Lefschetz fibrations and symplectic homology

Lefschetz fibrations and symplectic homology

Mark McLean

We show that for each there are infinitely many finite type Stein manifolds diffeomorphic to Euclidean space which are pairwise distinct as symplectic manifolds.



1. Introduction

This paper is about the symplectic topology of Stein manifolds. If we have a symplectic manifold , then we say it carries a Stein structure if there exists a complex structure and an exhausting (i.e. proper and bounded from below) plurisubharmonic function such that , where is defined by . The triple is called a Stein manifold. We say that a Stein manifold is of finite type if only has finitely many critical points, each of which is non-degenerate.

We define an equivalence relation on Stein manifolds by: if there exists a sequence of Stein manifolds such that

  1. and

  2. is either symplectomorphic or Stein deformation equivalent to . (Stein deformation is defined later in Definition LABEL:defn:steindeformation.)

The aim of this paper is the following theorem:

Theorem 1.1.

Let . There exists a family of finite type Stein manifolds diffeomorphic to indexed by such that

We also have the following corollary:

Corollary 1.2.

Let be a compact manifold of dimension or higher. There exists a family of finite type Stein manifolds diffeomorphic to indexed by such that

We will prove this at the end of this introduction. The results of Paul Seidel and Ivan Smith in [SS:rama] show that there exists a finite type Stein manifold diffeomorphic to but not symplectomorphic to . They show this by constructing an affine variety which has a Lagrangian torus which cannot be moved off itself by a Hamiltonian isotopy. In fact they show that none of these Stein manifolds can be embedded in a subcritical Stein manifold. Also, they have an argument (explained in [Seidel:biasedview]) that shows that the symplectic homology groups of these varieties are non-trivial. In this paper we strengthen this result in two ways:

  1. we give examples in all even dimensions (not just in dimension where );

  2. we also show there are countably many pairwise distinct examples in each of these dimensions.

(1) is straightforward but (2) is much harder and involves various new ideas. We show in Corollary LABEL:corollary:embeddinginsubcritical that these manifolds cannot be embedded in a subcritical Stein manifold.

There is no analogue of our result in dimension 4 because any finite type Stein manifold diffeomorphic to is symplectomorphic to (see the introduction to [SS:rama]). Having said that, Gompf in [Gompf:handlebody] constructs uncountably many non-finite type Stein manifolds which are homeomorphic to , but are pairwise not diffeomorphic to each other. We hope to address the question of whether Theorem 1.1 holds in dimension 6, and whether we can distinguish the contact boundaries of these manifolds up to contactomorphism, in future work.

We will now construct an example of a family of Stein manifolds as in Theorem 1.1 in dimension 8. Let and consider a smooth point, say . Let be the blowup of at . Then is a Stein manifold where is the proper transform of . The variety is called the Kaliman modification of (see [Kaliman:eisenman]). We will think of this modification in two stages:

  1. Cut out the hypersurface in to get .

  2. Blow up at infinity to get .

Operation (2) attaches a -handle along a knot which is transverse to the contact structure. All our Stein manifolds in this family will be constructed from . If we have two Stein manifolds and , then it is possible to construct their end connected sum (see Theorem LABEL:thm:endconnectsum). Roughly what we do here is join and with a -handle, and then extend the Stein structure over this handle. Finally, is our family of Stein manifolds.

1.1. Sketch of the proof of the main theorem

We will now give an outline of the proof. We will only consider the examples in dimension 8 constructed above, as the higher dimensional examples are similar. For each Stein manifold with a trivialisation of the canonical bundle, we have an integer graded commutative algebra where is called symplectic homology 111With our convention, the pair-of-pants product makes (and not ) a unital ring.. The reason why we write instead of is because we want the unit to be in degree and not in degree . If and are Stein manifolds with and , then (see [Seidel:biasedview, Section 7]). The reason why we need is because symplectic homology is only known to be invariant up to exact symplectomorphism. For each Stein manifold , we can define another invariant which is the number of idempotents of (this invariant might be infinite). Hence all we need to do is show that for , . The next fact we need is that for any two Stein manifolds and , . This means that , and hence . So all we need to do is show that . If , since we have and ; but since can a priori be infinite dimensional in each degree, finiteness of is much harder. Most of the work in this paper involves proving .

