Contact Manifolds with Flexible Fillings

Abstract

We prove that all flexible Weinstein fillings of a given contact manifold with vanishing first Chern class have isomorphic integral cohomology. As an application, we show that in dimension at least 5 any almost contact class that has an almost Weinstein filling has infinitely many different contact structures. We also construct the first known infinite family of almost symplectomorphic Weinstein domains whose contact boundaries are not contactomorphic. These contact structures are distinguished by positive symplectic homology, which we prove is a contact invariant for flexibly-filled contact structures. The key step is a procedure for increasing the degrees of Reeb chords of loose Legendrians.

Appendix: Scaling Map Lemma

In this section, we prove the scaling map Lemma 5.6. This lemma was used in Section 5 to prove Lemma 5.7 and Proposition 5.8, which were in turn needed to prove Theorem 5.4, loose Legendrians are ADC. The scaling lemma is a result about contactomorphisms of 1-jet spaces, i.e. neighborhoods of Legendrians, and can be stated reference to the ambient contact manifold or looseness. We recall the notation we used. We fix a metric on $\Lambda$. As before, we let $U^\varepsilon = U^{\varepsilon }(\Lambda ) \subset (J^1(\Lambda ), \alpha _{std})$ denote $\{ \Vert y\Vert< \varepsilon , |z| < \varepsilon \}$; here $\alpha _{std} = dz - \sum _{i=1}^n y_i dx_i$, where $x_i$ are any local coordinates on $\Lambda$ and $y_i$ are the dual coordinates on $T^*\Lambda$. For $c>0$, let $s_c$ be the scaling contactomorphism of $J^1(\Lambda )$

$$\begin{aligned} s_c(x,y,z) = (x,cy, cz). \end{aligned}$$

(6.1)

Note that we can multiply the y coordinate by c since it is the coordinate on the vector space fibers of $T^*\Lambda$. Also, note that $s_c^*\alpha = c\alpha$. In the following lemma, we construct a contactomorphism supported in $U^1$ that scales $U^{1/2}$ down an arbitrary amount using $s_c$ but does not change the contact form by more than a fixed bounded amount. To do so, we first construct such a contactomorphism without the correct support and then modify this contactomorphism using contact Hamiltonians. The following restates the scaling map Lemma 5.6.

Lemma 6.1

For any positive $\varepsilon$ and $\delta$ such that $\delta \le \varepsilon$, there exists a contactomorphism $f_{\delta }$ of $(J^1(\Lambda ), \alpha )$ supported in $U^\varepsilon$ such that $f_\delta |_{U_{\varepsilon /2}}= s_\delta$ and $f_\delta ^* \alpha < 4 \alpha$.

Remark 6.2

• (1) By cutting off the contact Hamiltonian for $s_c$ with a bump function, it is easy to construct such f with $f_\delta ^* \alpha < C_\delta \alpha$ for some constant $C_\delta$ depending on $\delta$. However it is unclear what effect this cutting off has on $C_\delta$. The point is that there is a bound on $f_\delta ^* \alpha$ independent of $\delta$, which we will show by explicit computation.

• (2) Note that $(f_\delta |_{U_{\varepsilon /2}})^*\alpha = \delta \alpha$. However, this condition by itself is not enough for the proof of Lemma 5.7; we will actually need $f_\delta |_{U_{\varepsilon /2}}= s_\delta$.

Proof of Lemma 6.1

It is enough to prove the case $\varepsilon = 1$. If we have constructed $f_\delta$ for $\varepsilon =1$, in general we can take as our contactomorphism $s_\varepsilon \circ f_\delta \circ s_\varepsilon ^{-1}$. This has the desired support, $s_\varepsilon \circ f_\delta \circ s_\varepsilon ^{-1}|_{U^{\varepsilon /2}} = s_\delta$, and $(s_\varepsilon \circ f_\delta \circ s_\varepsilon ^{-1})^*\alpha = \varepsilon (f_\delta \circ s_\varepsilon ^{-1})^* \alpha \le 4 \varepsilon (s_\varepsilon ^{-1})^* \alpha = 4 \alpha$.

