Arborealization II: Geomorphology of Lagrangian ridges
GGEOMORPHOLOGY OF LAGRANGIAN RIDGES
DANIEL ´ALVAREZ-GAVELA, YAKOV ELIASHBERG, AND DAVID NADLER
Abstract.
We prove an “h-principle without pre-conditions” for the elimination of tangencies ofa Lagrangian submanifold with respect to a Lagrangian distribution. The main result says thatthe tangencies can always be completely removed at the cost of allowing the Lagrangian to developcertain non-smooth points, called
Lagrangian ridges , modeled on the corner { p = | q |} ⊂ R togetherwith its products and stabilizations. This result will play an essential role in our forthcoming paperon the arborealization program [AGEN]. Contents
1. Introduction 12. Formal solution 73. Alignment of ridges 154. Integrable solution 225. Adapted version 27References 291.
Introduction
Lagrangian ridges and ridgy isotopies.
Let L be a smooth compact Lagrangian submanifoldof a symplectic manifold ( M, ω ). The goal of this paper is the simplification of singularities of tangencyof L with respect to a field of Lagrangian planes γ ⊂ T M . When γ is tangent to the fibres of aLagrangian fibration M → B these tangencies are the same as singular points of the smooth map L → B . If ω = dλ and L is exact, then the tangencies are also the same as the singular points of theLegendrian front (cid:98) L → B × R , where (cid:98) L is the Legendrian lift of L in the contactization M × R . Theimage Σ ⊂ B × R of the singular locus is known as the caustic in the literature.Our viewpoint is local on L . As the deformation will always be done in a neighborhood of the givenLagrangian L we can assume that the symplectic manifold M is the cotangent bundle T ∗ L endowedwith the standard symplectic form ω = d ( pdq ). All considered Lagrangians Λ ⊂ T ∗ L will be exact,i.e. pdq | Λ = dh , and hence could be lifted to Legendrian submanifolds (cid:98) Λ = { ( x, h ( x )) , x ∈ L } ⊂ T ∗ L × R = J ( L ), where J ( L ) is endowed with the standard contact structure ξ = { dz − pdq = 0 } .Even if γ is integrable, C ∞ -generic Lagrangian tangency singularities are in general non-classifiable,see [AGV85]. However, if certain homotopical conditions given in terms of the homotopy class ofthe Lagrangian plane field γ | L are met, then by a C -small Hamiltonian isotopy the singularities canbe reduced to the simplest ones of the so-called fold type, see Section 1.3 below. In the presence YE is partially supported by NSF grant DMS-1807270.DN is partially supported by NSF grant DMS-1802373. a r X i v : . [ m a t h . S G ] D ec DANIEL ´ALVAREZ-GAVELA, YAKOV ELIASHBERG, AND DAVID NADLER of homotopical obstructions the higher Lagrangian tangency singularities cannot not be removed bymeans of a Hamiltonian isotopy, so any attempt at removing them must allow for deformations of theLagrangian more dramatic than a Hamiltonian isotopy.Note that one can trade Lagrangian fold singularities for corner singularities of the Lagrangian itself.Namely, if { q = p } is a fold with respect to the vertical distribution γ = { dq = 0 } , we can replace itwith a corner { q = | p |} which is transverse to γ , see Figures 5 and 6. Our main result 1.4 shows that ina similar way, by creating certain standard combinatorial singularities called ridges , one can make theLagrangian transverse to a Lagrangian distribution γ even without any homotopical pre-conditions.These ridges are built out of the corner by taking products and stabilizations. Note that a posterioriall the results can be reformulated back in the smooth category by smoothing the ridges.We now define ridgy Lagrangians and ridgy isotopies. In the standard symplectic R = T ∗ R consider the subset R = { pq = 0; q ≥ , p ≥ } . This is the model ridge of order 1. The model ridgeof order k in the standard symplectic R n = T ∗ R n is defined to be the product R k,n = R k × R n − k ⊂ ( T ∗ R ) k × T ∗ R n − k , 0 ≤ k ≤ n , i.e the ( n − k )-fold stabilization of R k = R × · · · × R ( k times). Figure 1.
The model ridge R ⊂ T ∗ R . Note that R is the union of the half-line { p = 0 , q ≥ } together with the inner conormal { q = 0 , p ≥ } of its boundary point q = 0. The model R is simplectomorphic to { p = | q |} ⊂ T ∗ R . Example 1.1.
The order n ridge R n,n ⊂ T ∗ R n is the union to all the inner conormals of the faces of aquadrant in R n , hence is the union of the 2 n linear Lagrangians { p j = q k = 0 , q j , p k ≥ , j ∈ I, k (cid:54)∈ I } ,where I ⊂ { , . . . , n } . See Figure 2. Definition 1.2. An n -dimensional ridgy Lagrangian in a symplectic manifold M is a closed subset L ⊂ M which is covered by open subsets U ⊂ M such that ( U, U ∩ L ) is symplectomorphic to some( B, B ∩ R k,n ), for B ⊂ R n a ball centered at the origin.A ridgy Lagrangian has a natural stratification L = R ⊃ R ⊃ · · · ⊃ R n , where R k is the locusof ridges of order ≥ k . Note that the stratum R k \ R k +1 is a smooth (open) isotropic submanifold ofdimension n − k . EOMORPHOLOGY OF LAGRANGIAN RIDGES 3
Figure 2.
The order 2 ridge R , ⊂ T ∗ R is the union of the inner conormals tothe faces of the quadrant { q ≥ , q ≥ } . In black { p = p = 0 , q , q ≥ } ,in blue { p = q = 0 , q , p ≥ } , in green { p = q = 0 , q , p ≥ } and in red { q = q = 0 , p , p ≥ } . The model R , is symplectomorphic to { p = | q |} × { p = | q |} ⊂ T ∗ R × T ∗ R = T ∗ R . Figure 3.
A 2-dimensional ridgy Lagrangian has order 1 ridges along a union ofsimple closed curves which intersect each other at order 2 ridges.
Definition 1.3.
We now define the notion of a ridgy isotopy of a smooth Lagrangian submanifold L in a symplectic manifold M .(1) Let N , . . . , N m ⊂ L be co-oriented separating hypersurfaces defined by equations φ j = 0 forsome C ∞ -functions φ j : L → R without critical points on N j . We assume that the N j areco-oriented by the outward transversals to the domains { φ j ≤ } . Denote φ + j = max( φ j , θ j which is equal to 1 on N j and to 0 outside a neighborhood of N j . Define a function Φ : L → R (which is C and piecewise C ∞ ) by the formulaΦ := m (cid:88) j =1 θ j (cid:0) φ + j (cid:1) . An earthquake isotopy with faults N j is defined as a family of Lagrangians L t given by thehomotopy of generating functions t Φ, i.e. L t = { p = td Φ } , t ≥
0, see Figure 4.
DANIEL ´ALVAREZ-GAVELA, YAKOV ELIASHBERG, AND DAVID NADLER (2) A ridgy isotopy is an earthquake isotopy followed by an ambient Hamiltonian isotopy.Of course, the earthquake isotopy can be realized by an ambient Hamiltonian homotopy beginningfrom any t > Figure 4.
An earthquake isotopy. Note that in general the hypersurfaces N j mayintersect each other.1.2. Main results.
We can now state our main result. Recall that L is a smooth, compact Lagrangiansubmanifold of a symplectic manifold M . Theorem 1.4.
For any Lagrangian distribution γ there exists a ridgy isotopy L t of L such that L (cid:116) γ . Theorem 1.4 will play an essential role in our forthcoming paper [AGEN] on the arborealizationprogram [N15, N17, S18].
Remark 1.5.
Theorem 1.4 also holds(1) in C -close form; namely, we can demand that the ridgy isotopy L t is C -small. This meansthat given a fixed but arbitrary Riemannian metric on M , for any ε > x, f t ( x )) < ε for f t : L → L t the parametrization of the ridgy isotopy L t which isgraphical during the earthquake isotopy and then is given by the ambient Hamiltonian isotopy.In particular L t stays within a Weinstein neighborhood of L in M .(2) in relative form; namely, if L (cid:116) γ on Op ( A ) for A ⊂ L a closed subset then we can demandthat L t = L on Op ( A ). Here and below we use Gromov’s notation Op ( A ) for an arbitrarilysmall but non-specified open neighborhood of A . EOMORPHOLOGY OF LAGRANGIAN RIDGES 5
We will also prove an adapted version of Theorem 1.4 which is necessary for our applications, seeSection 5 below, where L has a boundary and corner structure and γ is itself adapted to that structure. Remark 1.6.
By definition, the local geometry of a ridgy Lagrangian L is given by the linear models R k,n . The space of linear fields γ transverse to some R k,n has interesting moduli without evidentcanonical representatives.1.3. h-principle for removing higher Lagrangian tangency singularities. The problem of sim-plifying the tangency locus of a smooth Lagrangian submanifold L ⊂ M with respect to a Lagrangianplane field γ ⊂ T M was first studied by Entov [E97], who used the method of surgery of singular-ities to establish an h-principle for the class of Σ -nonsingular plane fields, i.e. those γ for whichdim( T L ∩ γ ) <
2. In [AG18a] and [AG18b] the methods of holonomic approximation and wrinklingwere used by the first author to extend this h-principle to arbitrary Lagrangian plane fields. Thesimplest version of the h-principle can be formulated as follows.
Theorem 1.7.
Suppose that γ is homotopic through Lagrangian plane fields to a Lagrangian field (cid:98) γ which is transverse to L . Then L is Hamiltonian isotopic to a smooth Lagrangian submanifold (cid:98) L whosetangency singularities with respect to γ consist only of folds. The fold is the simplest type of singularity. In the case where γ is integrable, the germ is givenby (a stabilization of) the local model { q = p } ⊂ T ∗ R , where the Lagrangian field is the verticaldistribution γ = { dq = 0 } . If the tangency locus of L with respect to γ consist only of folds, thendim( T L ∩ γ ) ≤ { dim( T L ∩ γ ) = 1 } is a transversely cut out smoothhypersurface in L . Moreover, the line field (cid:96) = T L ∩ γ is transverse to Σ inside T L . These propertiescharacterize the fold (also in the non-integrable case).
