Flexible Lagrangians
Abstract
We introduce and discuss notions of regularity and flexibility for Lagrangian manifolds with Legendrian boundary in Weinstein domains. There is a surprising abundance of flexible Lagrangians. In turn, this leads to new constructions of Legendrians submanifolds and Weinstein manifolds. For instance, many closed manifolds of dimension can be realized as exact Lagrangian submanifolds of with possibly exotic Weinstein symplectic structures. These Weinstein structures on , infinitely many of which are distinct, are formed by a single handle attachment to the standard ball along the Legendrian boundaries of flexible Lagrangians. We also formulate a number of open problems.
1 Liouville and Weinstein cobordisms
The main goal of the paper is a discussion of two new notions of regularity and flexibility for exact Lagrangian cobordisms with Legendrian boundaries in Weinstein cobordisms, see Sections 2 and 3. In particular, we prove an existence principle for flexible Lagrangian cobordisms (Theorem 4.2), explore applications to Lagrangian and Legendrian embeddings and exotic Weinstein structures, and formulate throughout the paper numerous open problems.
A Liouville cobordism between contact manifolds is a dimensional cobordism ) equipped with a pair of a symplectic form and an expanding (Liouville) vector field for , i.e. , which is outward pointing along and inward pointing along , such that the contact structure induced by the Liouville form on coincides with . If in addition we are given a Morse function that is defining for and Lyapunov for , i.e. it attains its minimum on , its maximum on and has no critical points on , and satisfies the inequality for some , then the triple is called a Weinstein cobordism structure on between contact manifolds and , see [14, 36, 15, 10]. A cobordism with will be referred as a Weinstein domain.
Stable manifolds of zeroes of for a Weinstein cobordism structure are necessarily isotropic, see [15], and in particular the indices of all critical points of are always . A cobordism is called subcritical if there are are no critical points of index .
Every Liouville or Weinstein cobordism can be canonically completed by adding cylindrical ends and with and equal to up to an additive constant. Here we denote by the coordinate corresponding to the first factor. An important feature of completed Liouville cobordisms, see [15, 10], is that if , is a homotopy of completed Liouville cobordisms then there exists an isotopy such that , , which preserves the Liouville field at infinity. In other words, the symplectic structure of a completed Liouville cobordism remains unchanged up to isotopy when one deforms the Liouville structure.
Usually we will not distinguish in the notation between a Weinstein cobordism and its completion. Moreover, we will allow contact manifolds to have boundaries and will require in this case the cobordisms to be trivial over boundaries of the contact manifolds. Alternatively, a Weinstein cobordism between manifolds with boundary can be viewed as a sutured manifold with corner along the suture, see Fig. 1.1 (taken from [18]). More precisely, we assume that the boundary is presented as a union of two manifolds and with common boundary , along which it has a corner. Of course, in this case the function cannot be chosen constant on and .
Given a dimensional Weinstein cobordism we consider in it exact Lagrangian cobordisms with Legendrian boundary . We will additionally require to be tangent to near . This condition can always be achieved by a small isotopy of fixed on . The boundary components will always assumed to be closed and contained in . Sometimes we will talk about a parameterized Lagrangian cobordism, i.e. a diffeomorphism of a smooth cobordism onto an exact Lagrangian cobordism in between Legendrian manifolds in .
A Lagrangian cobordism can be canonically completed to a submanifold with cylindrical ends in the completion of . An isotopy between two exact Lagrangian cobordisms with Legendrian boundaries will be always understood in this class, i.e. as a Hamiltonian isotopy of the completions which at infinity is required to preserve the Liouville vector field . We note that any exact Lagrangian isotopy with Legendrian boundary lifts to a Hamiltonian isotopy of completions.
A Morse decomposition for the Lyapunov function yields an equivalent definition of a Weinstein cobordism as a Weinstein handlebody, formed by attaching handles with symplectically isotropic core discs along contactly isotropic sphere in the regular contact level sets of . A Weinstein cobordism of dimension is called flexible if it can be presented as a Weinstein handlebody so that all critical (i.e. of index ) handles are attached along loose Legendrian links, see [10, 26] for precise definitions and discussion. A flexible Weinstein structure is a choice of such a presentation.
2 Regular Lagrangians
Let be a Weinstein cobordism, and
a Lagrangian cobordism with Legendrian boundary.
Definition 2.1.
We say is regular if can be deformed to a Weinstein structure through Weinstein structures for which remains Lagrangian, such that the new Liouville vector field is tangent to . This is equivalent to the condition , where is the corresponding Liouville form.
We call such a tangent Weinstein structure to the regular Lagrangian . It follows that all critical points of are global critical points of , and the local models near such can be described by a “coupled handle attachment” picture. Indeed, let be the index of a critical point of and the index of as a critical point of . We have . A Weinstein handle of index is isomorphic to the subset , where we denoted by and the canonical coordinates in and . The handle contains a Lagrangian subhandle of index which is the intersection with of the total space of the conormal bundle to
When passing through the critical level of the critical point the Weinstein handle is attached to along , and the Lagrangian handle is attached to along .
It turns out that given a regular , any Weinstein cobordism structure tangent to can be further adjusted. Let us call a tangent to Weinstein cobordism structure special if there exists a regular value of the function such that

