ScalingStacks

We start by reviewing the construction of skein lasagna modules from [27, Section 5.2].

Following [27] and [25], for a framed link $L\subset{\mathbb{R}}^{3}$, we write

$$\operatorname{KhR}_{N}(L)=\bigoplus_{i,j\in{\mathbb{Z}}}\operatorname{KhR}_{N}^{i,j}(L)$$

for the $\mathfrak{gl}_{N}$ version of Khovanov-Rozansky homology. Here, $i$ denotes the homological grading and $j$ denotes the quantum grading.

If we have an oriented manifold $S$ diffeomorphic to the standard $3$-sphere $S^{3}$, and a framed link $L\subset S$, we can define a canonical invariant $\operatorname{KhR}_{N}(S,L)$ as in [27, Definition 4.12]. We sometimes drop $S$ from the notation and simply write $\operatorname{KhR}_{N}(L)$.

Given a framed cobordism $\Sigma\subset S^{3}\times[0,1]$ from $L_{0}$ to $L_{1}$, there is an induced map

$$\operatorname{KhR}_{N}(\Sigma)\colon\operatorname{KhR}_{N}(L_{0})\to\operatorname{KhR}_{N}(L_{1})$$

which is homogeneous of bidegree $(0,(1-N)\chi(\Sigma))$.

Let $W$ be a four-manifold and $L\subset\partial W$ a framed link. A lasagna filling $F=(\Sigma,\{(B_{i},L_{i},v_{i})\})$ of $W$ with boundary $L$ consists of

• •

  A finite collection of disjoint $4$-balls $B_{i}$ (called input balls) embedded in the interior or $W$;

• •

  A framed oriented surface $\Sigma$ properly embedded in $W\setminus\cup_{i}B_{i}$, meeting $\partial W$ in $L$ and meeting each $\partial B_{i}$ in a link $L_{i}$; and

• •

  for each $i$, a homogeneous label $v_{i}\in\operatorname{KhR}_{N}(\partial B_{i},L_{i}).$

The bidegree of a lasagna filling $F$ is

$$\deg(F):=\sum_{i}\deg(v_{i})+(0,(1-N)\chi(\Sigma)).$$

If $W$ is a $4$-ball, we can define a cobordism map

$$\operatorname{KhR}_{N}(\Sigma)\colon\bigotimes_{i}\operatorname{KhR}_{N}(\partial B_{i},L_{i})\to\operatorname{KhR}_{N}(\partial W,L)$$

and an evaluation

$$\operatorname{KhR}_{N}(F):=\operatorname{KhR}_{N}(\Sigma)(\otimes_{i}v_{i})\in\operatorname{Kh}(\partial W,L).$$

We define the skein lasagna module as the bigraded abelian group

$$\mathcal{S}_{0}^{N}(W;L):={\mathbb{Z}}\{\text{lasagna fillings $F$ of $W$ with boundary $L$}\}/\sim$$

where $\sim$ is the transitive and linear closure of the following relation:

1. (a)

   Linear combinations of lasagna fillings are set to be multilinear in the labels $v_{i}$;

2. (b)

   Furthermore, two lasagna fillings $F_{1}$ and $F_{2}$ are set to be equivalent if $F_{1}$ has an input ball $B_{i}$ with label $v_{i}$, and $F_{2}$ is obtained from $F_{1}$ by replacing $B_{i}$ with another lasagna filling $F_{3}$ of a $4$-ball such that $v_{i}=\operatorname{KhR}_{N}(F_{3})$, followed by an isotopy rel $\partial W$ (where the isotopy is allowed to move the input balls):

For future reference, here is a useful lemma.

Skein lasagna modules decompose according to relative homology classes, as noted in [25, Section 2.3]:

(1)

$\displaystyle\mathcal{S}_{0}^{N}(W;L)=\bigoplus\limits_{\alpha\in H_{2}(W,L;{\mathbb{Z}})}\mathcal{S}_{0}^{N}(W;L,\alpha).$

