Area preserving isotopies of self transverse immersions of S 1 in R 2

Transcription

1 U.U.D.M. Project Report 2010:17 Area preserving isotopies of self transverse immersions of S 1 in R 2 Cecilia Karlsson Examensarbete i matematik, 30 hp Handledare och examinator: Tobias Ekholm December 2010 Department of Mathematics Uppsala University

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3 Area preserving isotopies of self transverse immersions of S 1 in R 2. Cecilia Karlsson December 20, 2010 Abstract Let C and C be two smooth self transverse immersions of S 1 into R 2. Both C and C subdivide the plane into a number of disks and one unbounded component. An isotopy of the plane which takes C to C induces a 1-1 correspondence between the disks of C and C. An obvious necessary condition for there to exist an area preserving isotopy of the plane taking C to C is that there exists an isotopy for which the area of every disk of C equals that of the corresponding disk of C. In this paper we show that this is also a sufficient condition. 1 Introduction Let C be a smooth self transverse immersion of S 1 into the plane R 2 (by Sard s theorem any immersion is self transverse after arbitrarily small perturbation). Then C subdivides the plane into a number of bounded connected components and one unbounded component. The bounded components are topological disks and we call them the disks of C. Let C be another self transverse immersion of S 1 into R 2 such that there exists an isotopy of the plane taking C to C. Then the isotopy induces a 1-1 correspondence between the disks of C and the disks of C. In this paper we study the existence of area preserving isotopies of the plane taking C to C, where, if dx dy denotes the standard area form on R 2, we say that an isotopy φ τ : R 2 R 2, 0 τ 1 is area preserving if φ τ (dx dy) = dx dy for every τ [0, 1]. Since φ τ area preserving implies that area(φ τ (U)) = area(u) for any measurable U R 2, an obvious necessary condition for the existence of an area preserving isotopy φ τ taking C to C is that the area of any disk D of C satisfies area(d) = area(d ) (1.1) 1

4 where D is the disk of C which corresponds to D under φ τ. We call an isotopy which satisfies (1.1) disk area preserving. The main result of the paper shows that this is also a sufficient condition. More precisely, we have the following result. Theorem 1.1. Let C and C be two self transverse immersions of S 1 into R 2 and assume that there is a disk area preserving isotopy ψ τ, 0 τ 1 of R 2 taking C to C (i.e., ψ 0 = id, ψ 1 (C) = C, and for every disk D of C, area(ψ 1 (D)) = area(d)). Then there exists an area preserving isotopy φ τ, 0 τ 1, of R 2 with φ 0 = id and φ 1 (C) = C. Theorem 1.1 is proved in Section 5. Problems related to the existence of a topological isotopy (without area condition) taking C to C were studied by many authors, see e.g. [8, 3, 6]. From the point of view of symplectic geometry C is an immersed Lagrangian submanifold, and area preserving isotopies are Hamiltonian isotopies. For related questions in higher dimensions see e.g. [7, 5, 4]. In short outline, our proof of Theorem 1.1 is as follows. First, we construct an isotopy χ τ which takes C to C and such that for every disk D of C we have area(χ τ (D)) = area(d) for all τ. We call such an isotopy semi-area preserving with respect to C. The semi-area preserving isotopy is constructed from the disk area preserving isotopy ψ τ by first composing it with a time dependent scaling so that the resulting isotopy shrinks the area of each disk of C for all times. The resulting isotopy is then modified: we introduce segments connecting the boundary of each disk to the unbounded component and push the curve along these segments in order to increase the area of any disk until the isotopy becomes semi-area preserving, see Section 3. Second, we subdivide the semi-area preserving isotopy into small time steps and use a cohomological argument to show the existence of an area preserving isotopy, see Section 4. For simpler notation below, we assume that all maps are smooth and that all immersions are self transverse. Acknowledgements I would like to thank my advisor Tobias Ekholm for introducing me to the subject, and for guiding me through it. I would also like to thank my friends and teachers for always supporting and encouraging me. 2

