Weak and strong fillability of higher dimensional contact manifolds

Abstract

For contact manifolds in dimension three, the notions of weak and strong symplectic fillability and tightness are all known to be inequivalent. We extend these facts to higher dimensions: in particular, we define a natural generalization of weak fillings and prove that it is indeed weaker (at least in dimension five), while also being obstructed by all known manifestations of “overtwistedness”. We also find the first examples of contact manifolds in all dimensions that are not symplectically fillable but also cannot be called overtwisted in any reasonable sense. These depend on a higher dimensional analogue of Giroux torsion, which we define via the existence in all dimensions of exact symplectic manifolds with disconnected contact boundary.

Appendix: Cotamed complex structures: existence and convexity

1.1 A.1 Contractibility of the space of cotamed almost complex structures

To go from the linear situation to global existence results on a manifold we will need the following result.

Proposition 2.1

(Sévennec)

The space of complex structures on a finite dimensional vector space tamed by two given symplectic forms is either empty or contractible.

Using the fact that the space of complex structures tamed by a symplectic form is nonempty (which follows for instance by the linear Darboux theorem), and applying the proposition above twice to the same symplectic form, we recover as a special case the classical result of Gromov that states that the space of tamed complex structures is contractible. The proof of the proposition uses the following two lemmas, of which the first is more or less standard.

Lemma A.1

(Cayley, Sévennec)

Let V be a real finite dimensional vector space and ${\mathcal{J}}(V)$ the space of complex structures on V. We can define for any fixed $J_{0} \in {\mathcal{J}}(V)$ a map

$$ \mu_{J_0}\colon J \mapsto (J + J_0)^{-1} \cdot (J - J_0) $$

which is a diffeomorphism from

$$ {\mathcal{J}}_{J_0}^*(V) := \bigl\{ J \in {\mathcal{J}}(V)\ |\ J + J_0 \in \operatorname{GL}(V)\bigr\} $$

to

$$ \mathcal{A}_{J_0}^*(V) := \bigl\{ A \in \operatorname{End}(V)\ |\ A J_0 = -J_0 A\ \mbox{\textit{and}}\ A - I\in \operatorname{GL}(V)\bigr\} . $$

The inverse of this map is given by $\mu_{J_{0}}^{-1} \colon A \mapsto (A- I)J_{0}(A - I)^{-1}$.

Proof

One can view $\mathcal{A}^{*}_{J_{0}}(V)$ as the set of J ₀-complex antilinear maps that do not have any eigenvalue equal to 1. Using the equations (J−J ₀)J ₀=−J(J−J ₀) and (J+J ₀)J ₀=J(J+J ₀), one sees that the image of $\mu_{J_{0}}$ consists of J ₀-complex antilinear maps, and $\mu_{J_{0}}(J) - I = -2 (J + J_{0})^{-1} J_{0}$ is invertible. □

Lemma A.2

(Sévennec)

Let (V,ω) be a finite dimensional symplectic vector space and denote by ${\mathcal{J}}_{t}(\omega) \subset {\mathcal{J}}(V)$ the space of complex structures tamed by ω. Choosing any $J_{0} \in {\mathcal{J}}_{t}(\omega)$, it follows that ${\mathcal{J}}_{t}(\omega)$ lies in ${\mathcal{J}}_{J_{0}}^{*}(V)$, and the image of ${\mathcal{J}}_{t}(\omega)$ under the associated map $\mu_{J_{0}}$ is a convex domain in $\mathcal{A}_{J_{0}}^{*}(V)$.

We first explain how to prove Proposition 2.1 using the above lemma. Suppose there is a complex structure J ₀ tamed by ω ₀ and ω ₁. The space of cotamed complex structures ${\mathcal{J}}_{t}(\omega_{0}) \cap {\mathcal{J}}_{t}(\omega_{1})$ is then diffeomorphic under the map $\mu_{J_{0}}$ to the intersection of the convex subsets given by the lemma. This intersection is again convex and hence contractible.

Proof of Lemma A.2

For any complex structure J tamed by ω, the endomorphism J+J ₀ is invertible because for any nonzero w, we have ω(w,(J+J ₀)w)>0, so in particular (J+J ₀)w is not zero. This proves the first part of the lemma.

