Subsection 3.2. (0NH7) · ScalingStacks

3.2. Three-handles

In [25, Proposition 2.1] the following result was shown:

Proposition 3.4. Let $i\colon W\to W^{\prime}$ be the inclusion of a four-manifold $W$ into $W^{\prime}$ . Then we have a natural map

i_{*}\colon\mathcal{S}_{0}^{N}(W;\emptyset)\to\mathcal{S}_{0}^{N}(W^{\prime},\emptyset).

If $W^{\prime}$ is the result of a $k$ -handle attachment to $W$ , then $i_{*}$ is a surjection for $k=3$ and an isomorphism for $k=4$ .

0NFW

Corollary 3.5. We have $\mathcal{S}_{0}^{N}(S^{4})\cong{\mathbb{Z}}$ , concentrated in bidegree zero.

In this section we focus on the case of 3-handle attachments. We will generalize the statement of Proposition 3.4 to 3-handle attachments in the presence of boundary links and explicitly describe the kernel of the resulting maps on $\mathcal{S}_{0}^{N}$ .

Consider the following setting. Let $W$ be a four-manifold with a framed link $L\subset Y=\partial W$ and an embedded $2$ -dimensional sphere $S\subset Y$ , disjoint from $L$ . Let $Z$ be the cobordism given by attaching a 3-handle to $W$ along $S$ , and let

W^{\prime}=W\cup Z.

Let $Y^{\prime}=\partial W^{\prime}$ be the outgoing boundary of $Z$ , so that $\partial Z=(-Y)\cup Y^{\prime}$ . Inside $Z$ we have the two-dimensional annular cobordism $A=I\times L$ , from $L=\{0\}\times L$ to a new link $L^{\prime}=\{1\}\times L$ . Given $\alpha^{\prime}\in H_{2}^{L}(W^{\prime};{\mathbb{Z}})\cong H_{2}^{L}(W;{\mathbb{Z}})/([S])$ , let us consider the set of all $\alpha\in H_{2}^{L}(W;{\mathbb{Z}})$ whose equivalence class modulo $[S]$ is $\alpha^{\prime}$ :

\langle\alpha^{\prime}\rangle:=\{\alpha\in H_{2}^{L}(W;{\mathbb{Z}})\mid\alpha\text{ mod }[S]=\alpha^{\prime}\}.

We obtain a cobordism map as in (6):

\Psi_{Z;A,\alpha}:\mathcal{S}_{0}^{N}(W;L,\alpha)\to\mathcal{S}_{0}^{N}(W^{\prime};L^{\prime},\alpha^{\prime}).

Let

\Psi_{Z;A,\alpha^{\prime}}:=\sum_{\alpha\in\langle\alpha^{\prime}\rangle}\Psi_{Z;A,\alpha}:\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\mathcal{S}_{0}^{N}(W;L,\alpha)\to\mathcal{S}_{0}^{N}(W^{\prime};L^{\prime},\alpha^{\prime}).

0NFX

Remark 3.6. When $L=\emptyset$ (and therefore $A=\emptyset$ ), then $\Psi_{Z;\emptyset}$ is exactly the map $i_{*}$ from Proposition 3.4.

Let $J$ be the equator of $S$ (which is an unknot in $Y$ ). Equip $J$ with an arbitrary orientation. By pushing a hemisphere of $S$ slightly from $Y=\{0\}\times Y$ into the cylinder $I\times Y$ , and taking its union with $I\times L$ , we obtain a properly embedded cobordism in $I\times Y$ , going from $L\cup J$ to $L$ . There are two such hemispheres, which produce two cobordisms, denoted $\Delta_{+}$ and $\Delta_{-}\subset I\times Y$ . We orient $\Delta_{+}$ and $\Delta_{-}$ so that their boundary orientation is the one on $J$ . (Note that they are therefore “oppositely oriented,” in the sense that they do not match up to produce an orientation on $S$ .) Let us identify $W\cup(I\times Y)$ with $W$ itself using a standard collar neighborhood. Then, the cobordism maps associated to $\Delta_{+}$ and $\Delta_{-}$ take the form

\Psi_{I\times Y;\Delta_{+},\alpha}\colon\mathcal{S}_{0}^{N}(W;L\cup J,\alpha+[\Delta_{+}])\to\mathcal{S}_{0}^{N}(W;L,\alpha),

\Psi_{I\times Y;\Delta_{-},\alpha}\colon\mathcal{S}_{0}^{N}(W;L\cup J,\alpha+[\Delta_{-}])\to\mathcal{S}_{0}^{N}(W;L,\alpha).

From here we get direct sum maps

\Psi_{I\times Y;\Delta_{+},\alpha^{\prime}}:=\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\Psi_{I\times Y;\Delta_{+},\alpha}

and

\Psi_{I\times Y;\Delta_{-},\alpha^{\prime}}:=\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\Psi_{I\times Y;\Delta_{-},\alpha}.

Observe that these two maps have the same domain

\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\mathcal{S}_{0}^{N}(W;L\cup J,\alpha+[\Delta_{+}])=\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\mathcal{S}_{0}^{N}(W;L\cup J,\alpha+[\Delta_{-}])

and the same range $\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\mathcal{S}_{0}^{N}(W;L,\alpha).$ Let

f:=\Psi_{I\times Y;\Delta_{+},\alpha^{\prime}}-\Psi_{I\times Y;\Delta_{-},\alpha^{\prime}}.