5 2 Background In this section we introduce notation and discuss background material on Hamiltonian vector fields on surfaces, for proofs see e.g. [1, 2]. Let M be a surface and let v : M T M a vector field with compact support. We write Φ t v : M M for the time t flow of v. Let ω be a symplectic form on M and write I : T M T M for the isomorphism defined through the equation α(η) = ω(η, I(α)) for all α T M, η T x M. Let H : M R be a smooth function with compact support. The vector field X H = I(dH) is the Hamiltonian vector field of H and its flow is area preserving. Let C be an immersion of S 1 into the plane and let ϕ : S 1 R 2 be a ( parametrization ) of C. Write e(s) for the unit vector field along C such that dϕ ds (s), e(s) is a positively oriented basis of R 2 for all s S 1. Then for all sufficiently small ǫ > 0 the map Φ : S 1 ( ǫ, ǫ) R 2, Φ(s, t) = ϕ(s) + te(s). (2.1) parameterizes a neighborhood C ǫ of C. Let dx dy be the standard symplectic form on R 2 and consider coordinates (s, t) on S 1 R = (R/2πZ) R with the corresponding symplectic form ds dt. The following lemma is a special case of Moser s lemma, see e.g. [2] for a proof. Lemma 2.1. Let C be an immersion of S 1 in R 2 and let Φ be as in (2.1). Then there exists δ > 0 and a diffeomorphism ϑ : S 1 R S 1 R with ϑ(s, 0) = (s, 0) such that for all t < δ. (Φ ϑ) dx dy = ds dt, Below we will often combine Lemma 2.1 with a Hamiltonian isotopy χ τ of S 1 R with support inside S 1 ( δ, δ) to construct an area preserving isotopy χ τ of R 2 defined by χ τ (x) = { Φ ϑ χτ ((Φϑ) 1 (x)) if x Φ ϑ(s 1 ( ǫ, ǫ)), x otherwise. 3

6 In the following lemma we use this argument to construct area preserving isotopies between nearby curves C and C which agree near double points. We will use the following terminology: For C R 2 an immersed circle, we call an arc A C a maximal smooth arc of C if A {x i } n i=1 = {x i, x j } = A, where {x i } n i=1 C are the double points of C. Lemma 2.2. Let C be an immersion of S 1 into R 2 and let ξ : S 1 ( ǫ, ǫ) R 2 be an area preserving parameterization of a neighborhood C ǫ of C as in Lemma 2.1. Assume that C is an immersion of S 1 into R 2 which coincides with C in a neighborhood U x of every double point x of C and such that there is a function g : S 1 ( ǫ, ǫ) with C = ξ(γ), where Γ is the graph of g. If there exists a disk area preserving isotopy taking C to C then there exists an area preserving isotopy of the plane taking C to C. Proof. Shrink C ǫ so that we still have C C C ǫ, but so that the parameterization is 1-1 outside U x, i.e. so that C ǫ U x consists of a number of simply connected components V A where each component corresponds to a maximal smooth arc A of C. Let W C ǫ be a neighborhood of C C so that V A W and U x W are simply connected for all V A, U x. Let G : S 1 R be defined by G(s) = s 0 g(s )ds, and let G : R 2 R be a function satisfying G((ξ 1 ) 1 (x)) for x W V A G(x) = G((ξ 1 ) 1 (x )) for x W U x 0 for x / C ǫ where ξ 1 = ((ξ 1 ) 1, (ξ 1 ) 2 ). Then G is a well-defined function: Suppose that ξ(s 1, t 1 ) = ξ(s 2, t 2 ) for s 1 s 2. Then we have ξ(s 1, t 1 ) U x for some x. Since G is constant in U x we can assume that ξ(s 1, t 1 ) = x. But ξ((s 1, s 2 ) {0}) is a 1-chain, so it bounds a number of disks of C. Since every disk of C has the same area as the corresponding disk of C we have s2 s 1 g(s)ds = 0, so G(s 1 ) = G(s 2 ). The Hamiltonian vector field of G in the parameterization of C ǫ is X G = g(s) t for (s, t) W V A and X G = 0 in U x W. Hence its time 1-flow takes (s, g(s)) to (s, 0) for all s and we get an area preserving isotopy of the plane taking C to C. 3 Construction of a semi-area preserving isotopy with respect to C In this section we construct a semi-area preserving isotopy from a disk area preserving isotopy. 4