We first explain how to construct a contactomorphism of $J^1(\Lambda )$ with all the desired properties except support in $U^1$. For $t\in [0,1)$, we consider the diffeomorphism $h_t: [0,1] \rightarrow [0, 1]$ that is the identity map near $\{1\}$ and is obtained by smoothing the piecewise-linear map whose graph in $\mathbb {R}^2$ connects the points

• (0, 0) and $(1/2, (1-t)/2)$ with a line of slope $1-t$,

• $(1/2, (1-t)/2)$ and (1, 1) with a line of slope $1+t$.

Fig. 6

Graph of $h_t$ and $\frac{dh_t}{dz}$.

See Figure 6. Note that $h_0 = Id$ and as t approaches 1, the graph of $h_t$ approaches a line of slope 0 and a line of slope 2. Note that $1-t, 1+t < 2$ for $t < 1$. In particular, we can assume that $h_t(z) = (1-t) z$ for $z\in [0,1/2]$ and $\frac{d h_t}{d z} \le 2$ everywhere; also $\frac{d h_t}{d z} > 0$ since $h_t$ is a diffeomorphism. Extend $h_t$ to a diffeomorphism $h_t: \mathbb {R} \rightarrow \mathbb {R}$ satisfying $h_t(-z) = -h_t(-z)$ and $h_t(z) = z$ for $|z|> 1$. Let $\varphi _t: J^1(\Lambda )\rightarrow J^1(\Lambda )$ be the contactomorphism defined by $\varphi _t(x, y, z) = (x, \frac{d h_t}{dz}y, h_t(z))$; because $\frac{d h_t}{dz} > 0$, this is a diffeomorphism. In particular, we have $\varphi _t|_{U_{1/2}}= s_{1-t}$. Also, note that $\varphi _t^* \alpha = \frac{d h_t}{d z} \alpha$ and so $\varphi _t^* \alpha \le 2\alpha$ for all $t \in [0,1)$. Then $\varphi _{1-\delta }$ satisfies all conditions except it is not supported in $U^{1}$; the issue is that if $\frac{d h_t}{d z}(z) \ne 1$, then $(x, \frac{d h_t}{d z}y, h_t(z)) \ne (x,y,z)$ for all$y \in \mathbb {R}^n \backslash \{0\}$.

We now explain how to modify $\varphi _{t}$ to get the correct support while preserving the other properties. We first describe $h_t$ more carefully. In particular, we will use the following lemma about smooth functions, which we prove at the end of this section.

Lemma 6.3

There exists a smooth family of diffeomorphisms $h_t= h(t, \cdot ): \mathbb {R} \rightarrow \mathbb {R}$ defined for $t\in [0,1)$ such that $h_0(z) = z$, $h_t(z) = (1-t)z$ for $z\in [-\frac{1}{2}, \frac{1}{2}]$, $h_t(z) = z$ for $|z| > 1$ and

$$\begin{aligned} \max _{z\in \mathbb {R}} \left( \frac{\partial h_t}{\partial z} \right) ^{-1} \frac{\partial ^2 h_t}{\partial t \partial z} \le \frac{5}{4} \end{aligned}$$

for all $t\in [0,1)$.

Remark 6.4

In fact, we can replace $\frac{5}{4}$ by any number bigger than 1.

Our old $h_t$ satisfies all conditions except for possibly the last one and the $h_t$ constructed in this lemma look very much like the $h_t$ we constructed earlier.

Again we consider the contact isotopy $\varphi _t$ of $J^1(\Lambda )$ with $\varphi _t(x,y,z) =$$(x, \frac{\partial h_t }{\partial z}y, h_t(z))$. The vector field $X_t = (\frac{d\varphi _t}{dt}) \circ \varphi _t^{-1}$ is a time-dependent contact vector field whose flow is defined for all $t\in [0,1)$; for example, see Section 2 of [MP15]. Therefore, there exists a time-dependent contact Hamiltonian $H_t: J^1(\Lambda ) \rightarrow \mathbb {R}$ such that the corresponding contact vector field is precisely $X_t$; we do not need an explicit formula for $H_t$ but note that $H_t = \alpha (X_t)$. Because $\varphi _t$ is the identity map for $|z|>1$, $X_t$ and hence $H_t$ vanish for $|z|>1$. Let $b: \mathbb {R}^+ \rightarrow [0,1]$ be a smooth non-increasing function supported on [0, 1) such that $b=1$ on [0, 3/4]. Let $G_t: J^1(\Lambda ) \rightarrow \mathbb {R}$ be defined by $G_t(x,y,z) = b(\Vert y\Vert ^2)H_t(x,y,z)$, where we use the metric on $\Lambda$ to define $\Vert y\Vert ^2$. Note that $G_t$ is supported in $U^1$. Let $\psi _t$ be the contact isotopy obtained by integrating the contact vector field $Y_t$ of $G_t$. Since $G_t$ is supported in $U^1$, so is $\psi _t$ and therefore $\psi _t$ is defined for all $t\in [0,1)$. We will show that $\psi _{1-\delta }$ is the desired contactomorphism. Since $G_t = H_t$ in $U^{1/2}$, we have $Y_t = X_t$ in $U^{1/2}$. Furthermore, $\varphi _t(U^{1/2}) \subseteq U^{1/2}$ for all t and so $\psi _t|_{U^{1/2}} = \varphi _t|_{U^{1/2}} = s_{1-t}$. Therefore $\psi _{1-\delta }|_{U^{1/2}} = s_\delta$. It remains to show that $\psi _{1-\delta }^*\alpha < 4\alpha$; we will show that $\psi _t^*\alpha < 4 \alpha$ for all $t\in [0,1)$.

