ScalingStacks

Example 3.8. Let $W=S^{2}\times D^{2}$ and $S$ the sphere $S^{2}\times\{p\}$, where $p\in\partial D^{2}$. Then attaching the 3-handle gives $W^{\prime}=B^{4}$. Let us see what Theorem 3.7 gives in this case. For simplicity, we ignore the decomposition into relative homology classes.

The skein lasagna module of $W$ has the structure of a commutative algebra over ${\mathbb{Z}}$, with the multiplication given by putting lasagna fillings side-by-side, in the decomposition

$$(S^{2}\times D^{2})\cup_{S^{2}\times I}(S^{2}\times D^{2})\cong S^{2}\times D^{2},$$

where $I\subset\partial D^{2}$ is an interval. As a ${\mathbb{Z}}$-algebra, $\mathcal{S}_{0}^{N}(W;\emptyset)$ was computed in [25, Theorem 1.2] to be

$$\mathcal{S}_{0}^{N}(W;\emptyset)\cong{\mathbb{Z}}[A_{1},\dots,A_{N-1},A_{0},A_{0}^{-1}]$$

where $A_{i}$ comes from the lasagna filling corresponding to the closed surface $S^{2}\times\{0\}$, equipped with the standard orientation, and marked with $N-1-i$ dots. (As mentioned in Section 2.2, this is equivalent to introducing one input ball intersecting $S^{2}\times\{0\}$ in an unknot labeled $X^{N-1-i}$.)

The cobordism maps

$$\Psi_{I\times Y;\Delta_{+}},\ \Psi_{I\times Y;\Delta_{-}}:\mathcal{S}_{0}^{N}(W;J)\to\mathcal{S}_{0}^{N}(W;\emptyset)$$

are as follows. The unknot $J$ is contained in a ball in the boundary of $W$ (say, a neighborhood of the disk $\Delta_{+}$). Then, according to [25, Corollary 1.5], we have

$$\mathcal{S}_{0}^{N}(W;J)\cong\mathcal{S}_{0}^{N}(W)\otimes_{{\mathbb{Z}}}\operatorname{KhR}_{N}(J)\cong\mathcal{S}_{0}^{N}(W)\otimes_{{\mathbb{Z}}}\bigl({\mathbb{Z}}[X]/(X^{N})\bigr).$$

(Strictly speaking, Corollary 1.5 in [25] is phrased for coefficients in a field $\mathbbm{k}$, due to the fact that its proof requires choosing a basis of $\operatorname{KhR}_{N}(J)$. In our case, $J$ is the unknot, so $\operatorname{KhR}_{N}(J)$ is free over ${\mathbb{Z}}$, and therefore the same argument applies with coefficients in ${\mathbb{Z}}$.)

Both maps $\Psi_{I\times Y;\Delta_{+}}$ and $\Psi_{I\times Y;\Delta_{-}}$ correspond to capping the unknot by disks. The first map acts only on the factor $\operatorname{KhR}_{N}(J)$ and is given by

$$\Psi_{I\times Y;\Delta_{+}}(v\otimes X^{N-1-i})=\begin{cases}v&\text{if }i=0,\\ 0&\text{if }i=1,\dots,N-1.\end{cases}$$

A useful picture to have in mind is that we can represent $X^{N-1-i}$ by a dotted disk (with the number of dots specified by the exponent of $X$), which is completed by $\Delta_{+}$ to a dotted sphere that bounds a ball in $W$, and hence can be evaluated to a scalar as shown above. To compute the action of $\Psi_{I\times Y;\Delta_{-}}$, on the other hand, note that the disk $\Delta_{-}$ completes the dotted disk to a homologically essential dotted sphere, corresponding to a generator in $\mathcal{S}_{0}^{N}(W;\emptyset)$:

$$\Psi_{I\times Y;\Delta_{-}}(v\otimes X^{N-1-i})=v\cdot A_{i}.$$

Therefore, taking the coequalizer of the two maps as in Theorem 3.7 boils down to setting

$$A_{0}=1,\ \ A_{1}=\dots=A_{N-1}=0$$

in $\mathcal{S}_{0}^{N}(W;\emptyset)$. We deduce that

$$\mathcal{S}_{0}^{N}(W^{\prime};\emptyset)\cong{\mathbb{Z}}[A_{1},\dots,A_{N-1},A_{0},A_{0}^{-1}]/(A_{1},\dots,A_{n-1},A_{0}-1)\cong{\mathbb{Z}},$$

which is the known answer for the skein lasagna module of $B^{4}$; see [27, Example 4.6].