ScalingStacks

Pick an oriented link $L$ and a component $L^{\prime}$ of $L.$ Let $L_{0}$ be $L$ with the orientation of $L^{\prime}$ reversed and let $l$ be the linking number of $L^{\prime}$ and $L\setminus L^{\prime}.$ Fixing a plane diagram $D$ of $L$, we count $l$ as half the number of double intersection points in $D$ of $L^{\prime}$ with $L\setminus L^{\prime}$ with weights $+1$ or $-1$ according to the following convention

Denote by $D_{0}$ the diagram $D$ with the reversed orientation of $L^{\prime}.$ Since $D_{0}$ and $D$ are the same as unoriented diagrams, $\overline{C}(D_{0})=\overline{C}(D).$ Also

$$x(D_{0})=x(D)-2l,y(D_{0})=y(D)+2l$$

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We obtain

Let $K,K_{1}$ be oriented knots and $(-K)$ be $K$ with orientation reversed. In a similar fashion we deduce

Let $D$ be a diagram of an oriented link $L$ and denote by ${\mathrm{cm}}(L)$ the number of connected components of $L.$ Then it is easy to see that $C^{i}_{j}(D)=0$ if parities of $j$ and ${\mathrm{cm}}(L)$ differ. This observation implies

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

	$\displaystyle\cdots\longrightarrow 0$	$\displaystyle\longrightarrow$	$\displaystyle\overline{C}(D(\ast 00))\stackrel{{\scriptstyle\partial^{0}}}{{\longrightarrow}}\overline{C}(D(\ast 01))\{-1\}\oplus\overline{C}(D(\ast 10))\{-1\}\stackrel{{\scriptstyle\partial^{1}}}{{\longrightarrow}}$
		$\displaystyle\stackrel{{\scriptstyle\partial^{1}}}{{\longrightarrow}}$	$\displaystyle\overline{C}(D(\ast 11))\{-2\}\longrightarrow 0\longrightarrow\cdots$

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

	$\displaystyle\cdots\longrightarrow 0$	$\displaystyle\longrightarrow$	$\displaystyle\overline{C}(D_{0})\otimes A\stackrel{{\scriptstyle\partial^{0}}}{{\longrightarrow}}\overline{C}(D_{0})\{-1\}\oplus\overline{C}(D_{0})\{-1\}\stackrel{{\scriptstyle\partial^{1}}}{{\longrightarrow}}$
		$\displaystyle\stackrel{{\scriptstyle\partial^{1}}}{{\longrightarrow}}$	$\displaystyle\overline{C}(D_{1})\{-2\}\longrightarrow 0\longrightarrow\cdots$

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

$$0\longrightarrow C(D_{0})\otimes\mathbf{1}\stackrel{{\scriptstyle\partial^{0}}}{{\longrightarrow}}C(D_{0})\{-1\}\longrightarrow 0$$

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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

	$\displaystyle u$	$\displaystyle:$	$\displaystyle\overline{C}(D(\ast 00))\longrightarrow\overline{C}(D_{0})$		(139)
	$\displaystyle l$	$\displaystyle:$	$\displaystyle\overline{C}(D(\ast 00))\longrightarrow\overline{C}(D_{0})$		(140)
	$\displaystyle w$	$\displaystyle:$	$\displaystyle\overline{C}(D_{0})\longrightarrow\overline{C}(D_{1})$		(141)

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

$$v:\overline{C}(D_{0})\longrightarrow\overline{C}(D_{0})\otimes A=\overline{C}(D(\ast 00))$$

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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

$$u_{X}=u\circ v,\hskip 21.68121ptl_{X}=l\circ v.$$

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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

	$\displaystyle 0$	$\displaystyle\longrightarrow$	$\displaystyle\overline{C}(D_{0})\{k-1\}\stackrel{{\scriptstyle\partial^{0}}}{{\longrightarrow}}\overline{C}(D_{0})\{k-3\}\stackrel{{\scriptstyle\partial^{1}}}{{\longrightarrow}}\dots$
		$\displaystyle\stackrel{{\scriptstyle\partial^{k-3}}}{{\longrightarrow}}$	$\displaystyle\overline{C}(D_{0})\{3-k\}\stackrel{{\scriptstyle\partial^{k-2}}}{{\longrightarrow}}\overline{C}(D_{0})\{1-k\}\stackrel{{\scriptstyle\partial^{k-1}}}{{\longrightarrow}}\overline{C}(D_{1})\{-k\}\longrightarrow 0$

where

	$\displaystyle\partial^{k-1}$	$\displaystyle=$	$\displaystyle w$
	$\displaystyle\partial^{k-2}$	$\displaystyle=$	$\displaystyle u_{X}-l_{X}$
	$\displaystyle\partial^{k-3}$	$\displaystyle=$	$\displaystyle u_{X}+l_{X}$
	$\displaystyle\partial^{k-4}$	$\displaystyle=$	$\displaystyle u_{X}-l_{X}$
		$\displaystyle\dots$
	$\displaystyle\partial^{0}$	$\displaystyle=$	$\displaystyle u_{X}-(-1)^{k}l_{X},$

i.e.

$$\partial^{k-i}=u_{X}-(-1)^{i}l_{X},\hskip 21.68121pt\mbox{ for }2\leq i\leq k.$$

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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

