# From Rulings to Augmentations

This is part III of a post on the relationship between augmentations and rulings. If you missed parts I and II, you can find them here and here.

Now we’ll work on going from rulings to augmentations. Fuchs does this using what he calls “splashes” in diagrams, but I find it easier to see this using Sabloff’s method of defining an augmentation for the dipped diagram as one can then get an augmentation for the original diagram.

Given a ruling for the original diagram in plat position, we will define an augmentation for the dipped diagram. First, augment $c_k$ if the ruling is switched at $c_k$, augment $a_{rs}$, the crossing of strands $r$ and $s$ in the $a$-lattice, if strands $r$ and $s$ are paired between $c_k$ and $c_{k+1}$, (this is what we called Property (R) in part II), and augment $b_{rs}$, the crossing of strands $r$ and $s$ in the $b$-lattice, if one of the following holds:

• ruling looks like (a) at the previous crossing $c_k$ and strands $r$ and $s$ are crossing strands,
• ruling looks like (b) or (c) at the previous crossing $c_k$ and strands $r$ and $s$ are crossing or companion strands,
• ruling looks like (e) or (f) at the previous crossing $c_k$ and strands $r$ and $s$ are companion strands.

Recall the various crossing configurations.

(From Sabloff’s paper.)

Let’s check for a couple of these cases that this gives an augmentation of the dipped diagram. In other words, check that for each crossing in the dipped diagram there are an even number of totally augmented disks in the diagram with positive corner at that crossing.

First, check the left end of the diagram. Since, in the ruling strands $2k$ and $2k-1$ are paired at the left, we know the crossing in the first $a$-lattice of strands $2k$ and $2k-1$ is augmented for $1\leq k\leq m$.

We then see that we have the totally augmented disks depicted.
So $\epsilon'\circ\partial=0$ on this portion of the diagram.

Most of the crossings in the dips I will leave for you to check, but we will check the dip after a crossing of configuration (c). Thus the ruling is switched at that crossing. Suppose strands $i$ and $i+1$ cross at the crossing $c_k$  and that strand $i$ is paired with $L$ and strand $i+1$ is paired with $K$.
Since the ruling is switched at the crossing, we know the crossing is augmented. We also see that the following other crossings are augmented as well, from the pairing of the strands in the ruling.
To check whether $\epsilon'\partial=0$ on the crossings in the dip after the original crossing, look for totally augmented disks.

Clearly there aren’t any totally augmented disks with positive corner at $c_k$, so $\epsilon'\partial c_k=0$.

We see that there are two totally augmented disks contributing to $\epsilon'\circ\partial$ of the crossing $b_1$ in the $b$-lattice of strands $K$ and $i$ and so $\epsilon'\partial(b_1)=1+1=0$. (Recall that we are working mod 2.)
We see that there are two totally augmented disks contributing to $\epsilon'\circ\partial$ of the crossing $b_2$ in the $b$-lattice of strands $L$ and $i$ and so $\epsilon'\partial(b_1)=1+1=0$.
Similarly, we have disks for crossings $b_3$ and $b_4$ in the $b$-lattice of, respectively, strands $K$ and $i+1$ and strands $L$ and $i+1$.
None of the remaining crossings in the $a$– or $b$-lattice have totally augmented disks, so we have checked that on this region of the dipped diagram, $\epsilon'$ is an augmentation.

Now, let’s look at the right end of the dipped diagram. Since we have the ruling of the original diagram, we know that at the right end of the diagram, strands $2k$ and $2k-1$ are paired in the ruling for $1\leq k\leq m$. Following our algorithm, this means that the crossings in the $a$-lattice of strands $2k$ and $2k-1$ are augmented for $1\leq k\leq m$.
Thus we have the following totally augmented disks for $q_k$.
Again, we see that there are two totally augmented disks with positive corner at $q_k$, so $\epsilon'\circ\partial(q_k)=1+1=0$.

Thus, with some checking of the remaining cases, we have shown that the augmentation of the dipped diagram we defined, is in fact an augmentation and so, given a way to define an augmentation of the dipped diagram of a knot from a ruling of the knot.