7 Let C and C be two immersions of S 1 into R 2 such that there exists a disk area preserving isotopy φ t taking C to C. Let B r be the open disk of radius r centered at 0, B r,p the open disk of radius r centered at p. Without loss of generality we can assume that φ t has support in some B r. We start with bumping out each disk of C respectively C in a semi-area preserving way so that they all intersect R 2 B r+1. So, for each disk D of C pick a maximal smooth arc A D in D and let L D be a line segment going from B r+1 to A D. Let L ν D be a simply connected neighborhood of L D. Choose the lines and the neighborhoods in such a way so that L ν D 1 L ν D 2 = for D 1 D 2, so that L ν D does not contain any double points of C, so that L D and L ν D intersects C transversely at each point of intersection and so that L ν D B r is simply connected for all r [r, r + 1]. We also require that for A a maximal smooth arc of C with L D A = n we have that L ν D A = 2n, and also that L D D = 1. See Figure 1. B r+1 D 3 D 1 D 2 D 4 L D1 L D2 L D3 L D4 Figure 1: An immersion C with line-segments L D connecting each disk of C with R 2 B r+1. Let D 1,..., D k be the disks of C and let and let each L ν D i be so thin so that a = min i=1,2,..,k {area(d i )} area( k i=1 L ν D i ) < 1 a. (3.1) 4 Now isotope C with a semi-area preserving isotopy γ t, 0 t 1, with respect to C so that γ 1 (C) contains all the lines L D, with γ t having support 5

8 inside i Lν D i. See Figure 2. Isotope C with a semi-area preserving isotopy γ t, 0 t 1, with respect to C so that γ 1 (C ) contains all the arcs φ 1 (L D ), and so that γ t has support in k i=1 φ 1 (L ν D i ). Let C 1 = γ 1 (C), C 1 = γ 1 (C ) and let ζ = γ 1 φ 1γ 1 1. Also, let 0 < δ < 1 and choose γ t in such a way that ζ(c 1 B r+δ ) B r+δ, ζ C1 B r+δ = id. (3.2) L ν D 3 B r+1 D 4 D 2 D 4 L D3 D 2 D 3 γ 0 (C L ν D 3 ) γ 1/2 (C L ν D 3 ) γ 1 (C L ν D 3 ) Figure 2: Example of how we can let γ t act in L ν D 3 in Figure 1. Now we want to shrink B r and then take the shrunken version of C 1 to the shrunken C 1 in a semi-area preserving way. To do this, choose an ǫ > 0 so that area(d) πǫ 2 (r + δ) 2 > 1 2 a (3.3) for every disk D of C. Let ψ t, 0 t 1, be an isotopy with support in B r+1 satisfying ψ t (x, y) = (ψ 1 t (x, y), ψ 2 t (x, y)) = (x + tx(ǫ 1), y + ty(ǫ 1)) for (x, y) B r+δ and so that (ψt 1 1 (x, y), ψ 2 t 1 (x, y)) (x, y) (ψt 1 2 (x, y), ψ 2 t 2 (x, y)) (x, y) whenever t 1 < t 2 for (x, y) D γ 1(L ν D ) (B r+1 B r+δ ). Also let φ t (x) = (ǫφ 1 t (x/ǫ), ǫφ2 t (x/ǫ)), an isotopy of the plane with support in B ǫr. Lemma 3.1. There exists an isotopy ξ t, 0 t 1, of the plane such that φ 1 ψ 1 (x) = ξ 1 ψ 1 ζ(x) for all x C 1. 6

9 We will prove this lemma later on, but first let χ t, 0 t 1, be an isotopy acting as the composition of ψ t and φ t and let χ t, 0 t 1, be an isotopy acting as the composition of ψ t and ξ t. Then both χ t and χ t have support in B r+1, and we wish to extend them outside B r+1 to isotopies χ t, χ t being semi-area preserving with respect to C 1 respectively C 1, and so that χ 1 (x) = χ 1 ζ(x) for all x C 1. If we can do this we get a semi-area preserving isotopy with respect to C taking C to C, namely the isotopy F t : R 2 R 2, t [0, 4], given by γ t (x) for t [0, 1] χ t 1 (x) for t [1, 2] F t (x) = χ (3.4) 3 t (x) for t [2, 3] (x) for t [3, 4]. See Figure 3. γ 4 t C φ 1 C γ 1 C 1 ζ C 1 γ 1 χ 1 ψ 1 ψ 1 (C 1 ) ψ 1 ψ 1 (C 1 ) χ 1 φ 1 χ 1 (C 1 ) id ξ 1 χ 1 (C 1 ) Figure 3: We want to extend χ t and χ t outside B r+1. To prove Lemma 3.1 we first notice the following: Proposition 3.2. For each disk D of C we have that φ 1 ψ 1 γ 1 (L ν D) = ψ 1 ζγ 1 (L ν D). Moreover, for each x C 1 ( γ 1 (L ν D ) B r+δ) it holds that φ 1 ψ 1 (x) = ψ 1 ζ(x). Proof. Since γ t respectively γ t have support in the disjoint unions L ν D respectively φ 1 (L ν D ) and since ψ t, φ t and φ t all have support in B r+1 we 7