all critical points of in the sublevel set lie on and the indices of these critical points for and coincide;

there are no critical points of on .
In other words, has the following handlebody presentation. First, one attaches handles corresponding to critical points of and then the remaining handles, so that their attaching spheres do not intersect .
For a special tangent to Weinstein cobordism we set and view as a Weinstein subcobordism of with the induced Weinstein structure. We call the complementary Weinstein cobordism to and note that determines a presentation , where is endowed with its canonical Weinstein structure. The following lemma asserts that up to homotopy of Weinstein structures for which remains Lagrangian, the existence of such a presentation is equivalent to regularity:
Lemma 2.2.
Let be an exact Lagrangian subcobordism in a Weinstein cobordism tangent to . Then there is a homotopy of tangent to Weinstein structures such that is special.
Proof.
Suppose first that for each critical point of the function its index on coincides with its index as a critical point of on the whole .^{1}^{1}1In this case the modification of the Lagrangian after passing through the corresponding critical value coincides with the ambient Legendrian surgery defined by G. Dimitroglou Rizell in [30]. Then for any critical point of its stable manifold is contained in , and hence for any critical point there are no trajectories converging to at the negative direction, and to at the positive one. Hence, using Lemma 9.45 from [10] we can deform without changing and (and hence keeping Weinstein structure tangent to ) so that the critical values corresponding to critical points from are all smaller than the critical values corresponding to critical points of which are not in . Then an intermediate regular value has the required properties, and thus the Weinstein structure is special.
Suppose now that the index of a critical point of the function is less than the index of for on the whole . Let be the stable disc of on and the stable disc of for the function . Note that that there exists a function which coincides with on , has a critical point of index at and two additional critical points on of indices and , respectively. Hence the attaching of one handle of index corresponding to the point can be replaced by attaching of three handles of indices , and corresponding to the points , and . Moreover, only the first handle intersects the Lagrangian . Hence, the claim follows from the already considered case.∎
The following proposition characterizes regular Lagrangian discs.
Proposition 2.3.
Let be a Lagrangian disc with Legendrian boundary. It is regular if and only if there is a Weinstein handlebody representation of the cobordism for which coincides with the cocore Lagrangian disc of one of the index handles.
Proof.
If is a cocore disc for a Weinstein structure, then this structure is tangent to it, and hence is regular. Conversely, if is regular for a Weinstein structure , then by Lemma 2.2 this structure (after Weinstein homotopy of tangent to Weinstein structures) admits a Weinstein handlebody consisting of the ball with Lagrangian equatorial disc and other Weinsteins handles glued to . One can deform the Weinstein structure on by creating two additional critical points of index and such that serves as the cocore disc of the corresponding index handle. It remains to note that using Proposition 10.10 from [10] one can reorder critical points of a Lyapunov function so that the handle corresponding to the critical point is the last one to attach.
∎
The regularity property for Lagrangian submanifolds also has (at least conjecturally) a Lefschetz fibration characterization. E. Giroux and J. Pardon have suggested to us that the following can probably be proven along the lines of [19], adapting results of [7]: for any regular Lagrangian submanifold with Legendrian boundary there exists a Lefschetz fibration over which projects to a ray in . Of course, the converse statement is true: a Lagrangian with such a Lefschetz presentation is regular.
There are many other natural examples of regular Lagrangians, including the zero section and cotangent fibers of a cotangent bundle, and more generally smooth loci of Lagrangian skeleta or ascending Lagrangian cocores of the flow of . A necessary condition for the regularity of a closed , or more generally of a cobordism with , is given by the following:
Lemma 2.4.