Remark 1.8.
Even in the smooth (as opposed to symplectic) category, the elimination of folds is notusually possible. Moreover, while in the smooth category the only non-trivial constraints are on thetopology of the image of the fold, see [G09, G10], where the only thing which matters is that the foldlocus is non-empty, see [E70], in the symplectic case there are also constraints on the topology of folds(e.g. the number of its components) in the source Lagrangian, see [E98]. See also [FP98, FP06] forfurther constrains on the caustic locus.The fold is closely related to the order 1 ridge. More precisely, observe that the 1-dimensionalLagrangian model { q = p } has a fold type tangency to the vertical Lagrangian distribution γ = { dq = 0 } , while the ridgy Lagrangian { q = ε | p |} is transverse to γ . Let us take a cut-off function σ : [0 , ∞ ) → [0 ,
1] which is equal to 1 on [0 , ], equal to 0 outside [0 ,
1] and has non-positive derivative.Define the function φ ε ( p ) = 13 (cid:18) − σ (cid:18) | p | ε (cid:19)(cid:19) p + ε σ (cid:18) | p | ε (cid:19) sign( p ) p and set L ε := (cid:26) q = ∂φ ε ( p ) ∂p (cid:27) . Then L = (cid:8) q = p (cid:9) is a smooth Lagrangian with the fold tangency singularity to γ , while L ε for any ε > γ . Since we define the deformation at the level offunctions, exactness is automatic, see Figure 5. DANIEL ´ALVAREZ-GAVELA, YAKOV ELIASHBERG, AND DAVID NADLER
Figure 5.
A fold (blue) becomes a ridge (red). Exactness means that the area of theregion bounded by the blue and the red curves is zero when counted with sign.Note that if the deformation is performed close enough to the fold point, then the resulting ridgyLagrangian is transverse to the Lagrangian plane field with respect to which the smooth Lagrangianhad a fold. Hence Theorem 1.4 is an immediate consequence of Theorem 1.7 when γ is homotopic toa Lagrangian plane field transverse to L , see Figure 6. Figure 6.
A fold tangecy can be replaced with an order 1 transverse ridge.Note that the above relation between folds and ridges only holds for order 1 ridges, i.e. stabiliza-tions of the standard 1-dimensional ridge R ⊂ T ∗ R . Higher order ridges carry subtler homotopicalinformation corresponding to the higher singularities Σ k and are necessary to overcome the homotopytheoretic obstruction to the simplification of singularities. Thus Theorem 1.4 shows the best one cando if nothing is known about the homotopy class of γ .1.4. Structure of the article.
We begin our proof of Theorem 1.4 by showing existence of a formalsolution, which is established in Section 2 by working one rank 1 form at a time. The resulting formal
EOMORPHOLOGY OF LAGRANGIAN RIDGES 7 solution is then deformed to an integrable solution in two steps. First, in Section 3 we align the ridgedirections to the homotopy class necessary for integrability. Then in Section 4 we integrate our formalsolution and finish the proof of our main theorem. Finally, in Section 5 we state and prove a versionadapted to Lagrangians with a boundary and corner structure.1.5.
Acknowledgements.
The authors are very grateful to John Pardon and Laura Starkston forhelpful discussions. 2.
Formal solution
Tectonic fields.
We begin by introducing the notion of a tectonic field, which is the formalanalogue of a ridgy Lagrangian. Recall that a polarization of a symplectic vector space V consistsof a pair of transverse linear Lagrangian subspaces τ, ν ⊂ V . For a fixed polarization ( τ, ν ) there isa bijective correspondence between graphical linear Lagrangian subspaces of V (i.e. transverse to ν )and quadratic forms on τ . Indeed, both can be thought of as symmetric linear maps τ → τ ∗ , whereby symmetric we mean equal to its own transpose under the canonical isomorphism τ ∗∗ (cid:39) τ .We will repeatedly go back and forth between the two viewpoints. Note that given two graphicallinear Lagrangian subspaces λ , λ ⊂ V we have dim( λ ∩ λ ) = dim ker( λ − λ ). In particular, λ and λ are transverse if and only if λ − λ is a nonsingular quadratic form on τ . Given a smooth manifold L , for any x ∈ L there is a canonical polarization of T x ( T ∗ L ) given by τ = T x L and ν = T ∗ x L . Hencewe can identify graphical linear Lagrangian subspaces of T x ( T ∗ L ) with quadratic forms on T x L . Viathis identification, graphical Lagrangian plane fields on T ∗ L defined along the zero section L form amodule over C ∞ ( L ). Remark 2.1.
By a Lagrangian plane field on L we mean a field of Lagrangian planes in T ∗ L definedalong the zero section. Similarly, by a field of quadratic forms on L we will always mean a smoothfamily λ x of quadratic forms on T x L , x ∈ L . This is the same as a graphical Lagrangian plane field. Definition 2.2.
Suppose we are given dividing, co-oriented embedded hypersurfaces N , . . . , N k ⊂ L .We assume that the N j are mutually transverse, i.e. each N j is transverse to all possible intersectionsof the other N i , i (cid:54) = j . A tectonic field λ over L with faults along N j is a collection of fields of quadraticforms λ C over the closures C of the components C ⊂ L \ (cid:83) j N j such that there exist non-vanishing1-forms (cid:96) j on T L | N j , j = 1 , . . . , k, with the following property: · for any point point x ∈ N j \ (cid:83) j N i we have λ C + − λ C − = (cid:96) j , where we denote by C ± the components of L \ (cid:83) j N j adjacent to x and where the co-orientationof N j points into C + . Remark 2.3.
The hyperplane fields τ j = ker( (cid:96) j ) are co-oriented by the choice of the defining 1-forms (cid:96) j , but the co-orientation is not determined by the tectonic field λ and co-orientation of N j as we couldreplace (cid:96) j with − (cid:96) j .The hypersurfaces N j are called faults , the connected component of L \ (cid:83) j N j are called plates and the hyperplane fields τ j are called ridge directions . We will moreover demand that the followingtransversality condition is satisfied. DANIEL ´ALVAREZ-GAVELA, YAKOV ELIASHBERG, AND DAVID NADLER
Figure 7.
A tectonic field. The discontinuity of λ along the green arrow is the rank1 form µ j = (cid:96) j corresponding to the hypersurface N j . Note that the hyperplane fields τ j = ker( (cid:96) j ) need not be tangent to the N j . · Along each intersection N j ∩ · · · ∩ N j m the ridge directions τ j s , s = 1 , . . . , m , are transverseto all possible intersections of the other ridge directions τ j r , r (cid:54) = s . Remark 2.4.
Tectonic fields do not form a module over C ∞ ( L ), but they can be multiplied byfunctions which are positive on (cid:83) j N j and can be added when the union of the corresponding collectionsof faults and ridge directions satisfies the transversality conditions. For example this is vacuouslysatisfied when one of the tectonic fields is actually a smooth Lagrangian field.2.2. Formal transversalization.
The main goal of Section 2 is to prove the following transversal-ization result, which is the formal version of our main Theorem 1.4.
Theorem 2.5.
For any Lagrangian field γ there exists a tectonic field ζ such that ζ (cid:116) γ . In fact we will prove the following more general extension result with C -control. Theorem 2.6.
Let γ be a Lagrangian field and let ζ be a tectonic field. For any two disjoint closedsubsets K , K ⊂ L there exists a tectonic field (cid:98) ζ such that the following properties hold. · (cid:98) ζ is C -close to ζ . · (cid:98) ζ (cid:116) γ on Op ( K ) . · (cid:98) ζ = ζ on Op ( K ) . The C -closeness part of the statement means the following. Given a fixed but arbitrary Riemannianmetric on L , for any ε > ζ and (cid:98) ζ is smaller than ε . Note thatTheorem 2.6 implies Theorem 2.5 in its stronger relative form: if ζ (cid:116) γ on Op ( A ) for A ⊂ L a closedsubset, then we can demand that (cid:98) ζ = ζ on Op ( A ) . To see this take K = A and K = L \ Op ( A ).In Section 5 we will prove a version of Theorem 2.6 for the case where L is a manifold with boundaryand corners and γ is adapted to the corner structure. EOMORPHOLOGY OF LAGRANGIAN RIDGES 9
Inductive step.
The key ingredient in the proof of the formal transversalization theorem is thefollowing inductive procedure, in which we only deal with a rank 1 form at a time.
Lemma 2.7.
Let λ, η be smooth fields of quadratic forms on L , with η = α(cid:96) for a field of linear forms (cid:96) and a real valued function α . Let ζ be a tectonic field which is transverse to λ . Then there exists a C -small tectonic field ζ (cid:48) such that ζ + ζ (cid:48) is a tectonic field transverse to λ + η .Proof. Denote by N , . . . , N k the faults, by τ , . . . , τ k the ridge directions and by Q , . . . , Q m the platesof the tectonic field ζ . Note that η has rank ≤ λ + η − ζ has rank ≥ n −
1. Let Σ ⊂ L denote the locus where the rank of λ + η − ζ is exactly n −
1, i.e. Σ = { det( λ + η − ζ ) = 0 } , where hereand below we fix an arbitrary Riemannian metric on L to compute the determinant. Set Σ j = Q j ∩ Σ.Our first goal is to reduce Lemma 2.7 to the case where the following properties hold.(A) Σ j is a smooth codimension 1 submanifold with boundary and corners of Q j .(B) Σ j is transverse to all intersections of the faults N i , i (cid:54) = j .(C) τ = ker( η ) is transverse to all possible intersections of the ridge directions τ j , . . . , τ j m alongthe intersection of N j ∩ · · · ∩ N j m with Σ. Figure 8.