Observe that in the case where $L$ is not null-homologous in $W$ (i.e. $[L]\neq 0\in H_{1}(W;{\mathbb{Z}})$), then there are no lasagna fillings, so $\mathcal{S}_{0}^{N}(W;L)=0$. When $[L]=0\in H_{1}(W;{\mathbb{Z}})$, consider the boundary map in the long exact sequence of the pair $(W,L)$:

$$\partial:H_{2}(W,L;{\mathbb{Z}})\to H_{1}(L;{\mathbb{Z}}).$$

The only classes $\alpha\in H_{2}(W,L;{\mathbb{Z}})$ that can contribute non-trivially are those that map to the fundamental class $[L]\in H_{1}(L;{\mathbb{Z}})$ under $\partial$. Let us introduce the notation

$$H_{2}^{L}(W;{\mathbb{Z}}):=\partial^{-1}([L])\subseteq H_{2}(W,L;{\mathbb{Z}}).$$

Note that, using the long exact sequence of the pair, the difference of two classes in $H_{2}^{L}(W;{\mathbb{Z}})$ can be identified with an element of $H_{2}(W;{\mathbb{Z}})$. Thus, $H_{2}^{L}(W;{\mathbb{Z}})$ is a torsor over $H_{2}(W;{\mathbb{Z}})$; it can be identified with the latter group after choosing a base element in $H_{2}^{L}(W;{\mathbb{Z}})$.

The decomposition  (1) becomes

(2)

$\displaystyle\mathcal{S}_{0}^{N}(W;L)=\bigoplus\limits_{\alpha\in H_{2}^{L}(W;{\mathbb{Z}})}\mathcal{S}_{0}^{N}(W;L,\alpha).$

We will use the decomposition (2) in the case of a general link $L$; when $[L]\neq 0$, we have $H_{2}^{L}(W;{\mathbb{Z}})=\emptyset$ and $\mathcal{S}_{0}^{N}(W;L)=0$.

Let us consider two four-manifolds $W$ and $Z$ that have some part $Y$ of their boundaries in common, as follows:

$$\partial W=Y\amalg Y_{0},\ \ \partial Z=(-Y)\amalg Y_{1},$$

where $\amalg$ denotes disjoint union. We can glue $W$ and $Z$ along $Y$ to form a new four-manifold $W\cup Z$ with boundary $Y_{0}\amalg Y_{1}$. Suppose we are also given links $L_{0}\subset Y_{0}$, $L_{1}\subset Y_{1}$ and $L\subset Y$. Let $\overline{L}\subset-Y$ denote the mirror reverse of $L$. Then, we have a map

(3)

$$\Psi:\mathcal{S}_{0}^{N}(W;L\cup L_{0})\otimes\mathcal{S}_{0}^{N}(Z;\overline{L}\cup L_{1})\to\mathcal{S}_{0}^{N}(W\cup Z;L_{0}\cup L_{1})$$

obtained by gluing lasagna fillings along $L$:

$$[F]\otimes[G]\mapsto[F\cup G].$$

It is easy to see that if two lasagna fillings $F_{1}$ and $F_{2}$ are equivalent in $W$, and $G_{1}$ and $G_{2}$ are equivalent in $Z$, then $F_{1}\cup G_{1}$ and $F_{2}\cup G_{2}$ are equivalent in $W\cup Z$, so (3) is well-defined.

Starting from here, we see that skein lasagna modules are functorial under inclusions, in the following sense. We consider the case when $Y_{0}=\emptyset$, and we fix a lasagna filling $G$ of $Z$ with boundary $\overline{L}\cup L_{1}$. We can think of $Z$ as a cobordism from $Y=\partial W$ to $Y_{1}$. Then, there is an induced cobordism map

(4)

$$\Psi_{Z;G}=\Psi(\cdot\otimes[G]):\mathcal{S}_{0}^{N}(W;L)\to\mathcal{S}_{0}^{N}(W\cup Z;L_{1}).$$

Observe that the maps (4) behave well with respect to compositions:

(5)

$$\Psi_{Z^{\prime};G^{\prime}}\circ\Psi_{Z;G}=\Psi_{Z\cup Z^{\prime};G\cup G^{\prime}}.$$

Furthermore, in terms of the decompositions (2), given $\alpha\in H^{L}_{2}(W;{\mathbb{Z}})$, by attaching to it the class of $G$ in $H_{2}^{L\cup L_{1}}(Z;{\mathbb{Z}})$ we get a class $\alpha_{1}\in H_{2}^{L_{1}}(W\cup Z;{\mathbb{Z}})$. Then, $\Psi_{Z;G}$ maps $\mathcal{S}_{0}^{N}(W;L,\alpha)$ to $\mathcal{S}_{0}^{N}(W\cup Z;L_{1},\alpha_{1})$. We let

(6)

$$\Psi_{Z;G,\alpha}:\mathcal{S}_{0}^{N}(W;L,\alpha)\to\mathcal{S}_{0}^{N}(W\cup Z;L_{1},\alpha_{1})$$

denote the restriction of $\Psi_{Z;G}$.