Now fix a nonzero vector v∈V, and let C _v be the set of $A\in \operatorname{End}(V)$ that anticommute with J ₀, and that satisfy

$$ \omega\bigl((A - I) v, (A- I) J_0v\bigr) = -\omega\bigl((A - I) v, J_0(A+ I) v\bigr) > 0 . $$

We now prove that $C_{v}\subset \operatorname{End}(V)$ is convex. Every segment A _s =(1−s)A ₀+sA ₁ with s∈[0,1] for arbitrary A ₀,A ₁∈C _v defines a polynomial of degree 2

$$ P(s) = -\omega\bigl((A_s - I) v, J_0(A_s + I) v\bigr) , $$

and the above inequality corresponds to checking that P(s) is positive for all values s∈[0,1]. The leading coefficient −ω((A ₁−A ₀)v,J ₀(A ₁−A ₀)v) of P(s) is never positive, because J ₀ tames ω, so that P(s) is either a line or a parabola facing downward. In both cases P(s)≥min{P(0),P(1)}>0 for all s∈(0,1) so the inequality holds for the whole segment A _s .

Note that C _v ≠∅ since 0∈C _v . Define the intersection

$$ C^* := \bigcap_{v\ne 0} C_v , $$

which is a nonempty convex subset of $\operatorname{End}(V)$. In fact, one has $C^{*} \subset \mathcal{A}_{J_{0}}^{*}(V)$, because if there were a matrix A∈C ^∗ with det(A−I)=0, then A would have an eigenvector w∈V with eigenvalue 1, but then −ω((A−I)w,J ₀(A+I)w)=0 so that A∉C _w .

Since C ^∗ lies in the domain of $\mu_{J_{0}}^{-1}$ and ${\mathcal{J}}_{t}(\omega)$ lies in the domain of $\mu_{J_{0}}$, we have $C^{*} = \mu_{J_{0}}({\mathcal{J}}_{t}(\omega))$, so that the image of the complex structures tamed by ω is convex as we wanted to show. □

1.2 A.2 Existence of a cotamed complex structure

In this appendix, we prove Proposition 2.2, which we now recall:

Proposition 2.2

Let V be a finite dimensional real vector space equipped with two symplectic forms ω ₀ and ω ₁. The following properties are equivalent:

• (1) the segment between ω ₀ and ω ₁ consists of symplectic forms

• (2) the ray starting at ω ₀ and directed by ω ₁ consists of symplectic forms

• (3) there is a complex structure J on V tamed both by ω ₀ and by ω ₁.

The equivalence between (1) and (3) was explained to us by Jean-Claude Sikorav. It relies on the simultaneous reduction of symplectic forms. Specifically, we need [38, Theorem 9.1] which we shall state (in a slightly weakened form) and reprove (in its full force) below as Proposition A.3, since the very general context of [38] makes it hard to read for people interested only in the symplectic case.

Recall that according to the linear Darboux theorem, any symplectic form on a 2n-dimensional vector space is represented in some basis by the standard matrix

We now want to understand what can be said for a pair of symplectic structures. Below we give an approximate normal form which is sufficient for our purposes and more pleasant to state than the precise result (cf. [38, Theorem 9.1]), though the precise result can also be extracted from the proof that we will give at the end of this section.

Proposition A.3

Let ω ₀ and ω ₁ be symplectic forms on a finite dimensional vector space V. There exists a matrix A ₁ that splits into blocks of the form

for λ,ν≠0 with the following property: for any ε>0, there is a basis of V such that ω ₀ is represented by a block diagonal matrix with standard blocks Ω _2k , and ω ₁ is represented by a matrix which is ε-close to A ₁.

If the linear segment between ω ₀ and ω ₁ consists of symplectic forms, then the coefficients λ in the 2×2-blocks of A ₁ described above cannot be negative.

The relation with cotamed complex structures will come from the following.

Proposition A.4

• (a) Let V=ℝ² with two antisymmetric bilinear forms ω ₀ and ω ₁ defined by ω _j (v,w)=v ^t A _j w, where

  

  If λ>0, then is tamed by both forms.

