ScalingStacks

Setting $c=0$ and taking $\mathbb{Z}$ instead of $R=\mathbb{Z}[c]$ as the base ring, everything we did in Sections 2, 4 and 5 goes through in exactly the same manner. The role of the ring $A$ will be played by the free graded abelian group $\cal A$ of rank $2$ with generators $\mathbf{1}$ and $X$ in degrees $1$ and $-1$ correspondingly. $\cal A$ has commutative algebra and cocommutative coalgebra structures:

	$\displaystyle\mathbf{1}^{2}=\mathbf{1},\hskip 7.22743pt\mathbf{1}X=X\mathbf{1}=X,\hskip 7.22743ptX^{2}=0$		(149)
	$\displaystyle\Delta(\mathbf{1})=\mathbf{1}\otimes X+X\otimes\mathbf{1},\hskip 7.22743pt\Delta(X)=X\otimes X$		(150)

and the identity (16) holds. By abuse of notations, we use $m$ and $\Delta$ to denote multiplication and comultiplication in ${\cal A}.$ Earlier $m$ and $\Delta$ were used to denote multiplication and comultiplication in $A.$ As in Section 2.3, we construct a functor $\cal F$ from the category $\mathcal{M}$ of closed one-manifolds and cobordisms between them to the category of graded abelian groups and graded homomorphisms. To a disjoint union of $k$ circles functor $\cal F$ assigns the group ${\cal A}^{\otimes k}.$ To elementary surfaces $S_{2}^{1},S_{1}^{2},S_{0}^{1},S_{1}^{0},S_{2}^{2}$ and $S_{1}^{1}$ (see Section 2.3) functor $\cal F$ assigns maps $m,\Delta,\iota,\epsilon,\mbox{Perm}$ and Id between suitable tensor powers of ${\cal A}.$ The maps $\iota:\mathbb{Z}\to{\cal A}$ and $\epsilon:{\cal A}\to\mathbb{Z}$ are given by

$$\iota(1)=\mathbf{1},\hskip 14.45377pt\epsilon(\mathbf{1})=0,\hskip 14.45377pt\epsilon(X)=1,$$

(151)

while Perm is just the permutation map ${\cal A}\otimes{\cal A}\to{\cal A}\otimes{\cal A}.$

To a diagram $D$ of an oriented link $L$ we can then associate a commutative $\mathcal{I}$-cube ${\cal V}_{D}$ of graded abelian groups and grading-preserving homomorphism, by the same procedure as the one described in Section 4.2, using the functor $\cal F$ instead of $F$. In particular, for $\mathcal{L}\subset\mathcal{I}$ we have ${\cal V}_{D}(\mathcal{L})={\cal F}(D(\mathcal{L}))\{-|\mathcal{L}|\}$ where $\{k\}$ shifts the grading down by $k.$

Let $\cal{AB}$ be the category of graded abelian groups and grading-preserving homomorphisms. Let ${\cal E}_{\mathcal{I}}$ be the skew-commutative $\mathcal{I}$-cube $E_{\mathcal{I}}\otimes_{R}\mathbb{Z}$ over $\cal{AB}.$

Tensoring ${\cal V}_{D}$ with ${\cal E}_{\mathcal{I}}$ over $\mathbb{Z},$ we get a skew commutative $\mathcal{I}$-cube ${\cal V}_{D}\otimes{\cal E}_{\mathcal{I}}$ over the category $\cal{AB}.$ From this skew-commutative $\mathcal{I}$-cube we get a complex $\overline{{\cal C}}({\cal V}_{D}\otimes{\cal E}_{\mathcal{I}})$ of graded abelian groups with a grading-preserving differential (Section 3.4). Denote this complex by $\overline{\cal C}(D)$ and by ${\cal C}(D)$ the shifted complex

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

(152)

If we consider $\mathbb{Z}$ as a graded $R$-module, concentrated in degree $0,$ so that $c\mathbb{Z}=0,$ then

$$\overline{\cal C}(D)=\overline{C}(D)\otimes_{R}\mathbb{Z}\hskip 14.45377pt\mbox{ and }\hskip 14.45377pt{\cal C}(D)=C(D)\otimes_{R}\mathbb{Z}.$$

(153)

To a plane diagram $D$ of an oriented link $L$ we thus associate a complex of graded abelian groups ${\cal C}(D)$. Denote the $i$-th cohomology group of the $j$-th graded summand of ${\cal C}(D)$ by ${\cal H}^{i,j}(D).$ These cohomology groups are finitely-generated abelian groups. For each diagram $D$ as we vary $i$ and $j$ over all integers, only a finite number of these groups are non-zero.