When the lasagna filling $G$ consists of a surface $S$ (an embedded cobordism $S\subset Z$ from $L$ to $L_{1}$) with no input balls, we will simply write $\Psi_{Z;S,\alpha}$ for $\Psi_{Z;G,\alpha}$. Furthermore, we could decorate $S$ with $n$ dots at a chosen location, for $0\leq n\leq N-1$, as usual in $\mathfrak{gl}_{N}$ foams; cf. [27, Example 2.3]. This corresponds to constructing a lasagna filling $S(n\bullet)$ with $n$ input balls intersecting $S$ along unknots, each decorated with the generator

$$X\in\operatorname{KhR}_{N}(U)\cong{\mathbb{Z}}[X]/(X^{N}).$$

(This filling is equivalent to one where we consider a single input ball intersecting $S$ in an unknot, decorated with $X^{n}$.) When the chosen location of the dot placement is clear from the context, then we denote the corresponding map by

(7)

$$\Psi_{Z;S(n\bullet),\alpha}:\mathcal{S}_{0}^{N}(W;L,\alpha)\to\mathcal{S}_{0}^{N}(W\cup Z;L_{1},\alpha_{1}).$$

Let $W$ be a smooth, oriented, connected, compact four-manifold (possibly with boundary). By standard Morse theory, $W$ can be decomposed into $k$-handles for $k=0,\dots,4$, arranged according to their index $k$. Furthermore, without loss of generality, we can arrange so that there is a unique 0-handle, and the number of 4-handles is either $0$ or $1$, according to whether $W$ has empty boundary or not.

Denote the numbers of $1$-, $2$- and 3-handles by $m$, $n$ and $p$, respectively. After attaching the 1-handles to the 0-handle we get the handlebody $\natural^{m}(S^{1}\times B^{3})$, with boundary $\#^{m}(S^{1}\times S^{2})$. (Here, $\natural$ denotes the boundary connected sum, and $\#$ the usual interior connected sum.) The attaching circles for the 2-handles form a link

$$K\subset\#^{m}(S^{1}\times S^{2}),$$

with components $K_{1},\dots,K_{n}$. The link also has a framing, which specifies how the 2-handles are attached. Once these are attached, the boundary of the resulting manifold must be of the form $Y\#p(S^{1}\times S^{2})$. Attaching the 3-handles gets rid of the $p$ summands of $S^{1}\times S^{2}$, so the resulting boundary is some $3$-manifold $Y$. In the case $\partial W\neq\emptyset$, we stop here and we have $\partial W=Y$. In the case where $W$ is closed, we must have $Y=S^{3}$ and we attach the 4-handle (a four-ball) to $S^{3}$ at the last step to eliminate the boundary.

The handle decomposition allows us to represent $W$ by a Kirby diagram. This consists of drawing $\#^{m}(S^{1}\times S^{2})$ as $m$ pairs of spheres in ${\mathbb{R}}^{3}$, where we think of the spheres in each pair as identified to produce a 1-handle (and we also add the point at infinity to ${\mathbb{R}}^{3}$). We then draw a picture of the attaching link $K$ for the 2-handles, where the link can go through the 1-handles. The framing of $K$ can be specified by drawing parallel copies of the components of $K$. (The components that don’t go through the 1-handles can be viewed as living in $S^{3}$; for those, an alternative way to specify the framing is by an integer, which is the difference between the given framing and the Seifert framing.) To determine $W$, in principle we should also specify the attaching spheres for the 3-handles. These are usually not drawn in the Kirby diagram. In the case where $\partial W=\emptyset$, this leaves no ambiguity, because there is a unique way to fill $\#^{p}(S^{1}\times S^{2})$ by 3-handles and then by a 4-handle.

For example, we show here a Kirby diagram of $W=\mathbb{CP}^{2}\#\mathbb{CP}^{2}$ with one 1-handle and three 2-handles. For the attaching curve of the 2-handle that goes through the 1-handle, we specified the framing by drawing a parallel copy by a dashed curve; for the other 2-handles, we used numbers:

For more details about the subject, we refer to the book [10].