ScalingStacks

Given a link $L$ in $\mathbb{R}^{3}$ we can take its generic projection on the plane. A generic projection is the one without triple points and double tangencies. An isotopy class of such projections is called a plane diagram of $L,$ or, simply, a diagram. The following four types of transformations of plane diagrams preserve the isotopy type of the associated link.

I. Addition/removal of a left-twisted curl:

II. Addition/removal of a right-twisted curl:

III. Tangency move:

IV. Triple point move:

$\square$

Fix a plane diagram $D$ with $n$ double points of an oriented link $L.$ Denote by $\mathcal{I}$ the set of double points of $D.$ To $D$ we will associate an $\mathcal{I}$-cube $V_{D}$ over the category $R{\mbox{-mod}_{0}}$ of graded $R$-modules. This cube will not depend on the orientation of components of $L.$

Given a double point of a diagram $D$, it can be resolved in two possible ways:

Let us call the resolution on the left the $0$-resolution, and the one on the right the $1$-resolution. A resolution of $D$ is a resolution of each double point of $D.$ Thus, $D$ admits $2^{n}$ resolutions. There is a one-to-one correspondence between resolutions of $D$ and subsets $\mathcal{L}$ of the set $\mathcal{I}$ of double points. Namely, to $\mathcal{L}\subset\mathcal{I}$ we associate a resolution, denoted $D(\mathcal{L}),$ by taking $1$-resolution of each double point that belongs to $\mathcal{L}$ and $0$-resolution if the double point does not lie in $\mathcal{L}.$

A resolution of a diagram $D$ is always a collection of simple disjoint curves on the plane and is thus a $1$-manifold embedded in the plane. Now the functor $F$ from $(1+1)$ cobordisms to $R$-modules (see Section 2.3) comes into play. To a union of $k$ circles it assigns the $k$-th tensor power of $A.$ The functor $F$, applied to the diagram $D(\mathcal{L}),$ considered as a one-dimensional manifold, produces a graded $R$-module $A^{\otimes k}$ where $k$ is the number of components of $D(\mathcal{L}).$ We raise the grading of $A^{\otimes k}$ by $|\mathcal{L}|,$ the cardinality of $\mathcal{L},$ and assign the $R$-module $F(D(a))\{-|\mathcal{L}|\}$ to the vertex $V_{D}(\mathcal{L})$ of the cube $V_{D}:$

$$V_{D}(\mathcal{L})=F(D(\mathcal{L}))\{-|\mathcal{L}|\}$$

(40)

(Recall from Section 3.1 that the automorphism $\{1\}$ of the category $R{\mbox{-mod}_{0}}$ lowers the grading by $1$.) Let us now define maps between vertices of $V_{D}.$ Choose $(\mathcal{L},a)\in r(\mathcal{I}).$ We want to have a map

$$\xi_{a}^{V_{D}}(\mathcal{L}):V_{D}(\mathcal{L})\longrightarrow V_{D}(\mathcal{L}a).$$

(41)

The diagrams $D(\mathcal{L})$ and $D(\mathcal{L}a)$ differ only in the neighborhood of the double point $a$ of $D,$ as an example below (for $n=1$, so that $D$ has one double point, $\mathcal{I}=\{a\}$) demonstrates ($D$ is the leftmost diagram, $D(\emptyset)$ is the diagram in the center and $D(a)$ is depicted on the right):

Take the direct product of the plane $\mathbb{R}^{2}$ and the interval $[0,1].$ We identify the diagram $D(\mathcal{L})$ (resp. $D(\mathcal{L}a)$) with a one-dimensional submanifold of $\mathbb{R}^{2}\times\{0\}$ (resp $\mathbb{R}^{2}\times\{1\}$.) We can choose a small neighbourhood $U$ of $a$ such that $D(\mathcal{L})$ and $D(\mathcal{L}a)$ coincide outside $U$ and inside they look as follows:

The boundary of $U$ is depicted by a dashed circle, the central picture shows the intersection of $D(\mathcal{L})$ and $U,$ the rightmost picture shows the intersection of $D(\mathcal{L}a)$ and $U.$ Let $S$ be a surface properly embedded in $\mathbb{R}^{2}\times[0,1]$ such that

1. 1.

   The boundary of $S$ is the union of the diagrams $D(\mathcal{L})$ and $D(\mathcal{L}a).$

2. 2.

   Outside of $U\times[0,1]$ surface $S$ is the direct product of $D(\mathcal{L})\cap(\mathbb{R}^{2}\setminus U)$ and the interval $[0,1].$

3. 3.

   The connected component of $S$ that has a nonempty intersection with $U\times[0,1]$ is homeomorpic to the two-sphere with three holes.