Let be a regular Lagrangian cobordism with . Then the inclusion is injective. Here the homology is taken with integer coefficients if is orientable and with coefficients otherwise. Moreover, in the orientable case the image of the (relative) fundamental class of in is indivisible.
Proof.
Assuming the Weinstein structure is special, we observe that a generic fiber of the cotangent bundle has boundary which does not intersect the attaching spheres of any of the additional Weinstein handles, and hence represents a homology class . But , and hence the class is indivisible. ∎
The results of [17] and [27] show that this injectivity condition does not necessarily hold when is loose, or when but is overtwisted, and these therefore provide examples of nonregular Lagrangian cobordisms.
However, we do not know any counterexample to the positive answer to the following problem:
Problem 2.5.
Suppose (or more generally when is tight and is not loose). Is every Lagrangian cobordism regular? In particular, does the conclusion of Lemma 2.4 hold for such ? For instance, is the image of the fundamental class of a closed Lagrangian manifold in a Weinstein manifold necessarily indivisible (and in particular nonzero)?
3 Flexible Lagrangians
Definition 3.1.
We say a Lagrangian cobordism is flexible if it is regular with a special tangent Weinstein structure for which the complementary Weinstein cobordism is flexible.
In the case when one can equivalently characterize flexibility in terms of tangent but not necessarily special Weinstein structures.
Lemma 3.2.
A Lagrangian cobordism with is flexible if and only it is regular with a tangent to Weinstein cobordism structure which admits a partition into elementary cobordisms such that links of attaching spheres of index handles are loose in the complement of .
Proof.
The proof repeats the steps of the proof of Lemma 2.2. If for each critical point of the function its index on coincides with its index as a critical point of on , then we modify the Weinstein structure into one which is special and tangent to without changing and . For the resulting Weinstein structure the cobordism is automatically flexible.
Suppose that the index of a critical point point of the function is strictly less than the index of for on . Letting be the stable disc of on and the stable disc of for the function , we modify, as in the proof of Lemma 2.2 the Weinstein structure by changing the index of for to at the expense of creating two new critical points of index and respectively on the stable .
Finally we observe that if , then the index handle corresponding to the point is attached along a loose Legendrian by assumption. If , the second index handle corresponding to the point is in canceling position with the index handle corresponding to and hence is also attached along a loose knot. This implies the flexibility of .
The opposite implication is straightforward. ∎
The next proposition gives two fundamental examples of flexible Lagrangian submanifolds. Recall that a product of completed Weinstein cobordisms and is again a (completed) Weinstein cobordism (of manifolds with boundary). For a Weinstein cobordism structure we denote and observe that the the structure is tangent to the Lagrangian diagonal , and hence is regular for .
Proposition 3.3.

Let be a flexible Weinstein cobordism and let denote the result of attaching an handle to along a loose Legendrian knot . Then the cocore disc of the attached handle is flexible.

Let be a flexible Weinstein cobordism structure. Then the diagonal is flexible for .
Proof.
(i) This is immediate from Lemma 3.2.
(ii) To show that is flexible, we must show that all index handles determined by are attached along Legendrians which are loose in the complement of . The proof of this fact is essentially identical to the proof that the product of a flexible Weinstein manifold with any other Weinstein manifold is always flexible (and moreoever, in the case of , one can ensure by construction that the loose charts stay away from ). This folkloric statement was known for a while to several specialists and recently was proven by MurphySiegel [28, Proposition 3.7]. ∎
Problem 3.4.
Is the converse to each of the statements in Proposition 3.3 true? In other words, is it true that

If a Lagrangian cocore of an handle is flexible, then its attaching Legendrian sphere is loose? Moreover, can a flexible Weinstein cobordism remain flexible after attaching an handle along a nonloose Legendrian knot?