We now recall some properties of contact Hamiltonians. Suppose $\varphi _t$ is a contact isotopy induced by an arbitrary contact Hamiltonian $H_t$. Then $\varphi _t^*\alpha = \lambda _t \alpha$ for some $\lambda _t$. We have the following explicit formula for $\lambda _t$

$$\begin{aligned} \lambda _t = e^{\int _0^t \mu _s ds }, \end{aligned}$$

(6.2)

where $\mu _t= dH_t(R_\alpha )\circ \varphi _t$. For completeness, we review the proof of this formula, which is given in [Gei08], p. 63. Note that

$$\begin{aligned} \frac{d\lambda _t}{dt}\alpha = \frac{d\varphi _t^*\alpha }{dt} = \varphi _t^*L_{X_t} \alpha = \varphi _t^* (dH_t(R_\alpha ) \alpha ) = \mu _t \lambda _t \alpha \end{aligned}$$

where the second-to-last equality follows from the definition of the contact Hamiltonian. So we get $\frac{d\lambda _t}{dt} = \mu _t \lambda _t$, which proves Equation 6.2. In particular, to bound $\lambda _t$ it is sufficient to bound $dH_t(R_\alpha )$. Solving for $dH_t(R_\alpha )\circ \varphi _t$ in terms of $\lambda _t$, we get

$$\begin{aligned} dH_t(R_\alpha )\circ \varphi _t = \mu _t = \lambda _t^{-1} \frac{d \lambda _t}{dt}. \end{aligned}$$

In our situation, we have $\lambda _t = \frac{d h_t(z)}{d z}$ and so

$$\begin{aligned} dH_t(R_\alpha )\circ \varphi _t = \left( \frac{\partial h_t}{\partial z} \right) ^{-1} \frac{\partial ^2 h_t}{\partial t \partial z}. \end{aligned}$$

By the last condition of Lemma 6.3, $\underset{J^1(\Lambda )}{\max } \ dH_t(R_\alpha ) = \underset{J^1(\Lambda )}{\max } \ dH_t(R_\alpha ) \circ \varphi _t \le 5/4$ for all $t\in [0,1)$.

We are interested in bounding the function $\gamma _t$ defined by $\psi _t^*\alpha = \gamma _t\alpha$. Recall that $\psi _t$ is generated by $G_t = b(\Vert y\Vert ^2)H_t$. Since b is independent of z and $R_\alpha = \partial _z$, we have that $dG_t(R_\alpha ) = b(\Vert y\Vert ^2) dH_t(R_\alpha )$; the fact that we can just factor out b and not consider derivatives of b is key. Let $\nu _t = dG_t(R_\alpha )\circ \psi _t$. We have $\underset{J^1(\Lambda )}{\max } \ dH_t(R_\alpha ) \ge 0$ because $H_t = 0$ for $|z| > 1$ for all t. Also, $0 \le b(\Vert y\Vert ^2) \le 1$. Therefore

$$\begin{aligned} \max _{J^1(\Lambda )}\nu _t = \max _{J^1(\Lambda )}{dG_t(R_\alpha )} \le \max _{J^1(\Lambda )}{dH_t(R_\alpha )} =\max _{J^1(\Lambda )}\mu _t \le \frac{5}{4}. \end{aligned}$$