4. 4.

   The projection $S\longrightarrow[0,1]$ onto the second component of the product $\mathbb{R}^{2}\times[0,1]$ has only one critical point – the saddle point that lies inside $U\times[0,1].$

Example: Let $D$ be a diagram with two double points, $\mathcal{I}=\{a,b\},$ diagram $D(a),$ respectively $D(ab),$ consists of 2, respectively 3 simple curves:

The boundary of the neighbourhood $U$ of the double point $a$ is depicted by the dashed circle on the diagram above. Then the surface $S$ looks like

Recall that earlier we defined $V_{D}(\mathcal{L})$ to be $F(D(\mathcal{L}))$ for $\mathcal{L}\subset\mathcal{I},$ with the degree raised by $|\mathcal{L}|.$ Now define the map

$$\xi_{a}^{V_{D}}(\mathcal{L}):V_{D}(\mathcal{L})\longrightarrow V_{D}(\mathcal{L}a)$$

to be given by

$$F(S):F(D(\mathcal{L}))\longrightarrow F(D(\mathcal{L}a)).$$

Note that the degree of $F(S)$ is equal to $-1,$ the Euler characteristic of the surface $S$ (Proposition 3). But $|\mathcal{L}a|=|\mathcal{L}|+1,$ so, with degrees shifted:

	$\displaystyle V_{D}(\mathcal{L})$	$\displaystyle=$	$\displaystyle F(D(\mathcal{L}))\{-|\mathcal{L}|\}$		(42)
	$\displaystyle V_{D}(\mathcal{L}a)$	$\displaystyle=$	$\displaystyle F(D(\mathcal{L}a))\{-|\mathcal{L}|-1\}$		(43)

and the map $\xi_{a}^{V_{D}}(\mathcal{L})$ is a grading-preserving map of graded $R$-modules.

The proof consists of verifying commutativity relations (28) for maps $\xi_{a}^{V_{D}}(\mathcal{L}).$ They follow immediately from the functoriality of $F.$

$\square$

Example: Let $D$ be the diagram

The four resolutions of this diagram are given below

Applying the functor $F$, we get

	$\displaystyle F(D(\emptyset))=A^{\otimes 2},\hskip 21.68121ptF(D(a))=A,$
	$\displaystyle F(D(b))=A,\hskip 21.68121ptF(D(ab))=A^{\otimes 2}.$

The cube $V_{D}$ has the form

$$\begin{CD}A^{\otimes 2}@>{m}>{}>A\{-1\}\\ @V{}V{m}V@V{}V{\Delta}V\\ A\{-1\}@>{\Delta}>{}>A^{\otimes 2}\{-2\}\end{CD}$$

$\square$

Let us now go back to our construction. So far, to a plane diagram diagram $D$ with the set $\mathcal{I}$ of double points we associated a $\mathcal{I}$-cube $V_{D}$ over the category $R{\mbox{-mod}_{0}}$ of graded $R$-modules. We would like to build a complex of graded $R$-modules out of $V_{D}.$ We know how to build a complex from a skew-commutative $\mathcal{I}$-cube (see Section 3.4). To make a skew-commutative $\mathcal{I}$-cube out of an $\mathcal{I}$-cube $V_{D}$ we put minus sign in front of some structure maps $\xi^{V_{D}}$ of $V_{D}$ so that for any commutative square of $V_{D}$ an odd number out of the four maps constituting the square change signs. A more intrinsic way to do this is to tensor $V_{D}$ with the skew-commutative $\mathcal{I}$-cube $E_{\mathcal{I}},$ defined at the end of Section 3.3.

To the skew-commutative $\mathcal{I}$-cube $V_{D}\otimes E_{\mathcal{I}}$ there is associated the complex $\overline{C}(V_{D}\otimes E_{\mathcal{I}})$ of graded $R$-modules (see Section 3.4). Denote this complex by $\overline{C}(D):$

$$\overline{C}(D)\stackrel{{\scriptstyle\mbox{\scriptsize def}}}{{=}}\overline{C}(V_{D}\otimes E_{\mathcal{I}})$$

(44)

Thus, $\overline{C}(D)$ is a complex of graded $R$-modules and grading-preserving homomorphisms. It does not depend on the orientations of the components of the link $L.$

Recall (Section 3.1) that the category $\mbox{Kom}(R{\mbox{-mod}_{0}})$ has two commuting automorphisms: $[1],$ which shifts a complex one term to the left; and $\{1\},$ which lowers the grading of each component of the complex by $1.$

To the diagram $D$ of link $L$ we associated (Section 2.4) two numbers, $x(D)$ and $y(D).$ Define a complex $C(D)$ by

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

(45)

Define $H^{i}(D)$ as the $i$-th cohomology group of $C(D).$ It is a finitely-generated graded $R$-module.

