A knot grid diagram is admissible

Regard the grid torus as the standard Heegaard torus separating the two solid tori of $S^3$, with horizontal circles meridians of the first and vertical circles meridians of the second. The $n$ parallel meridian disks in either solid torus cut it into $n$ balls, one for each row or column annulus. Push the row $O$-to-$X$ arcs into the first solid torus and the column $X$-to-$O$ arcs into the second. Their pushing disks lie in disjoint row or column annuli and are disjoint from the compression disks, so each arc system is simultaneously boundary parallel as required by [F2]. The over/under convention is vertical over horizontal, exactly [F1]'s link. Thus these data represent the same oriented knot and satisfy the topological part of the multipointed definition; it remains to prove its admissibility condition.

Let $p_{ij}$ be the coefficient of cell $R_{ij}$ in a periodic domain $P$. Because $\mathrm{\partial }P$ is a sum of whole horizontal and vertical circles, the coefficient jump across a vertical circle is independent of the row, and the jump across a horizontal circle is independent of the column. Fix reference indices $i_0,j_0$ and set $a_i=p_{i,j_0}$ and $b_j=p_{i_0,j}-p_{i_0,j_0}$. Summing the constant jumps from the reference row and column gives $p_{ij}=a_i+b_j$ for every cell. Since the marking permutation in [F1] is one cycle, number its basepoint pairs in reverse order of knot travel so that row region $A_i$ contains $w_i,z_i$ and column region $B_i$ contains $w_i,z_{i+1}$, as allowed by the convention in [F2]. Then $P={\sum }_i{a}_i{A}_i+{\sum }_i{b}_i{B}_i$, with local multiplicities $n_{w_i}(P)=a_i+b_i$ and $n_{z_i}(P)=a_i+b_{i-1}$.

Suppose $P$ is periodic and all $n_{w_i}(P)=0$. Then $b_i=-a_i$, so $n_{z_i}(P)=a_i-a_{i-1}$ and ${\sum }_i{n}_{z_i}(P)=0$. If one $n_{z_i}(P)$ is nonzero, these integer multiplicities include both a positive and a negative value; each is a region coefficient at the corresponding $X$-cell, so $P$ has both signs.

If all $n_{z_i}(P)$ also vanish, then all $a_i$ are the same integer $c$ and all $b_i=-c$. Hence $P=c({\sum }_i{A}_i-{\sum }_i{B}_i)=0$, because each of the two sums is the whole torus with coefficient one. Thus a nonzero periodic domain with zero $O$-multiplicity cannot have all $X$-multiplicities zero, and step 2.1 proves it has both signs. By [F2], the grid diagram is admissible.