ScalingStacks

Let $R=\mathbb{Z}[c]$ denote the ring of polynomials with integral coefficients. Introduce a $\mathbb{Z}$-grading on $R$ by

$${\mathrm{deg}}(1)=0,\hskip 21.68121pt{\mathrm{deg}}(c)=2.$$

(7)

Denote by $R{\mbox{-mod}_{0}}$ the abelian category of graded $R$-modules. Denote the $i$-th graded component of an object $M$ of $R{\mbox{-mod}_{0}}$ by $M_{i}.$ Morphisms in the category $R{\mbox{-mod}_{0}}$ are grading-preservings homomorphisms of modules. For $n\in\mathbb{Z}$ denote by $\{n\}$ the automorphism of $R{\mbox{-mod}_{0}}$ given by shifting the grading down by $n.$ Thus for a graded $R$-module $N=\oplus_{i}N_{i},$ the shifted module $N\{n\}$ has graded components $N\{n\}_{i}=N_{i+n}.$

In this paper we will sometimes consider graded, rather than just grading-preserving, maps. A map $\alpha:M\to N$ of graded $R$-modules is called graded of degree $i$ if $\alpha(M_{j})\subset N_{i+j}$ for all $j\in\mathbb{Z}.$

Let $R{\mbox{-mod}}$ be the category of graded $R$-modules and graded maps between them. This category has the same objects as the category $R{\mbox{-mod}_{0}},$ but more morphisms. It is not an abelian category.

A graded map $\alpha$ is a morphism in the category $R{\mbox{-mod}_{0}}$ if and only if the degree of $\alpha$ is $0.$ At the end we will favor grading-preserving maps and when at some point we look at a graded map $\alpha:M\to N$ of degree $i,$ later we will make it grading-preserving by appropriately shifting the degree of one of the modules. For example, $\alpha$ gives rise to a grading-preserving map $M\to N\{i\},$ also denoted $\alpha.$

Let $M$ be a finitely-generated graded $R$-module. As an abelian group, $M$ is the direct sum of its graded components: $M={\mathop{\oplus}\limits_{j\in\mathbb{Z}}}M_{j},$ where each $M_{j}$ is a finitely-generated abelian group. Define the graded Euler characteristic $\widehat{\chi}(M)$ of $M$ by

$$\widehat{\chi}(M)=\sum_{j\in\mathbb{Z}}{\mathrm{dim}}_{\mathbb{Q}}(M_{i}\otimes_{\mathbb{Z}}\mathbb{Q})q^{j}$$

(8)

Since

$$\widehat{\chi}(R)=1+q^{2}+q^{4}+\dots=\frac{1}{1-q^{2}},$$

(9)

$\widehat{\chi}(M)$ is not, in general, a Laurent polynomial in $q$, but an element of the Laurent series ring. Moreover, for any $M$ as above, there are Laurent polynomials $a,b\in\mathbb{Z}[q,q^{-1}]$ such that

$$\widehat{\chi}(M)=a+\frac{b}{1-q^{2}}$$

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Let $A$ be a free graded $R$-module of rank $2$ spanned by $\mathbf{1}$ and $X$ with

$${\mathrm{deg}}(\mathbf{1})=1,\hskip 21.68121pt{\mathrm{deg}}(X)=-1$$

(11)

We equip $A$ with a commutative algebra structure with the unit $\mathbf{1}$ and multiplication

$$\mathbf{1}X=X\mathbf{1}=X,X^{2}=0.$$

(12)

We denote by $\iota$ the unit map $R\to A$ which sends $1$ to $\mathbf{1}.$ This map is a graded map of graded $R$-modules and it increases the degree by $1.$

We equip $A$ with a coalgebra structure with a coassociative cocommutative comultiplication

		$\displaystyle\Delta(\mathbf{1})$	$\displaystyle=\mathbf{1}\otimes X+X\otimes\mathbf{1}+cX\otimes X$		(13)
		$\displaystyle\Delta(X)$	$\displaystyle=X\otimes X$		(14)

and a counit

$$\epsilon(\mathbf{1})=-c,\hskip 36.135pt\epsilon(X)=1.$$

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$A,$ equipped with these structures, is not a Hopf algebra. Instead, the identity

$$\Delta\circ m=(m\otimes\mbox{Id})\circ(\mbox{Id}\otimes\Delta)$$

(16)

holds.