Proof: Set $c=0$ in the proof of Theorem 1. $\square$

For a diagram $D$ of the link $L,$ denote the isomorphism classes of ${\cal H}^{i,j}(D)$ by ${\cal H}^{i,j}(L).$

Denote by $\overline{{\cal C}}^{i}(D)$ (respectively, ${\cal C}^{i}(D)$) the $i$-th group of the complex $\overline{{\cal C}}(D)$ (respectively, ${\cal C}(D)$) and by $\overline{{\cal C}}^{i}_{j}(D)$ (respectively, by ${\cal C}^{i}_{j}(D)$) the $j$-th graded component of $\overline{{\cal C}}^{i}(D)$ (respectively, ${\cal C}^{i}(D)$), so that $\overline{{\cal C}}^{i}(D)={\mathop{\oplus}\limits_{j\in\mathbb{Z}}}\overline{{\cal C}}^{i}_{j}(D),$ respectively, ${\cal C}^{i}(D)={\mathop{\oplus}\limits_{j\in\mathbb{Z}}}{{\cal C}}^{i}_{j}(D).$ For an diagram $D$ denote by ${\cal H}^{i}(D)$ the graded abelian group ${\mathop{\oplus}\limits_{j\in\mathbb{Z}}}{\cal H}^{i,j}(D).$ In other words, ${\cal H}^{i}(D)$ is the $i$-th cohomology group of ${\cal C}(D).$ Denote by $\overline{{\cal H}}^{i}(D)$ the $i$-th cohomology group of the complex $\overline{\cal C}(D)$ and by $\overline{{\cal H}}^{i,j}(D)$ the $j$-th graded component of $\overline{{\cal H}}^{i}(D),$ so that $\overline{{\cal H}}^{i}(D)={\mathop{\oplus}\limits_{j\in\mathbb{Z}}}\overline{{\cal H}}^{i,j}(D).$

The Kauffman bracket of an oriented link $L$ is equal to the graded Euler characteristic of the cohomology groups ${\cal H}^{i,j}(L),$ as stated in the following proposition.

The proof is completely analogous to that of formula (47). $\square$

The statements and proofs of Propositions 22-24 transfer without change to the case of cohomology groups ${\cal H}^{i,j},$ as indicated below.

Let $L$ be an oriented link and $L^{\prime}$ a component of $L.$ Denote by $l$ the linking number of $L^{\prime}$ with its complement $L\setminus L^{\prime}$ in $L.$ Let $L_{0}$ be the link $L$ with the orientation of $L^{\prime}$ reversed.

Similarly to Proposition 24 we can prove

Let $L$ be an oriented link and denote by $L^{!}$ the mirror image of $L$. Let $D$ be a diagram of $L$ with $n$ crossings, $\mathcal{I}$ the set of these crossings, and $D^{!}$ the corresponding diagram of $L^{!}$:

If $M$ is a graded abelian group, $M={\mathop{\oplus}\limits_{j\in\mathbb{Z}}}M_{j},$ define the dual graded abelian group $M^{\ast}$ by $(M^{\ast})_{j}=\mbox{Hom}(M_{-j},\mathbb{Z}).$ The dual map $f^{\ast}:N^{\ast}\to M^{\ast}$ of a map $f:M\to N$ is defined as the dual of $f$ in the sense of linear algebra.

For $\mathcal{L}\subset\mathcal{I}$ denote by $\widetilde{\mathcal{L}}$ the complement $\mathcal{I}\setminus\mathcal{L}.$

Let ${\cal V}$ be a commutative $\mathcal{I}$-cube over the category $\cal{AB}$ of graded abelian groups and grading-preserving homomorphisms. Define the dual cube ${\cal V}^{\ast}$ by

$${\cal V}^{\ast}(\mathcal{L})=({\cal V}(\widetilde{\mathcal{L}}))^{\ast}$$

(158)

and the structure map

$$\xi_{a}^{{\cal V}^{\ast}}:{\cal V}^{\ast}(\mathcal{L})\longrightarrow{\cal V}^{\ast}(\mathcal{L}a)$$