• (b) Let V=ℝ⁴, and let ω ₀ and ω ₁ be antisymmetric bilinear forms defined by the matrixes

  

  with μ≠0. Then there exists a complex structure J on ℝ⁴ that is tamed by both forms.

Proof

We only need to prove (b). For simplicity write V as ℂ², and the matrices A ₀ and A ₁ as

with z=λ+iμ=re ^iψ. The matrices

define complex structures on V, and it follows that is positive definite if cosϕ<0, and is positive definite if cos(ψ−ϕ)<0. As long as ψ≠π (which we have excluded by requiring that μ≠0), it follows that we can choose ϕ such that ϕ∈(π/2,3π/2) and ϕ−ψ∈(π/2,3π/2)+2πℤ. □

Proof of Proposition 2.2

We first explain the easy equivalence between (1) and (2). The (open) ray starting at ω ₀ and directed by ω ₁ and the open interval between ω ₀ and ω ₁ span the same cone in the space of antisymmetric bilinear forms. Since being symplectic is invariant under nonzero scalar multiplication, we have the equivalence.

The implication (3) $\implies$ (1) is also direct because, for any t∈[0,1], we have

$$ \big((1 - t) \omega_0 + t \omega_1\big)(v, Jv) = (1 - t) \omega_0(v, Jv) + t \omega_1(v, Jv) , $$

which is positive whenever v∈V is nonzero. So in particular, no such v can be in the kernel of an element of the segment between ω ₀ and ω ₁.

To prove (1) $\implies$ (3), we use the fact that by Proposition A.3, there is a matrix $A_{1}'$ that splits into certain standard blocks, such that we can find for any ε>0 a basis of V for which ω ₀ is in canonical form, and for which ω ₁ is represented by a matrix that is ε-close to $A_{1}'$.

If condition (1) holds, then the blocks of $A_{1}'$ correspond to the ones described in Proposition A.4, and we obtain the existence of a complex structure J on V that is tamed both by the standard symplectic form and by $A_{1}'$. By choosing ε>0 sufficiently small, it follows that J is also tamed by ω ₀ and ω ₁, because tameness is an open condition. □

Proof of Proposition A.3

The proof will proceed in several steps.

Decomposition into generalized eigenspaces. In the first step we shall decompose V into suitable subspaces that are both ω ₀- and ω ₁-orthogonal.

Let φ _r :V→V ^∗ for r=0,1 be the isomorphisms defined by φ _r (v):=ω _r (v,⋅). We consider the endomorphism $B = \varphi_{0}^{-1} \circ \varphi_{1}$ of V so that ω ₁(v,w)=ω ₀(Bv,w). The endomorphism B is invertible and it is ω ₀-symmetric since:

$$ \omega_0(Bv, w) = \omega_1(v, w) = -\omega_1(w, v) = -\omega_0(Bw, v) = \omega_0(v, Bw) . $$

To define the generalized eigenspaces of B, complexify the vector space V to obtain V ^ℂ, and extend the ω _r to sesquilinear forms $\omega_{r}^{{\mathbb {C}}}$. A computation analogous to the preceding one shows that B is $\omega_{0}^{{\mathbb {C}}}$-symmetric and we still have $\omega_{0}^{{\mathbb {C}}}(v,Bw) = \omega_{1}^{{\mathbb {C}}}(v,w)$.

The characteristic polynomial of B splits over ℂ as $P(X) = \prod_{\lambda}(X - \lambda)^{m_{\lambda}}$, so we can decompose V ^ℂ into generalized eigenspaces

$$ V^{\mathbb {C}}= \bigoplus_{\lambda \in Sp(B)} E^{\mathbb {C}}_\lambda;\quad E^{\mathbb {C}}_\lambda = \ker (B - \lambda)^{m_\lambda} . $$

Lemma A.5

If λ and μ are eigenvalues of B such that $\lambda \neq \bar{\mu}$, then $E^{{\mathbb {C}}}_{\lambda}$ and $E^{{\mathbb {C}}}_{\mu}$ are both $\omega^{{\mathbb {C}}}_{0}$- and $\omega^{{\mathbb {C}}}_{1}$-orthogonal.

Proof

We prove by induction on k and l that ker(B−λ)^k and ker(B−μ)^l are orthogonal.