10 get φ 1 ψ 1 γ 1 (L ν D ) (R2 B r+1 ) = L ν D (R2 B r+1 ) = ψ 1 γ 1φ 1 (L ν D) (R 2 B r+1 ) = ψ 1 ζγ 1 (L ν D) (R 2 B r+1 ). We also notice that for x B r+δ we have ) φ 1 ψ 1 (x) = φ 1 (ǫx) = (ǫφ 11 (ǫxǫ ), ǫφ21 (ǫxǫ ) = ψ 1 φ 1 (x). (3.5) Thus φ 1 ψ 1 γ 1 (L ν D) B r+1 = φ 1 ψ 1 (L ν D (B r+1 B r )) φ 1 ψ 1 (L ν D B r ) = ψ 1 (L ν D (B r+1 B r )) ψ 1 φ 1 (L ν D B r ) ψ 1 ζγ 1 (L ν D ) B r+1 = ψ 1 γ 1 φ 1(L ν D (B r+1 B r )) ψ 1 γ 1 φ 1(L ν D B r) = ψ 1 (L ν D (B r+1 B r )) ψ 1 φ 1 (L ν D B r). The first statement of the claim follows. For the second one, first consider the case when x C 1 γ 1 (L ν D ). Then x = γ1 1 (x) C L ν D B r and φ 1 γ1 1 (x) B r φ 1 (L ν D ), so φ 1 γ1 1 (x) = γ 1 φ 1γ1 1 (x). Hence, by (3.5) we get φ 1 ψ 1 (x) = φ 1 ψ 1 γ 1 1 (x) = ψ 1φ 1 γ 1 1 (x) = ψ 1γ 1 φ 1γ 1 1 (x) = ψ 1ζ(x). Next, if x C 1 B r+δ the statement follows from (3.2) and from the fact that φ t has support in B ǫr. Proof of Lemma 3.1. To simplify the notation let φ 1 ψ 1 γ 1 (L ν D ) = L ν D. By Proposition 3.2 it is enough to find an isotopy ξ t with support in Lν D B ǫ(r+δ) such that ξ 1 ψ 1 ζ(x) = φ 1 ψ 1 (x) for all x C 1 ( L ν D B r+δ). For each disk D of C we have that C 1 L ν D B r+δ consists of n disjoint arcs A 1, A 2,..., A n and clearly φ 1 ψ 1 (C 1 ) L ν D B ǫ(r+δ) = n i=1 φ1 ψ 1 (A i ). By (3.2) we have ψ 1 ζ(a i ) L ν D B ǫ(r+δ) for all arcs A i, and Proposition 3.2 implies that φ 1 ψ 1 (A i ) = ψ 1 ζ(a i ) for all i. Hence, since L ν D B ǫ(r+δ) is simply connected we can find an isotopy ξt D so that ξ1 Dψ 1ζ(x) = φ 1 ψ 1 (x) for all x C 1 L ν D B r+δ. See Figure 4. After a small perturbation of ( L ν D B ǫ(r+δ)) at the intersection points of φ 1 ψ 1 (C 1 ) if necessary, we can choose ξt D to have support inside L ν D B ǫ(r+δ). Composing the ξt D :s we get an isotopy ξ t of the plane as in the statement of the lemma. 8

11 L ν D B ǫ(r+δ) Figure 4: An example of the intersection of φ 1 ψ 1 (C 1 ) and ψ 1 ζ(c 1 ) with L ν D B ǫ(r+δ). The solid arcs belong to φ 1 ψ 1 (C 1 ) and the dashed arcs belong to ψ 1 ζ(c 1 ). Clearly we can find an isotopy taking the dashed arcs to the solid arcs. Now, for each disk D of C we have that χ 1 γ 1 (D) B r+1 = γ 1 (D) B r+1 = χ 1ζγ 1 (D) B r+1, which follows from (3.2) and from the fact that χ t, χ t have support in B r+1. After a small bending of the tubes L ν D outside B r+1 we can make them as long as we wish, while still having that two different tubes does not intersect each other. To compensate the area change of γ 1 (D) respectively ζγ 1 (D) under χ t respectively χ t we define an isotopy Ft D, 0 t 1, with support in the extended tube L ν D B r+1. Choose Ft D so that the function β D (t) = area(f D t (γ 1 (D) ( L ν D B r+1))) is strictly increasing in t, and so that β D (1) = area(d). Also let Ft D be such that F0 D = id and such that the area of all other disks of C 1 (and C 1 ) are constant under Ft D. Let and consider b D = area(γ 1 (D) ( L ν D B r+1)) c D = area(γ 1 (D) ( L ν D i B r+1 )) D i D α D (t) = area(d) area(χ t (γ 1 (D) B r+1 )) c D α D (t) = area(d) area(χ t (ζγ 1(D) B r+1 )) c D. 9