(159)

of ${\cal V}^{\ast}$ being the dual of the structure map

$$\xi_{a}^{{\cal V}}:{\cal V}(\widetilde{\mathcal{L}}\setminus a)\longrightarrow{\cal V}(\widetilde{\mathcal{L}})$$

(160)

of ${\cal V}.$

Denote by $\{s\}$ the automorphism of the category $\cal{AB}$ that shifts the grading down by $s.$ If ${\cal V}$ is a commutative $\mathcal{I}$-cube over $\cal{AB},$ denote by ${\cal V}\{s\}$ the commutative $\mathcal{I}$-cube ${\cal V}$ with the grading of each group ${\cal V}(\mathcal{L})$ shifted down by $s.$

Proof: Introduce a basis $\{\mathbf{1}^{\ast},X^{\ast}\}$ in the abelian group ${\cal A}^{\ast}=\mbox{Hom}({\cal A},\mathbb{Z})$ by

$$\mathbf{1}^{\ast}(\mathbf{1})=0,\mathbf{1}^{\ast}(X)=1,X^{\ast}(\mathbf{1})=1,X^{\ast}(X)=0$$

(161)

Denote by $m^{\ast},\Delta^{\ast}$ maps dual to $\Delta$ and $m$ respectively:

	$\displaystyle m^{\ast}$	$\displaystyle:$	$\displaystyle{\cal A}^{\ast}\otimes{\cal A}^{\ast}\longrightarrow{\cal A}^{\ast}$
	$\displaystyle\Delta^{\ast}$	$\displaystyle:$	$\displaystyle{\cal A}^{\ast}\longrightarrow{\cal A}^{\ast}\otimes{\cal A}^{\ast}$

Then, in the basis $\{\mathbf{1}^{\ast},X^{\ast}\}$ these maps are

	$\displaystyle m^{\ast}(\mathbf{1}^{\ast}\otimes X^{\ast})=m^{\ast}(X^{\ast}\otimes\mathbf{1}^{\ast})=X^{\ast}$
	$\displaystyle m^{\ast}(\mathbf{1}^{\ast}\otimes\mathbf{1}^{\ast})=\mathbf{1}^{\ast}$
	$\displaystyle m^{\ast}(X^{\ast}\otimes X^{\ast})=0$
	$\displaystyle\Delta^{\ast}(\mathbf{1}^{\ast})=\mathbf{1}^{\ast}\otimes X^{\ast}+X^{\ast}\otimes\mathbf{1}^{\ast}$
	$\displaystyle\Delta^{\ast}(X^{\ast})=X^{\ast}\otimes X^{\ast}$

Hence, under the isomorphism $\mu:{\cal A}\to{\cal A}^{\ast}$ of graded abelian groups, given by $\mu(\mathbf{1})=\mathbf{1}^{\ast}$ and $\mu(X)=X^{\ast},$ maps $m,\Delta$ become $m^{\ast},\Delta^{\ast}.$

Note that the $\widetilde{\mathcal{L}}$-resolution $D(\widetilde{\mathcal{L}})$ of diagram $D$ and the $\mathcal{L}$-resolution $D^{!}(\mathcal{L})$ of $D^{!}$ are isomorphic. Let $k$ be the number of circles in $D(\widetilde{\mathcal{L}}).$ Then

$${\cal F}(D(\widetilde{\mathcal{L}}))={\cal F}(D^{!}(\mathcal{L}))={\cal A}^{\otimes k}$$

(162)

and, via $\mu,$ we can identify

$${\cal F}(D(\widetilde{\mathcal{L}}))=({\cal A}^{\ast})^{\otimes k}=({\cal F}(D^{!}(\mathcal{L})))^{\ast}$$

(163)

Since $\mu$ maps $m,\Delta$ to $m^{\ast},\Delta^{\ast},$ we see that after suitable shifts (recall that ${\cal V}_{D}(\mathcal{L})$ is equal to ${\cal F}(D(\mathcal{L}))$ shifted up by $|\mathcal{L}|$) the identification (163) extends to an isomorphism of $\mathcal{I}$-cubes ${\cal V}_{D^{!}}\{-n\}$ and $({\cal V}_{D})^{\ast}.$

$\square$

Given a complex $C$ of graded abelian groups and grading-preserving homomorphisms

$$\cdots\longrightarrow C^{i}\stackrel{{\scriptstyle d^{i}}}{{\longrightarrow}}C^{i+1}\longrightarrow\cdots$$

(164)

define the dual complex $C^{\ast}$ by $(C^{\ast})^{i}=(C^{-i})^{\ast}$ and the differential $(d^{\ast})^{i}$ being the dual of the differential $d^{-i-1}$ of $C.$

From the last proposition we easily obtain

This implies

Note that this corollary provides a necessary condition for a link to be amphicheiral.