To start the induction, note that if v _λ ∈ker(B−λ), and v _μ ∈ker(B−μ), then

$$ (\bar{\lambda}- \mu) \omega_0^{\mathbb {C}}(v_\lambda, v_\mu) = \omega_0^{\mathbb {C}}\bigl((B - \bar{\mu}) v_\lambda, v_\mu\bigr) = \omega_0^{\mathbb {C}}\bigl(v_\lambda, (B-\mu) v_\mu\bigr) = 0 , $$

thus since $\lambda \ne \bar{\mu}$, it follows that $\omega_{0}^{{\mathbb {C}}}(v_{\lambda}, v_{\mu}) = 0$. Similarly, $\omega_{1}^{{\mathbb {C}}}(v_{\lambda}, v_{\mu}) = \omega_{0}^{{\mathbb {C}}}(v_{\lambda}, B v_{\mu}) = \mu \omega_{0}^{{\mathbb {C}}}(v_{\lambda}, v_{\mu}) = 0$.

Assume now it has already been shown for the integers k and l that ker(B−λ)^k and ker(B−μ)^l are both $\omega_{0}^{{\mathbb {C}}}$- and $\omega_{1}^{{\mathbb {C}}}$-orthogonal. Choose a vector $v_{\lambda}' \in \ker (B - \lambda)^{k+1}$ and use the fact that $Bv_{\lambda}' = \lambda v_{\lambda}' + w$ for some w∈ker(B−λ)^k. Then we obtain for any v _μ ∈ker(B−μ)^l,

and also $\omega_{1}^{{\mathbb {C}}}(v_{\lambda}', v_{\mu}) = \omega_{0}^{{\mathbb {C}}}(B v_{\lambda}', v_{\mu}) = \bar{\lambda}\omega_{0}^{{\mathbb {C}}}(v_{\lambda}', v_{\mu}) + \omega_{0}^{{\mathbb {C}}}(w, v_{\mu}) = 0$, which proves the induction step from (k,l) to (k+1,l). Since λ and μ have completely symmetric roles, this also explains how to go to (k,l+1). □

We now relate this decomposition of V ^ℂ to the initial real vector space V. For a real eigenvalue λ, the intersection $V\cap E_{\lambda}^{{\mathbb {C}}}$ defines a real subspace E _λ with $\dim_{{\mathbb {R}}}E_{\lambda}= \dim_{{\mathbb {C}}}E_{\lambda}^{{\mathbb {C}}}$. Complex conjugation defines an isomorphism $E_{\lambda}^{{\mathbb {C}}}\to E_{\bar{\lambda}}^{{\mathbb {C}}}, v_{\lambda}\mapsto \bar{v}_{\lambda}$, and we can write $V\cap (E_{\lambda}^{{\mathbb {C}}}\oplus E_{\bar{\lambda}}^{{\mathbb {C}}})$ for λ∈ℂ∖ℝ as the direct sum of real subspaces $E_{\{\lambda,\bar{\lambda}\}} = \{v+\bar{v} \mid v\in E_{\lambda}^{{\mathbb {C}}}\} \oplus \{i (v-\bar{v}) \mid v\in E_{\lambda}^{{\mathbb {C}}}\}$.

This way we find a decomposition of V into pairwise ω ₀- and ω ₁-orthogonal subspaces

$$ E_{\mu_1}\oplus \dotsm \oplus E_{\mu_k} \oplus E_{\{\lambda_1,\bar{\lambda}_1\}} \oplus \dotsm \oplus E_{\{\lambda_l,\bar{\lambda}_l\}} $$

with μ ₁,…,μ _k ∈ℝ∖{0}, and λ ₁,…,λ _l ∈ℂ∖ℝ.

Blocks with real eigenvalue. For the following considerations, we restrict to one of the subspaces $E_{\lambda_{j}}$ with λ _j ∈ℝ, and denote λ _j for simplicity just by λ. We will construct a basis of E _λ such that ω ₀ and ω ₁ have the particularly nice form described in the proposition. Note that ω ₀ and ω ₁ are both nondegenerate on E _λ .