0NFY

Theorem 3.7. The map $\Psi_{Z;A,\alpha^{\prime}}$ associated to a 3-handle addition from $W$ to $W^{\prime}$ is surjective, and its kernel is exactly the image of $f$ . Therefore, $\mathcal{S}_{0}^{N}(W^{\prime},L^{\prime},\alpha^{\prime})$ is isomorphic to

\Bigl(\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\mathcal{S}_{0}^{N}(W,L,\alpha)\Bigr)/\operatorname{im}(f),

that is, to the coequalizer of the maps $\Psi_{I\times Y;\Delta_{+},\alpha^{\prime}}$ and $\Psi_{I\times Y;\Delta_{-},\alpha^{\prime}}$ .

0NFZ

Proof. We first show that $\Psi_{Z;A,\alpha^{\prime}}$ vanishes on the image of $f$ , that is,

\Psi_{Z;A,\alpha^{\prime}}\circ\Psi_{I\times Y;\Delta_{+},\alpha^{\prime}}=\Psi_{Z;A,\alpha^{\prime}}\circ\Psi_{I\times Y;\Delta_{-},\alpha^{\prime}}.

Indeed, from the composition law (5) we see that the left hand side is associated to the surface cobordism $\Delta_{+}\cup A$ and the right hand side to $\Delta_{-}\cup A$ . However, inside the 3-handle $Z$ , the sphere $S$ gets filled with a core $B^{3}$ , and therefore $\Delta_{+}$ and $\Delta_{-}$ are isotopic rel boundary. It follows that the two cobordism maps are the same.

Therefore, $\Psi_{Z;A,\alpha}$ factors through a map

\Phi\colon\Bigl(\bigoplus_{\alpha\in\langle\alpha^{\prime}\rangle}\mathcal{S}_{0}^{N}(W,L,\alpha)\Bigr)/\operatorname{im}(f)\to\mathcal{S}_{0}^{N}(W^{\prime},L^{\prime},\alpha^{\prime}).

We need to prove that $\Phi$ is bijective. For this, we construct its inverse $\Phi^{-1}$ . Given a lasagna filling $F^{\prime}$ of $W^{\prime}$ with boundary $L^{\prime}$ , observe that the cocore of the 3-handle $Z$ is one-dimensional, and therefore we can isotope $F^{\prime}$ to be disjoint from this cocore; after this, we can push it into $W$ , to obtain a lasagna filling there, called $F$ , with boundary $L$ . We set

\Phi^{-1}[F^{\prime}]=[F].

To see that $\Phi^{-1}$ is well-defined, we need to check that if two lasagna fillings $F^{\prime}_{0}$ and $F^{\prime}_{1}$ are equivalent in $W$ , then the corresponding fillings $F_{0}$ and $F_{1}$ differ (up to equivalences in $W$ ) by an element of $\operatorname{im}(f)$ . We use Lemma 2.1, in which we fix balls $R_{i}\subset W$ away from the 3-handle, and consider the equivalences listed in the lemma (with the ball replacements happening in $R_{i}$ ). Then, the equivalences in $W^{\prime}$ give rise to equivalences in $W$ , with one exception: an isotopy of the surfaces may intersect the one-dimensional cocore of $Z$ (which is an interval). Generically, this happens in a finite set of points, each point at a different time during the isotopy. Every time the isotopy meets the cocore, the corresponding surfaces in $W$ differ by replacing a hemisphere of $S$ (with boundary some closed curve $\gamma$ ) with its complement in $S$ . Up to an isotopy supported near $S$ , we can assume that $\gamma$ is the equator $J$ with its chosen orientation. (For example, if $\gamma$ is $J$ with the opposite orientation, we can rotate it by $\pi$ about a transverse axis to get $J$ with the original orientation.) Then, the hemispheres being interchanged are $\Delta_{+}$ and $\Delta_{-}$ and hence the classes of $F_{0}$ and $F_{1}$ differ by an element in the image of $f$ .

This shows that $\Phi^{-1}$ is well-defined, and its definition makes it clear that it is an inverse to $\Phi$ . It follows that $\Phi$ is bijective, and the conclusions follow. ∎

0NG0

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].

0NG1

Remark 3.9. Example 3.8 gives an alternate formula for 3-handle attachments. Let us go back to the general setting in this section, with a 3-handle attached to an arbitrary four-manifold $W$ along a sphere $S$ to produce $W^{\prime}$ , and a framed link $L\subseteq\partial W$ away from $S$ . Observe that $\mathcal{S}_{0}^{N}(W,L)$ is naturally a module over the algebra $\mathcal{S}_{0}^{N}(S^{2}\times D^{2};\emptyset)$ , with the module action being given by attaching fillings in a neighborhood of the sphere $S$ . It follows from the definitions that

\mathcal{S}_{0}^{N}(W^{\prime};L^{\prime})\cong\mathcal{S}_{0}^{N}(W;L)\otimes_{\mathcal{S}_{0}^{N}(S^{2}\times D^{2};\emptyset)}\mathcal{S}_{0}^{N}(B^{3}\times I;\emptyset).

Here, the algebra $\mathcal{S}_{0}^{N}(S^{2}\times D^{2};\emptyset)$ is the free polynomial ring in $A_{1},\dots,A_{N-1},A_{0},A_{0}^{-1}$ and $\mathcal{S}_{0}^{N}(B^{3}\times I;\emptyset)=\mathcal{S}_{0}^{N}(B^{4})$ is ${\mathbb{Z}}$ as a module over that algebra, where $A_{0}$ acts by $1$ and the other $A_{i}$ by $0$ . We conclude that

\mathcal{S}_{0}^{N}(W^{\prime};L^{\prime})\cong\mathcal{S}_{0}^{N}(W;L)/(A_{0}-1,A_{1},\dots,A_{N}).

Ciprian Manolescu, Kevin Walker, Paul Wedrich

Original source: arXiv:2206.04616v2