Pick diagrams $D_{1},D_{2}$ of oriented links $L_{1},L_{2}$ and consider a diagram $D_{1}\sqcup D_{2}$ of the disjoint union $L_{1}\sqcup L_{2}.$ We then have an isomorphism of cochain complexes

$${\cal C}(D_{1}\sqcup D_{2})={\cal C}(D_{1})\otimes{\cal C}(D_{2})$$

(167)

of free graded abelian groups. From the Künneth formula we derive

Let $D_{1},D_{2}$ be diagrams of oriented knots $K_{1},K_{2}.$ We picture $D_{1}$ and $D_{2}$ schematically as

Consider diagrams $D_{3},D_{4}$ and $D_{5},$ depicted below, of oriented links $K_{1}\sqcup K_{2},$ $K_{1}\#K_{2},$ and $K_{1}\#(-K_{2})$:

By resolving the central double point of $D_{5},$ we get a short exact sequence of complexes of graded abelian groups

$$0\longrightarrow\overline{{\cal C}}(D_{3})[-1]\{-1\}\longrightarrow\overline{{\cal C}}(D_{5})\longrightarrow\overline{{\cal C}}(D_{4})\longrightarrow 0$$

(168)

After shifts, we obtain an exact sequence

$$0\longrightarrow{\cal C}(D_{3})[-1]\{-1\}\longrightarrow{\cal C}(D_{5})[-1]\{-2\}\longrightarrow{\cal C}(D_{4})\longrightarrow 0$$

(169)

which induces, for every integer $j,$ a long exact sequence of cohomology groups

$$\begin{array}[]{ccccccccc}\cdots&\to&{\cal H}^{i-1,j-1}(D_{3})&\to&{\cal H}^{i-1,j-2}(D_{5})&\to&{\cal H}^{i,j}(D_{4})&\to&\\ &\to&{\cal H}^{i,j-1}(D_{3})&\to&{\cal H}^{i,j-2}(D_{5})&\to&{\cal H}^{i+1,j}(D_{4})&\to&\cdots\end{array}$$

(170)

Since the diagrams $D_{3},D_{4}$ and $D_{5}$ represent oriented links $K_{1}\sqcup K_{2},K_{1}\#K_{2}$ and $K_{1}\#(-K_{2}),$ respectively, then in view of Proposition 29, we obtain

Let $D$ be an plane diagram of a link. In this section we construct a spectral sequence whose $E_{1}$-term is made of groups ${\cal H}^{i,j}(D)$ and which converges to cohomology groups $H^{i,j}(D).$

Due to the direct sum decomposition $R={\mathop{\oplus}\limits_{k\geq 0}}c^{k}\mathbb{Z}$ of abelian groups, we have an abelian group decomposition

$$C^{i}_{j}(D)={\mathop{\oplus}\limits_{k\geq 0}}{\cal C}^{i}_{j-2k}(D)$$

(172)

where, recall, $C^{i}_{j}(D)$ and ${\cal C}^{i}_{j}(D)$ were defined in Sections 4.2 and 7.1 respectively. Let us fix a $j\in\mathbb{Z}.$ Denote by $d$ the differential in the weight $j$ subcomplex of the complex $C(D)$:

$$\cdots\stackrel{{\scriptstyle d}}{{\longrightarrow}}C^{i-1}_{j}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}C^{i}_{j}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}C^{i+1}_{j}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}\cdots$$

(173)

and by $\partial$ the differential in the complex

$$\cdots\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i-1}_{j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i}_{j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i+1}_{j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}\cdots$$

(174)

where we suppress the dependence of $\partial$ on $k$. Under the identification (172) differential $d$ becomes a differential of the complex

$$\cdots\stackrel{{\scriptstyle d}}{{\longrightarrow}}{\mathop{\oplus}\limits_{k\geq 0}}{\cal C}^{i}_{j-2k}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}{\mathop{\oplus}\limits_{k\geq 0}}{\cal C}^{i+1}_{j-2k}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}\cdots$$

(175)

