ScalingStacks

Given a plane diagram $D$, a straighforward computation of cohomology groups $H^{i}(D)$ is daunting. These groups are cohomology groups of the graded complex $C(D)$ and the ranks of the abelian groups $C^{i}_{j}(D)$ grow exponentially in the complexity of $D.$ Probably there is no fast algorithm for computing $H^{i}(D)$, since these groups carry full information about the Jones polynomial, computing which is $\#P$-hard ([JVW]).

Yet, one can try to reduce $C(D)$ to a much smaller complex, albeit still exponentially large, but more practical for a computation. In this section we provide an example by simplifying $C(D)$ in the case when $D$ contains a chain of positive half-twists and apply our result by computing cohomology groups of $(2,n)$ torus links.

Let $D$ be a plane diagram with $n$ crossings and suppose that $D$ contains a subdiagram pictured below

Four possible resolutions of these two double points of $D$ produce diagrams $D(\ast 00),D(\ast 01),D(\ast 10),D(\ast 11)$:

Note that diagrams $D(\ast 01)$ and $D(\ast 10)$ are isomorphic and $D(\ast 00)$ is isomorphic to a union of $D(\ast 01)$ and a simple circle. The complex $\overline{C}(D)$ is isomorphic to the total complex of the bicomplex

where the differentials $\partial^{0}$ and $\partial^{1}$ are determined by the structure maps of the skew $\mathcal{I}$-cube $V_{D}\otimes E_{\mathcal{I}}$ (where $\mathcal{I}$ is the set of crossings of $D$). Denote this bicomplex by $C.$ To simplify notation we denote the diagram $D(\ast 01)$ by $D_{0}$ and $D(\ast 11)$ by $D_{1}.$

Then the bicomplex $C$ becomes

Clearly, the differential $\partial^{0},$ if restricted to the subcomplex $\overline{C}(D_{0})\otimes\mathbf{1}$ of $\overline{C}(D_{0})\otimes A,$ is injective, and so the total complex of the subbicomplex

of $C$ is acyclic. Denote this subbicomplex by $C_{s}$ and the quotient bicomplex by $C/C_{s}.$ The total complexes ${\mathrm{Tot}}(C)$ and ${\mathrm{Tot}}(C/C_{s})$ of $C$ and $C/C_{s}$ are quasi-isomorphic, so to compute the cohomology of $\overline{C}(D)={\mathrm{Tot}}(C)$ it suffices to find the cohomology of ${\mathrm{Tot}}(C/C_{s}).$

We next give a precise description of the bicomplex ${\mathrm{Tot}}(C/C_{s}).$ Let $u,l,w$ be maps of complexes

induced by surfaces

and

respectively. Note that each of these maps have degree $-1,$ and to make them homogeneous we need to shift gradings of our complexes appropriately. We will use the same notations for shifted maps since it will always be clear what the shifts are.

Let

be the map of complexes $v(t)=t\otimes X,t\in\overline{C}(D_{0}).$ The map $v$ has degree $-1.$ Denote by $u_{X}$ and $l_{X}$ the compositions

These are degree $-2$ maps of complexes and for each $i$ they induce degree $0$ maps $\overline{C}(D_{0})\{i\}\longrightarrow\overline{C}(D_{0})\{i-2\},$ also denoted $u_{X}$ and $l_{X}.$

We skip the proof which is a simple linear algebra. $\square$

We thus see that the cohomology $\overline{H}^{i}(D)$ of the diagram $D$ can be computed via the quotient complex ${\mathrm{Tot}}(C/C_{s})$ of $\overline{C}(D).$ The quotient complex is smaller than the original one and computing its cohomology requires less work. This reduction is not drastic since ranks of homogeneous components of complexes $\overline{C}(D)$ and ${\mathrm{Tot}}(C/C_{s})$ have the same order of magnitude, but a similar reduction (described next, when $D$ contains a long chain of positive twists) leads to an effective computation of $H^{i}(D)$ for certain diagrams $D.$

Suppose that a diagram $D$ contains a chain of $k$ positive half-twists

As before, denote by $D_{0}$ and $D_{1}$ diagrams that are suitable resolutions of the $k$-chain of $D.$

From our previous discussion we retain degree $-2$ maps $u_{X},l_{X}$ and degree $-1$ map $w$ between (appropriately shifted) complexes $\overline{C}(D_{0})$ and $\overline{C}(D_{1}).$ Let $C^{\prime}$ be the bicomplex

where

i.e.

The proof goes by induction on $k,$ induction base $k=2$ being given by Corollary 10, and consists of finding a suitable acyclic subcomplex to quotient by. We omit the details. $\square$

We conclude this section by applying this proposition to compute cohomology groups of $(2,k)$ torus links. Fix $k>0$ and denote by $D$ the diagram

of the $(2,k)$ torus link $T_{2,k}.$

The diagram $D_{0}$ is isomorphic to a simple circle and $D_{1}$ to a disjoint union of two simple circles. Then $u_{X}=l_{X}$ is the operator $A\to A$ of multiplication by $X$ and the bicomplex $C^{\prime}$ becomes a complex

Recalling that $x(D)=k$ and $y(D)=0,$ we get