September | 2012 | The Electric Handle Slide

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.

While hoping to get some classification results about symplectic fillings, I ran into a problem with symplectic cut-and-paste. Essentially there are two rather different compatibility conditions one can consider on the boundaries of two symplectic manifolds which ensure that they can be glued together to form a closed symplectic manifold. These two notions do not usually coincide so having one condition on one side, and the other condition on the other side does not mean they will glue together.

The first condition is the standard notion used when considering symplectic cobordisms: one boundary must be concave and the other convex. A concave or convex boundary inherits a contact structure, and a concave boundary will glue to a contactomorphic convex boundary to give a closed symplectic manifold. More precisely, a compact hypersurface of a symplectic manifold (e.g. the boundary) has contact type if there is a symplectic dilation (a.k.a. Liouville vector field) (a vector field v such that $\mathcal{L}_v(\omega)=\omega$ ) defined in a neighborhood of the hypersurface that is everywhere transverse to the hypersurface. Equivalently, a hypersurface N has contact type of there exists $\alpha\in \Omega^1(N)$ such that $d\alpha =i^*\omega$ and $\alpha$ is nowhere zero on $\ker(\omega|_N)$ . Here $\alpha$ will be a contact form on N, inducing a contact structure $\xi$ , oriented by $d\alpha$ . If the boundary of a symplectic manifold has contact type and the symplectic dilation is pointing in the outward normal direction, we say the boundary is convex; if the symplectic dilation points to the inward normal direction we say the boundary is concave. Equivalently, if we use the defintion involving the 1-form $\alpha$ , $\alpha\wedge (d\alpha)^{n-1}$ is a volume form on the hypersurface and thus induces an orientation. If that orientation agrees with the boundary orientation, we say the boundary is convex, and if they disagree it is concave.

The second condition that allows gluing of symplectic manifolds, is the existence of a fixed point free $S^1$ action on the boundary, such that the orbits of the $S^1$ action are tangent to the kernel of the restriction of the symplectic form to the boundary, $\ker(\omega|_{\partial W})$ . If two symplectic manifolds have orientation-reversing diffeomorphic boundaries, each with an $S^1$ action as above, and the symplectic reductions of the boundaries are symplectomorphic, then the manifolds can be glued together to get a closed symplectic manifold. The symplectic reduction here is the quotient of the boundary by the $S^1$ action, with the induced symplectic form (which remains non-degenerate since we are modding out by the kernel directions). A number of authors have used such $S^1$ actions for symplectic constructions including McDuff, Guillemin-Sternberg, McCarthy-Wolfson, etc. This $S^1$ action can be used to prove Gompf’s symplectic normal sum construction (here the $S^1$ action is has orbits which are concentric circles in the normal fibers to the symplectic surface we are doing the symplectic sum along).

Directionality

One significant difference between these two structures, is that one is inherently directional — a symplectic manifold with convex boundary cannot be viewed as a symplectic manifold with concave boundary by simply reversing some orientations, or negating the symplectic structure (and vice versa). So to glue, we can’t start with two symplectic manifolds which both have convex ends. Symplectic manifolds whose boundaries have $S^1$ action satisfying the correct conditions, are similarly considered either positive or negative, and to glue them together we need one postive side and one negative side. However it is possible to take a positive side and turn it into a negative side simply by reversing the orientation of the $S^1$ action, so we can glue two positive sides together simply by reversing the $S^1$ action on one of them to turn it into a negative side. Here is my pictorial analogy via puzzle pieces:

Here’s the mathematical justification. First consider a separating hypersurface which has contact type, so we have a Liouville vector field transverse to the hypersurface, $\mathcal{L}_v\omega = \omega$ . One side has convex boundary and the other has concave boundary depending on the direction of the transverse vector field. Suppose we wanted to cut out the side with convex boundary and glue to another symplectic manifold with convex boundary, so we wish we could realize the same manifold up to diffeomorphism as a symplectic manifold with concave boundary. Usually this is not possible. Replacing the symplectic dilation v with -v, makes it no longer a symplectic dilation for $\omega$ since $\mathcal{L}_{-v}\omega=-\mathcal{L}\omega =-\omega$ . So we cannot just switch the direction of v. If we modify the symplectic structure, by replacing $\omega$ by $-\omega$ , we keep the same symplectic dilation, $\mathcal{L}_v(-\omega) = -\omega$ , the orientation on the symplectic manifold is unchanged since $(-\omega)\wedge(-\omega)=\omega\wedge \omega$ gives the positive volume form. The contact structure on the boundary reverses its orientation, but the orientation induced on the boundary by the contact form is unchanged: $-\alpha\wedge(-d\alpha)=\alpha\wedge d\alpha$ . Therefore the boundary cannot easily be changed from convex to concave or vice versa. Typically the possible diffeomorphism types of convex fillings do not agree with the possible diffeomorphism types of concave fillings.

