From Augmentations to Rulings

This is part II of a post on the relationship between augmentations and rulings. Here is part I if you missed it.

Given a front diagram of a Legendrian knot in plat position, we will use Sabloff’s method of dips to show that if there exists an augmentation of the Chekanov-Eliashberg DGA, then there exists a normal ruling of the diagram. We will extend the augmentation \epsilon of the diagram to an augmentation \epsilon' of the dipped diagram satisfying Property (R) which will simultaneously build a normal ruling of the original front diagram.

We will extend \epsilon to \epsilon' by adding dips to the diagram from left to right and extending the augmentation and ruling as we go as follows: If a crossing c_k is augmented in the original front diagram, then augment c_k in the dipped diagram. In a dip, we decide whether a crossing in the a-lattice is augmented by following Property (R).

Property (R): At the dip between c_k and c_{k+1}, a crossing in the a-lattice is augmented iff the two strands which cross there are paired in the ruling of the original front diagram between c_k and c_{k+1}.

A crossing b in the b-lattice between crossings c_k and c_{k+1} is augmented based on whether the strands which cross at b are crossing or companion strands in the ruling of the original front diagram and which diagram below c_k looks like .

abcdef

(Figure from Sabloff’s paper.)

In particular, if c_k has configuration ____ and is augmented, then b is augmented if the strands crossing at b are _____ strands. (Use the following to fill in this statement.)

(a), crossing
(b) or (c), crossing and companion
(d), none
(e) or (f), companion

Note that this definition relies on simultaneously building a ruling for the original front diagram. So, let us describe how to build the ruling. If we look at the left end of the ruling of a front diagram in plat position, the ruling will have strand 2k paired with strand 2k-1 for all 1\leq k\leq m. To extend the ruling over a crossing c_k, if c_k is augmented and it looks like configuration (a), (b), or (c) just to the left of c_k, then switch the ruling at c_k. Otherwise, don’t switch the ruling.

One can check that this gives us an augmentation of the dipped diagram. For example, suppose c_k is augmented and has configuration (c) just to the left. We will denote augmented crossings by a large dot. Suppose strands i and i+1 cross at c_k and that strand i is paired with strand L and strand i+1 is paired with strand K between crossings c_k and c_{k+1}.

Since c_k is augmented and looks like configuration (c) to the left, our ruling is switched at c_k.

cAug

(I apologize for the crude drawings, but Inkscape wasn’t cooperating.)

Now let us use this diagram to compute \epsilon'\circ\partial. Recall that the only disks that contribute to \epsilon'\circ\partial(c) are totally augmented disks, disks with a positive corner at c and with all negative corners augmented by \epsilon'. Each such disk will contribute 1 to \epsilon'\circ\partial(c), so as long as there are a even number of such disks, \epsilon'\circ\partial(c)=0.

There are no such disks for c_k, so \epsilon'\circ\partial(c)=0. A more interesting computation is the crossing b_1 in the b-lattice of strands i and K. We see that there are two such disks.

b-K-i-Disks

So \epsilon'\circ\partial(b_1)=0.

We also have the following disks for the crossing b_2 in the b-lattice of strands i+1 and L. So \epsilon'\circ\partial(b_2)=0.

b-L-i+1-Disks

A crossing with more standard disks appearing, is the crossing b_3 in the $b$-lattice where strands i+1 and K cross. It has the following disks.

b-K-i+1-Disks

There are no such disks for any crossing in the a-lattice, so \epsilon'\circ\partial=0 on the a-lattices. One can check that \epsilon'\circ\partial=0 for all remaining crossings.

While the dipped diagram makes differentials much easier, they tend to be a pain to deal with. Luckily, there’s a shortcut for finding the ruling associated to an augmentation of the original front diagram, without going through the dipped diagram:

We will modify our augmentation of the crossings of the plat position diagram while extending the ruling from left to right. Start the ruling at the left cusps like usual, and extend as follows. If a crossing c_k looks like configuration (a), (b), or (c) to the left and the crossing is augmented, then switch the ruling at the crossing and update the augmentation on the crossings to the right based on the number of disks to the right which contribute Qc_kR and involve the crossing strands.

shortcutDisk

Otherwise do not switch at the crossing. However, if a crossing c_k looks like configuration (e), or (f) to the left and c_k is augmented, then update the augmentation of the crossings to the right based on the number of disks to the right which contribute Qc_kR and involve the companion strands.

This is easier to see in an example. Let’s look at the trefoil where all crossings of degree 0 are augmented.

shortcutTrefoil1Start the ruling at the left.

shortcutTrefoil2The first crossing is augmented and the left of the crossing matches (a), so switch the ruling at c_1.

shortcutTrefoil3Now update the augmentation by looking for the disks QaR. We only find one.

shortcutTrefoil4So we get the diagram so far, with updated augmentation.

shortcutTrefoil5As c_2 is not augmented, the ruling is not switched and no updating of the augmentation is necessary.

shortcutTrefoil6The final crossing does not match (a), (b), or (c) to the left, so do not switch the ruling at c_3.

shortcutTrefoil7

It turns out the augmentation where only c_1 and c_2 are augmented will give the same ruling. Ng and Sabloff’s paper The Correspondence Between Augmentations and Rulings for Legendrian Knots discusses how this is a many-to-one map and gives the number of \rho-graded augmentations going to the same \rho-graded ruling for \rho\vert2r(K) and \rho=0 or \rho odd. (You’ll have to read up on what a \rho-graded augmentation is in the paper.)

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One response to “From Augmentations to Rulings

  1. Pingback: From Rulings to Augmentations | The Electric Handle Slide

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