-graded $R$-modules, $\tau_{2}$ is grading-preserving.

Let $\delta_{2}$ be the map of complexes

associated to the surface

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

where $d_{2}$ stands for the differential of $\overline{C}(D_{2}).$

Proof: Parts 1 and 2 of this proposition are proved similarly to Proposition 18. The isomorphism $X_{1}\cong Y_{1}$ comes from the diagram isomorphisms

These diagram isomorphisms induce isomorphisms of complexes

which allow us to identify $x$ in the definition (125) of $X_{1}$ with $x$ in the definition (128) of $Y_{1}$ and, similarly, identify $y$’s. An isomorphism $X_{1}\cong Y_{1}$ of complexes is then given by

$\square$

The above isomorphism of complexes $X_{1}$ and $Y_{1}$ induces a quasi-isomorphism of $\overline{C}(D_{1})$ and $\overline{C}(D_{2}).$ Note that $x(D_{1})=x(D_{2})$ and $y(D_{1})=y(D_{2}).$ Therefore, the complexes $C(D_{1})$ and $C(D_{2})$ are quasi-isomorphic and the cohomology groups $H^{i}(D_{1})$ and $H^{i}(D_{2})$ are isomorphic as graded $R$-modules. Q.E.D.

This finishes the proof of Theorem 1.