	$\displaystyle 0$	$\displaystyle\longrightarrow$	$\displaystyle A\{k-1\}\longrightarrow A\{k-3\}\longrightarrow\dots$
		$\displaystyle\stackrel{{\scriptstyle 0}}{{\longrightarrow}}$	$\displaystyle A\{5-k\}\stackrel{{\scriptstyle 2X}}{{\longrightarrow}}A\{3-k\}\stackrel{{\scriptstyle 0}}{{\longrightarrow}}A\{1-k\}\stackrel{{\scriptstyle\Delta}}{{\longrightarrow}}A\otimes A\{-k\}\longrightarrow 0$

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

In this section by a surface $S$ in $\mathbb{R}^{4}$ we mean an oriented, compact surface $S,$ possibly with boundary, properly embedded in $\mathbb{R}^{3}\times[0,1].$ The boundary of $S$ is then a disjoint union

$$\partial S=\partial_{0}S\sqcup-\partial_{1}S$$

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of the intersections of $S$ with two boundary components of $\mathbb{R}^{3}\times[0,1]$:

	$\displaystyle\partial_{0}S$	$\displaystyle=$	$\displaystyle(S\cap\mathbb{R}^{3}\times\{0\})$
	$\displaystyle-\partial_{1}S$	$\displaystyle=$	$\displaystyle(S\cap\mathbb{R}^{3}\times\{1\})$

Note that $\partial_{0}S$ and $\partial_{1}S$ are oriented links in $\mathbb{R}^{3}.$

The surface $S$ can be represented by a sequence $J$ of plane diagrams of oriented links where every two consecutive diagrams in $J$ are related either by one of the Reidemeister moves I-IV (Section 4.1) or by one of the four moves depicted below (see [CS] where such representations by sequences of plane diagrams are studied in detail).

Following Carter-Saito [CS] and Fisher [Fs], we call these moves birth, death, fusion ([CS] and [Fs] deal with the non-oriented version of these moves). We call $J$ a representation of $S.$ The first diagram in the sequence $J$ is necessarily a diagram of oriented link $\partial_{0}S,$ and the last diagram is a diagram of $\partial_{1}S.$

The birth move consists of adding a simple closed curve to a diagram $D.$ Denote the new diagram by $D_{1}.$ Then $C(D_{1})=C(D)\otimes_{R}A$ and the unit map $\iota:R\to A$ of the algebra $A$ induces a map of complexes $C(D)\to C(D_{1}).$ This is the map we associate to the birth move.

The death move consists of removing a simple circle from a diagram $D_{1}$ to get a diagram $D.$ In this case the counit $\epsilon:A\to R$ induces a map of complexes $C(D_{1})\to C(D).$

Finally, to a fusion move between diagrams $D_{0}$ and $D_{1}$ we associate a map $C(D_{0})\to C(D_{1})$ corresponding to the elementary surface with one saddle point in the manner discussed in Section 4.3.

In Section 5 to each Reidemeister move between diagrams $D_{0}$ and $D_{1}$ we associated a quasi-isomorphism map of complexes $C(D_{0})\longrightarrow C(D_{1}).$

Given a representation $J$ of a surface $S$ by a sequence of diagrams, denote the first and last diagrams of $J$ by $J_{0}$ and $J_{1}$ respectively. Then to $J$ we can associate a map of complexes

$$\varphi_{J}:C(J_{0})\to C(J_{1})$$

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which is the composition of maps associated to elementary transformations between consecutive diagrams of $J.$ The map $\varphi_{J}$ induces a map of cohomology groups

$$\theta_{J}:H^{i,j}(J_{0})\to H^{i,j+\chi(S)}(J_{1}),\hskip 21.68121pti,j\in\mathbb{Z}.$$

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We now ready to state our main conjecture.

In other words, we conjecture that, after a suitable $\mathbb{Z}_{2}$ extension of the link cobordism category, our construction associates honest cohomology groups $H^{i}(L)$ to oriented links $L$ in $\mathbb{R}^{3}$ (and not just isomorphism classes of groups) and associates homomorphisms between these groups to isotopy classes of oriented surfaces embedded in $\mathbb{R}^{3}\times[0,1].$ In the categorical language, we expect to get a functor from the category of ($\mathbb{Z}_{2}$-extended) oriented link cobordisms to the category of bigraded $R$-modules and module homomorphisms.

Suppose that the above conjecture is true. Then, in the case of a closed oriented surface $S$ embedded in $\mathbb{R}^{4},$ the map $\theta_{S}$ of cohomology groups is a homomorphism from $R$ to itself (since $\partial S=\emptyset$ and the cohomology of the empty link is equal to the ground ring $R$). This homomorphism has degree $\chi(S)$ and is automatically $0$ when $\chi(S)<0.$ Thus, the conjectural invariants are zero whenever $S$ has empty boundary and the Euler characteristic of $S$ is negative. If, again, $\partial S=\emptyset$ and the Euler characteristic of $S$ is nonnegative (when $S$ is connected, $S$ is then necessarily a 2-sphere or a 2-torus), the homomorphism $\theta_{S}:R\to R$ is determined by $\theta_{S}(1)=kc^{\frac{\chi(S)}{2}}$ and amounts to an integer number $k.$ Hence, we expect to have integer-valued invariants of closed oriented surfaces with non-negative Euler characteristic, embedded in $\mathbb{R}^{4}.$