Consider a bigraded abelian group

$$C={\mathop{\oplus}\limits_{k\geq 0,i\in\mathbb{Z}}}{\cal C}^{i}_{j-2k}(D)$$

(176)

where we set the grading of ${\cal C}^{i}_{j-2k}(D)$ to $(i,-k).$ We thus have a bigraded abelian group $C$ and two maps, $d$ and $\partial$, from $C$ to $C.$ Map $\partial$ is bigraded of degree $(1,0)$ while $d$ is only graded relative to the first grading. However, we can decompose

$$d=\partial+\widetilde{\partial}$$

(177)

where $\widetilde{\partial}$ has grading $(1,-1)$ and satisfies

	$\displaystyle\widetilde{\partial}\widetilde{\partial}$	$\displaystyle=$	$\displaystyle 0,$		(178)
	$\displaystyle\partial\widetilde{\partial}+\widetilde{\partial}\partial$	$\displaystyle=$	$\displaystyle 0$		(179)

Besides, since $\partial$ is a differential, $\partial\partial=0.$ Therefore, $H^{i,j}(D)$ is equal to the $i$-th cohomology group of the total complex of the bicomplex $(C,\partial,\widetilde{\partial}).$ Group ${\cal H}^{s,j-2k}(D)$ is equal to the $s$-th cohomology group of the subcomplex (relative to the differential $\partial$)

$$\cdots\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i-1,j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i,j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}C^{i+1,j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}\cdots$$

(180)

of $C.$ Therefore, for each $j\in\mathbb{Z}$ we get a spectral sequence whose $E_{1}$-term is given by cohomology groups ${\cal H}^{s,j-2k}(D)$, $s\in\mathbb{Z},k\geq 0$ and which converges to cohomology groups $H^{i,j}(D).$ In the few cases where we managed to compute cohomology groups, we have $H^{i,j}(D)={\mathop{\oplus}\limits_{k\geq 0}}{\cal H}^{i,j-2k}(D)$ and,consequently, the spectral sequence degenerates at $E_{1}.$ We have no idea whether this is true for any diagram $D.$

Perhaps the graded groups ${\cal H}^{i}(D)$ are easier to compute than $H^{i}(D).$ The latter are computed via the complex $C(D)$ of free graded $R$-modules, and a complex for ${\cal H}^{i}(D)$ is obtained by tensoring $C(D)$ with $\mathbb{Z}$ over $R,$ so that free graded $R$-modules become free abelian groups of the same rank. In practice, the computation of ${\cal H}^{i}(D)$ faces the problem of effectively simplifying complexes of abelian groups of exponentially high rank. The shortcut for the computation of $H^{i}(D)$ described in Section 6.2 works equally well for groups ${\cal H}^{i}(D),$ with Proposition 25 generalized to complexes $\overline{{\cal C}}(D).$ This easily leads to a computation of cohomology groups ${\cal H}^{i,j}(T_{2,k})$ of the $(2,k)$ torus link $T_{2,k},$ oriented as in Section 6.2.

These definitions are from [LT] and [T].

Proof: The differential $\partial^{0}:\overline{\cal C}^{0}(D)\to\overline{\cal C}^{1}(D)$ is not injective and, hence, $\overline{{\cal H}}^{0}(D)\not=0$ if and only if $D$ is $-$adequate. Similarly for $\overline{{\cal H}}^{n}(D)$ and $+$adequate diagrams (groups $\overline{{\cal H}}^{i}(D)$ were defined at the end of Section 7.1). $\square$

Denote by $c(L)$ the crossing number of $L$. It is the minimal number of crossings in a plane diagram of $L.$

Proof: Let $D$ be a diagram of $L$ with $c(L)$ crossings. Then $\overline{\cal C}^{i}(D)=0$ for $i<0$ and for $i>{\mathrm{hl}}(L).$ Consequently, $\overline{{\cal H}}^{i}(D)=0$ for $i<0$ and $i>{\mathrm{hl}}(L).$

$\square$

Proof: By Proposition 36 $\overline{{\cal H}}^{0}(D)\not=0$ and $\overline{{\cal H}}^{n}(D)\not=0.$ Therefore, $c(L)\geq{\mathrm{hl}}(L)\geq n.$ But since $D$ is an $n$-crossing diagram of $L,$ the crossing number of $L$ is $n.$ $\square$

Corollary 13 was originally obtained by Thistlethwaite (Corollary 3.4 of [T]) through the analysis of the $2$-variable Kauffman polynomial (not to be confused with the Kauffman bracket).