If the diagonal in the product is flexible, then is flexible.
Problem 3.5.
(around the nearby Lagrangian conjecture)

Are all regular closed Lagrangians in Hamiltonian isotopic?

Let be a flexible closed Lagrangian in a Weinstein domain . Are all (regular) closed Lagrangian submanifolds in Hamiltonian isotopic to ? To tie this to (i), we note that the section in a cotangent bundle is tautologically flexible.
4 Existence and classification of flexible Lagrangians
By a formal parameterized Lagrangian cobordism in we mean a pair , where is a smooth embedding of an dimensional cobordism , and , , is a homotopy of injective homomorphisms such that

(i) ;

(ii) is a Lagrangian homomorphism, i.e. is a Lagrangian subspace for all ;

(iii) ;

(iv) ; and

(v) is transverse to for all .
A genuine parameterized Lagrangian cobordism can be viewed as formal by setting , . Two formal parameterized Lagrangian cobordisms are formally Lagrangian isotopic if they are isotopic through formal parameterized Lagrangian cobordisms. Note that a formal parameterized Lagrangian cobordism has well defined formal Legendrian classes of its positive and negative boundaries. If is a subcobordism and is the inclusion map then we will drop the word “parameterized” from the term “formal parameterized Lagrangian cobordism”.
Remark 4.1.
One can define a weaker notion of a formal parameterized Lagrangian cobordism by dropping the transversality condition (v) from the definition. We note, however, that every homotopy class of weak formal Lagrangians contains a unique homotopy class of strong ones. Indeed, the obstructions to deform a weak to a strong one lie in the homotopy groups , where . But for , and hence obstructions vanish.
Theorem 4.2.

(Existence) In a flexible Weinstein cobordism , any formal Lagrangian cobordism with nonempty positive boundary of each of its components is formally Lagrangian isotopic to a flexible genuine Lagrangian cobordism.

(Uniqueness) Let
be two flexible Lagrangian cobordisms. Then given diffeomorphisms and such that and is homotopic to via a homotopy of nondegenerate not necessarily closed forms vanishing on , there exists an isotopy of to a symplectomorphism such that .
Remark 4.3.

Note that the condition in (i) is essential. For example, a flexible Weinstein manifold has no compact exact Lagrangians, even though it may have many formal compact Lagrangians.

If the formal embedding in Theorem 4.2(i) is genuine Lagrangian near then the formal homotopy can be constructed fixed near .