By contrast, with the $S^1$ action, condition the directionality is more superficial. I’ll explain the relation of the $S^1$ action and the symplectic structure in more detail first and then show the directionality can easily be reversed. One way a circle action can arise is as the flow of a Hamiltonian vector field. If we start with a Hamiltonian $H:M\to \mathbb{R}$ , and consider a regular level set $N=H^{-1}(c)$ , then the Hamiltonian vector field $v_H$ defined by $dH(\cdot)=\omega(v_H,\cdot)$ lies tangent to N, since $dH(v_H) = \omega(v_H,v_H)=0$ and $TN=\ker(dH)$ . So the flow of $v_H$ preserves the level sets of H. Additionally, $v_H$ is nowhere zero on N, since the zeros of $v_H$ correspond to critical points of H. Therefore if the orbits of $v_H$ are compact, we get a fixed point free $S^1$ action on N. Notice that the span of $v_H$ is the kernel of $\omega|_N$ since for any $w\in TN$ , $\omega(v_H,w) = dH(w) = 0$ .

Theorem [McDuff (free action case), McCarthy-Wolfson (fixed point free generalization), converse is Duistermaat-Heckman]: If I is an interval of regular values containing c, the symplectic form on $H^{-1}(I)$ is uniquely determined by the manifold $N=H^{-1}(c)$ , the $S^1$ action on N, and a family of symplectic forms $\{\tau_t\}$ on the (orbifolds) $H^{-1}(t)/S^1$ , where $\tau_{t_2}-\tau_{t_1}=(t_1-t_2)c$ (c is the Chern class of a principal $S^1$ bundle over the orbifold which is finitely branched covered by N — in the case the circle action on N is free, this principal bundle is just N).

The idea is that the symplectic forms $\tau_t$ are the symplectic reduction of $\omega$ on $N/S^1$ , and the symplectic form in the other directions are determined by the $S^1$ invariance. It turns out that this is the model for any hypersurface of a symplectic manifold with a compatible $S^1$ action. If N is any hypersurface with a fixed point free $S^1$ action such that $\ker(\omega|N)$ is tangent to the orbits, then there exists a neighborhood of N, symplectomorphic to a standard model: $(N\times (-\varepsilon,\varepsilon),\omega_0)$ coming from a Hamiltonian circle action. Call an embedding of N into $(M,\omega)$ , $\omega$ -compatible if the image of the $S^1$ orbits is tangent to $latex\ker(\omega|N)$.

McCarthy and Wolfson prove and use this theorem to show that one can cut two closed symplectic manifolds along such an N, and switch the top pieces and reglue to get two new symplectic manifolds. The “top” or + piece is determined as follows. The circle action is oriented and is tangent to the vector field $v_H$ . Since the symplectic form on $(M,\omega)$ is nondegenerate, and $v_H\in \ker(\omega|_N)$ , there must be a vector field X transverse to N, which pairs positively with $v_H, \; \omega(v_H,X)>0$ . The piece of the complement of N for which this vector field X points inward is the top, $M^+$ side, and the piece for which this vector field points outward is the bottom, $M^-$ side. McCarthy and Wolfson state their gluing theorem as follows:

Theorem: Suppose $(M_i^{2n},\omega_i)$ for i=1,2 are symplectic manifolds, $N^{2n-1}$ is a manifold with a fixed point free $S^1$ action, and there are $latex\omega_i$-compatible embeddings $j_i: N\to (M_i\omega_i)$ such that the image of N is a separating hypersurface. Also suppose that the symplectic reductions $(j_i(N)/S^1,\pi_*(\omega|{j_i(N)}))$ are symplectomorphic. Then there is a symplectic structure $\omega$ on $M=M_1^-\cup_N M_2^+$ , which agrees with the original symplectic structures, on a complement of a neighborhood of N.