Therefore by Equation 6.2

$$\begin{aligned} \gamma _t = e^{\int _0^t \nu _s ds } \le e^{\int _0^t \frac{5}{4} ds} = e^{\frac{5}{4}t}< e^\frac{5}{4} < 4 \end{aligned}$$

for all $t\in [0,1)$ as desired. $\square$

Remark 6.5

We knew a priori that $\int _0^t \mu _s ds = \ln \lambda _t \le \ln 2$ for all points in $J^1(\Lambda )$ but this is not enough to conclude that $\int _0^1 \max _{J^1(\Lambda )}\mu _t dt < \infty$, as required for the last part of the proof of Lemma 6.1. This is why we needed to us Lemma 6.3 to bound $\max _{J^1(\Lambda )}\mu _t$.

Fig. 7

Graph of g.

Proof of Lemma 6.3

Now we explain how to construct the desired $h_t$. Let $g:[0,1] \rightarrow [-1, \frac{5}{4}]$ be a smooth function such that g equals $-1$ on [0, 1/2], vanishes near 1, and $\int _0^1 g(x) dx = 0$; see Figure 7. Define $h_t: [0,1] \rightarrow [0,1]$ by $h_t(z) = \int _0^z (tg(s)+1)ds$. Since $\frac{\partial h_t}{\partial z} = tg(z)+1 > 0$ for $t < 1$, $h_t$ is a smooth family of increasing (in z) functions. Also, $h_0(z) =z, h_t(z) = (1-t)z$ for $z\in [0,1/2],$ and $h_t(z) = z$ near 1. We need to bound

$$\begin{aligned} \left( \frac{\partial h_t}{\partial z}\right) ^{-1} \frac{\partial ^2 h_t}{\partial t \partial z} = (tg(z)+1)^{-1} \frac{\partial (tg(z)+1)}{\partial t} = \frac{g(z)}{tg(x)+1} \end{aligned}$$

in z for fixed $t<1$. If $g(z)\le 0$, then because $tg(z)+1 >0$ for $t<1$, we have that $g(z)/(tg(z)+1) \le 0$. If $g(z)>0$, then $tg(z)+1 \ge 1$ and so $g(z)/(tg(z)+1)) \le g(z) \le \frac{5}{4}$. So $g(z)/(tg(z)+1) \le \frac{5}{4}$ for all $z\in [0,1]$ and $t < 1$, as desired. Finally, we extend $h_t$ to a diffeomorphism $h_t: \mathbb {R} \rightarrow \mathbb {R}$ such that $h_t(z)= -h_t(-z)$ and $h_t(z) = z$ for $|z|>1$. Since $\frac{\partial h_t}{\partial z}(z)= \frac{\partial h_t}{ \partial z}(-z)$, this bound holds for all $z\in \mathbb {R}$. $\square$

Remark 6.6

Lemma 6.1 also holds for $(J^1(\Lambda ) \times E, \alpha )$. Here E is a trivial conformal symplectic bundle over $\Lambda$ and $\alpha = dz - y dx - vdu$, where (x, y, z, u, v) are local coordinates on $J^1(\Lambda ) \times E$. In this case, we take $s_c(x,y,z,u,v) := (x, cy, cz, \sqrt{c}u, \sqrt{c}v)$ and $U^\varepsilon = \{\Vert y\Vert , |z|, |u|, |v| < \varepsilon \}$ so that $s_c^*\alpha = c\alpha$ and $s_c(U^\varepsilon ) \subset U^{\varepsilon \sqrt{c}}$. We modify the proof a bit by using $\varphi _t(x,y,z,u,v) = (x,\frac{\partial h_t}{\partial z}y,h_t(z), \sqrt{\frac{\partial h_t}{\partial z}}u, \sqrt{\frac{\partial h_t}{\partial z}} v)$, which is smooth since $\frac{\partial h_t}{\partial z}$ is never zero, and $G_t = b(\Vert y\Vert ^2) b(|u|^2)b(|v|^2)H_t$. Note that any isotropic submanifold with trivial symplectic conormal bundle has a neighborhood that is strictly contactomorphic to a neighborhood of $\Lambda$ in $(J^1(\Lambda ) \times E, \alpha )$. When $\Lambda$ is a sphere, this is precisely the neighborhood of $\Lambda$ used in subcritical surgery.