The formal Lagrangian isotopy in Theorem 4.2(i) need not be small, because the proof relies on the flexibility of the ambient structure. The situation here is similar to that of Theorem 7.19 of [10]: any formal Legendrian embedding in an overtwisted contact manifold is formally Legendrian isotopic to a genuine Legendrian embedding, but the isotopy need not be small.
Proof of Theorem 4.2.
To prove the existence part, let be a formal Lagrangian submanifold and denote a tubular neighborhood of in with its canonical symplectic structure . There exist an extension of to an embedding and a homotopy , , of fiberwise isomorphisms extending , such that and is a symplectic bundle isomorphism . The homotopy (:= ) of nondegenerate but not necessarily closed forms on extends to a homotopy , , of nondegenerate forms on such that and . In particuar, is a genuine symplectic structure on a neighborhood of and is Lagrangian. We denote by the complement of with its induced formal symplectic structure. We wish to now show that admits a Weinstein structure in the same almost symplectic class as relative boundary.
The condition for each component of implies that for . Indeed, the pair is connected because the cobordism is Weinstein. Hence, any relative spheroid extends for to a spheroid , where we denoted
and identified the the upperhalf sphere with the disc . Assuming without loss of generality that is transverse to we conclude that if then , and hence the homotopy class is trivial, and if the image intersects transversely in finitely many points. Hence, is generated by small spheres linked with and transported to the base point in by some paths in . Moreover, the condition for each connected component of allows us to choose and the condition provides a homotopy of the connecting path to , i.e. the generating relative spheroid is trivial in . Hence, the pair is connected. Then the classical WhiteheadSmale’s handle exchange argument, see [32], allows us to construct a defining function on the cobordism without critical points of index .
Using Theorem 13.1 from [10], which holds for cobordisms with corners, we construct a flexible Weinstein cobordism structure on which agrees with the standard symplectic structure on (the boundary of) the given neighborhood of the Lagrangian . Together with the canonical subcritical Weinstein structure on , it yields a flexible Weinstein structure on which is in the same almost symplectic homotopy class as the original symplectic structure on . Using Theorem 14.3 and Proposition 11.8 from [10], we can construct a diffeotopy connecting the identity with a symplectomorphism . Then the parameterized Lagrangian cobordism is in the prescribed formal class.
To prove the second part denote and . Let us observe that the Lagrangian neighborhood theorem allows us to assume that is symplectic on . Let , and denote the given Weinstein structures on and . By assumption the Weinstein structures and restricted to are in the same relative to almost symplectic class, and hence according to Theorem 14.3 and Proposition 11.8 from [10] is isotopic to a symplectomorphism via an isotopy fixed on , and in particular, we have . ∎
Remark 4.4.
An interesting aspect of Theorem 4.2 is that when , the positive Legendrian boundaries of a flexible Lagrangians necessarily cannot be loose in the sense of [26] (as they are filled by exact Lagrangians), and indeed must have nontrivial holomorphic curve invariants. For instance, the wrapped Floer homology of must be 0 by i.e., Lemma 6.3, or more directly, one can note that is flexible, hence and is a unital module over . It follows [11] that there is an isomorphism between the (linearized) Legendrian contact homology and the relative homology . See Problem 4.13 for further discussion.
Corollary 4.5.
Let be an manifold with nonempty boundary, equipped with a fixed trivialization of its complexified tangent bundle . Then there exists a flexible Lagrangian embedding with Legendrian boundary where is the standard symplectic ball, realizing the trivialization . In particular, any manifold with boundary can be realized as a flexible Lagrangian submanifold of with Legendrian boundary in .
Proof.
With respect to a reference trivialization , the trivialization is equivalent to the data of a map such that . Given such data, Gromov’s hprinciple for Lagrangian immersions produces a Lagrangian immersion transverse to . Moreover, using Whitney’s cancellation technique and the fact that we can regularly (but not symplectically) homotope to an embedding . The resulting embedding inherits a formal Lagrangian structure from the immersion . We then complete the proof using Theorem 4.2.
∎
To explore further consequences of the above constructions we first recall a theorem of Michèle Audin. Given a connected closed ndimensional manifold and an immersion with transverse double points we denote by the algebraic count of double points. This is an integer if is orientable and is even and an element of otherwise. is an invariant of the regular homotopy class of which vanishes if and only if the class contains an embedding. For a closed connected manifold of dimension we denote by Kervaire’s semicharacteristic
A relationship between and is given (in nice cases) by the following result:
Theorem 4.6.
[M. Audin, [6]] Let be a closed manifold of odd dimension . Then for any Lagrangian immersion with transverse double points, we have at least in the following cases:

is stably parallelizable;

and is orientable;

and is spin.
Let us call an dimensional, , connected closed manifold with trivial complexified tangent bundle admissible if at least one of the following conditions holds:

;

is even, is orientable and ; .

