{0,1} Maslov potentials on Legendrian knots

Question

A Legendrian knot is a curve in $\mathbb{R}^3$ on which $dz - ydx$ vanishes identically. Its projection to the $x,z$ plane is called a front diagram; as we can recover $y = dz/dx$ this determines the curve. Note that since $y$ is finite, we can never have vertical tangents; instead there are cusps in the diagram.

A $\mathbb{Z}$-Maslov potential is an assignment of integers to the arcs connecting cusps such that the number below a cusp is one less than the number above a cusp. The condition to admit such a thing is that the 'rotation number' -- which can be computed by counting the difference between the number of times you go from the top to the bottom of a cusp minus the number of times you go from the bottom to the top as you traverse the knot diagram -- be zero. Evidently a Maslov potential must take on at least two values.

Which rotation number zero Legendrian knots admit a diagram which admits a Maslov potential taking values in {$0,1$}?

For instance, the closure of any positive braid is such a knot.

Answer 1

I don't have anything close to a complete answer, but for many $tb$-maximizing knots we can get some obstructions from Legendrian contact homology. The generators for the DGA $A(K)$ associated to a given front diagram (as described by Ng) are right cusps, which have grading 1, and crossings, whose grading is the difference in Maslov potentials of the two strands through a crossing, so if the Maslov potential is valued in {0,1} then every generator has grading 0 or ±1. In particular, any 2-graded normal ruling of the front is actually just a graded normal ruling, and vice versa.

The classical invariants of $K$ satisfy the HOMFLY-PT bound $tb(K)+|r(K)| \leq -\deg_a P_K(a,z) - 1$, and Rutherford showed that $K$ has a 2-graded normal ruling if and only if $K$ achieves equality and $r(K)=0$. Thus if $K$ achieves the bound with $r(K)=0$ and $K$ has a {0,1} Maslov potential, we see that $K$ actually has a graded normal ruling. A theorem of Fuchs now says that $A(K)$ has an augmentation, so we can compute the linearized contact homology of $K$, which is generated over $\mathbb{Z}/2$ by crossings and right cusps whose gradings are bounded as above. In particular, if $K$ has a {0,1} Maslov potential and $tb(K) = -\deg_a P_K(a,z)-1$ then $A(K)$ must have augmentations, and the linearized homology for any augmentation has to be supported entirely in degrees -1,0,1.

We can use this on some examples in the Legendrian knot atlas as follows. The first $m(5_2)$ knot has linearized contact homology in degree 2, so it can't have a {0,1} Maslov potential, whereas the second $m(5_2)$ clearly does. The second $m(9_{45})$ knot has two different linearized contact homologies, and one of them is supported in degrees -1,0,1 but the other isn't, so it can't have a {0,1} Maslov potential either. Finally, the first three $7_4$ knots achieve the HOMFLY-PT bound ($tb=1$) but don't have any linearized contact homology, so they can't have {0,1} Maslov potentials while the fourth $7_4$ does.