# Symplectic Gluing

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.

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### 6 responses to “Symplectic Gluing”

1. I believe the answer the question is actually no, contrary to McCarthy-Wolfson’s claim that their procedure is a special case of convex-concave gluing.

The easiest counter-example is of $S^2 \times S^1 \subset S^2 \times S^2$. It is straightforward to find an $\omega$-compatible structure on $S^2 \times S^1$ but $S^2 \times D^2$ does not strongly fill the Stein fillable contact structure on $S^2 \times S^1$.

More generally, take Kahler manifolds $(X, \omega)$ and $(X \times S^2, \omega \oplus \omega_{FS})$ and the separating hypersurface $Y = X \times S^1$. Then there is an $\omega$-compatible structure on $Y$ but I believe there is no way to make it of contact type. If, instead, $Y$ was an $S^1$-bundle over $X$ with nonzero Euler number then it would admit both $\omega$-compatible and contact type structures.

• lstarkston

Thanks, that is what I suspected the answer would be. How do you know $S^2\times D^2$ does not strongly fill the tight contact structure on $S^2\times S^1$?

• Paolo

It follows from Yasha’s filling by holomorphic disks.

• A simpler reason is that the restriction of $\omega$ to $S^2 \times S^1$ is Poincare dual to the $S^2$ factor so it can’t be exact.

2. lstarkston

Some follow-up…

The filling by holomorphic disks argument shows that $S^2\times D^2$ is not a convex fillings of $S^2\times S^1$ but it does not show it is not a concave filling. (For others’ reference, the relevant paper is Eliashberg’s “Filling by holomorphic discs and its applications”. The classification of convex fillings of $S^2\times S^1$ follows by similar argument, some more details are given in this paper: http://arxiv.org/pdf/1104.1543.pdf by Geiges and Zehmisch.) We’d like an example with an $\omega$-compatible $S^1$ action on the boundary whose boundary is not contact type (convex or concave).

The specific symplectic structure on $S^2\times D^2$ given by the direct sum of area forms $\omega = \omega_{S^2}\times \omega_{D^2}$ gives a symplectic structure which is compatible with the obvious $S^1$ action on $S^1\times S^2$ and cannot have convex/concave boundary by Peter’s reason above, since it evaluates positively on an $S^2$, so I think with this specific symplectic form $S^2\times D^2$ is a counterexample.

There are however other exact symplectic forms on $S^2\times D^2$: embed $S^2\times D^2$ into $\mathbb{R}^4$ (you can clearly embed $S^2\times D^1$ into $\mathbb{R}^3$ and then cross with an interval) and restrict the standard (exact) symplectic form on $\mathbb{R}^4$. The boundary is probably not of contact type in this case but I’m not entirely sure (its certainly not convex, but technically could be concave).