For any Stein manifold , is graded by the Conley-Zehnder index taken with negative sign. The group has a ring structure making into a graded algebra. This ring is also a graded algebra. This means that as a vector space, it is of the form and if and then their product is in . Hence idempotents in are contained in and are a linear combination of elements with grading in the torsion part of (see Lemma LABEL:lemma:finitelymanyidempotents). The problem is that . In order to find out which elements of are idempotents, we will show that is isomorphic as a ring to where was defined above. Because is so much simpler than and , it is possible by a direct calculation to show that has finitely many idempotents.

Proving that relies on the following theorem. This theorem is the heart of the proof. We let , be Lefschetz fibrations, and (resp. ) be smooth fibres of (resp. ). Let and be Stein domains with a holomorphic and symplectic submanifold of .

Theorem 1.3.

Suppose and satisfy the following properties:

  1. is a subfibration of .

  2. The support of all the monodromy maps of are contained in the interior of .

  3. Any holomorphic curve in with boundary inside must be contained in .

Then .

Remark 1: There exist Lefschetz fibrations , with the above properties such that as convex symplectic manifolds, (resp. ) is convex deformation equivalent to (resp. ). This is because we can choose an algebraic Lefschetz fibration on where the closures of all the fibres pass through . Then blowing up at infinity (operation (2) of the Kaliman modification) is the same as blowing up each fibre at infinity and keeping the same monodromy. Hence .

Remark 2: Given varieties and in dimension such that is obtained from by blowing up at infinity, there are Lefschetz fibrations , satisfying properties (1) and (2) such that as convex symplectic manifolds, (resp. ) is convex deformation equivalent to (resp. ). These do not satisfy property (3) because is obtained from by filling in a boundary component of the fibres with a disc.

We will prove Theorem 1.3 in two stages. In stage (i), we construct a graded algebra related to a Lefschetz fibration and show it is equal to symplectic homology. This is covered in sections LABEL:section:lefschetzfibrationproofs and LABEL:section:bettercofinal. In stage (ii), we prove that . This is covered in section LABEL:section:transferisomorphism. In a little more detail:

(i) Let be a smooth fibre of and a disc in . In section LABEL:section:lefschetzcofinal, we show (roughly) that the chain complex for is generated by:

  1. critical points of some Morse function on ;

  2. two copies of fixed points of iterates of the monodromy map around a large circle;

  3. pairs where is a Reeb orbit on the boundary of and is either a Reeb orbit of or a fixed point in the interior of .

This is done in almost exactly the same way as the proof of the Künneth formula for symplectic homology [Oancea:kunneth]. The differential as usual involves counting cylinders connecting the orbits and satisfying the perturbed Cauchy -Riemann equations. The orbits in (1) and (2) actually form a subcomplex , and we define Lefschetz symplectic homology to be the homology of this subcomplex. Let be the map corresponding to the inclusion . We need to show that is an isomorphism. In order to do this we will show that the elements of in (3) have very high index. The orbits in (3) actually correspond to orbits of some Hamiltonian on the complex plane . We can ensure that this Hamiltonian has orbits of index as large as we like. In particular, we can assume that the index of is so large that the index of the pair is greater than any number we like (because the index of is the sum ). This means for any , the map is an isomorphism if we ensure the indices of the orbits are all greater than . Hence .

(ii) Let (resp. ) be the chain complex for the group (resp. ). The fibration is a trivial fibration . We have a short exact sequence where is generated by orbits of the form in . The orbit is a critical point of some Morse function on and is either a Reeb orbit of or a fixed point in the interior of . In this case the homology of the chain complex is actually a product where is the relative cohomology group . Because , we have that which gives us our isomorpism between and . Property (3) in Theorem 1.3 is needed here to ensure that the above exact sequence exists. If we didn’t have this property, then there would be some spectral sequence from (with an extra grading coming from the classes of these orbits) to .