is even, is not orientable and is even;

is odd, satisfies one of the conditions (i)–(iii) of Audin’s theorem and .
Theorem 4.7.
Let be a closed admissible dimensional manifold. Then there exists a Weinstein structure on in the same formal homotopy class as the standard one, which contains as a flexible Lagrangian submanifold in the homology class of the section (with coefficients in the nonorientable case). Moreover, infinitely many of the are distinct as Weinstein manifolds.
Remark 4.8.
Conversely, any closed regular Lagrangian in with a possibly exotic, but formally standard Weinstein structure must have a trivial complexified tangent bundle and realizes the generator homology class (with coefficients if is not orientable), see Lemma 2.4. Furthermore, if is even then if is orientable and is even otherwise. Indeed, we have and if is not orientable this holds . If is odd then one can deduce from Audin’s theorem that for all admissible , i.e. in all cases listed in that theorem the condition is also necessary.
The proof of Theorem 4.7 roughly will proceed as follows: we remove a disc from to obtain a manifold with spherical boundary. Corollary 4.5 produces a flexible Lagrangian embedding with parametrized Legendrian boundary ; if this Legendrian lies in the same formal Legendrian isotopy class as the standard unknot, then the result of attaching a handle to along (which contains as a regular Lagrangian) will be formally homotopic to . It will thus be necessary to understand the Legendrian isotopy class of the aforementioned Legendrian embedding.
Recall that the formal Legendrian isotopy class of a parameterized Legendrian sphere in the standard contact is determined by two invariants (see [26, 10, 33, 20]): the rotation class and the generalized ThurstonBennequin invariant . If is even then can be defined as the linking number between and its pushoff by the Reeb flow. If is odd the rotation class identically vanishes, while the above definition of always yields , where the sign depends only on dimension. When there is indeed only 1 formal Legendrian isotopy class of spheres. However, for all odd there are exactly two classes, see [26, 10]. They are distinguished by a modified ThurstonBennequin invariant, which we will continue to denote by , and which can be defined as follows, see [10].
The vanishing of allows us to connect with the Legendrian unknot by a regular Legendrian homotopy. Viewing the homotopy as an immersed cylinder in , and assuming that the immersion has transverse double points, we set , where is the number of double points. It turns out that this residue is independent of the choice of a regular homotopy.
In order to prove Theorem 4.7 we will need the following two Lemmas:
Lemma 4.9.
Let be a nonorientable manifold of dimension bounded by a sphere, and a parameterization of its boundary. Suppose that the complexified tangent bundle is trivial. Then for any there exists a Lagrangian embedding with Legendrian boundary such that and .
Proof.
Let be a Lagrangian embedding with Legendrian boundary provided by Corollary 4.5. Using the stabilization procedure, see [10], one can modify for any integer the Legendrian knot by a Legendrian regular homotopy to a Legendrian embedding with . Note that . Let be a Lagrangian immersion, corresponding to this regular homotopy which connects and . The algebraic number of double points of is equal to . Gluing the Lagrangian cylinder with the embedding we get a Lagrangian immersion of whose boundary Legendrian sphere has its ThurstonBennequin invariant equal to and has the same rotation class as .
Suppose now that has the same parity as . Since also has the same parity as , we have for some integer . Apply the above construction to . Then, using nonorientability of one can cancel all double points in pairs by a smooth (not necessarily Lagrangian) isotopy, thus obtaining a formal Lagrangian embedding with the required Legendrian boundary invariants. Applying again Corollary 4.5 we construct a genuine Lagrangian embedding with the prescribed invariants of the boundary.
∎
Lemma 4.10.
Suppose that is admissible, and is obtained from by removing an ball, . Suppose that the boundary is parameterized by a diffeomorphism . If is orientable then for any Lagrangian embedding with Legendrian boundary , the Legendrian embedding is in the formal Legendrian isotopy class of the Legendrian unknot . If is nonorientable then there exists a Lagrangian embedding with Legendrian boundary such that is in the formal Legendrian isotopy class of the Legendrian unknot .
Proof.
Now, suppose that for a Lagrangian embedding and a diffeomorphism . Then . We already noted that this is always the case when is odd. If is even and is orientable then the Hurewicz homomorphism is injective (see e.g. Theorem 20.9.6 in [21]). Hence if the bounding Lagrangian is orientable. But the same argument applies in the nonorientable case to .
