ScalingStacks

Proof. For a lasagna diagram $D$ (as in Figure 2) we define a homomorphism

$$\mathrm{KhR}_{N}(D):\bigotimes_{i}\mathrm{KhR}_{N}(S_{i},L_{i})\to\mathrm{KhR}_{N}(S,L),$$

as follows. We first choose points $q_{j}\in S_{j}\setminus L_{j}$ and $q\in S\setminus L$ and then a properly embedded 1-complex $T\subset B$, disjoint from $\Sigma$, such that the underlying graph of $T$ is a tree and the endpoints of the 1-complex are $\{q_{j},q\}$. Choose a small closed tubular neighborhood $N$ of $T$, also disjoint from $Y$. The complement of $N$ in $B\setminus\sqcup B_{i}$ is diffeomorphic to ${\mathbb{R}}^{3}\times[0,1]$ with some embedded surface $\Sigma^{\prime}$. We will view $\Sigma^{\prime}$ as a bordism between two links in two copies of ${\mathbb{R}}^{3}$. One copy is identified with $S^{q}:=S\setminus\{q\}$, which contains the link $L$. The other copy $X$ is the remainder of the boundary, and can be expressed as the boundary connect sum of the 3-balls $S_{i}^{q_{i}}:=S_{i}\setminus\{q_{i}\}$, connected along the tree $T$. The 3-ball $X$ contains the split disjoint union of the links $L_{i}$. Khovanov–Rozansky homology for links in 3-balls gives us a map

$$\mathrm{KhR}_{N}(\Sigma^{\prime}):\mathrm{KhR}_{N}(X,\sqcup_{i}L_{i})\to\mathrm{KhR}_{N}(S^{q},L)$$

which, together with the monoidality maps from §4.3, specifies a map

$\displaystyle\mathrm{KhR}_{N}(D):\bigotimes_{i}\mathrm{KhR}_{N}(S_{i},L_{i})$

$\displaystyle\cong\bigotimes_{i}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\stackrel{{\scriptstyle T}}{{\to}}\mathrm{KhR}_{N}(X,\sqcup_{i}L_{i})\to\mathrm{KhR}_{N}(S^{q},L)\cong\mathrm{KhR}_{N}(S,L).$

Here the first and last maps are the ‘evaluation’ isomorphisms discussed below Definition 4.12, and we highlight that the monoidality map depends on the tree $T$.

We must check that the overall map above does not depend on the choices of $q_{i}$ and $T$. This is straightforward, so we merely sketch the argument. Isotoping the points $q_{i}$ does not change the map, by the same argument that showed that $\mathrm{KhR}_{N}$ is well-defined for links in 3-spheres; see §4.2. Isotoping $T$ disjointly from $\Sigma$ clearly does not affect the map. Changing the combinatorics of the underlying tree of $T$ can be done in such a way that $N$ varies continuously and remains far from $\Sigma$, and so does not affect the map. Isotoping $T$ through $\Sigma$ does not affect the map, thanks to the sweep-around property (see Theorem 1.1 above). Thus $\mathrm{KhR}_{N}(\Sigma)$ is well-defined.

Next we must show compatibility with the operad composition. For this we consider three lasagna diagrams:

• •

  $D_{1}$ with output boundary $(S_{1},L_{1},q_{1})$, with input boundaries $(S_{i},L_{i},q_{i})_{i\in J}$, surface $\Sigma_{1}$, and tree $T_{1}$,

• •

  $D_{2}$ with output boundary $(S_{2},L_{2},q_{2})$, with input boundaries $(S_{1},L_{1},q_{1})$ along with $(S_{i},L_{i},q_{i})_{i\in K}$, surface $\Sigma_{2}$, and tree $T_{2}$

• •

  $D$, the result of gluing $D_{1}$ inside $D_{2}$, with outer boundary $(S_{2},L_{2},q_{2})$, input boundary $(S_{i},L_{i},q_{i})_{i\in J\cup K}$, surface $\Sigma=\Sigma_{1}\cup\Sigma_{2}$, and tree $T=T_{1}\cup T_{2}$.

