Integrality and rectangle changes of the grid Alexander grading

Each $I(P,Q)$ is an integer count. For $P=Q$, $J(P,P)=I(P,P)$, so the two self-pair terms in the formulas for $M_{\mathbb{O}}$ and $M_{\mathbb{X}}$ are integers; the cross terms occur doubled. Hence both Maslov functions are integer-valued.

Suppose a rectangle $r$ is contained in the chosen fundamental square. Write its southwest and northeast corners as $x_1,x_2\in \mathbf{x}$ and its northwest and southeast corners as $y_1,y_2\in \mathbf{y}$; all other state points are common. In $J(\mathbf{x},\mathbf{x})-J(\mathbf{y},\mathbf{y})$ the common-point terms cancel, the ordered corner pair contributes $1$, and each point of $\mathbf{x}$ strictly inside $r$ contributes twice, once against each changed corner. Thus this difference is $1+2\mathrm{\#}(\mathbf{x}\cap \mathrm{Int}r)$. In $2J(\mathbf{x},\mathbb{O})-2J(\mathbf{y},\mathbb{O})$ the same cancellation leaves twice the number of $O$-markings in $r$. Substituting these two counts into [F1] gives $M_{\mathbb{O}}(\mathbf{x})-M_{\mathbb{O}}(\mathbf{y})=1-2n_{\mathbb{O}}(r)+2\mathrm{\#}(\mathbf{x}\cap \mathrm{Int}r)$. Replacing every $O$ by $X$ gives the stated formula for $M_{\mathbb{X}}$.

For a marking set $P\in \{\mathbb{O},\mathbb{X}\}$, write $M_P(\mathbf{x})=J(\mathbf{x}-P,\mathbf{x}-P)+1$ and let $\mathrm{N}\mathrm{W}(P_i)$ denote the northwest grid vertex of the cell containing the $i$th mark, read on the bottom edge when that cell is in the top row. Let ${\mathbf{x}}_{\mathrm{N}\mathrm{W}P}$ be the state consisting of these vertices. Expanding the $J$ formula in [F1] gives $M_P({\mathbf{x}}_{\mathrm{N}\mathrm{W}P})=\mathrm{\#}\{(i,j):\mathrm{N}\mathrm{W}(P_i)<\mathrm{N}\mathrm{W}(P_j)\}-\mathrm{\#}\{(i,j):\mathrm{N}\mathrm{W}(P_i)<P_j\}-\mathrm{\#}\{(i,j):P_i<\mathrm{N}\mathrm{W}(P_j)\}+\mathrm{\#}\{(i,j):P_i<P_j\}+1.$ For distinct $i,j$ with neither mark in the top row, all four inequalities are equivalent and their signed contributions cancel. If $P_j$ is in the top row, the second and fourth inequalities are equivalent and the first and third are false; if $P_i$ is in the top row, the first and second inequalities are equivalent and the third and fourth are false. These contributions also cancel. For $i=j$, the only inequality that can hold is $\mathrm{N}\mathrm{W}(P_i)<P_i$, and it holds exactly for the unique mark in the top row. The displayed count is therefore $-1+1=0$, proving $M_{\mathbb{O}}({\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{O}})=M_{\mathbb{X}}({\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{X}})=0$.

For two complementary rectangles $r$ and $r^{\mathrm{\prime }}$ with the same vertical edges, let $v$ be their common width, counted in column strips. The one-$O$- and one-$X$-per-column conditions in [F2] give $n_{\mathbb{O}}(r)+n_{\mathbb{O}}(r^{\mathrm{\prime }})=v$ and $n_{\mathbb{X}}(r)+n_{\mathbb{X}}(r^{\mathrm{\prime }})=v$. The state has one point on each of the $v-1$ vertical circles strictly between those two edges, so $\mathrm{\#}(\mathbf{x}\cap \mathrm{Int}r)+\mathrm{\#}(\mathbf{x}\cap \mathrm{Int}{r}^{\mathrm{\prime }})=v-1$. If the formula of step 1.2 holds for $r$ from $\mathbf{x}$ to $\mathbf{y}$, these three equalities give the formula for $r^{\mathrm{\prime }}$ from $\mathbf{y}$ to $\mathbf{x}$ by subtraction. The same argument applies to complementary rectangles sharing horizontal edges, counting rows instead of columns. Of the four torus rectangles for the two states, one lies in the fundamental square; successive horizontal or vertical complementations reach each of the other three, so the formulas hold for every torus rectangle.

Subtract the $M_{\mathbb{X}}$ change from the $M_{\mathbb{O}}$ change and divide by two. The constant terms and state-interior terms cancel, giving $\frac{M_{\mathbb{O}}(\mathbf{x})-M_{\mathbb{X}}(\mathbf{x})}{2}-\frac{M_{\mathbb{O}}(\mathbf{y})-M_{\mathbb{X}}(\mathbf{y})}{2}=n_{\mathbb{X}}(r)-n_{\mathbb{O}}(r)$. The common normalization $-(n-1)\mathrm{/}2$ in $A_G$ cancels, proving the rectangle-change formula and that all differences of $A_G$ are integers.

The row and column connections from $O$ to $X$ and back form one $n$-cycle by [F2]. Equivalently, the relative permutation between ${\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{X}}$ and ${\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{O}}$ is an $n$-cycle. Factor that cycle into $n-1$ transpositions and apply them successively to obtain grid states between the two states. Two states that differ by one transposition have four intersection corners: the two row circles and two column circles cut the torus into four rectangles, and one of those rectangles has the first state at its southwest and northeast corners and the second state at its northwest and southeast corners. Thus each transition is realized by a torus rectangle, whether empty or not. Each rectangle changes $M_{\mathbb{X}}$ by an odd integer by steps 1.2 and 2.1, so $M_{\mathbb{X}}({\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{O}})\equiv n-1(\mathrm{m}\mathrm{o}\mathrm{d}2)$, using the zero normalization at ${\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{X}}$ from step 1.3. Since $M_{\mathbb{O}}({\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{O}})=0$, the numerator $M_{\mathbb{O}}-M_{\mathbb{X}}-(n-1)$ defining twice $A_G({\mathbf{x}}_{\mathrm{N}\mathrm{W}\mathbb{O}})$ is even. Hence this value is integral.

Any two grid states correspond to permutations, and a finite product of transpositions carries one to the other. Each transposition gives a rectangle transition. Step 3.1 shows that $A_G$ changes by an integer at every transition; since one state has integral Alexander value by step 3.2, all grid states do.