Let k+1 be the nilpotency index of B−λ, i.e. (B−λ)^k+1=0 and (B−λ)^k≠0. Let v ₀ be an element of E _λ not in ker(B−λ)^k. We set v _j :=ε ^−j(B−λ)^j v ₀ to define a collection of vectors v ₀,…,v _k . Choose now a vector w _k ∈E _λ with ω ₀(v _k ,w _k )=1 and ω ₀(v _j ,w _k )=0 for every j≠k, and define inductively w _j−1:=ε ⁻¹(B−λ)w _j , or equivalently

$$ B w_j = \lambda w_j + \varepsilon w_{j-1} $$

for j≥1.

Lemma A.6

The vectors v ₀,…,v _k ,w ₀,…,w _k are linearly independent and satisfy the relations ω _r (v _j ,v _j′)=ω _r (w _j ,w _j′)=0 for all r=0,1, and j,j′, and

$$ \omega_0(v_j, w_{j'}) = \delta_{j,j'} \quad\mbox{\textit{and}}\quad \omega_1(v_j, w_{j'}) = \lambda \delta_{j,j'} + \varepsilon \delta_{j,j'-1} . $$

Proof

We start by proving ω _r (v _j ,v _j′)=0. For this we will use an induction on |j−j′|. If j−j′=0 then the statement follows directly from the antisymmetry of ω _r . Suppose that the claim is true for j−j′≤m and consider any j and j′ with j−j′=m+1 (in particular j≥1). We have

by the induction hypothesis. Using the fact that Bv _j′=εv _j′+1+λv _j′, we compute

The first term is zero by the induction hypothesis and the second one is zero because of the preceding computation. The proof of ω _r (w _j ,w _j′)=0 follows the same lines, and will be omitted.

Note that

and in particular this implies that v ₀,…,v _k ,w ₀,…,w _k are linearly independent vectors with respect to which ω ₀ has standard form.

The remaining relation for ω ₁ can be obtained by

 □

If we restrict ω ₀ and ω ₁ to the subspace $E = \operatorname{span}(v_{0},\dotsc,v_{k}, w_{0},\dotsc,w_{k})$ and represent them in this basis, we now find that ω ₀ is in standard form Ω _2k and ω ₁ is represented by a matrix ε-close to λΩ _2k .

To continue the proof, restrict ω ₀, ω ₁, and B to the ω ₀-symplectic complement E′ of the space E. Note that E′ is stable under B because for u∈E′,

$$ \omega_0(v_j,Bu) = \omega_0(Bv_j,u) = \lambda \omega_0(v_j,u) + \varepsilon \omega_0(v_{j-1},u) = 0 , $$

and similarly for ω ₀(w _j ,Bu)=0. We can thus proceed as before to reduce all eigenspaces E _λ with λ∈ℝ to ω ₀-symplectic blocks in normal form.

Blocks with complex eigenvalue. We proceed now to the generalized complex eigenspace $E^{{\mathbb {C}}}_{\lambda}$ with λ∈ℂ∖ℝ. Let k be the largest integer for which $E^{{\mathbb {C}}}_{\lambda}\ne \ker (B-\lambda)^{k}$, and construct as before a chain of vectors $v_{0},\dotsc,v_{k} \in E^{{\mathbb {C}}}_{\lambda}$ by starting with an element $v_{0} \in E^{{\mathbb {C}}}_{\lambda}\setminus \ker (B-\lambda)^{k}$, and defining inductively

$$ v_{j+1} := \varepsilon^{-1} (B-\lambda) v_j . $$

Using complex conjugation, we also find a chain $\bar{v}_{0},\dotsc, \bar{v}_{k}$ that lies in $E^{{\mathbb {C}}}_{\bar{\lambda}}$. Since B is the complexification of a real linear map, $\bar{v}_{j+1} := \varepsilon^{-1} (B- \bar{\lambda}) \bar{v}_{j}$ holds.