Compatibility with the operad composition now boils down to the claim:

$$\mathrm{KhR}_{N}(D)=\mathrm{KhR}_{N}(D_{2})\circ(\mathrm{KhR}_{N}(D_{1})\otimes\mathbf{1})$$

as maps

$$\left(\bigotimes_{i\in J}\mathrm{KhR}_{N}(S_{i},L_{i})\right)\otimes\left(\bigotimes_{i\in K}\mathrm{KhR}_{N}(S_{i},L_{i})\right)\to\mathrm{KhR}_{N}(S_{2},L_{2})$$

We compare these two homomorphisms on the level of 3-ball link homologies, that is, with respect to a fixed choice of basepoints $q_{i}$ and $q$, and we suppress associators. On this level $\mathrm{KhR}_{N}(D)$ is determined by the homomorphism

(5.1)

$$\left(\bigotimes_{i\in J}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)\otimes\left(\bigotimes_{i\in K}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)\xrightarrow{T}\mathrm{KhR}_{N}(X,\sqcup_{i}L_{i})\to\mathrm{KhR}_{N}(S_{2}^{q},L_{2})$$

where $X$ denotes the boundary connect sum of the 3-balls $S_{i}^{q_{i}}$ for $i\in J\cup K$ along the tree $T$, and the first map is provided by lax monoidality. On the other hand, the homomorphism $\mathrm{KhR}_{N}(D_{2})\circ(\mathrm{KhR}_{N}(D_{1})\otimes\mathbf{1})$ is determined by the composite

	$\displaystyle\left(\bigotimes_{i\in J}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)\otimes\left(\bigotimes_{i\in K}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)\xrightarrow{T_{1}\otimes\mathbf{1}}$	$\displaystyle\;\mathrm{KhR}_{N}(X_{J},\sqcup_{i\in J}L_{i})\otimes\left(\bigotimes_{i\in K}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)$
	$\displaystyle\xrightarrow{\mathrm{KhR}_{N}(\Sigma^{\prime}_{1})\otimes\mathbf{1}}$	$\displaystyle\;\mathrm{KhR}_{N}(S_{1}^{q_{1}},L_{1})\otimes\left(\bigotimes_{i\in K}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)$
	$\displaystyle\xrightarrow{T_{2}}$	$\displaystyle\;\mathrm{KhR}_{N}(X_{K},\sqcup_{i\in\{1\}\cup K}L_{i})$
	$\displaystyle\xrightarrow{\mathrm{KhR}_{N}(\Sigma^{\prime}_{2})}$	$\displaystyle\;\mathrm{KhR}_{N}(S_{2}^{q},L_{2}).$

Here we write $X_{J}$ for the boundary connect sum of the 3-balls $S_{i}^{q_{i}}:=S_{i}\setminus\{q_{i}\}$ for $i\in J$ that is determined by $T_{1}$, and $X_{K}$ for the boundary connect sum of the $S_{i}^{q_{i}}$ for $i\in\{1\}\cup K$ along $T_{2}$. After commuting the map induced by $\Sigma^{\prime}_{1}$ past the second monoidality map, we arrive at

(5.2)

$$\left(\bigotimes_{i\in J}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)\otimes\left(\bigotimes_{i\in K}\mathrm{KhR}_{N}(S_{i}^{q_{i}},L_{i})\right)\xrightarrow{T}\mathrm{KhR}_{N}(X,\sqcup_{i\in J\cup K}L_{i})\xrightarrow{\mathrm{KhR}_{N}(\Sigma^{\prime}_{2}\circ(\Sigma^{\prime}_{1}\cup\mathbf{1}))}\mathrm{KhR}_{N}(S_{2}^{q},L_{2}).$$

Since the link cobordism $\Sigma^{\prime}_{2}\circ(\Sigma^{\prime}_{1}\cup\mathbf{1})$ is isotopic to $\Sigma$, the functoriality of $\mathrm{KhR}_{N}$ implies that the maps in (5.1) and (5.2) are equal. This proves the claim. ∎