Grading ${\mathrm{deg}},$ given by (7),(11), induces a grading, also denoted ${\mathrm{deg}},$ on tensor powers of $A$ by

$${\mathrm{deg}}(a_{1}\otimes\dots\otimes a_{n})={\mathrm{deg}}(a_{1})+\dots+{\mathrm{deg}}(a_{n})\mbox{ for }a_{1},\dots,a_{n}\in A$$

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Here and further on all tensor products are taken over the ring $R$ unless specified otherwise.

We next describe the effect of the structure maps $\iota,m,\epsilon,\Delta$ on the gradings. We say that a map $f$ between two graded $R$-modules $V=\oplus V_{n}$ and $W=\oplus W_{n}$ has degree $k$ if $f(x)\in W_{n+k}$ whenever $x\in V_{n}.$

Consider the surfaces $S_{2}^{1},S_{1}^{2},S_{0}^{1},S_{1}^{0},S_{2}^{2}$ and $S_{1}^{1}$, depicted below

Each of these surfaces $S_{a}^{b}$ defines a cobordism from a union of $a$ circles to a union of $b$ circles. We denote by $\mathcal{M}$ the category whose objects are closed one-dimensional manifolds and morphisms are two-dimensional cobordisms between these manifolds generated by the above cobordisms. Specifically, objects of $\mathcal{M}$ are enumerated by nonnegative integers $\mbox{Ob}(\mathcal{M})=\{\overline{n}|n\in\mathbb{Z}_{+}\}.$ A morphism between $\overline{n}$ and $\overline{m}$ is a compact oriented surface $S$ with boundary being the union of $n+m$ circles. The boundary circles are split into two sets $\partial_{0}S$ and $\partial_{1}S$ with $\partial_{0}S$ containing $n$ and $\partial_{1}S$ containing $m$ circles. An ordering of elements of each of these two sets is fixed. The surface $S$ is presented as a concatenation of disjoint unions of elementary surfaces, depicted above. Morphisms are composed in the usual way by gluing boundary circles. Two morphisms are equal if the surfaces $S,T$ representing these morphisms are diffeomorphic via a diffeomorphism that extends the identification $\partial_{0}S\cong\partial_{0}T,\partial_{1}S\cong\partial_{1}T$ of their boundaries. $\mathcal{M}$ is a monoidal category with tensor product of morphisms defined by taking the disjoint union of surfaces.

Let us construct a monoidal functor from $\mathcal{M}$ to the category $R{\mbox{-mod}_{0}}$ of graded $R$-modules and graded module maps. Assign graded $R$-module $A^{\otimes n}$ to the object $\overline{n}$ and to the elementary surfaces $S_{2}^{1},S_{1}^{2},S_{0}^{1},S_{1}^{0},S_{2}^{2},S_{1}^{1}$ assign morphisms $m,\Delta,\iota,\epsilon,\mbox{Perm},\mbox{Id}$

$$\begin{array}[]{lll}F(S_{2}^{1})=m,&F(S_{1}^{2})=\Delta,&F(S_{0}^{1})=\iota,\\ F(S_{1}^{0})=\epsilon,&F(S_{2}^{2})=\mbox{Perm},&F(S_{1}^{1})=\mbox{Id},\\ \end{array}$$

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where $\mbox{Perm}:A\otimes A\to A\otimes A$ is the permutation map, $\mbox{Perm}(u\otimes v)=v\otimes u,$ and Id is the identity map $\mbox{Id}:A\to A.$

To check that $F$ is well-defined one must verify that for any two ways to glue an arbitrary surface $S$ in ${\mathrm{Mor}}(\mathcal{M})$ from copies of these six elementary surfaces, the two maps of $R$-modules, defined by these two decompositions of $S,$ coincide. This follows from the commutative algebra and cocommutative coalgebra axioms of $A$ and the identity (16).