12 : [0, 1] R are well- Proposition 3.3. The functions βd 1 α D, βd 1 defined. α D Proof. For each disk D of C we must show that α D (t), α D (t) [b D, area(d)] for t [0, 1]. Clearly α D (t), α D (t) < area(d) for all t. Suppose that χ 1/2 = ψ 1 = χ 1/2, so that χ t, χ t act as ψ 2t for t 1/2 and as φ 2t 1 respectively ξ 2t 1 when t 1/2. Then, for t < 1/2 we have that α D (t), α D (t) are non-decreasing in t, and since α D(0) = b D = α D (0) we thus have that βd 1 α D(t), βd 1 α D (t) are defined for t [0, 1/2]. Next consider the case when t > 1/2. By (3.2) and since both ξ t and φ t are supported in B ǫ(r+δ) we have that χ t (γ 1 (D) (B r+1 B r+δ )) = ψ 1 (γ 1 (D) (B r+1 B r+δ )) = χ t(ζγ 1 (D) (B r+1 B r+δ )). Hence, since ψ 1 (γ 1 (D) (B r+1 B r+δ )) ψ 1 (L ν D ) we get that area(χ t (γ 1 (D) B r+1 )) = = area(χ t (γ 1 (D) B r+δ )) + area(χ t (γ 1 (D) (B r+1 B r+δ ))) πǫ 2 (r + δ) 2 + area( ψ 1 (L ν D )) πǫ2 (r + δ) 2 + area( L ν D ) < πǫ 2 (r + δ) a where the second last inequality follows from the construction of ψ t and the last one follows from (3.1). So for these t we have that α D (t) = area(d) area(χ t (γ 1 (D) B r+1 )) c D > area(d) πǫ 2 (r + δ) a c D > 1 4 a c D > b D + c D c D = b D where the second last inequality follows from (3.3) and the last one from (3.1). Similar calculations show that α D (t) > b D for t > 1/2. Now extend χ t outside B r+1 to an isotopy χ t satisfying F D χ t (x) = (x) for x L ν β 1 D (α D(t)) D B r+1 x for x / B r+1 L D ν D 10

13 and extend χ t in a similar way to an isotopy χ t satisfying χ t (x) = F D β 1 D (α (t))(x) for x L ν D B r+1 D x for x / B r+1 L D ν D. Let F t : R 2 R 2, t [0, 4], be defined as in (3.4). Then we have the following: Theorem 3.4. The isotopy F t is a semi-area preserving isotopy with respect to C taking C to C. Proof. By construction we have that the composition of χ t and γ t is a semiarea preserving isotopy with respect to C and that the composition of χ t andγ t is a semi-area preserving isotopy with respect to C. So it sufficies to show that χ 1 γ 1 (x) = χ 1 γ 1 φ 1(x) for all x C, i.e. χ 1 (x) = χ 1ζ(x) for all x C 1. By Lemma 3.1 and since χ t, χ t have support in B r+1 we have that χ 1 (x) = χ 1 ζ(x) for all x C 1 B r+1. But this also implies that α D (1) = α D (1) for all disks D of C, so χ 1 R 2 B r+1 = χ 1 R 2 B r+1. Hence, for x C 1 B r+1 we get χ 1 (x) = χ 1 ζ(x) = χ 1ζ(x) where the first equality follows from (3.2). 4 Area preserving isotopies between nearby curves In this section we show that if C and C are two immersed circles in the plane such that there is a semi-area preserving isotopy ψ τ, 0 τ 1, with respect to C taking C to C, then there is an area preserving isotopy taking C to ψ τ0 (C) for τ 0 sufficiently small. Thus, by compactness arguments, we can find an area preserving isotopy taking C completely to C. First we prove that if C lies sufficiently close to C then there is an area preserving isotopy taking C to C. To this end we begin with finding a suitable parameterization of a neighborhood of C, and then we define what we mean by C being sufficiently close to C. Given C, let ν > 0 be so small that B ν,x1 B ν,x2 = for any double points x 1 x 2 of C. Let ξ : S 1 ( ǫ, ǫ) R 2 be an area preserving parameterization of a neighborhood C ǫ of C as in Lemma 2.1. Then at each double point x of C we get a double point of ξ, i.e. a subset U x C ǫ where C ǫ overlaps itself. Let ǫ > 0 be so small that U x is a disk contained in B ν,x and so that C U x consists of two smooth arcs L s, L t intersecting at x. Suppose that x = ξ(0, 0) and that L s = ξ([ s 1, s 1 ] {0}). Since L s intersects L t 11

