aa r X i v : . [ m a t h . S G ] S e p A Lagrangian Klein bottle you can’t squeeze
Jonny EvansSeptember 4, 2020
Abstract
Suppose you have a nonorientable Lagrangian surface L in a symplec-tic 4-manifold. How far can you deform the symplectic form before thesmooth isotopy class of L contains no Lagrangians? I solve this ques-tion for a particular Lagrangian Klein bottle. I also discuss some relatedconjectures. Here are two overlapping questions:
Question 1.1 (Minimal nonorientable genus) . Given a symplectic 4-manifold ( X, ω ) and a Z / -homology class β ∈ H ( X ; Z / , what is the minimal nonori-entable genus of a nonorientable Lagrangian surface L ⊂ X with [ L ] = β ? Question 1.2 (Nonsqueezing) . Given a symplectic 4-manifold ( X, ω ) and anonorientable Lagrangian surface L ⊂ X , how far can you deform ω in coho-mology before there is no Lagrangian smoothly isotopic to L ?If L is orientable then these questions are less interesting: the genus is deter-mined by [ L ] = − χ ( L ) and, in Question 1.2, it is necessary to deform ω subjectto the cohomological condition R L [ ω ] = 0 . By contrast, if L is nonorientable,we have H ( L ; R ) = 0 , which means that it is possible to deform ω , keeping L Lagrangian, in such a way that [ ω ] ranges over an open set in H ( X ; R ) .I will give some general discussion of these questions in turn, then give a concreteexample of a Lagrangian Klein bottle for which Question 1.2 can be answeredcompletely (Theorem 3.1).One running theme throughout the discussion is the use of visible and trop-ical Lagrangians in almost toric 4-manifolds: these provide a rich source ofLagrangian submanifolds coming respectively from straight lines and tropicalcurves in integral affine surfaces. I have found them useful for thinking aboutsome of the phenomena under discussion, and for formulating conjectures. Visi-ble Lagrangians were introduced in Symington’s work [16]; tropical Lagrangianswere introduced independently by Mikhalkin [9] and Matessi [8].1 .1 Acknowledgements
I would like to thank Ivan Smith, Emily Maw, Georgios Dimitroglou Rizell,and Daniel Cavey for helpful conversations around this topic. My research issupported by EPSRC Grant EP/P02095X/2.
Definition 2.1.
Define the nonorientable genus of the nonorientable surface k RP to be k . Proposition 1.1 of [4] shows that any Z / -homology class ina symplectic 4-manifold can be represented by some embedded nonorientableLagrangian, so Question 1.1 has a well-defined answer, which I will denote by N ( X, ω, β ) . Remark . Audin [1] showed that P ( β ) = χ ( L ) = 2 − k mod 4 , where P denotes the Pontryagin square operation and χ is the Euler character-istic. If you find a Lagrangian with nonorientable genus k then you can performa Hamiltonian finger move locally to introduce pairs of intersections with indexdifference 1 and then perform Polterovich surgery [13] on these self-intersectionsto get an embedded Lagrangian with nonorientable genus k + 4 . This meansthat the set of genera which can be realised is { N ( X, ω, β ) , N ( X, ω, β ) + 4 , . . . } . Remark . The quantity N ( X, ω, β ) is known in a small range of cases, thelower bound being the principal difficulty.1. When X satisfies [ ω ] · c ( X ) > , we know that N ( X, ω,
0) = 6 . Thisfollows from Givental’s construction [5] of a Lagrangian RP in the4-ball and from the fact, proved by Shevchishin [14] that X contains nonullhomologous Lagrangian Klein bottles (see also the beautiful papers byNemirovski [12, 11]).2. Let X a,b,c be the blow-up of the 4-ball in three subballs so that the sym-plectic areas of the exceptional spheres E , E , E are a, b, c . Shevchishinand Smirnov [15] show that E + E + E contains a Lagrangian RP ifand only if the following inequalities all hold a < b + c, b < c + a, c < a + b. They call these the symplectic triangle inequalities . This gives the lowerbound N ( X a,b,c , ω, E + E + E ) ≥ when a, b, c violate the triangleinequalities. ŋ is the International Phonetic Alphabet symbol for the “ng” sound. emark . After the fact, we see that there is a tropical or almost toric moti-vation for the Shevchishin-Smirnov triangle inequalities. The almost toric basediagram in Figure 1 depicts the blow-up X a,b,c ; the affine lengths a, b, c indi-cated correspond to the sizes of the exceptional spheres E , E , E . In red youcan see a tropical curve; using the ideas of Mikhalkin [9] and Matessi [8], wecan construct a Lagrangian submanifold L living over a (small thickening of a)tropical curve. This tropical Lagrangian is diffeomorphic to RP if and only ifthe inequalities all hold: the preimage of the point marked with cross-hairs is acircle in L whose neighbourhood is a Möbius strip. × ×⊕ cc a b × × cc a b Figure 1: Almost toric base diagrams for X a,b,c with a tropical curve in red. Left:The symplectic triangle inequalities and the associated tropical Lagrangian isdiffeomorphic to RP (with the core circle of a cross-cap living over the pointmarked by the cross-hair symbol). Right: The symplectic triangle inequalitiesare violated and the associated tropical Lagrangian is diffeomorphic to a disc. S × S Let X = S × S . Modulo an overall scale factor, any symplectic form on X isdiffeomorphic to one from the family λp ∗ σ + p ∗ σ , where p , p : X → S are thetwo projections and σ is an area form on S . We know that N ( X, ω,
0) = 6 , whichleaves two interesting Z / -homology classes up to diffeomorphism: β = [ ⋆ × S ] and the class ∆ of the diagonal. The Pontryagin squares are P ( β ) = 0 and P (∆) = 2 , so there is a chance to represent β by Lagrangian Klein bottles. Lemma 2.5. If λ < then β is represented by a Lagrangian Klein bottle.Proof. The rectangle in Figure 2 is the toric moment polygon for the standardHamiltonian torus action on S × S with symplectic form ω λ . There is aLagrangian Klein bottle living over the line ℓ (slope / ) in the diagram. To seethis, consider the two S factors sitting inside R and let ( p j , θ j ) be cylindricalcoordinates on the j th factor ( j = 1 , ). These are action-angle coordinates, so ω = P dp j ∧ dθ j . The line ℓ is given by p = p and the Lagrangian Klein3 λ ⊗ ⊗ Figure 2: A visible Lagrangian Klein bottle in ( S × S , ω λ ) for λ < . Thecores of two cross-caps are indicated with cross-hairs.bottle is cut out by this equation together with θ = − θ . This is certainlyLagrangian for this symplectic form. To see that L is a Klein bottle, notice thatthe regular level sets of p restricted to L are circles θ = − θ in the ( θ , θ ) -torus, which collapse -to- onto the circles of maxima and minima at p = ± λ (as the torus collapses to the circle with coordinate θ ). The projections of thesecircles are denoted with cross-hairs in Figure 2. Remark . This L is a visible Lagrangian in the sense of Symington [16] aswell as being a tropical Lagrangian in the sense of Matessi [8] and Mikhalkin[9]. This Klein bottle is well-known: it appears in [3] as a Hamiltonian minimalLagrangian, in [6] as a Hamiltonian suspension, and in [4] as a fibre connect-sumof RP s. It has minimal Maslov number 1 and has a monotone representativein its Lagrangian isotopy class if λ = 1 .If λ ≥ then the line ℓ does not fit into the rectangle. The following conjectureseems natural; while I cannot prove it, it inspired Theorem 3.1 below. Conjecture 2.7.
There is no Lagrangian Klein bottle in the class β if λ ≥ . It is interesting to consider what happens for large λ . We have essentiallyno tools to prove lower bounds when the Lagrangians are of high genus andmay be Floer-theoretically obstructed. The most pessimistic conjecture is thatLagrangians with high genus become flexible enough that: Conjecture 2.8. lim λ →∞ N ( X, ω λ , β ) < ∞ . The following lemma gives an upper bound on N ( X, ω λ , β ) , but it goes to infinitywith λ . Lemma 2.9.