The singular locus Σ ⊂ L .Suppose first that we know Lemma 2.7 to be true when (A) and (B) hold. Let λ, η and ζ as inthe statement of the lemma. By genericity of transversality we can find a C -small smooth field ofquadratic forms ϕ such that the hypersurfaces det( λ + ϕ + η − ζ ) = 0 are transversely cut out oneach plate of ζ and are transverse to all intersections of the faults. Then by assumption we can applyLemma 2.7 with λ + ϕ instead of λ (which is still transverse to ζ since ϕ is C -small), obtaining a C -small tectonic field ζ (cid:48) such that ζ + ζ (cid:48) is transverse to λ + ϕ + η . Hence ζ (cid:48)(cid:48) = ζ (cid:48) − ϕ is a C -smalltectonic field such that ζ + ζ (cid:48)(cid:48) is transverse to λ + η . It therefore suffices to prove Lemma 2.7 underthe assumption that (A) and (B) hold.Next, suppose that we know Lemma 2.7 to be true when (A), (B) and (C) hold. Let λ, η and ζ be as in the statement of the lemma and assume that (A) and (B) hold. Note that the transversalitycondition in (C) has codimension ≥ n − m and the intersection has codimension m + 1. Thereforeby genericity we can find a smooth field of rank ≤ (cid:101) η = (cid:101) α (cid:101) (cid:96) which is C -close to η and suchthat condition (C) holds if we replace η by (cid:101) η . Then by assumption we can apply Lemma 2.7 with (cid:101) η instead of η . The output is a C -small tectonic field ζ (cid:48) such that ζ + ζ (cid:48) is transverse to λ + (cid:101) η . Hence ζ (cid:48)(cid:48) = ζ (cid:48) + η − (cid:101) η is a C -small tectonic field such that ζ + ζ (cid:48)(cid:48) is transverse to λ + η . It therefore sufficesto prove Lemma 2.7 under the assumption that (A), (B) and (C) hold.We now proceed to prove Lemma 2.7 under the assumption that (A), (B) and (C) hold. Extend eachΣ j to a closed hypersurface (cid:98) Σ j ⊂ L , so that the collection N , . . . , N k , (cid:98) Σ , . . . , (cid:98) Σ k forms a transversesystem of hypersurfaces. This is possible because Σ j is defined by the equation det( λ + η − ζ ) | Q j = 0, soit suffices to extend the function ∆ j = det( λ + η − ζ ) | Q j to L . A generic extension provides the desiredtransversality. Note that Σ j is canonically co-oriented by the direction in which ∆ j is increasing andhence we can extend this co-orientation to (cid:98) Σ j using the extension of ∆ j .We will construct the tectonic field ζ (cid:48) inductively, working plate by plate. We begin with the firstplate Q . Fix a tubular neighborhood U = (cid:98) Σ × ( − ,
1) of (cid:98) Σ with coordinates ( x, u ) so that ∂ u agrees with the specified co-orientation of Σ Write η = α(cid:96) as in the statement of the lemma. Fix acutoff function ψ : [0 , → [0 ,
1] such that ψ = 1 near 0 and ψ = 0 near 1. Since λ − ζ is nonsingular,the restriction of λ + η − ζ to τ = ker( η ) is nonsingular. Let δ ∈ {± } be the sign of its determinanton Q . Pick ε > ζ ε given by ζ ε = − δψ ε ( u ) (cid:96) , ψ ε ( u ) = ε sign( u ) ψ ( | u | /ε ) . Remark 2.8.
Note that ζ ε is a tectonic field with fault (cid:98) Σ and C -norm proportional to ε . Claim 2.9. If ε is chosen small enough, then ζ + ζ ε is transverse to λ + η on Q .Proof of Claim 2.9. Fix an arbitrary point in Σ . Choose a local frame κ , . . . , κ n − of τ ∗ . The n ( n + 1) / (cid:96) , κ j , ( κ j + (cid:96) ) , ( κ j + κ k ) , i, j = 1 , . . . , n − i < j form a local frame. Byconsidering the symmetric matrix which corresponds to this frame we can compute the determinantdet( λ + η − ζ ) to be of the form A ( x, u ) f ( x, u ) + B ( x, u ), x ∈ Σ , u ∈ ( − , A is anon-vanishing function, namely the complementary minor corresponding to the forms ( κ j + κ i ) .That Σ is cut out transversely means ∂ u det( λ + η − ζ ) > u = 0, so by rescaling if necessarywe may assume that this folds for all u ∈ ( − ,
1) and hence det( λ + η − ζ ) is a strictly increasingfunction of u in the tubular neighborhood U , see Figure 9. Figure 9.
We may assume ∂ u det( λ + η − ζ ) > u ∈ ( − , EOMORPHOLOGY OF LAGRANGIAN RIDGES 11
Moreover, with respect to that same frame we can write det( λ + η − ζ − ζ ε ) in the form A ( x, u )( f ( x, u ) ± ψ ε ( u )) + B ( x, u ), where ± is the sign δ of A . Hence we havedet( λ + η − ζ − ζ ε ) = det( λ + η − ζ ) + | A ( x, u ) | ψ ε ( u ) , which is bounded away from zero, see Figures 10 and 11. (cid:3) Figure 10.
The function | A ( x, u ) | ψ ε ( u ), which has a discontinuity at u = 0. Figure 11.
The function det( λ + η − ζ − ζ ε ), which is the sum of the functionsdet( λ + η − ζ ) and | A ( x, u ) | ψ ε ( u ) illustrated in Figures 9 and 10.Before proceeding with the inductive process on the next plate we examine the new singular locusΣ ε j = det( λ + η − ζ − ζ ε ) = 0 on Q j for j >
1. If Σ j ∩ (cid:98) Σ = ∅ , then for ε small enough this singularlocus is just Σ j and nothing changes. Suppose however that P j = Σ j ∩ (cid:98) Σ (cid:54) = ∅ . After the additionof ζ ε to ζ the hypersurface (cid:98) Σ becomes a fault, which causes Σ j to disconnect along P j . The crucialobservation is the following. Claim 2.10.
The new singular locus Σ ε j is displaced in opposite directions on each side of the fault (cid:98) Σ and hence intersects (cid:98) Σ in two disjoint parallel copies of P j in (cid:98) Σ . Proof of Claim 2.10.
To verify the claim, choose a tubular neighborhood U j = Σ j × R of Σ j in Q j with coordinates ( x, u j ) such that ∂ u j agrees with the specified co-orientation of Σ j . Together withthe coordinate u of the tubular neighborhood U of (cid:98) Σ this gives us coordinates ( x, u , u j ) of atubular neighborhood of P j in Q j . Near P j we can write det( λ + η − ζ − ζ ε ) as before in the form A ( y, u , u j )( f ( y, u , u j ) ± ψ ε ( u )) + B ( y, u , u j ), y ∈ P j , u , u j ∈ ( − , A is a nonvanishingfunction and ± is the sign δ of A . In terms of our previous notation x = ( y, u j ). The hypersurface Σ ε j is cut out by the equation det( λ + η − ζ − ζ ε ) = 0which is equivalent to det( λ + η − ζ ) = −| A ( y, u , u j ) | ψ ε ( u ) . That Σ j is cut out transversely means that ∂ u j det( λ + η − ζ ) > j , so we may assume thatthis condition holds in the tubular neighborhood. Solving for u j , the implicit function theorem impliesthat on each side of (cid:98) Σ the above equation cuts out a smooth hypersurface which is graphical over Σ j .Moreover, the intersection of these hypersurfaces with (cid:98) Σ = { u = 0 } is given by the equationsdet( λ + η − ζ ) = | A ( y, , u j ) | ε , det( λ + η − ζ ) = −| A ( y, , u j ) | ε , coming from u < u > λ + η − ζ ) is a strictly increasing function of u j on U ∩ U j which vanishes at u j = 0, these solutions have strictly positive and strictly negative u j coordinates respectively. Let u + j ( y ) > u − j ( y ) < y ∈ P j .Then (cid:83) y y × × [ u − j ( y ) , u + j ( y )] is a tubular neighborhood of P j = (cid:83) j y × × (cid:98) Σ = { u = 0 } withboundary Σ ε j ∩ (cid:98) Σ , which was to be proved. (cid:3) We now reconnect Σ ε j back together along P j in the ( u , u j ) plane by parametrically closing up thefamily of broken curves cut out by det( λ + η − ζ − ζ ε ) = 0. For each fixed y ∈ P j we know that theinterval 0 × [ u − j ( y ) , u + j ( y )] is disjoint from the curve T y = Σ ε j ∩ ( y × ( − , u × ( − , u j ) ⊂ ( − , except at its endpoints { (0 , u ± j ( y )) } = ∂T y . Moreover, at these boundary points T y is transverse tothe vertical axis u = 0, see Figure 13.Consider a family of ‘S’ shaped curves (in fact, backwards ‘S’) S y ∈ ( − , u × ( − , u j whichreconnect the point (0 , u − j ( y )) with the point (0 , u + j (0)) in the complement of T y so that (cid:98) T y = T y ∪ S y is a smooth family of boundaryless curves, see Figure 14. This produces a smooth extension of Σ ε j in Q j to a smooth hypersurface which for generic S y satisfies the required transversality conditions withrespect to the faults.At the boundary of Q j this new hypersurface does not match up with the old extension (cid:98) Σ j of Σ j ,but there is a homotopically canonical way to let the ‘S’ curves die out just outside of Q j so that theydo match up away from Op ( Q j ), see Figures 15 and 16. Hence the reconnected Σ ε j extends to a closedembedded hypersurface in L which abusing notation we denote by the same symbol (cid:98) Σ j , and which wemay assume satisfies the required transversality conditions with respect to the faults.We now continue on to plate Q . Replace the tubular neighborhood U of the old (cid:98) Σ with a tubularneighborhood of the new (cid:98) Σ . Again we have coordinates ( x, u ). Pick ε (cid:28) ε and define ζ ε = ± ψ ε ( u ) (cid:96) , EOMORPHOLOGY OF LAGRANGIAN RIDGES 13
Figure 12.
The singular locus changes after the first step of the inductive process.
Figure 13.
The discontinuity of Σ ε j along (cid:98) Σ = { u = 0 } .where the sign is determined as above. If ε is chosen small enough, then ζ + ζ ε + ζ ε is still transverseto λ + η on Q , since transversality is an open condition. For the same reason, on the family of ‘S’shaped curves which we added to reconnect Σ ε we still have transversality. Along Σ ε itself we alsoachieve transversality by the computation carried out in the first step. Hence ζ + ζ ε + ζ ε is transverseto λ + η on Q . We have thus achieved transversality on Q ∪ Q .Observe that on Q j , j >
2, the new singular locus det( λ + η − ζ − ζ ε − ζ ε ) = 0 will split alongthe intersection of the old singular locus with (cid:98) Σ . We proceed just like before, reconnecting the newsingular locus using ‘S’ shaped curves and letting them die when they cross a fault (which now inaddition to N , . . . , N k also includes (cid:98) Σ ). We can then keep on going with the inductive process until Figure 14.