Next, we define two chains w ₀,…,w _k in $E^{{\mathbb {C}}}_{\bar{\lambda}}$ and $\bar{w}_{0},\dotsc, \bar{w}_{k}$ in $E^{{\mathbb {C}}}_{\lambda}$ by starting with a vector $w_{k} \in E^{{\mathbb {C}}}_{\bar{\lambda}}$ with $\omega^{{\mathbb {C}}}_{0}(v_{k},w_{k}) = 1$ and $\omega^{{\mathbb {C}}}_{0}(v_{j},w_{k}) = 0$ for every j≠k, and defining $w_{j-1} := \varepsilon^{-1} (B-\bar{\lambda}) w_{j}$, or equivalently

$$ B w_j = \bar{\lambda}w_j + \varepsilon w_{j-1} $$

for j≥1. Similarly, we obtain $\bar{w}_{j-1} = \varepsilon^{-1} (B- \lambda) \bar{w}_{j}$.

Lemma A.7

• (a) The space spanned by $v_{0},\dotsc, v_{k - 1}, \bar{v}_{0}, \dotsc, \bar{v}_{k - 1}$ and the one spanned by $w_{0},\dotsc, w_{k - 1}, \bar{w}_{0}, \dotsc, \bar{w}_{k - 1}$ are each isotropic with respect to both ω ₀ and ω ₁.

• (b) The $\omega_{0}^{{\mathbb {C}}}$-pairings for these vectors are given by

  

• (c) The $\omega_{1}^{{\mathbb {C}}}$-pairings for these vectors are given by

  

Proof

To prove (a) note that since $\lambda \ne \bar{\lambda}$, the spaces $E^{{\mathbb {C}}}_{\lambda}$ and $E^{{\mathbb {C}}}_{\bar{\lambda}}$ are both $\omega_{0}^{{\mathbb {C}}}$- and $\omega_{1}^{{\mathbb {C}}}$-isotropic, so we only need to show that $\omega_{r}^{{\mathbb {C}}}(\bar{v}_{j}, v_{j'}) = \omega_{r}^{{\mathbb {C}}}(\bar{w}_{j}, w_{j'}) = 0$ for all j,j′, and for r=0,1. If j=j′, we write v _j as v _x +iv _y , and we use sesquilinearity as follows:

By the same computation, $\omega_{1}^{{\mathbb {C}}}(\bar{v}_{j}, v_{j}) = 0$.

If the statement is true for j′−j=m≥0, then

and

which finishes the induction. The argument for $\omega_{r}^{{\mathbb {C}}}(\bar{w}_{j}, w_{j'})$ is identical.

To prove (b), note first that the second two equations are the complex conjugate of the first two. Since $v_{j}, \bar{w}_{j'} \in E_{\lambda}^{{\mathbb {C}}}$, it also follows immediately that $\omega_{0}^{{\mathbb {C}}}(\bar{v}_{j}, \bar{w}_{j'}) = 0$, so that we are only left with showing $\omega_{0}^{{\mathbb {C}}}(v_{j}, w_{j'}) = \delta_{j,j'}$, but the required computation is identical to the one used to show the analogous relation in the proof of Lemma A.6.

The equalities for (c) follow similarly. □

We will now intersect the complex subspace spanned by the chains defined above with the initial real vector space V to finish the proof of the proposition. For this, define for all j≤k the real vectors

and

which all lie in $E_{\lambda, \bar{\lambda}}$. Using the results deduced above, we obtain for all r=0,1, and j,j′ the equations $\omega_{r}(v_{j}^{+}, v_{j'}^{\pm}) = \omega_{r}(v_{j}^{-}, v_{j'}^{\pm}) = 0$ and $\omega_{r}(w_{j}^{+}, w_{j'}^{\pm}) = \omega_{r}(w_{j}^{-}, w_{j'}^{\pm}) = 0$, and finally

and similar computations for the other matrix elements, which prove the desired result with $\mu = \operatorname{Re}\lambda$ and $\nu = \operatorname{Im}\lambda$.

Sign of real eigenvalues. Assume that all 2-forms in the family

$$ \omega_t := (1-t) \omega_0 + t \omega_1 $$

for t∈[0,1] are nondegenerate. The λ-coefficients in the 2×2-blocks of $A_{1}'$ correspond to the real eigenvalues of the map B, so that if λ<0 with eigenvector v, then we have ω ₁(v,⋅)=ω ₀(Bv,⋅)=λω ₀(v,⋅), and it follows that ω _t (v,⋅)=(1−t+tλ)ω ₀(v,⋅) has to vanish for a certain value t ₀∈(0,1), so that $\omega_{t_{0}}$ is degenerate.  □