We have N ( X, ω λ , β ) ≤ ℓ + 2 when λ < ℓ + 2 .Proof. If λ < ℓ + 1 then there is a tropical Lagrangian in the class β withnonorientable genus ℓ + 2 . We show the tropical curve for ℓ = 2 in Figure 3below; for general ℓ we simply repeat the pattern between the vertical blue barsas often as required to get from the left-hand side to the right-hand side of therectangle.The edges of this tropical curve are: 4 ⊗ ⊗⊗ ⊗ pattern ⊗⊗ ⊗⊗ ⊗ Figure 3: A tropical curve giving a Lagrangian of genus ℓ +2 in the case ℓ = 2 .• internal edges parallel to either (3 , or (2 , − ,• external edges parallel to (2 , − or (1 , .The corresponding tropical Lagrangian intersects the horizontal spheres witheven multiplicity and the vertical spheres with odd multiplicity, so it inhabitsthe class β . The vertices of the tropical curve are not smooth : each has self-intersection equal to 2. By [9, Theorem 3.2], this tropical curve therefore yieldsan immersed Lagrangian with ℓ double points and ℓ cross-caps whereit hits the boundary (marked with cross-hairs in Figure 3). When we per-form Polterovich surgery at the double points, we obtain a Lagrangian whichis topologically a surface of genus ℓ with ℓ + 2 cross-caps. This has Eulercharacteristic − ℓ − ℓ − − ℓ , so the nonorientable genus is ℓ . Remark . It seems harder to make the genus significantly smaller usingtropical Lagrangians, but there is no reason to believe that tropical Lagrangiansshould give a sharp upper bound for N . For each connected open interval I ⊂ R (length | I | ), let C I denote the cylinder I × ( R / π Z ) with coordinates ( p, θ ) , equipped with the symplectic form π dp ∧ dθ ; this has total area | I | . Let S denote the 2-sphere equipped with its areaform σ satisfying R S σ = 2 .Let U I = S × C I . Note that U I is obtained from ( S × S , ω | I | ) by excising thespheres S × { n, s } , where n, s denote the poles of the second factor. Arguing asin Lemma 2.5, we see that if | I | > , the only nontrivial class β ∈ H ( U I ; Z / is represented by a Lagrangian Klein bottle (see Figure 4). At each vertex of a tropical curve, the outgoing edges v , v , v must sum to zero; if wewrite m for the determinant | v ∧ v | = | v ∧ v | = | v ∧ v | then the self-intersection of thisvertex is defined to be m − . Smoothness means all vertices have self-intersection zero. | I |⊗ ⊗ Figure 4: The visible Lagrangian Klein bottle in U I when | I | > . Theorem 3.1.
Suppose that | I | ≤ . If ι : K → U I is a Lagrangian embeddingof the Klein bottle in the class β then ι ∗ : Q = H ( K ; Q ) → H ( U I ; Q ) = Q isthe zero map.Remark . The proof of Theorem 3.1 will occupy the rest of the paper. Ituses SFT and neck-stretching.
Remark . Note that if | I | > then H ( L ; Q ) → H ( U I ; Q ) is an isomorphismfor the Lagrangian Klein bottle L coming from Lemma 2.5. To see this, takeeither one of the circles living over the points marked with cross-hairs in Figure4; this is a generator for both H ( L ; Q ) and H ( U I ; Q ) . We deduce: Corollary 3.4.
The Lagrangian Klein bottle in U (0 , ǫ ) from Lemma 2.5 cannotbe squeezed into U (0 , .Remark . To reduce Conjecture 2.7 to this result, you would need to producea pair of symplectic spheres in the class [ S × ⋆ ] which “link” your LagrangianKlein bottle in an appropriate way. Since this class has non-minimal symplecticarea, it is difficult to control the SFT limit of such spheres.We now proceed to the proof of Theorem 3.1. Pick a flat metric g on the Klein bottle. There is a contact form (the canoni-cal 1-form) on the unit cotangent bundle M ⊂ T ∗ K whose closed Reeb orbitscorrespond to closed geodesics on K . We will not distinguish notationally be-tween geodesics and the corresponding Reeb orbits and we will write − γ for thegeodesic obtained by reversing γ . There are two isolated simple geodesics γ , γ which are the core circles for two disjoint embedded Möbius strips in K . Anyisolated geodesic is a multiple cover of one of these and all other geodesics oc-cur in one-parameter families. We call the isolated geodesics odd and the othergeodesics even . Theorem 3.6 (Mohnke [10, Section 2.1]) . There exists an almost complex struc-ture J − on the cotangent bundle T ∗ K with the following properties:1. J − is cylindrical at infinity and suitable for neck-stretching. . For any geodesic γ there is a finite-energy J − -holomorphic cylinder f γ in T ∗ K asymptotic to γ and − γ .3. [10, Lemma 7(2)] Any J − -holomorphic cylinder in T ∗ K which intersectsthe zero-section is one of these f γ for some closed geodesic γ .Remark . If we let W := T ∗ K denote the compactification of the cotangentbundle obtained by gluing on its ideal contact boundary M then there is a well-defined intersection pairing H ( W, M ; Z / ⊗ H ( W ; Z / → Z / . The cylinders f γ define elements of H ( W, M ; Z / and we have [10, Lemma 7(3)] f γ · K = ( if γ is odd if γ is even. Remark . Note that there are also no finite energy planesin T ∗ K , nor in the symplectisation R × M , for any cylindrical almost complexstructure adapted to our chosen contact form. This is because there are nocontractible Reeb orbits, and a finite energy plane would provide a nullhomotopyof its asymptote. Let I = (0 , and ¯ I = [0 , . Suppose there is a Lagrangian Klein bottle K ⊂ U I such that Q = H ( K ; Q ) → H ( U I ; Q ) = Q is nonzero (in