The discontinuity from Figure 13 can be stitched up with an ‘S’ shapedcurve (because of our conventions we actually see a backwards ‘S’).
Figure 15.
To let the ‘S’ shaped curves die out first push the lobes of the ‘S’ inopposite directions so that the curves become graphical over the u axis. Figure 16.
Then one can linearly interpolate along the u j direction until the curvescollapse to the u axis.we get to the last stage, which results in a C -small tectonic field ζ = ζ ε + · · · + ζ ε m m satisfying therequired properties. This completes the proof. (cid:3) Remark 2.11.
The proof of Lemma 2.7 automatically gives the relative form: if η = 0 on Op ( A ) for A ⊂ L a closed subset, then we can demand that ζ (cid:48) = 0 on Op ( A ). EOMORPHOLOGY OF LAGRANGIAN RIDGES 15
Extension step.
In this section we use the inductive lemma 2.7 to prove the formal transver-salization theorem 2.6. The main point is that any quadratic form is a sum of rank 1 forms. First weprove a local version of the result, which we will then globalize.
Lemma 2.12.
Let γ be a smooth field of quadratic forms on the open unit ball B ⊂ R n , let ξ be atectonic field on B and let (cid:101) B ⊂ B be a smaller ball whose closure is contained in B . There exists a C -small tectonic field ζ such that ξ + ζ is a tectonic field transverse to γ on the closure of (cid:101) B and suchthat ζ = 0 near ∂B .Proof. Fix a smooth field of quadratic forms σ on B which is transverse to ξ . This is always possible,for instance we can take σ to be almost vertical. Write the difference γ − σ as a sum α (cid:96) + · · · + α N (cid:96) ,where the (cid:96) j are smooth fields of linear forms. For example we can use the linear forms X i + X j , where1 ≤ i ≤ j ≤ n . Then N = n ( n + 1) / X i X j = (cid:0) ( X i + X j ) − X i − X j (cid:1) ensuresthat such a decomposition exists. Let (cid:101) α j be a function which is equal to α j on the closure of (cid:101) B andis equal to zero near ∂B .We begin by applying Lemma 2.7 to λ = σ , η = (cid:101) α (cid:96) and ζ = ξ . We obtain a C -small tectonic field ζ such that σ + (cid:101) α (cid:96) − ξ − ζ is nonsingular. Next we apply Lemma 2.7 to λ = σ + (cid:101) α (cid:96) , η = (cid:101) α (cid:96) , and ζ = ξ + ζ . We obtain a C -small tectonic field ζ such that σ + (cid:101) α (cid:96) + (cid:101) α (cid:96) − ξ − ζ − ζ is nonsingular.We repeat this process inductively. When at the last step we apply Lemma 2.7, we obtain a C -smalltectonic field ζ = ζ + · · · + ζ N such that σ + (cid:80) Nj =1 (cid:101) α j (cid:96) j − ξ − ζ is nonsingular. In particular γ − ξ − ζ is nonsingular on (cid:101) B . Moreover, since ζ j = 0 near ∂B for each j = 1 , . . . , N , the same is true of ζ . (cid:3) Proof of Theorem 2.6.
For
C > C = {| det( γ ) | < C } ⊂ L . Choose C sufficiently large sothat ζ (cid:116) γ outside of Ω C . Let B , . . . , B m be a cover of Ω C ∩ K by open balls B j ⊂ Ω C \ K .In particular γ is graphical on each B j , hence can be thought of as a field of quadratic forms. Takeslightly smaller balls (cid:101) B j whose closure is contained in B j and such that the collection (cid:101) B , . . . , (cid:101) B m stillcovers Ω C ∩ K . We will construct the desired ζ inductively, one B j at a time.First apply Lemma 2.12 on B to γ and ξ = ζ , producing a C -small tectonic field ζ such that ζ = 0 near ∂B and such that ζ + ζ is transverse to γ on (cid:101) B . Suppose that we have constructed C -small tectonic fields ζ , . . . , ζ k supported on (cid:83) kj =1 B j such that ζ + (cid:80) kj =1 ζ j is transverse to γ on (cid:83) kj =1 (cid:101) B j . Apply Lemma 2.12 on B k +1 to γ and ξ = ζ + (cid:80) kj =1 ζ k to obtain a C -small tectonic field ζ k +1 such that ζ k +1 = 0 near ∂B k +1 and such that ζ + (cid:80) k +1 j =1 ζ j is transverse to γ on (cid:101) B k +1 . Sincetransversality is an open condition, by taking ζ k +1 to be sufficiently C -small we can ensure that ζ + (cid:80) k +1 j =1 ζ j is also transverse to γ on (cid:83) kj =1 (cid:101) B j . Hence ζ + (cid:80) k +1 j =1 ζ j is transverse to γ on (cid:83) k +1 j =1 (cid:101) B j andthe inductive procedure can continue.At the last stage of the inductive procedure we obtain a tectonic field ζ (cid:48) = ζ + (cid:80) mj =1 ζ j which is C -close to ζ , such that ζ (cid:48) (cid:116) γ on Ω C ∩ K and such that ζ (cid:48) = ζ outside of Ω C \ K . If ζ (cid:48) is sufficiently C -close to ζ then ζ (cid:48) (cid:116) γ also on Ω C \ Ω C , because we chose C > ζ (cid:116) γ in that region.Hence ζ (cid:48) (cid:116) γ on K and ζ (cid:48) = ζ on K . This completes the proof. (cid:3) Alignment of ridges
Aligned transversalization.
Let L be a smooth manifold and let Λ ⊂ T ∗ L be a ridgy La-grangian. Denote by R ⊂ Λ the ridge locus and let Λ \ R = P ∪ · · · ∪ P k be the decomposition into connected components (each of which is a smooth manifold with corners). Suppose that Λ is graphicalover L and denote by Q j the image of P j under the projection T ∗ L → L . Then Λ is given over Q j asthe graph of a closed 1-form β j . Assume for simplicity that Λ is exact, so that we can write β j = dh j for h j : Q j → R a smooth function. Set λ j = Hess( h j ) on Q j , where we use an auxiliary Riemannianmetric on L to write down the Hessian. Note that the λ j assemble to a tectonic field λ with plates Q j . Definition 3.1.
When a tectonic field λ arises in this way we say that it is integrable .A tectonic field provides the infinitesimal data to integrate a graphical ridgy Lagrangian. However,for the integration to be possible in a neighborhood of the fault locus we need the additional conditionthat the ridges are aligned with the faults. Definition 3.2.
We say that a tectonic field λ is aligned if τ j = T N j for every fault N j and corre-sponding ridge direction τ j , see Figure 17. Figure 17.
An aligned tectonic field.For a Lagrangian plane field γ in T ∗ L , the problem under consideration is to deform the zero section L by a ridgy isotopy so that it becomes transverse to γ . In the previous section we found a formalsolution to this transversalization problem, i.e. a tectonic field λ such that λ (cid:116) γ . In this sectionwe take a step towards integrability by upgrading our formal solution to an aligned solution. Moreprecisely, we have the following aligned version of Theorem 2.5. Theorem 3.3.
For any Lagrangian field γ there exists an aligned tectonic field ζ such that ζ (cid:116) (cid:98) γ for (cid:98) γ a Lagrangian field homotopic to γ . Note that we gain alignment of the tectonic field (cid:98) ζ at the cost of deforming the Lagrangian field γ to a homotopic field (cid:98) γ . Nevertheless, in the next section we show that it is possible to integratethe aligned solution (cid:98) ζ produced by Theorem 3.3 to obtain a ridgy Lagrangian which after an ambientHamiltonian isotopy is transverse to γ itself, thus proving our main result Theorem 1.4.The rest of the present section is devoted to the proof of Theorem 3.3. In fact, we prove below thefollowing more general extension result, which is the aligned analogue of Theorem 2.6. EOMORPHOLOGY OF LAGRANGIAN RIDGES 17
Theorem 3.4.
Let γ be a Lagrangian field and let ζ be an aligned tectonic field. For any two disjointclosed subsets K , K ⊂ L , there exists an aligned tectonic field (cid:98) ζ and a Lagrangian field (cid:98) γ homotopicto γ such that the following properties hold. · (cid:98) ζ is C -close to ζ . · (cid:98) ζ (cid:116) (cid:98) γ on Op ( K ) . · (cid:98) ζ = ζ on Op ( K ) .Moreover, we can assume that the homotopy between γ and (cid:98) γ is fixed on Op ( K ) . Homotopically aligned transversalization.
It will be useful to also consider the homotopicalversion of definition 3.2.
Definition 3.5.
We say that a tectonic field is homotopically aligned if there exists a homotopy oflinear isomorphisms Ψ t : T x L → T x L , x ∈ L , such that Ψ = id T x L and Ψ ( τ j ) = T N j .We call Ψ t the homotopical alignment and consider it part of the defining data of a homotopicallyaligned tectonic field. Note that Theorem 3.4 follows immediately from the analogous homotopicallyaligned statement. Theorem 3.6.
