The factorization of a marked MOY graph

Let $\mathrm{\Gamma }$ be a finite planar graph in a disk whose edges are either oriented arcs between marks (including boundary points) or wide (thick) edges, each wide edge bounded by four oriented edge-ends, as in the Khovanov-Rozansky diagrams. Place finitely many marks, with at least one on every internal edge and every circle of $\mathrm{\Gamma }$, place any finite number of marks (possibly none) on each boundary edge, and label all marks and boundary points by distinct variables $x_1,\dots ,x_r$; the boundary points carry orientations ${ϵ}_p\in \{1,-1\}$. With $C_c$ and $C_t$ the arc and wide-edge factorizations of Arc and wide-edge Khovanov-Rozansky factorizations, define

$C(\mathrm{\Gamma }):={⨂}_c{C}_c\otimes {⨂}_t{C}_t,$

the tensor product taken over $S=\mathbb{Q}[a,x_1,\dots ,x_r]$ with all variables shared (the tensor product of the two-term factorizations of the local pieces, with the Koszul sign convention), and view $C(\mathrm{\Gamma })$ as a factorization over the smaller polynomial ring $R=\mathbb{Q}[a,x_p:p\text{ a boundary point}]$: the variables at internal marks are internal and are forgotten. Its potential is

$w_{\mathrm{\Gamma }}=a{\sum }_p{ϵ}_p{x}_p\mathrm{.}$

The sum is over the boundary points: every internal label occurs at exactly two edge-ends with opposite signs, whose contributions $+a{x}_i$ and $-a{x}_i$ cancel, while a boundary label occurs at exactly one edge-end, and the local potentials of the arc and wide-edge factorizations add to $w_{\mathrm{\Gamma }}$.

If $\mathrm{\Gamma }$ is closed (no boundary points), then $w_{\mathrm{\Gamma }}=0$ and $C(\mathrm{\Gamma })$ is a $2$-periodic complex $C^0(\mathrm{\Gamma })\stackrel{d}{\to }{C}^1(\mathrm{\Gamma })\stackrel{d}{\to }{C}^0(\mathrm{\Gamma })$ of bigraded $\mathbb{Q}[a]$-modules, whose cohomology is denoted $H(\mathrm{\Gamma })$. Each term is a free bigraded $\mathbb{Q}[a]$-module, generally of infinite rank. Contractible summands may be removed without changing its cohomology, but no finite-rank representative over $\mathbb{Q}[a]$ is asserted: already a one-mark circle has cohomology $\mathbb{Q}[x]\{-1,1\}$. For a nonempty closed graph, the row reduction proved below shows that $a$ acts trivially on $H(\mathrm{\Gamma })$. For the empty graph the empty tensor product is $\mathbb{Q}[a]$ in even parity, and $a$ need not act trivially. Caveats: $C(\mathrm{\Gamma })$ has infinite rank as an $R$-module whenever internal marks are present; the marks are auxiliary data, and a marking change alters $C(\mathrm{\Gamma })$ by a chain homotopy equivalence (proved later on this page); the potential vanishes exactly for closed graphs, which is why $C(\mathrm{\Gamma })$ is a genuine complex in that case.