The proof of this theorem occupies Section 5, together with some preliminary material contained in Section 4.3.

Define $H^{i,j}(D)$ as the $i$-th cohomology group of the degree $j$ subcomplex of $C(D).$ Thus, $H^{i,j}(D)$ is the graded component of $H^{i}(D)$ of degree $j$ and we have a decomposition of abelian groups

$$H^{i}(D)={\mathop{\oplus}\limits_{j\in\mathbb{Z}}}H^{i,j}(D).$$

(46)

We denote by $H^{i}(L)$ the isomorphism class of $H^{i}(D)$ in the category of graded $R$-modules. For an oriented link $L$ only finitely many of $H^{i}(L)$ are non-zero as $i$ varies over all integers.

We next show that the Kauffman bracket is equal to a suitable Euler characteristic of these cohomology groups.

Proof: First notice that $\widehat{\chi}(M\{n\})=q^{-n}\widehat{\chi}(M)$ for a finitely-generated graded $R$-module $M.$ Given a bounded complex

$$M:\hskip 21.68121pt\dots\to M^{i}\to M^{i+1}\to\dots$$

(48)

of finitely-generated graded $R$-modules, define

$$\widehat{\chi}(M)=\sum_{i\in\mathbb{Z}}(-1)^{i}\widehat{\chi}(M^{i})$$

(49)

Since

$$\widehat{\chi}(C(D))=\sum_{i\in\mathbb{Z}}(-1)^{i}\widehat{\chi}(H^{i}(D)),$$

(50)

it is enough to prove

$$K(L)=(1-q^{2})\widehat{\chi}(C(D))$$

(51)

For three diagrams $D_{1},D_{2}$ and $D_{3}$ that differ as shown below

the complex $\overline{C}(D_{1})[1]$ is isomorphic, up to a shift, to the cone of a map of complexes $\overline{C}(D_{2})\to\overline{C}(D_{3})\{-1\}.$ Therefore,

$$\widehat{\chi}(\overline{C}(D_{1}))=\widehat{\chi}(\overline{C}(D_{2}))-\widehat{\chi}(\overline{C}(D_{3})\{-1\})=\widehat{\chi}(\overline{C}(D_{2}))-q\widehat{\chi}(\overline{C}(D_{3}))$$

(52)

On the other hand, for diagrams $D_{1},D_{2},D_{3}$ as above, we have

$$<D_{1}>=<D_{2}>-q<D_{3}>$$

(53)

(see Section 2.4, where $<D>$ is defined). If the diagram $D$ is a disjoint union of $k$ simple plane curves then

$$\widehat{\chi}(\overline{C}(D))=\widehat{\chi}(A^{\otimes k})=(q+q^{-1})^{k}\widehat{\chi}(R)=\frac{(q+q^{-1})^{k}}{1-q^{2}}$$

(54)

and $<D>=(q+q^{-1})^{k}.$ Therefore, for any diagram $D$

$$<D>=(1-q^{2})\widehat{\chi}(\overline{C}(D)).$$

(55)

Since

$$\widehat{\chi}(C(D))=\widehat{\chi}(\overline{C}(D))[x(D)]\{2x(D)-y(D)\}=(-1)^{x(D)}q^{y(D)-2x(D)}\widehat{\chi}(\overline{C}(D)),$$

(56)

and in view of (21), proposition follows. $\square$

Let $U$ be a closed disk in the plane $\mathbb{R}^{2}$ and $\dot{U}$ the interior of $U$ so that $U=\partial U\cup\dot{U}.$ Let $T^{\prime}$ be a tangle in $(\mathbb{R}^{2}\setminus\dot{U})\times[0,1]$ with $m$ points (where $m$ is even) on the boundary $\partial U\times[0,1]$ and $T$ a generic projection of $T^{\prime}$ on $\mathbb{R}^{2}\setminus\dot{U}.$ The intersection of $T$ with $\partial U$ consists of $m$ points. Denote them by $p_{1},\dots,p_{m}$ (see an example on the diagram below, $\partial U$ is shown by a dashed circle).