Let γ be a Lagrangian field and let ζ be a homotopically aligned tectonic field. Forany two disjoint closed subsets K , K ⊂ L , there exists a homotopically aligned tectonic field (cid:98) ζ and aLagrangian field (cid:98) γ homotopic to γ such that the following properties hold. · (cid:98) ζ is C -close to ζ . · (cid:98) ζ (cid:116) (cid:98) γ on Op ( K ) . · (cid:98) ζ = ζ on Op ( K ) .Moreover, we can assume that the homotopy between γ and (cid:98) γ and is fixed on Op ( K ) and that thehomotopical alignment (cid:98) Ψ t for (cid:98) ζ agrees with the homotopical alignment Ψ t of ζ on Op ( K ) .Proof of Theorem 3.4 assuming Theorem 3.6. We apply Theorem 3.6 in the case where Ψ t = id T x L .The output is (cid:98) γ and (cid:98) ζ , with homotopical alignment (cid:98) Ψ t . Let Φ t be the unique homotopy of linearsymplectic isomorphisms of T x ( T ∗ L ), x ∈ L , lifting the linear isomorphism (cid:98) Ψ t of T x L and fixing thecotangent fibre T ∗ x L . Then taking the aligned tectonic field Φ t ( (cid:98) ζ ) and concatenating the homotopybetween γ and (cid:98) γ with the homotopy Φ t ( (cid:98) γ ) we obtain the conclusion of Theorem 3.4. (cid:3) Therefore we have reduced the aligned formal transversalization theorem 3.4 to the homotopicallyaligned formal transversalization theorem 3.6. To prove the homotopically aligned formal transver-salization theorem 3.6 we will take the tectonic field (cid:98) ζ produced by the formal transversalizationtheorem 2.6, which may not be homotopically aligned, and perform a local modification to adjust thehomotopical condition obstructing alignment.3.3. Formal ridges.
We begin by introducing the notion of a formal ridge.
Definition 3.7. A formal k -ridge over an n -dimensional vector space V is the data of a quadraticform λ on V and an unordered collection of k rank 1 forms µ , . . . , µ k on V such that each of thehyperplanes H j = ker( µ j ) is transverse to all finite intersections of the other H i , i (cid:54) = j . Let λ be a tectonic field of a smooth n -dimensional manifold L . A point at which exactly k ofthe faults of λ meet determines a formal k -ridge. Indeed, the 2 k Lagrangian planes correspondingto the tectonic field λ at the point x are given by λ + (cid:80) j ∈ J µ j , where J ranges over subsets of { , , . . . , k } and λ is the plane corresponding to the quadrant which is initial with respect to thefault co-orientations. We get a formal k -ridge by considering λ together with the µ i . Note that withthis choice of λ we have that each µ i is the square of a linear form (cid:96) j . However, we could also take adifferent plane in λ as our λ and replace each of the corresponding µ i with − µ i . Then we get anotherformal k -ridge which has the same collection of 2 k Lagrangian planes associated to it. Note that thereis no canonical ordering on the forms µ i .Denote [ λ ] = span( λ ), which is a field of coisotropic subspaces of T ( T ∗ L ) | L . The dimension of[ λ ] varies and equal to n + k along the formal k -ridge locus. Given a Lagrangian field η along L wedenote by η [ λ ] the symplectic reduction of η ∩ [ λ ] in [ λ ] / [ λ ] ⊥ ω . Note that the transversality of η to λ is equivalent to transversality of η to [ λ ] and transversality of η [ λ ] to λ [ λ ] . Here λ [ λ ] consists of thecollection of symplectic reductions of the Lagrangian planes of λ . Lemma 3.8.
The projection η (cid:55)→ η [ λ ] defined on the space of Lagrangian fields transverse to [ λ ] hascontractible fibers.Proof. Consider the fibre over a formal k -ridge point. We factor the projection η (cid:55)→ η [ λ ] as the map η → η ∩ [ λ ] and η ∩ [ λ ] → η [ λ ] . The second map is defined on the space of ( n − k )-dimensional isotropicsubspaces of [ λ ]. Let τ ⊂ [ λ ] be an ( n − k )-dimensional isotropic subspace. Then the fibre of thesecond map over the reduction of τ can be identified with the space of linear maps τ → [ λ ] ⊥ ω , henceis contractible. For the first map, take an ( n − k )-dimensional isotropic subspace τ ⊂ [ λ ] and let η be a Lagrangian plane whose intersection with [ λ ] is τ . Then the fibre of the first map over τ can beidentified with the space of quadratic forms on η/τ , hence is also contractible. (cid:3) For an inductive argument below it will be convenient to consider formal k -ridges with a fixedordering of the rank 1 forms µ j . We call this an ordered formal k -ridge . Lemma 3.9.
Let λ and λ be two ordered formal k -ridges on V . There exists a linear symplecticisomorphism Φ of V × V ∗ which sends λ to λ . Moreover Φ is determined up to contractible choiceby its restriction to [ λ (cid:48) ] , where λ (cid:48) is the ordered formal ( k − -ridge obtained from λ by forgetting µ k . Remark 3.10.
That Φ sends λ to λ means that the image of the Lagrangian plane λ + (cid:80) j ∈ J µ j by Φ is λ + (cid:80) j ∈ J µ j for every J ⊂ { , . . . , k } . Proof.
We argue by induction on k = 0 , , . . . , n . For k = 0 the existence part follows from thefact that the symplectic group acts transitively on the Lagrangian Grasmannian. The uniquenessfollows from the fact that a linear symplectic isomorphism is determined by its restriction to a pair oftransverse Lagrangian planes, together with the fact that the space of Lagrangian planes transverse toa fixed Lagrangian plane is contractible. We spell out the details of an explicit argument which willbe easily adaptable to the case k >
0. First we reduce to the case V = R n , λ = λ = R n ⊂ C n andΦ | R n = id R n . Write the symplectic matrix M ∈ Sp(2 n ) representing Φ in the block form corresponding EOMORPHOLOGY OF LAGRANGIAN RIDGES 19 to C n = R n × i R n M = A BC D . Then Φ | R n = id R n is equivalent to A = I n and C = 0. That M is symplectic means M T Ω M = Ω forΩ = I n − I n , where I n is the n by n identity matrix. It follows that D = I n and B T = B . Hence Φ is uniquelydetermined up to the contractible choice of the symmetric matrix B . This completes the base case.For the inductive step, observe that as before it suffices to consider the case λ = R n ⊂ C n .Furthermore, up a linear change of coordinates in R n we may assume that the kernel of µ j is thecoordinate hyperplane { q j = 0 } ⊂ R n . Let Φ be the linear symplectic isomorphism obtained byapplying the inductive hypothesis to the ordered formal ( k − λ and λ after forgetting µ k and µ k respectively. Then by pulling λ back by Φ we reduce to the case λ = λ and µ i = µ i for i < k .In this case have [ λ (cid:48) ] = { p j = 0 , j ≥ k − } and [ λ ] = [ λ ] = { p j = 0 , j ≥ k } . Note that the productof a horizontal shear of the symplectic subspace ( q k , p k ) and the identity on the complementary R n − fixes [ λ (cid:48) ]. Since the group of horizontal shears ( x, y ) (cid:55)→ ( x, y + ax ), a ∈ R , acts transitively on thespace of lines in R transverse to the horizontal axis { y = 0 } ⊂ R , we can find a linear symplecticisomorphism which is the identity on [ λ ] and takes λ to λ . This proves the existence part.For the uniqueness part it suffices to show that a linear symplectic isomorphism Φ of R n whichrestricts to the identity on [ λ ] = { p j = 0 , j ≥ k } is unique up to contractible choice. With the samenotation as above, write the symmetric matrix B in block form B = X YY T W . Here X is a k by k matrix and W is an ( n − k ) by ( n − k ) matrix, which are both symmetric. Theconditions on Φ are equivalent to X = 0 and Y = 0 Hence Φ is uniquely determined up to thecontractible choice of the symmetric matrix W . (cid:3) The model.
Consider a tectonic field λ on L . Let N ⊂ L be one of its faults, which bounds adomain U such that outside of U the field λ differs by adding a rank one 1 quadratic form µ along N . Let Ω ⊂ U be a domain with boundary and corners, where we decompose ∂ Ω = F ∪ F for F and F smooth so that F = ∂ Ω ∩ N , ∂F = F ∩ N and the corner is precisely ∂ Ω = F ∩ F . Let ν be a field of rank 1 quadratic forms on Ω such that ν = µ near F . Consider the field (cid:98) λ which isdefined to be λ outside of Ω and λ + ν on Ω. Ater smoothing corners, (cid:98) λ becomes a tectonic field with (cid:98) N = ( N \ F ) ∪ F as one of its faults, see Figure 18.We now apply this construction in a standard local model. Consider a constant tectonic field λ over R n ⊂ T ∗ R n with faults along the first k coordinate hyperplanes Q j := { q j = 0 } , j = 1 , . . . , k < n .That λ is constant means that the discontinuities of λ across a fault Q j are given by constant rank 1quadratic forms µ j . Figure 18.
The modification λ (cid:55)→ (cid:98) λ .Take a sphere Σ ⊂ R n of radius 1 centered at a point a with coordinates q k = 2, q j = 0 , j (cid:54) = k .Denote A = { ≤ q k ≤ q j = 0 , j (cid:54) = k } , and denote by Ω a neighborhood of A ∪ Σ in { q k ≥ } .Thus ∂ Ω = ( F ∪ F ) ∪ F , where F = ∂ Ω ∩ Q k , F is a ( n − Q k and F is a( n − Q k . Figure 19.
The standard model in the case n = 3, k = 2.We use the notation Q J for the fault intersections: Q J = (cid:84) j ∈ J Q j , J ⊂ K for K = { , . . . , k } . Wealso enumerate the quadrants on R n by multi-indices I ⊂ K , namely C I = { q i ≥ , i ∈ I ; q j ≤ , j ∈ K \ I } , and write C JI = C I ∩ Q J . Note that on C JI the tectonic field λ is a fixed formal r -ridge λ JI ,where r ≤ k is the cardinality of J .Let ν be a field of rank 1 quadratic forms over Ω which agrees with µ k over F . We will additionallyassume that near Q J the field ν is independent of coordinates q j , j ∈ J . Proposition 3.11.
Suppose λ is transverse to a constant Lagrangian distribution γ . Then (cid:98) λ is trans-verse to a distribution (cid:98) γ which is homotopic to γ by a deformation fixed outside of a compact set. EOMORPHOLOGY OF LAGRANGIAN RIDGES 21
Proof.