14 transversely at x there is a t 1 > 0 so that L t (( s 1, s 1 ) ( t 1, t 1 )) coincides with the graph of a function g : ( t 1, t 1 ) ( s 1, s 1 ) in the parameterization of C ǫ. Let S = ( s 1, s 1 ) ( t 1, t 1 ) and let ϑ : S R 2 be defined by ϑ(s, t) = (s g(t), t) = (µ(s, t), η(s, t)). Then ϑ dµ dη = ds dt, and ϑ maps L s S to the µ-axis and L t S to the η-axis. Let D x = ξϑ 1 (( s 1, s 1 ) ( t 1, t 1 )) where s 1 > 0 is so small that ϑ 1 (( s 1, s 1 ) ( t 1, t 1 )) S. Definition 4.1. We call the data {C ǫ, D x } a regular neighborhood of C. So a regular neighborhood of C consists of an immersed annulus C ǫ = ξ(s 1 ( ǫ, ǫ)), and also a parameterization of a neighborhood of each double point of C so that in this parameterization we have that C coincides with the coordinate axes of R 2. See Figure 5. t S 1 R ξ 1 (D x) C ǫ 1 s ξ ǫ C ǫ L t η L s ϑ L t ξ ϑ 1 D x µ L s Figure 5: An example of a regular neighborhood. 12

15 Now let C C ǫ be an immersion such that there exists a disk area preserving isotopy taking C to C. Let Q r be the open square with sides of length 2r centered at 0, and Q r,p the open square with sides of length 2r centered at p. Let δ > 0 be so small that for every double point x C we have that Q δ,x is contained in the parameterization of D x. Further, for each double point x C, let x be the corresponding double point of C, and let L s, L t C be the arcs corresponding to L s respectively L t in C. Suppose that C D x L s L t and that x Q δ,x in the parameterization of D x. Also suppose that L s D x respectively L t D x is a graph of a function g µ respectively g η over the µ- respectively η-axis in the parameterization of D x, satisfying g µ, g η, dgµ dgη ds, ds < δ. If this holds for all double points of C, and if C is a graph of a function g : S 1 ( δ, δ) in the parameterization of C ǫ satisfying dg ds < δ, we say that C is δ-close to C in {C ǫ, D x }. The following result shows that if C is sufficiently close to C in the above sense, then there is an area preserving isotopy taking C to C. Lemma 4.2. Let C be an immersion of S 1 in R 2 and let {C ǫ, D x } be a regular neighborhood of C. Then there exists a δ > 0 such that for every immersion C which is δ-close to C in {C ǫ, D x } there is an area preserving isotopy taking C to C. Before we prove this, let ψ τ be a semi-area preserving isotopy with respect to C taking C to C. For each τ 0 [0, 1], given a regular neighborhood of ψ τ0 (C) and a δ as in the lemma, if we have ψ τ (C) being δ-close to ψ τ0 (C) for 0 < τ τ 0 < ν then there is an area preserving isotopy taking ψ τ (C) to ψ τ0 (C). Thus, if this is the case for all τ 0 [0, 1] we can gradually replace the semi-area preserving isotopy with an area preserving isotopy and in the end get an area preserving isotopy taking C completely to C. In the following lemma we show that we always can find such a ν. Lemma 4.3. Let C be an immersion of S 1 in R 2 and let {C ǫ, D x } be a regular neighborhood of C. Let δ > 0 be such that Q δ,x D x in the parameterization of each D x, and suppose that χ τ is a semi-area preserving isotopy of the plane with respect to C. Then there exists a ν > 0 so that χ τ (C) is δ-close to C in {C ǫ, D x } for all 0 τ < ν. Proof. Let ν > 0 be so small so that χ τ (C) = C τ C ǫ for τ < ν. Let ξ : S 1 ( ǫ, ǫ) R 2 be an area preserving parameterization of C ǫ as in Lemma 2.1. To have C τ to be a graph of a function g τ : S 1 R satisfying 13