particular,it is injective). Think of K sitting inside U ¯ I and make symplectic cuts to U ¯ I at p = 0 , to obtain a Lagrangian Klein bottle K living in the manifold X = S × S equipped with the product symplectic form giving the factorsareas and respectively. Crucially, the symplectic cut introduces symplecticspheres S and S (at the p = 0 , cuts respectively) which are disjoint from K .Pick a sequence of almost complex structures J t , t ∈ R , on X with the followingproperties:• on a Weinstein neighbourhood of K , J t coincides with Mohnke’s almostcomplex structure J − ;• on a neck-stretching region ( a t , b t ) × M around K , J t is a neck-stretchingsequence;• the spheres S , S are J t -holomorphic for all t ∈ R .Pick a point k on K which does not lie on any of the cylinders f γ for anodd geodesic γ . Let u t : S → X be a J t -holomorphic curve representing theclass α = [ ⋆ × S ] and such that u t (0) = k ; there is a unique such u t up toreparametrisation by a theorem of Gromov [7, 2.4.C], since α is a minimal areasphere class in X .By the SFT compactness theorem [2], there is a sequence t i such that u t i con-verges (after reparametrisations) to a holomorphic building with components7n T ∗ K (the completion of the Weinstein neighbourhood of K ), components in R × M (the completion of the neck) and components in X \ K (the completionof the complement of the Weinstein neighbourhood). The components v , . . . , v n of the SFT limit building living in X \ K can becompactified, yielding topological surfaces in X with boundary on K ; we willstill denote these by v , . . . , v n . The sum of the ω -areas of the v i (weighted bymultiplicities if the SFT limit involves a branched cover) equals the ω -area of α , which is . Lemma 3.9.
There must be at least two planar components amongst the v i ,possibly geometrically indistinct (i.e. having the same image).Proof. First note that the limit building intersects K because u t (0) = k ∈ K for all t . It also necessarily has at least one component in X \ K because T ∗ K isexact and so contains no closed holomorphic curves. A genus zero holomorphicbuilding with at least two levels must have two planar components (just fortopological reasons) though these could be geometrically indistinct. Any planarcomponents live in X \ K . Lemma 3.10.
There are two components v , v of the limit building such that v i · S j = δ ij . These components are planar and there are no further componentsof the limit building in X \ L .Proof. Since α intersects S and S there must be components of the limitbuilding which intersect S and S . By positivity of intersections, either:(A) there is one component v which hits both S and S once transverselyand all other components are disjoint from S , S .(B) there are two components v , v such that v intersects S once trans-versely and is disjoint from S and vice versa for v .Moreover, each of these components occurs with multiplicity one in the SFTlimit in order to get the correct intersection numbers α · S , α · S .If v is a component which does not intersect S or S then it defines a class in H ( U I , K ; Z ) . By assumption, the kernel of the map Z ⊕ Z / H ( K ; Z ) → H ( U I ; Z ) = Z is precisely the torsion part. Therefore the long exact sequence · · · → H ( U I ; Z ) → H ( U I , K ; Z ) → H ( K ; Z ) → H ( U I ; Z ) → · · · splits off a sequence · · · → Z → H ( U I , K ; Z ) → Z / → . H ( U I , K ; Z ) are half-integer multiplesof the area of the generator β ∈ H ( U I ; Z ) , which is . Therefore v has integerarea.The area of α is , so the (weighted) sum of the ω areas of the v i equals 1. Since v has positive area, v must have positive area strictly less than , but this isnot possible if v has integer area. Therefore there cannot be any component v disjoint from S and S .By Lemma 3.9, there are at least two planar components (or one planar com-ponent with multiplicity two) in the limit building. This is not compatible withCase (A), so we must be in Case (B) and v , v must additionally be planes. Lemma 3.11.
1. All the remaining parts of the limit building are cylinders.2. At least one of these cylinders lives in T ∗ K and has the form f γ for anodd geodesic γ .3. There are no other cylindrical components of the SFT limit building in T ∗ K .Proof.
1. If a component has three or more punctures then the limit buildingmust contain at least three planar components (counted with multiplicity)but we have seen that all the planar components must live in X \ K (Remark 3.8) and that there are precisely two such components (Lemma3.10).2. Since u t (0) = k for all t , the limit building contains a component in T ∗ K ,which must be a cylinder of the form f γ by Theorem 3.6(3). At least one ofthese cylindrical components must correspond to an odd geodesic because α has odd intersection with K in H ( X ; Z / and the intersection numberpicks up contributions from each component of the building inside T ∗ K ,which are nontrivial if and only if γ is odd (Remark 3.7).3. If there are two or more cylindrical components in T ∗ K then there must bea further cylindrical component in R × M which connects the asymptotesof two of these cylinders. Since this cylinder has no positive asymptote,this cannot exist by the maximum principle. Proof of Theorem 3.1.
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