We construct (cid:98) γ inductively over the dimension n − | J | of the strata C JI intersecting Σ. Thesmallest dimensional stratum is C = C ,...,k − ,...,k . For every point x ∈ C ∩ Ω consider a linear symplecticisomorphism Φ ,x of R n which sends (cid:98) λ = λ to (cid:98) λ x and is the identity on [ λ (cid:48) ]. Here 0 denotes theorigin in R n , λ x is the formal k -ridge of λ at x and λ (cid:48) is the formal ( k − λ byforgetting µ k . According to Lemma 3.9 there exists a homotopically unique continuous family of suchisomorphisms. We can assume that Φ ,x is the identity if (cid:98) λ x = (cid:98) λ . Let us define (cid:98) γ x = Φ ,x ( γ ), whichis transverse to (cid:98) λ . Using Lemma 3.8 we can extend (cid:98) γ to C keeping fixed its reduction (cid:98) γ [ λ (cid:48) ] = γ [ λ (cid:48) ] and making it equal γ outside a neighborhood of Ω. Again applying Lemma 3.8 we conclude that theconstructed field (cid:98) γ is homotopic to γ via a homotopy γ t with a fixed reduction γ [ λ (cid:48) ] t .Next, we extend (cid:98) γ to a neighborhood of C = C { ,...,k − }{ ,...,k } so that it is independent of the coordinates q j , j ≤ k −
1. For any stratum C JI of codimension k − C we choose a point y ∈ C JI ∩ Ω ina neighborhood U of C where (cid:98) γ is already defined. We note that in this neighborhood there exists afamily of linear isomorphisms Φ y,x fixing [ λ IJ ] which maps (cid:98) λ y to (cid:98) λ x and (cid:98) γ y to (cid:98) γ x , x ∈ C JI ∩ Ω ∩ U . Weextend the family to all x ∈ C JI ∩ Ω and define (cid:98) γ x := Φ y,x ( (cid:98) γ y ). Next we extend it to C JI keeping fixedits reduction (cid:98) γ [ λ IJ ] , and making it equal γ outside a neighborhood of Ω. The same lemma implies thatthe constructed field (cid:98) γ is homotopic to γ via a homotopy γ t with a fixed reduction γ [ λ IJ ] t . Continuingthis process we construct the required distribution (cid:98) γ . (cid:3) Changing the homotopy class of the ridge directions.
Finally we show how the local modelconstructed above can be used to prove Theorem 3.6.
Proposition 3.12.
Let λ be a tectonic field over a manifold L which is transverse to a Lagrangian dis-tribution γ . Then there exists a homopically aligned tectonic field (cid:98) λ which is transverse to a Lagrangiandistribution (cid:98) γ homotopic to γ .Proof. We align the ridge directions inductively over the strata of the fault locus of λ . In fact wefirst refine the stratification so that (cid:83) j N j = (cid:83) n − k =0 ∆ k for ∆ k a union of k -dimensional simplices suchthat each simplex of ∆ k is contained in the locus where exactly r of the faults N j intersect for some r ≤ n − k (which depends on the simplex).Begin by defining the homotopical alignment Ψ t in an arbitrary way at the 0-simplices. For theinductive step, suppose that the homotopical alignment is defined on the ( k − C be a k -simplex in ∆ k where exactly r ≤ n − k of the faults intersect. We must extend thehomotopical alignment from ∂C (cid:39) S k − to C (cid:39) D k . Since the ridge directions are co-orientable, byconsidering the vectors normal to the ridge directions we can equivalently think in terms of the Stiefelmanifold of r -frames V r,n in R n . Since π k ( V s,n ) = 0 for s < n − k we can always homotopically alignthe ridge directions if r < n − k , and if r = n − k we can align all but one. The obstruction to aligningthat last ridge direction lies in π k ( V n − k,n ), which is Z if n − k is even or k = 1 and Z / n − k is oddand k >
1. We claim that one can change this obstruction by ± ν = c(cid:96) in this proposition in such a way that the hyperplane τ = { (cid:96) = 0 } is tangent to Σ with the center of the sphere on a ( k − C (cid:48) adjacent to C . Then we create a new spherical fault with its ridge aligned with its tangent plane fieldand we change the homotopy class of the ridge field on the component C by ± C (cid:48) , see Figure 20. Figure 20.
The local modification in the case n = 2, k = 1. Note that the homotopyclass of the line field changes by ± (cid:98) λ which according to Proposition 3.11 istransverse to a Lagrangian distribution homotopic to γ . (cid:3) Remark 3.13.
From the proof we see that Proposition 3.12 holds in relative form. This meansthat if there exists a homotopy of linear isomorphisms Ψ t : T x L → T x L such that Ψ = id T x L andΨ ( τ j ) = T N j on Op ( A ) for A ⊂ L closed, then we can demand that (cid:98) λ = λ on Op ( A ) and that thehomotopical alignment of (cid:98) λ agrees with Ψ t on Op ( A ). Moreover, we can demand that the homotopyof γ is constant on Op ( A ). Proof of Theorem 3.6.
Consider the tectonic field (cid:98) ζ which is produced by the formal transversalizationTheorem 2.6. Then by applying the relative form of Proposition 3.12 to (cid:98) ζ and γ with A = K weobtain the desired homotopically aligned tectonic field. (cid:3) Integrable solution
Holonomic approximation of ridges.
We now turn to the proof of our main theorem 1.4. Ourfirst task will be to solve the transversalization problem near the ridge locus, where the homotopicalinformation is concentrated. Since the ridge locus is a stratified subset of codimension 1, we canapply the method of holonomic approximation. It will be convenient to use the notion of a tangentialrotation. Recall that the Gauss map G Λ of a Lagrangian submanifold Λ of a symplectic manifold( M, ω ) is the map which to each q ∈ L assigns the linear Lagrangian subspace G Λ ( q ) = T q Λ ⊂ T q M . Definition 4.1.
A tangential rotation of a Lagrangian submanifold Λ of a symplectic manifold (
M, ω )is a homotopy of Lagrangian plane fields G t ( q ) ⊂ T q M along L , starting at the Gauss map G = G Λ .The above definition makes sense even when Λ is singular, for example ridgy, though in this case G Λ is only piecewise continuous. However, in this case we will always assume that the tangential rotation G t is constant in a neighborhood of the singular locus. The goal of the present section is to establishthe following proposition. EOMORPHOLOGY OF LAGRANGIAN RIDGES 23
Proposition 4.2.
Let γ be a Lagrangian field on L . There exists a C -small ridgy isotopy L t ⊂ T ∗ L of L = L such that L (cid:116) γ in a neighborhood of the ridge locus and such that there exists a tangentialrotation G t of L , fixed in that same neighborhood, for which G (cid:116) γ everywhere. Remark 4.3.
The relative version is as follows: if L (cid:116) γ on Op ( A ) for A ⊂ L a closed subset, thenwe can demand that L t = L and G t = G L on Op ( A ).As a first step towards Proposition 4.2 we have the following lemma. Lemma 4.4.
There exists a C -small integrable tectonic field (cid:101) ζ on L which is transverse to a La-grangian field (cid:101) γ homotopic to γ .Proof. We begin by invoking Theorem 3.3, which produces an aligned tectonic field ζ on L such that ζ (cid:116) (cid:98) γ for (cid:98) γ a Lagrangian distribution homotopic to γ . The tectonic field ζ jumps discontinuously alonga fault N j by a family of rank 1-forms µ j . Since ker( µ j ) = T N j , we can write µ j ( x ) = f j ( x ) du j where( x, u j ) ∈ N j × ( − ε, ε ) are tubular neighborhood coordinates for N j = N j × L and f j : N j → R is asmooth function. By reversing the orientation of ( − ε, ε ) if necessary we may assume that ζ +1 = ζ − + η j ,where ζ +1 and ζ − are the extensions of ζ | { u j < } and ζ | { u j > } to N j respectively. Consider the function h j ( x, u j ) = 12 ψ ( u j ) f j ( x ) u j , where ψ : ( − ε, ε ) → [0 ,
1] is a cutoff function such that ψ = 1 near 0 and ψ = 0 near ± ε . Hencealong N j we have Hess( h j ) = µ j . Note that h j is compactly supported in the tubular neighborhood N j × ( − ε, ε ). Consider next the function r j ( x, u j ) = h j ( x, u j ) u j ≥ , − h j ( x, u j ) u j ≤ , which also has compact support in the tubular neighborhood N j × ( − ε, ε ). Set r = (cid:80) j r j : L → R , apiecewise C function which generates an integrable tectonic field (cid:101) ζ . Note that if ζ was C -small then (cid:101) ζ is also C -small. Moreover, we claim that (cid:101) ζ (cid:116) (cid:101) γ for (cid:101) γ a Lagrangian distribution homotopic to γ .Consider the tectonic field λ = ζ − (cid:101) ζ . It is in fact a smooth Lagrangian field, because the discontinu-ities of ζ are exactly canceled by those of (cid:101) ζ . Observe that (cid:101) ζ is graphical, hence det( (cid:101) ζ ) is bounded. LetΩ C = {| det( (cid:98) γ ) | < C } . It follows that for C > (cid:101) ζ (cid:116) (cid:98) γ outside of Ω C . Moreoverfor C > (cid:98) γ is homotopic to a Lagrangian plane field (cid:101) γ which is equal to (cid:98) γ − λ on Ω C and transverse to (cid:101) ζ outside of Ω C . Note that the expression (cid:98) γ − λ makes sense on Ω C because (cid:98) γ isgraphical on Ω C . The claim, and therefore also the Lemma, now follow. Indeed, the condition (cid:101) ζ (cid:116) (cid:101) γ on Ω C is equivalent to the nonsingularity of the form (cid:101) γ − (cid:101) ζ = ( (cid:98) γ − λ ) − ( ζ − λ ) = (cid:98) γ − ζ , which is inturn equivalent to ζ (cid:116) (cid:98) γ , which is true. (cid:3) Remark 4.5.
In the relative version where L (cid:116) γ and ζ = 0 on Op ( A ) for A ⊂ L a closed set, wedemand that the homotopy between γ and (cid:101) γ is constant on Op ( A ).We also need the following elementary fact. Lemma 4.6.