Remark: Suppose that a surface $S\in\mathcal{M}$ contains a punctured genus two surface as a subsurface. Then $F(S)$ is the zero map. Indeed, we only need to check this when $S$ has genus two and one boundary component. Then that $F(S)=0$ follows from $m\circ\Delta\circ m\circ\Delta=0.$

Maps $F(S_{2}^{1}),F(S_{1}^{2}),F(S_{0}^{1}),F(S_{1}^{0}),F(S_{2}^{2})$ and $F(S_{1}^{1})$ between tensor powers of $A$ are graded relative to ${\mathrm{deg}}$ with degrees

	$\displaystyle{\mathrm{deg}}(F(S_{2}^{1}))$	$\displaystyle=$	$\displaystyle{\mathrm{deg}}(F(S_{1}^{2}))=-1,$
	$\displaystyle{\mathrm{deg}}(F(S_{0}^{1}))$	$\displaystyle=$	$\displaystyle{\mathrm{deg}}(F(S_{1}^{0}))=1,$
	$\displaystyle{\mathrm{deg}}(F(S_{2}^{2}))$	$\displaystyle=$	$\displaystyle{\mathrm{deg}}(F(S_{1}^{1})=0$

From this we deduce

In this section we review the Kauffman bracket and its relation to the Jones polynomial following Kauffman [Ka]. Fix an orientation of the 3-space $\mathbb{R}^{3}$. A plane projection $D$ of an oriented link $L$ in $\mathbb{R}^{3}$ is called generic if it has no triple intersections, no tangencies and no cusps. In this paper by a plane projection we will mean a generic plane projection. Given a plane projection $D,$ we assign a Laurent polynomial $<D>\in\mathbb{Z}[q,q^{-1}]$ to $D$ by the following rules

1. 1.

   A simple closed loop evaluates to $q+q^{-1}:$

   
2. 2.

   Each over and undercrossing is a linear combination of two simple resolutions of this crossing:

   
3. 3.

   $<D_{1}\bigsqcup D_{2}>=<D_{1}><D_{2}>$ where $<D_{1}\bigsqcup D_{2}>$ stands for the disjoint union of the diagrams $D_{1}$ and $D_{2}.$

From these rules we deduce that

Curves of the diagram $D$ inherit orientations from that of $L.$ Let $x(D)$ be the number of double points in the diagram $D$ that look like

and $y(D)$ the number of double points that look like

Then the quantity

$$K(D)=(-1)^{x(D)}q^{y(D)-2x(D)}<D>$$

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does not depend on the choice of a diagram $D$ of the oriented link $L$ and is an invariant of $L.$ We denote this invariant by $K(L).$ Up to a simple normalization, $K(L)$ is the Kauffman bracket of link $L$ and equal to the Jones polynomial of $L.$ The Kauffman bracket, $f[L],$ as defined in [Ka], is a Laurent polynomial in an indeterminate $A$ (this $A$ has no relation to the algebra $A$ in Section 2.2 of this paper). One easily sees that setting our $q$ to $-A^{-2}$ and dividing by $(-A^{2}-A^{-2})$ we get $f[L]$:

$$K(L)_{(q=-A^{-2})}=(-A^{2}-A^{-2})f[L]$$

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In this paper we will call $K(L)$ the scaled Kauffman bracket.

Let $L_{1},L_{2}$ and $L_{3}$ be three oriented links that differ as shown below.

The rules for computing the Kauffman bracket imply

$$q^{-2}K(L_{1})-q^{2}K(L_{2})=(q^{-1}-q)K(L_{3})$$

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Moreover, $K(L)=q+q^{-1}$ if $L$ is the unknot.

The Jones polynomial $V(L)$ of an oriented link $L$ is determined by two properties:

1. 1.

   The Jones polynomial of the unknot is $1.$

2. 2.

   For oriented links $L_{1},L_{2},L_{3}$ as above

   $$t^{-1}V(L_{1})-tV(L_{2})=(\sqrt{t}-\frac{1}{\sqrt{t}})V(L_{3})$$

   (24)

Therefore, the scaled Kauffman bracket and the Jones polynomial are related by

$$V(L)_{\sqrt{t}=-q}=\frac{K(L)}{q+q^{-1}}$$

(25)