Chekanov-Eliashberg DGA: Why d^2 = 0

The Chekanov-Eliashberg DGA is an invariant of Legendrian knots consisting of a dg-algebra whose differential is determined by counting rigid, punctured holomorphic disks in the plane with exactly one positive puncture and with boundary on the Lagrangian projection of a knot L.  Due to the Riemann mapping theorem, these counts of holomorphic disks can be determined diagrammatically and the entire dg-algebra structure can be obtained combinatorially.  This invariant fits into the broad framework of Symplectic Field Theory which gives invariants of objects across the contact and symplectic topology spectrum, such as symplectic homology and contact homology for closed manifolds, chain maps corresponding to symplectic cobordisms, and relative versions for Legendrians (resp. Lagrangians) in contact (resp. symplectic) manifolds.  In all cases, the invariants conjecturally are determined by counting punctured holomorphic disks of all genera and with arbitrarily many positive punctures.

However, the relevant analytic theorems are hard and the theories are not always defined in all cases.  Legendrian Contact Homology (or relative contact homology) is a generalization of DGA to Legendrians in all contact manifolds of any dimension and has been rigorously defined by Etnyre, Ekholm and Sullivan when the ambient contact manifold is of the form P \times \mathbb{R}, for P an exact symplectic manifold, such as 1-jet spaces and affine space.  Here, the differential only counts holomorphic disks with a single positive puncture.

I’m interested in understanding moduli spaces of curves with multiple positive punctures and as an introduction, I read Lenny Ng’s paper (Rational SFT for Legendrian Knots) where he works out a Legendrian knot invariant that includes information about holomorphic disks with arbitrarily many positive punctures.  “Rational” here refers to the fact that he is only considering holomorphic disks and not punctured curves of higher genus.

In this case, one has to account for two phenomena that cause lots of problems, bubbling and multiple covers, which can ruin the equation \partial^2 = 0 if they are not accounted for properly.  Ng follows an approach of Cieliebak, Latschev and Mohnke by using ideas from String Topology to account for these problems.  I’ll explain how he does this in a future post but before I get to that, I want to review basics about why counting holomorphic curves should even give a differential.

Why is \partial^2 = 0?

Take as an example the right-handed trefoil

whose differential is

\partial q_1 = 1 + q_2 q_5 q_6 + q_2 + q_4 + q_4 q_5 q_6 + q_6

\partial q_7 = 1 + q_6 q_5 q_2 + q_2 + q_4 + q_6 q_5 q_4 + q_6

\partial q_2 = q_3

\partial q_4 = q_3

\partial q_3 = \partial q_5 = \partial q_6 = 0

Now, conceptually identify each monomial in the differential with the disk that contributes it to the differential.  We can find \partial^2 by applying the Leibnitz rule.  For instance

\partial^2 (q_1) = (\partial q_2) q_5 q_6 + (\partial q_2)+ (\partial q_4) + (\partial q_4) q_5 q_6 =

$latex q_3 q_5 q_6 + q_3 + q_3  + q_3 q_5 q_6 = 0$ mod 2

Before canceling terms, think of what each monomial in \partial^2 represents.  The second q_3 q_5 q_6 term is determined by a holomorphic disk with one positive puncture at q_1 and a holomorphic disk with one positive puncture at $q_4$.  You should think of the monomial q_3 q_5 q_6 as representing the disk obtained by gluing these two disks together:

This is still a punctured holomorphic disk with one positive puncture, but it is not rigid and has an obtuse corner at q_3.  When computing the differential, I look for holomorphic curves with acute corners but in general I can find holomorphic curves with obtuse corners and the dimension of the moduli space in which it lives is the number of obtuse corners.  The reason is because at an obtuse corner, I can degenerate in two directions.  A local model of the puncture is that, traversing the boundary of the disk counter-clockwise, I get to the crossing, jump down a strand and immediately continue off to my right.

But I could also jump down, turn left and go a little ways, before turning around and going the other way.  Or I can first go past the crossing, before turning around, jumping down and heading to (what is now) my left.

Again these are still punctured holomorphic curves with boundary on L.  If you need a local picture to convince yourself that it’s OK to turn around, think of what happens at the origin to the imaginary axis under the holomorphic map z \mapsto z^2.

I can degenerate along each of these arcs in my knot projection until it intersects the boundary of my disk.  My moduli space is not compact, but it limits to a broken curve where my holomorphic curve breaks into two pieces, each of which is a holomorphic curve.  This happens in both directions and the standard picture is the heart.

Broken curves are, in some sense, the opposite of gluing.  As you can see from the differential, I get two copies of the monomial q_3 q_5 q_6, one of which corresponds to gluing at q_2 and the other from gluing at q_4.  Since we are working over \mathbb{F}_2, these terms cancel; if we kept track of signs they would still cancel.  So, whenever I apply the differential twice, I get monomials which each corrspond to an end of a 1-dimensional moduli space and so these monomials cancel in pairs.

In general, you can show that \partial^2 = 0 by showing

  • broken curves glue together
  • 1-dimensional moduli spaces can be compactified with broken curves

This proves that every broken curve is the boundary of some 1-dimensional moduli space and since the boundary of a compact 1-manifold is an even number of points, broken curves must cancel is pairs.  Since broken curves correspond to monomials in \partial^2 this proves that the map is in fact a differential.

A similar situation occurs in Morse homology.  The differential counts rigid gradient flow lines and applying the differential twice means I jump descend twice along rigid flow lines.

Now, if I perturb a little, there is in fact a flow line that starts and ends at the same critical points but jumps the middle critical point.  It lives in a 1-dimensional moduli space of flow lines and at the other end is another pair of rigid flow lines.  The endpoints are called broken trajectories because it breaks into two pieces at its limit.

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2 responses to “Chekanov-Eliashberg DGA: Why d^2 = 0

  1. Pingback: What does a moduli space look like? | The Electric Handle Slide

  2. Pingback: Morse Homotopy and A-infinity; Part 1 | The Electric Handle Slide

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