Let f t : Λ → M be an exact regular homotopy of Lagrangian embeddings of a compactmanifold Λ into a symplectic manifold M and for i = 0 , , let γ i ⊂ T M be a Lagrangian plane field along f i which is transverse to df i ( T Λ) . Then there exists a compactly supported Hamiltonian isotopy ϕ t : M → M such that ϕ t ◦ f = f t and dϕ ( γ ) = γ .Proof. By taking a family of Weinstein neighborhoods for f t we reduce to the case M = T ∗ Λ, f t = id Λ and γ = ν (the vertical distribution). Since γ is transverse to the zero section, we can think of γ asfamily of quadratic forms on the fibres λ q : T ∗ q Λ → R . Then the required Hamiltonian isotopy ϕ t isgiven by the quadratic Hamiltonian H ( q, p ) = λ q ( p ), cut off at infinity. (cid:3) Remark 4.7.
From the proof we also deduce the relative version: if f t = f and γ = γ on Op ( A )for A ⊂ Λ a closed subset, then we can demand that ϕ t = id M on Op ( A ).We are now ready for the proof of Proposition 4.2. Proof of Proposition 4.2.
Let Λ ⊂ T ∗ L be the graphical ridgy Lagrangian corresponding to theintegrable tectonic field (cid:101) ζ produced by Lemma 4.4. Denote by R ⊂ Λ the ridge locus. LetΛ \ R = Q ∪ Q ∪ · · · ∪ Q m , where each Q j is a smooth Lagrangian submanifold with corners.We will prove by induction that for each k = 0 , , . . . , m there exists a ridgy Lagrangian Λ k ⊂ T ∗ L satisfying the following properties.(a) Λ k is Hamiltonian isotopic to Λ(b) Λ k (cid:116) γ in a neighborhood of ∂Q k ∪ · · · ∪ ∂Q kk , where Q kj ⊂ Λ k corresponds to Q j ⊂ Λ underthe Hamiltonian isotopy.(c) There exists a homotopy γ kt of γ k = γ , fixed in a neighhborhood of ∂Q k ∪ · · · ∪ ∂Q kk , such thatΛ k (cid:116) γ k everywhere. Base case ( k = 0). In this case we take Λ = Λ so condition (a) is vacuous. Since k = 0, so is (b).Finally, condition (c) is the conclusion of Lemma 4.4. Inductive step ( k ⇒ k + 1). We apply the holonomic approximation theorem for 1-holonomicsections to the Lagrangian submanifold Q kk +1 and the stratified subset ∂Q kk +1 . Technically one shouldslightly enlarge Q k +1 so that ∂Q k +1 sits in its interior but this will not affect the proof. The preciseresult we need is the h-principle for Lagrangian embeddings which are D -directed along a stratifiedsubset, which is Theorem 1.20 in [AG18a]. For our application we take D to consist of all Lagrangianplanes transverse to γ . Condition (c) of the inductive hypothesis implies that the Gauss map of Q kk +1 ishomotopic to a map with image in D , moreover this homotopy can be taken relative to a neighborhoodof ∂Q k ∪ · · · ∪ ∂Q kk .The output of the h-principle is a Hamiltonian isotopy ϕ t : T ∗ L → T ∗ L such that ϕ ( Q kk +1 ) (cid:116) γ ina neighborhood of ϕ ( ∂Q kk +1 ). Moreover, by the parametric version of the h-principle we may assumethat ϕ t ( Q kk +1 ) is transverse to γ k − t for all t ∈ [0 , dϕ ( γ k ) = γ along ∂Q kk +1 ). Hence we have ϕ (Λ) (cid:116) γ in a neighborhood of ϕ ( ∂Q kk +1 ), i.e. also on the otherside of the ridges outside of Q kk +1 . Moreover, by the relative version of the holonomic approximationtheorem we can demand that ϕ t = id T ∗ L in a neighborhood of ∂Q k ∪ · · · ∪ ∂Q kk . It follows that if weset Λ k +1 = ϕ (Λ k ), then conditions (a) and (b) are satisfied for k + 1 instead of k .It remains to verify condition (c). Consider the homotopy γ k +1 t of γ which is given by the con-catenation of first γ kt and then dϕ t ( γ k ). The result is transverse to Λ k +1 = ϕ (Λ k ) because γ k istransverse to Λ k . Although this homotopy is constant in a neighborhood of ∂Q k +11 ∪ · · · ∪ ∂Q k +1 k , itis not constant near ∂Q k +1 k +1 . To fix this we recall that dϕ ( γ k ) = γ along ∂Q kk +1 and that ϕ t (Λ k ) EOMORPHOLOGY OF LAGRANGIAN RIDGES 25 is transverse to γ k − t along ∂Q k +1 k . Since the space of linear Lagrangian planes transverse to a fixedlinear Lagrangian plane is contractible, by parametrically interpolating between dϕ t ( γ k ) and γ k − t wecan cancel out the concatenations using a cutoff function and thus deform the homotopy so that it isconstant in a neighborhood of ∂Q k +1 k +1 . This completes the proof of the inductive step.To finish the proof of Proposition 4.2 we define the desired ridgy isotopy L t ⊂ T ∗ L of the zerosection L = L as the concatenation of two ridgy isotopies. The first one is a ridgy isotopy between L and Λ, the graphical ridgy Lagrangian obtained by integrating (cid:101) ζ . The second one is the Hamiltonianisotopy which one gets at the last stage k = m of the above inductive process. For the resulting ridgyLagrangian L = Λ m ⊂ T ∗ L we know that γ is homotopic, relative to a neighborhood of the ridge locusof L , to a Lagrangian plane field transverse to L . Take a family of symplectic bundle isomorphismsΦ t covering this homotopy of Lagrangian plane fields. We can take Φ t to be constant in a neighborhoodof the ridge locus of L . Then G t = Φ − t ( T L ) gives the required tangential rotation. (cid:3) Remark 4.8.
Since the h-principle holds in C -close form, at each step of the proof we can ensure C -closeness to the previous step and thus obtain that the ridgy isotopy is C -small.4.2. Ridgification of wrinkles.
To conclude the proof of Theorem 1.4, we need to take the ridgyLagrangian produced by Proposition 4.2 and further deform it in the complement of the ridge lo-cus so that it becomes transverse to γ . To achieve this we use a wrinkling technique, namely thetransversalization theorem for wrinkled Lagrangian embeddings. Theorem 4.9 (Theorem 5.1 of [AG18b]) . Let G t be a tangential rotation of a Lagrangian Λ ⊂ M suchthat G (cid:116) γ for γ ⊂ T M a Lagrangian distribution. Then there exists an exact homotopy of wrinkledLagrangian embeddings Λ t of Λ such that Λ (cid:116) γ . A wrinkled Lagrangian embedding is a smooth Lagrangian embedding outside of a disjoint unionof codimension 1 contractible spheres S ⊂ Λ. Along each sphere S the embedding has cuspidalsingularities of the form { p = q } × R n − ⊂ T ∗ R × T ∗ R n − , see Figure 21, with a codimension 1equatorial sphere Σ ⊂ S where the cuspidal singularities experience birth/death. The precise modelnear the equator is not important since as in Section 6.1 of [AG18b] we can surger away the birth/deathsingularities. More precisely, one can open up each sphere S along its equator into two parallel spheresso that Λ becomes a Lagrangian submanifold which is smooth away from the disjoint union of finitelymany pairs of contractible parallel spheres S ∪ S where the Lagrangian has cuspidal singularities.Moreover, this can be achieved while maintaining exactness and transversality to γ . We note that thiscuspidal Lagrangian can be smoothed so that the resulting smooth Lagrangian has fold tangencies on S ∪ S with opposite Maslov co-orientations, known as double folds . By performing this smoothingone deduces the h-principle for the simplification of caustics. However, it will be easier for us to workwith cuspidal Lagrangians directly. Remark 4.10.
Even though this will not be important, we note that the pairs of spheres S ∪ S ⊂ Λcould be nested, in the sense that we could have A ⊂ A for A i (cid:39) S n − × [0 , i = 1 ,
2, thecodimension 0 annuli in Λ corresponding to two pairs of parallel spheres.We deduce the following consequence of Theorem 4.9.
Figure 21.
A cuspidal Lagrangian singularity is a stabilization of the standard semi-cubical cusp { p = q } ⊂ T ∗ R . Corollary 4.11.
Let G t be a tangential rotation of a Lagrangian Λ ⊂ M such that G (cid:116) γ for γ ⊂ T M a Lagrangian distribution. Then there exists a ridgy isotopy Λ t of Λ such that Λ (cid:116) γ .Proof. We need to modify the wrinkled Lagrangian embedding Λ produced by Theorem 4.9 to make itridgy. We first resolve the equators of the wrinkles into cuspidal singularities as above while maintainingexactness and transversality to γ . Next, by a local interpolation at the level of generating functionswe can replace these cuspidal singularities with stabilizations of order 1 ridges { q = | p |} while stillmaintaining exactness and transversality to γ .Explicitly, take a cut-off function σ : [0 , ∞ ) → [0 ,
1] which is equal to 1 on [0 , ], equal to 0 outside[0 ,
1] and has non-positive derivative. Define the function φ ε ( p ) = 32 (cid:18) − σ (cid:18) | p | ε (cid:19)(cid:19) p / + 12 ε σ (cid:18) | p | ε (cid:19) sign( p ) p and set R ( ε ) := (cid:26) q = ∂φ ε ( p ) ∂p (cid:27) . For ε = 0 we have R (0) = { p = q } and for any ε > R ( ε ) is a ridgy Lagrangian with aridge at the origin, which gets infinitely sharp as ε →
0. If C ⊂ Λ is a component of the cuspidallocus, we can apply the above interpolation parametrically over x ∈ C to obtain a deformation Λ ( ε )of Λ (0) = Λ such that for any ε > ( ε ) is a ridgy Lagrangian. Moreover, by taking ε > ( ε ) is C -close to that of Λ . Hence for ε > ( ε ) is still transverse to γ .Therefore by parametrically implanting a 1-dimensional model we can replace the exact homotopyof wrinkled Lagrangian embeddings Λ t with a ridgy isotopy which at time 1 is transverse to γ . Thisridgy isotopy consists of an earthquake isotopy on the preimage of the cuspidal locus of Λ in Λ followedby an ambient Hamiltonian isotopy, the existence of which is guaranteed because we ensured exactnessat every stage by working at the level of functions. (cid:3) Remark 4.12.