Notice that when N is a 3-manifold, it is easy to rescale the symplectic structures on the symplectic reductions to ensure they are symplectomorphic, so this hypothesis is easily satisfied if one is willing to rescale the symplectic structure on the 4-manifold.

At initial glance, one would think that the $M^-$ and $M^+$ sides correspond to the concave and convex fillings. However, it is not hard to show that there is nothing dictating that the $M^-$ side cannot act as an $M^+$ side. By simply negating the Hamiltonian, $H\mapsto -H$ , we reverse the Hamiltonian vector field $v_H\mapsto -v_H$ , and thus the $S^1$ action. This does not change whether the hypotheses of the theorem are satisfied: the $S^1$ action with reversed orientation is still tangent to $\ker(\omega|{j_i(N)})$ , and the symplectic reductions are unchanged. However if before $\omega(v_H,X)>0$ and X pointed into $M^+$ and out of $M^-$ , then now $\omega(-v_H,-X)>0$ , and $-X$ points into $M^-$ and out of $M^+$ , so $M^+$ and $M^-$ have exchanged roles. In other words, the property of being a symplectic manifold whose boundary has a fixed point free $S^1$ is not an inherently sided/directional condition like a boundary of contact type is. The + and – sides look the same after a $180^{\circ}$ rotation of the vectors giving the circle and normal directions.

Trying to relate the two conditions:

So this $\omega$ -compatible $S^1$ action condition, is very different in character from the contact type (concave/convex) condition. Are these two conditions related at all? Certainly not every hypersurface of contact type has an $\omega$ -compatible $S^1$ action. A necessary condition for the existence of such an action, is that the contact structure on the hypersurface be transverse to the $S^1$ -action. This is because in a neighborhood of a hypersurface, N of contact type, the symplectic form looks like the symplectization of $(N,\ker\alpha)$ : $(\mathbb{R}\times N, d(t\alpha))$ . The kernel of the restriction of this to $\{1\}\times N$ is

$\ker((d(t\alpha))|_{\{1\}\times N}) = \ker((dt\wedge \alpha+td\alpha)|_{\{1\}\times N}) = \ker(d\alpha)$

The kernel of $d\alpha$ is the span of the Reeb vector field and is necessarily transverse to $\xi=\ker\alpha$ . Therefore if the $S^1$ action lies in the kernel of the symplectic form on N, it is necessarily transverse to the contact structure. The existence and classification of such contact structures has been studied in certain cases (e.g. Lisca-Matic). So not all convex/concave boundaries will have an $\omega$ -compatible $S^1$ action.

Question: Do all $\omega$ -compatibly embedded separating hypersurfaces have contact type (are concave/convex boundaries)?

It is mentioned in the introduction to McCarthy-Wolfson’s paper that this is the case, but it is not clear to me how to find a Liouville vector field transverse to the hypersurface in the standard model of the symplectic structure on $N\times(-\varepsilon,\epsilon)$ , which is suitably compatible with the circle action. Equivalently one could look for a contact form $\alpha$ on N such that $d\alpha=\omega|_N$ . Any ideas? Here is what happens when you try the vector field in the $(-\varepsilon,\varepsilon)$ direction, it doesn’t seem to work out, but I’m probably missing something.

For simplicity lets assume the circle action is free, so we don’t need to deal with orbifolds and branched covers. Then fix a symplectic structure $\tau_0$ on the quotient $(\{0\}\times N)/S^1$ , and define $\tau_t =\tau_0+tc_1(N)$ , where we view $\pi: N\to N/S^1$ as a principle $S^1$ bundle so its Chern class makes sense. Then the unique symplectic form on $N\times (-\varepsilon,\varepsilon)$ compatible with the $S^1$ action, is given by $\omega = \pi^*(\tau_t)+\alpha_t\wedge dt$ where $\alpha_t$ is dual to the tangent vector field to the $S^1$ action. We may guess that $\partial_t$ is the symplectic dilation transverse to N, so we need to check if $\mathcal{L}_{\partial_t}\omega=\omega$ :

$\mathcal{L}_{\partial_t}(\omega) = d\circ \iota_{\partial_t}(\pi^*(\tau_t)+\alpha_t\wedge dt)=d\alpha_t$
Its not clear to me why we should have $\omega = d\alpha_t$ . With any other transverse vector field, we will have some non-zero component of $\partial_t$ , so maybe we can add something to make this a Liouville vector field. Right now it seems unclear to me though. Any thoughts are welcome.