Rectangle-boundary pairings and grading changes in grid complexes

For two successive rectangles $r_1,r_2$, let $\psi =r_1+r_2$ be their cellwise sum as a two-chain, and let $\mathbf{x},\mathbf{z}$ be the initial and terminal states. The state permutation changes by one transposition at each rectangle. If $\mathbf{x}\mathrm{\ne }\mathbf{z}$, their relative permutation is either a product of two disjoint transpositions, so four points move, or a $3$-cycle, so three points move. These are the only nonidentity products of two transpositions.

For a counted rectangle $r$ the interior contains no state point, so [F3] gives $M_{\mathbb{O}}(\mathbf{x})-M_{\mathbb{O}}(\mathbf{y})=1-2n_{\mathbb{O}}(r)$. The coefficient ${\prod }_i{U}_i^{O_i(r)}$ lowers Maslov degree by $2n_{\mathbb{O}}(r)$ by [F2]; hence the full term has degree $-1$.

In the four-point case, the two transpositions are disjoint and commute. Exchanging their order gives the alternate intermediate state; the first rectangle in one decomposition has the same support as the second rectangle in the other, and conversely. Thus both decompositions have the same two rectangle interiors and are empty together. No third decomposition is possible, because the four distinct changing corners must be paired into the same two disjoint transpositions fixed by the endpoint permutation and the domain boundary.

In the three-point case, following the boundary of $\psi$ gives a six-corner region with five right-angle corners and one reflex corner of angle ${270}^{\circ }$; every cell coefficient is zero or one. At the reflex corner the boundary can be cut in exactly two ways into two rectangles. Each cut gives the required intermediate grid state, and the resulting interiors contain no additional state point because any such point would already lie in the interior of one of the original empty rectangles. The six-corner boundary has no other concave corner at which to make a cut, so these are exactly the two decompositions.

If $\mathbf{x}=\mathbf{z}$, the two rectangles reverse the same state transposition and their union is a periodic annulus. If the annulus were not thin, its transverse width would include at least two grid strips; the interior would then contain a state point on an intervening grid circle, contradicting emptiness. Hence every returning domain is one of the $n$ horizontal thin annuli, each occupying one row strip, or one of the $n$ vertical thin annuli, each occupying one column strip. Conversely, the two state points on the boundary circles of any one-row strip split that strip into two empty rectangles, and the same holds for every one-column strip, so all $2n$ listed annuli occur as returning composites. Each contains exactly one $O$ and one $X$ by the row and column marking conditions, so it is excluded in tilde and associated-graded minus modes. In the full filtered minus mode, the horizontal annulus through the row of $O_i$ contributes $U_i$, and the vertical annulus through its column also contributes $U_i$. Every $O_i$ therefore occurs twice among all returning domains, and the total is zero over ${\mathbb{F}}_2$.

In either non-annular case the two decompositions have the same cellwise domain $\psi$. A forbidden $O$- or $X$-marking occurs in one rectangle of a decomposition exactly when its cell has positive coefficient in $\psi$; hence a mode excluding that marking accepts either both decompositions or neither. For each $O_i$, the exponent in the product coefficient is its multiplicity in $\psi$, so the two accepted terms have the same monomial. They cancel over ${\mathbb{F}}_2$.

The Alexander change in [F3] is $A_G(\mathbf{x})-A_G(\mathbf{y})=n_{\mathbb{X}}(r)-n_{\mathbb{O}}(r)$. Multiplication by the coefficient lowers filtration by $n_{\mathbb{O}}(r)$, so the output filtration is $A_G(\mathbf{x})-n_{\mathbb{X}}(r)$. Thus the full filtered term drops by exactly $n_{\mathbb{X}}(r)$, and the tilde and associated-graded modes, which count only rectangles with $n_{\mathbb{X}}(r)=0$, preserve Alexander grading.