The ridge locus of the ridgy Lagrangian Λ produced by Corollary 4.11 thereforeconsists of a disjoint union of parallel contractible spheres, which may be nested.The relative version also holds: if G t is constant on Op ( A ) for A ⊂ Λ a closed subset, so that inparticular Λ (cid:116) γ on Op ( A ), then we can demand that ϕ t = id M on Op ( A ). This follows immediatelyfrom the corresponding relative version of Theorem 4.9. EOMORPHOLOGY OF LAGRANGIAN RIDGES 27
Proof of the main theorem 1.4.
We apply Proposition 4.2 to L and γ | L inside a Weinstein neighbor-hood U of L in M . Indeed, U is symplectomorphic to T ∗≤ δ L = {|| p || < δ } , where δ > L . If the resulting ridgy Lagrangian is sufficiently C -close to the zerosection then it will remain in this Weinstein neighborhood, hence can be viewed as a ridgy Lagrangianin M . We then apply the relative version of Theorem 4.11 to the output of Proposition 4.2. (cid:3) Remark 4.13.
The ridge locus of the ridgy Lagrangian solving the transversalization problem consistsof the fault locus of the formal solution together with a union of parallel contractible spheres.5.
Adapted version
Adapted transversalization.
Let L be a compact manifold with boundary and corners. Wedenote by ∂ k L ⊂ ∂L the locus of k -fold corners, so that ∂L = ∂ L (cid:116) · · · (cid:116) ∂ n L . Note that ∂ k L is an( n − k )-dimensional smooth manifold with closure (cid:83) j ≥ k ∂ j L .We now explain a version of our results which is tailored to this situation. Near each point of a k -fold corner x ∈ ∂ k L we have fixed collar coordinates x = ( z, t ) for z ∈ ∂ k L and t = ( t , . . . , t k ) ∈ I k ,where I is the germ of [0 ,
1) at 0. Note that near x we have ∂ k L = { t = · · · = t k = 0 } and for j > k the strata of ∂ j L whose closure contains x are given by setting exactly k − j of the coordinates t i tozero. We demand compatibility in the sense that the remaining j coordinates t i give the fixed collarstructure ∂ j L × I j near x . We call such a compatible collection of collar coordinates a collar structure on L . The definitions and theorems which we write below are all relative to a fixed but otherwisearbitrary collar structure.Note that we get an induced decompositions T ∗ L (cid:39) T ∗ ( ∂ k L ) × ( T ∗ I ) k in a neighborhood of eachpoint x ∈ ∂ k L . In T ∗ I we have two distinguished Lagrangian distributions, the horziontal distribution τ which is tangent to the zero section and the vertical distribution ν which is tangent to the cotangentfibres. We will want our Lagrangian fields and submanifolds to interact compatibly with this structure. Definition 5.1.
A Lagrangian field λ on L (possibly tectonic) is said to be: · horizontally adapted if λ = λ k × τ k ⊂ T ∗ ( ∂ k L ) × ( T ∗ I ) k near each x ∈ ∂ k L , k = 1 , , . . . , n . · vertically adapted if λ = λ k × ν k ⊂ T ∗ ( ∂ k L ) × ( T ∗ I ) k near each x ∈ ∂ k L , k = 1 , , . . . , n . Definition 5.2.
A Lagrangian submanifold (possibly ridgy) Λ ⊂ T ∗ L is said to be adapted if Λ =Λ k × I k ⊂ T ∗ ( ∂ k L ) × ( T ∗ I ) k near each x ∈ ∂ k L , k = 1 , , . . . , n . A (possibly ridgy) isotopy ofLagrangian submanifolds Λ t ⊂ T ∗ L is said to be adapted if each Λ t is adapted. Remark 5.3.
If Λ ⊂ T ∗ L is adapted, then T Λ is horizontally adapted.We can now state the adapted version of our main theorem 1.4.
Theorem 5.4.
For any vertically adapted Lagrangian field γ ⊂ T ∗ L there exists an adapted ridgyisotopy L t of the zero section L = L such that L (cid:116) γ . Theorem 5.4 also holds in C -close and relative forms. The proof of Theorem 5.4 proceeds just likein the unadapted case: first we construct a formal solution, then we align it and finally we integrateit. We must argue that the same proof works while ensuring that all the objects are adapted to thecollar structure at each step. Adapted formal transversalization.
The adapted version of the formal transversalizationtheorem 2.5 reads as follows.
Theorem 5.5.
For any vertically adapted Lagrangian field γ there exists a horizontally adapted tectonicfield λ such that λ (cid:116) γ . The extension form 2.6 of the result also has its adapted version.
Theorem 5.6.
Let γ be a vertically adapted Lagrangian field and ζ a horizontally adapted tectonicfield. For any two disjoint compact subsets K , K ⊂ L there exists a horizontally adapted tectonicfield (cid:98) ζ such that the following properties hold. · (cid:98) ζ is C -close to ζ , · (cid:98) ζ (cid:116) γ on Op ( K ) . · (cid:98) ζ = ζ on Op ( K ) . To prove Theorem 5.6 one inductively applies Lemma 2.12, just as in the proof of Theorem 2.6.The only difference is that before building (cid:98) ζ in the interior of L one builds (cid:98) ζ in a neighborhood of ∂L , inductively over the strata ∂ k L . Start with the deepest stratum ∂ n L where there is nothing toprove. At each step of the induction one has a horizontally adapted tectonic field (cid:98) ζ defined over aneighborhood of (cid:83) j ≥ k ∂ j L which satisfies the required properties. To continue with the induction onechooses a cover of ∂ k − L by balls and applies Lemma 2.12 in the manifold ∂ k − L , one ball at a time.Multiplying the resulting tectonic field by the horizontal distribution in the collar direction providesthe extension and so the induction can continue. Once the horizontally adapted tectonic field (cid:98) ζ hasbeen built in a neighborhood of ∂L it can be extended to the rest of L as in the unadapted case.5.3. Adapted aligned formal transversalization.
The adapted version of the aligned analogueTheorem 3.3 of Theorem 2.5 reads as follows.
Theorem 5.7.
For any vertically adapted Lagrangian field γ there exists a horizontally adapted alignedtectonic field λ such that λ (cid:116) γ . More generally, we have the adapted version of the aligned extension result Theorem 3.4.
Theorem 5.8.
Let γ be a vertically adapted Lagrangian field and ζ a horizontally adapted alignedtectonic field. For any two disjoint compact subsets K , K ⊂ L there exists a horizontally adaptedaligned tectonic field (cid:98) ζ and a vertically adapted Lagrangian field (cid:98) γ homotopic to γ such that the followingproperties hold. · (cid:98) ζ is C -close to ζ , · (cid:98) ζ (cid:116) (cid:98) γ on Op ( K ) . · (cid:98) ζ = ζ on Op ( K ) .Moreover, we can assume that the homotopy between γ and (cid:98) γ is through vertically adapted fields andis constant on Op ( K ) . Theorem 5.8 follows from the same local model for changing the homotopy class of the ridge direc-tions which we used to align the ridge directions in the unadapted case. Indeed, as in Section 3.5 wefirst reduce to the homotopically aligned version of Theorem 5.8. To prove the homotopically aligned
EOMORPHOLOGY OF LAGRANGIAN RIDGES 29 version we can start making the necessary local modifications to the (cid:98) ζ produced by Theorem 5.6 alongthe boundary ∂L first, then once we have a horizontally adapted solution near ∂L we can extend tothe interior as in the unadapted case. To construct the horizontally adapted solution (cid:98) ζ near ∂L wework inductively over the strata ∂ k L , starting with the deepest one ∂ n L in which there is nothing toprove. Whenever we need to adjust the homotopy class of the ridge directions for the tectonic field λ k in ∂ k L , we choose a domain Ω ⊂ L and a form ν as in the model 3.4 which are adapted to thecollar structure. Then not only is the modified tectonic field still adapted, but the homotopy of γ isby construction through vertically adapted fields.5.4. Adapted integration.
Finally we show how to integrate the λ produced by Theorem 5.7 so thatthe resulting ridgy Lagrangian remains adapted. In fact this follows easily from the parametric versionsof the holonomic approximation and wrinkling results which are used in the unadapted case. Firstobserve that since the aligned tectonic field λ is horizontally adapted, the introduction of integrableridges in Lemma 4.4 can be achieved with respect to coordinates that are compatible with the collarstructure. Hence the resulting integrable tectonic field is horizontally adapted.Next we turn to the adapted analogue of Proposition 4.2. We recall that the holonomic approxima-tion lemma for 1-holonomic sections [AG18a] holds in parametric form, and moreover holds relativeto a closed subset of the parameter space. Hence we can apply this result inductively over the strata ∂ k L so that at each stage of the induction the conclusion of the proposition holds in a neighbor-hood of (cid:83) j ≥ k ∂ j L and moreover such that the resulting ridgy Lagrangian and tangential rotation areadapted in this neighborhood. At the last stage of this inductive process we obtain the desired adaptedridgy Lagrangian in a neighborhood of ∂L , which can then be extended to the interior of L as in theunadapted case.To conclude we turn to the application of Theorem 4.9 in the adapted setting and finish the proofof Theorem 5.4. The C -approximation result for wrinkled Lagrangian embeddings also holds inparametric form, but only relative to a subset where the embedding is smooth. Therefore, whenapplying the result in a component of the stratum ∂ k L one will need to let the cuspidal singularitiesdie out as you move away from this component. After replacing these cuspidal singularities with ridges,this means that at the next stage of the induction there will be some ridges present. So in order tocontinue the inductive process we must first achieve transversality in a neighborhood of these ridges.This can be achieved using holonomic approximation just as in Section 4.1. We can then use wrinklingas before in the complement. References [AG18a] D. ´Alvarez-Gavela, Refinements of the holonomic approximation lemma,
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Department of Mathematics, Princeton Univeristy, Princeton, NJ 08540
E-mail address : [email protected] Department of Mathematics, Stanford University, Stanford, CA 94305
E-mail address : [email protected] Department of Mathematics, University of California, Berkeley, Berkeley, CA 94720-3840
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