If is even and is orientable then . This is proven in [20] but here is another argument. Consider a vector field tangent to such that agrees the Liouville vector field . Then the pushoff of along intersects at points. But is the Reeb vector field, so the linking number entering the definition of is equal to up to sign. If is not orientable then the above argument implies only that . But in that case Lemma 4.9 allows us to modify the embedding to ensure that . Suppose now that is odd and . Then we can use Audin’s Theorem 4.6 to deduce that coincides with the Kervaire semicharacteristic for all admissible . Indeed, let be a Lagrangian immersion realizing a regular Legendrian homotopy connecting with the Legendrian unknot . Such immersion exists for sufficiently large , see [16]. It has intersection points. In turn, the unknot bounds an immersed Lagrangian disc with 1 intersection point. Gluing together the Lagrangian embedding with Lagrangian immersions and we get a Lagrangian immersion of to with intersection points, and hence the claim follows from Audin’s Theorem 4.6. ∎
Proof of Theorem 4.7.
Using Corollary 4.5 we realize as a flexible Lagrangian submanifold with Legendrian boundary in the standard symplectic ball . According to Lemma 4.10 the gluing diffeomorphism viewed as a Legendrian embedding is formally Legendrian isotopic to the standard Legendrian unknot. Hence, by attaching to the ball a Weinstein handle of index along using we get a Weinstein domain diffeomorphic to the disk cotangent bundle with its symplectic structure in the standard formal symplectic homotopy class. The Weinstein domain contains as a closed flexible Lagrangian submanifold in the homology class of the section.
To prove the second part of the theorem, which concerns with infinitely many symplectomorphism types of the resulting Weinstein domains , we first observe that the Viterbo transfer map on symplectic homology is an isomorphism preserving symplectic cohomology’s TQFT operations and BV algebra structure, see e.g. [29] (the part about compatibility of Viterbo transfer map with the BV operator is folkloric). In turn, as a BV algebra is isomorphic to , the homology of the free loop space , with the ChasSullivan string topology BV algebra structure at least whenever is Spin, see [34, 31, 4, 5, 1]. Since and , there is a canonical grading on ; if , then is gradingpreserving. In particular, assuming are spin with , the Weinstein domains and are not symplectomorphic whenever the BV algebras and are nonisomorphic. Thus, the above construction provides a rich source of exotic symplectic structures on ; it is at least as rich as the collection of string topology BV algebra structure on various manifolds.
Hence, to get infinitely many nonsymplectomorphic structures it suffices to find for any infinitely many closed stably parallelizable manifolds with , for even, for odd and , and different . Since the rank of equals the number of conjugacy classes of , it suffices to find with fundamental groups with different numbers of conjugacy classes; if we furthermore assume that is finite, then automatically since is always torsionfree. For we can take the collection of Lens spaces . To get examples for , consider the CW complex which has one 0, 1, and 2cell such that the attaching map for the 2cell wraps times around the 1cell so that and . We can then embed into , , and take a regular neighborhood of . Then the closed dimensional manifold satisfies all the required conditions. Indeed, it is stably parallelizable, for even, for odd, and . ∎
Remark 4.11.
There are by now many constructions of exotic Weinstein structures on , for instance [23, 25, 24, 2]. A notable feature of the examples given above is that they are each constructed from a single handle attachment on standard . Hence they have minimal complexity, meaning the defining function for the resulting Weinstein structure can be chosen with exactly two critical points (in particular, the Weinstein geometry of each is entirely determined by the Legendrian isotopy class ; recall ). In addition, each example contains an exact Lagrangian ; and hence has nonvanishing symplectic homology with any coefficients (at least whenever is Spin).
It is conceivable that the examples are as diverse as diffeomorphism types of manifolds :
Problem 4.12.
Does symplectic topology of remember the diffeomorphism type of , or even of ?
As Remark 4.11 recalls, the symplectic topology of is entirely determined by the differing Legendrian topology of embeddings ; in particular the examples in Theorem 4.7 produce infinitely many nonisotopic Legendrians in the same formal isotopy class. Hence, one can recast Problem 4.12 in terms of more general questions about the richness of the Legendrian topology of boundaries of flexible Lagrangians . For instance,
Problem 4.13.
Does the Legendrian boundary of a flexible Lagrangian remember the topology of the filling? For instance, when there exists a unique formal class of Legendrian 2spheres in contact . Does the genuine Legendrian isotopy class of remember the topology of ?</