16 dgτ ds < δ in the parameterization of Cǫ we require that d ds (ξ 1 ) 1 (χ τ ξ(s, 0)) 0, d ds (ξ 1 ) 2 (χ τ ξ(s, 0)) d ds (ξ 1 ) 1 (χ τ ξ(s, 0)) < δ for all s S 1, where ξ 1 = ((ξ 1 ) 1, (ξ 1 ) 2 ). For each τ [0, 1] let (χ τ ) be the differential of the diffeomorphism χ τ. By Taylor expansion we get (χ τ ) = E + τ( d dτ (χ τ) τ=0 ) + O(τ 2 ) = E + O(τ) where E is the 2 2 unit matrix and O(τ) denotes a 2 2 matrix with entries of size O(τ). Thus (ξ 1 χ τ ξ) = (ξ 1 ) (E + O(τ))ξ = E + O(τ), so d ds (ξ 1 ) 1 (χ τ ξ(s, 0)) = 1 + O(τ) d ds (ξ 1 ) 2 (χ τ ξ(s, 0)) d ds (ξ 1 ) 1 (χ τ ξ(s, 0)) = O(τ) 1 + O(τ) = O(τ). Hence, for τ sufficiently small we indeed have that C τ is a graph of a function g τ : S 1 ( δ, δ) satisfying dgτ ds < δ. Next let x be a double point of C and consider D x. By continuity of χ τ we clearly have that χ τ (x) Q x,δ and that C τ D x χ τ (L s L t ) for τ small. And just as above we can keep χ τ (L s ), χ τ (L t ) as graphs of some functions g τ,µ, g τ,η over the µ- respectively η-axis in D x for τ small, satisfying g τ,µ, g τ,η, dgτ,µ dgτ,η ds, ds < δ. So, since there is only a finite number of double points of C we can find a ν > 0 as in the statement of the lemma. Proof of Lemma 4.2. Let σ > 0 be so small so that we in each parameterized D x can find a square Q σ,x = Q σ, where x corresponds to (0, 0) in the parameterization. Let σ > δ > 0 be sufficiently small so that there exists a smooth cut-off function ψ : R R satisfying ψ(y) = { 1 for y ( δ, δ) 0 for y / ( σ, σ) with ψ 1, dψ dy, d2 ψ dy 2 < δ. Now let C be an immersion which is δ-close to C in {C ǫ, D x }, and let x C be a double point. We start with showing that if δ is sufficiently 14

17 small then there is a neighborhood U of x and an area preserving isotopy χ τ, 0 τ 1, with support in D x so that χ 1 (C ) U coincides with C U and so that χ 1 (C ) is a graph over S 1 in C ǫ. By finding one such isotopy for each double point of C and then use Lemma 2.2 we get an area preserving isotopy taking C completely to C. So given a double point x C, first consider L s. We know that L s coincides with the graph of a function g µ : ( σ, σ) ( δ, δ) in Q σ. Let δ be so small that we can find an exact function g : R ( δ, δ) with support in ( σ, σ), whose graph coincides with L s in Q δ and which satisfies d g dµ < δ. Let G(µ) = µ σ g(µ )dµ, and consider the Hamiltonian H(µ, η) = G(µ)ψ(η) with corresponding vector field X H = G(µ) dψ dη (η) µ g(µ)ψ(η) η. Then the Hamiltonian isotopy Φτ X H = χ τ, 0 τ 1, takes L s to the µ-axis in Q δ, and has support in Q σ. Similar to the proof of Lemma 4.3 we want to find (χ 1 ) to make sure that χ 1 (C ) is still a graph over S 1 in the parameterization of C ǫ. Divide [0, 1] into N intervals of lenght 1/N. By Taylor expansion we have, for τ 1/N, that and χ 1 τ µ = χ1 0 µ + τ d χ 1 0 dτ µ + O(τ 2 ) = 1 + τ µ (G(µ)dψ dη (η)) + O(τ 2 ) = = 1 + τ g(µ) dψ dη (η) + O(1/N 2 ) χ 1 1/N+τ µ = χ1 1/N µ + τ d χ 1 1/N dτ µ + O(τ 2 ) = = (1 + 1/N g(µ) dψ dη (η) + O(1/N 2 )) + τ g(µ) dψ dη (η) + O(1/N 2 ). If we continue like this we get χ 1 N 1 1 µ = 1 + 1/N g(µ(n/n)) dψ dη (η(n/n)) + NO(1/N 2 ) n=0 = 1 + O(δ 2 ) + O(1/N) since g, dψ dη < δ. Hence for N big enough, depending on C, we get χ O(δ), where O(δ) depends on C, C ǫ and D x. Similarly we have χ 1 N 1 1 η = 0 + 1/NG(µ(n/N)) d2 ψ (η(n/n)) + O(1/N) = O(δ) dη2 n=0 15 µ =