Let $\mathcal{I}$ be the set of double points of $T.$ Pick two systems $Q_{0}$ and $Q_{1}$ of $\frac{m}{2}$ simple disjoints arcs in $U$ with ends in points $p_{1},\dots p_{m}$:

Then $Q_{0}\cup T$ and $Q_{1}\cup T$ (here and further on we denote them by $P_{0}$ and $P_{1}$ respectively) can be considered as two plane diagrams of links in $\mathbb{R}^{3}:$

To $P_{0}$ and $P_{1}$ there are associated $\mathcal{I}$-cubes $V_{P_{0}}$ and $V_{P_{1}}.$

Let $S$ be a compact oriented surface in $U\times[0,1]$ such that the boundary of $S$ is the union of $Q_{0}\times\{0\},Q_{1}\times\{1\}$ and $(p_{1}\cup\dots\cup p_{m})\times[0,1].$ To $S$ we associate an $\mathcal{I}$-cube map

$$\psi_{S}:V_{P_{0}}\longrightarrow V_{P_{1}}$$

as follows. For each $\mathcal{L}\subset\mathcal{I}$ we must construct a map

$$\psi_{S,\mathcal{L}}:V_{P_{0}}(\mathcal{L})\{-|\mathcal{L}|\}\longrightarrow V_{P_{1}}(\mathcal{L})\{-|\mathcal{L}|\}$$

(57)

and check the commutativity of diagrams (29).

To $\mathcal{L}$ there is associated a resolution $T(\mathcal{L})$ of double points of $T.$ Thus $T(\mathcal{L})$ is a collection of simple closed curves and arcs in $\mathbb{R}^{2}\setminus\mbox{\.{U}}$ with ends in $p_{1},\dots,p_{m}.$ Then (by (40)

	$\displaystyle V_{P_{0}}(\mathcal{L})$	$\displaystyle=$	$\displaystyle F(T(\mathcal{L})\cup Q_{0})\{-|\mathcal{L}|\}$		(58)
	$\displaystyle V_{P_{1}}(\mathcal{L})$	$\displaystyle=$	$\displaystyle F(T(\mathcal{L})\cup Q_{1})\{-|\mathcal{L}|\}$		(59)

where $F$ is the functor described in Section 2.3 ( $T(\mathcal{L})\cup Q_{0}$ and $T(\mathcal{L})\cup Q_{1}$ are collections of simple closed curves on the plane, so that we can apply functor $F$ to them).

Let $S^{\prime}$ be a surface in $\mathbb{R}^{2}\times[0,1]$ which is $S$ inside $U\times[0,1]$ and $T(\mathcal{L})\times[0,1]$ outside $U\times[0,1].$ Map $F(S^{\prime})$

$$F(S^{\prime}):F(T(\mathcal{L})\cup Q_{0})\longrightarrow F(T(\mathcal{L})\cup Q_{1})$$

(60)

is a graded map of $R$-modules of degree $\chi(S^{\prime})=\chi(S)-\frac{m}{2}.$ Define $\psi_{S,\mathcal{L}}$ as this map, shifted by $|\mathcal{L}|$:

$$\psi_{S,\mathcal{L}}=F(S^{\prime})\{-|\mathcal{L}|\}:V_{P_{0}}(\mathcal{L})\{-|\mathcal{L}|\}\longrightarrow V_{P_{1}}(\mathcal{L})\{-\mathcal{L}|\}$$

(61)

The commutativity condition (29) is immediate. We sum up our result as

Everything in this section extends to the case when the diagrams $Q_{0}$ and $Q_{1}$ are allowed to have simple closed circles in addition to $\frac{m}{2}$ simple disjoint acts joining points $p_{1},\dots p_{m}.$ For instance, $Q_{0}$ may look like

In this more general case to each compact oriented surface $S$ in $U\times[0,1]$ such that the boundary of $S$ is the union of $Q_{0}\times\{0\},Q_{1}\times\{1\}$ and $(p_{1}\cup\dots\cup p_{m})\times[0,1],$ in exactly the same fashion as before, we associate an $\mathcal{I}$-cube map

$$\psi_{S}:V_{P_{0}}\longrightarrow V_{P_{1}}$$

(63)

This map is a graded map of cubes over $R{\mbox{-mod}_{0}}$ of degree equal to the Euler characteristic of $S$ minus $\frac{m}{2}.$

Tensoring the map $\psi_{S}$ with the identity map of the skew-commutative $n$-cube $E_{\mathcal{I}}$ and passing to associated complexes, we obtain a map of complexes of graded $R$-modules

$$\psi^{\prime}_{S}:\overline{C}(P_{0})\longrightarrow\overline{C}(P_{1})$$

(64)

In general this map is not a morphism in the category $\mbox{Kom}(R{\mbox{-mod}_{0}})$ of complexes of graded $R$-modules and grading-preserving homomorphism, as it shifts the grading by $\chi(S)-\frac{m}{2},$ but $\psi^{\prime}_{S}$ becomes a morphism in $\mbox{Kom}(R{\mbox{-mod}_{0}})$ when the grading of $\overline{C}(P_{0})$ or $\overline{C}(P_{1})$ is appropriately shifted.