Lefschetz symplectic homology was partially inspired by Paul Seidel’s Hochshild homology conjectures [Seidel:hochschildhomology], which also relate symplectic homology to Lefschetz fibrations. His conjectures would in particular prove Theorem 1.3.


of Corollary 1.2. There is a standard Stein structure on such that is a non-trivial finite dimensional vector space. By Lemma LABEL:lemma:finitelymanyidempotents this means that . Also is an idempotent which means that . We let be defined as in the proof of the main theorem 1.1. We define

Then . These numbers are all different as and for . ∎

Acknowledgements: I would like to thank my supervisor Ivan Smith for checking this paper and for giving many useful suggestions. I would also like to thank Paul Seidel, Kai Cieliebak, Alexandru Oancea, Frédéric Bourgeois, Domanic Joyce, Burt Totaro, Jonny Evans and Jack Waldron for giving useful comments.

1.2. Notation

Throughout this paper we use the following notation:

  1. are manifolds (with or without boundary).

  2. is the boundary of .

  3. are exact Lefschetz fibrations (See Definition LABEL:defn:lefschetzfibration).

  4. If we have some data associated to (resp. ), then
    are data associated to
    (resp. ). For instance is the boundary of .

  5. is a symplectic form on or .

  6. is a -form such that .

  7. is an exact symplectic manifold.

  8. is an almost complex structure compatible with .

  9. If is a compact convex symplectic manifold (See Definition 2.1), then is the completion of (Lemma 2.4). Similarly by Definition LABEL:defn:halfconvexlefschetzcompletion, can be completed to (we leave and as they are by abuse of notation).

  10. is a convex symplectic or Stein deformation.

  11. will denote a smooth fibre of .

  12. If we have some subset of a topological space, then we will let be some open neighbourhood of .

2. Background

2.1. Stein manifolds

We will define Stein manifolds as in [SS:rama]. We let be a manifold and a -form where is a symplectic form.

Definition 2.1.

is called a compact convex symplectic manifold if is a compact manifold with boundary and the -dual of is transverse to and pointing outwards. A compact convex symplectic deformation is a family of compact convex symplectic manifolds parameterized by . We will let be the vector field which is -dual to .

Usually, a compact convex symplectic manifold is called a convex symplectic domain. We have a natural contact form on , and hence we call this the contact boundary of .

Definition 2.2.

Let be a manifold without boundary. We say that is a convex symplectic manifold if there exist constants tending to infinity and an exhausting function such that is a compact convex symplectic manifold for each . Exhausting here means proper and bounded from below. If the flow of exists for all positive time, then is called complete. If there exists a constant such that for all , is a compact convex symplectic manifold, then we say that is of finite type.

Definition 2.3.

Let be a smooth family of convex symplectic manifolds with exhausting functions . Suppose that for each , there are constants tending to infinity and an such that for each in and , is a compact convex symplectic manifold. Then is called a convex symplectic deformation.

The constants mentioned in this definition depend on but not necessarily in a continuous way. The nice feature of convex symplectic manifolds is that we have some control over how they behave near infinity. That is, the level set is a contact manifold for all . Also note that if we have a convex symplectic manifold of finite type then it has a cylindrical end. That is, there exists a manifold with a contact form such that at infinity, is symplectomorphic to ( is a coordinate on ).

Lemma 2.4.

A compact convex symplectic manifold can be completed to a finite type complete convex symplectic manifold .

This is explained for instance in [Viterbo:functorsandcomputations, section 1.1]. The proof basically involves gluing a cylindrical end onto . Let be a complete convex symplectic manifold. Let be a compact convex symplectic manifold which is a codimension exact submanifold of (i.e. for some smooth function on ).

Lemma 2.5.

We can extend the embedding to an embedding .

Comments 0
Request Comment
You are adding the first comment!
How to quickly get a good reply:
  • Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.
  • Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.
  • Your comment should inspire ideas to flow and help the author improves the paper.

The better we are at sharing our knowledge with each other, the faster we move forward.
The feedback must be of minimum 40 characters and the title a minimum of 5 characters
Add comment
Loading ...
This is a comment super asjknd jkasnjk adsnkj
The feedback must be of minumum 40 characters
The feedback must be of minumum 40 characters

You are asking your first question!
How to quickly get a good answer:
  • Keep your question short and to the point
  • Check for grammar or spelling errors.
  • Phrase it like a question
Test description