18 since G < σδ, d2 ψ dη 2 < δ, and χ 2 N 1 1 µ = 0 1/N d g (µ(n/n))ψ(η(n/n)) + O(1/N) = O(δ) dµ n=0 χ 2 N 1 1 η = 1 1/N g(µ(n/n)) dψ (η(n/n)) + O(1/N) = 1 + O(δ). dη Thus we get that n=0 (χ 1 ) = E + O(δ). (4.1) Now let ϑ = (ϑ 1, ϑ 2 ) : D x D x be a change of coordinates from (µ, η) to (s, t) C ǫ. In (s, t)-coordinates we by assumption have that L s D x = {(s, g(s))} for s (σ 1, σ 2 ), say, and g satisfies g, dg ds < δ. By (4.1) we have d ds ϑ1 (χ 1 ϑ 1 (s, g(s))) = 1 + O(δ) for all s (σ 1, σ 2 ), so χ 1 (L s ) is a graph of a function α in the parameterization of C ǫ if we let δ be small enough. Furthermore, for the slope of α we get that α d ds = ϑ2 (χ 1 ϑ 1 (s, g(s))) d ds ϑ1 (χ 1 ϑ 1 (s, g(s))) = O(δ) 1 + O(δ) = O(δ). Similar calculations show that χ 1 (L t) is a subset of both a graph over S 1 in the parameterization of C ǫ and a graph over the η-axis in the parameterization of D x for δ sufficiently small. Moreover, the slope of these graphs are of order δ. Now we want to find an isopoty χ t, 0 τ 1, taking χ 1 (L t ) to L t in a neighborhood of x, and such that χ 1 (χ 1 (L s )) still coincides with L s here. Since we by assumption had x Q δ, where x C is the double point corresponding to x, we have that χ 1 (x ) ( δ, δ) {0}. Hence we can find a 0 < δ < δ so that χ 1 (L t) coincides with the graph of an exact function f : R ( δ, δ) in ( δ, δ) ( δ, δ ), that is, χ 1 (L t) (( δ, δ) ( δ, δ )) = {(f(η), η)}. In addition we can choose f so that df dη < δ for all η R and so that f(η) = 0 for η > σ. Let F(η) = η σ f(η )dη. Then the isotopy Φ τ X H = χ τ, 0 τ 1, obtained from the Hamiltonian H(µ, η) = ψ(µ)f(η) takes χ 1 (L t ) to the η-axis in ( δ, δ) ( δ, δ ), and we have that χ 1 χ 1 (L s ) still coincides with the µ-axis in a neighborhood of (0, 0) = χ 1 χ 1 (x ). As before we get that ( χ 1 ) = E + [ dψ dµ f ψ df dη d 2 ψ dµ F dψ 2 dµ f ] 16 + O(1/N) = E + O(δ)

19 for N large. So for χ 1 χ 1 (L s) in C ǫ D x we have, with χ 1 (C ) = {(s, α(s))} here, that [ ] d ds ϑ χ 1ϑ (s, α) = 0 α + O(δ). Hence χ 1 χ 1 (L s) will be a subset of a graph over S 1 for δ small enough, and similarly we get that χ 1 χ 1 (L t) is a subset of a graph over S 1 in the parameterization of C ǫ too. By doing the same thing at all double points of C we get an area preserving isotopy taking C to C in a neighborhood of every double point of C, and so that the time-1 image of C is still a graph over S 1 in C ǫ. So by Lemma 2.2 there is an area preserving isotopy taking C completely to C. 5 Proof of Theorem 1.1 Now if we compose Lemma 3.4 with Lemma 4.3 and Lemma 4.2 we can prove our theorem: Proof of Theorem 1.1. By Lemma 3.4 there is a semi-area preserving isotopy φ τ, 0 τ 1, with respect to C taking C to C. Let C τ = φ τ (C) for τ [0, 1], and for each τ 0 [0, 1] let {Cτ ǫ 0, D τ 0 x } be a regular neighborhood of C τ0. By Lemma 4.2 we can find a δ τ0 > 0 so that for every C τ which is δ τ0 -close to C τ0 there exists an area preserving isotopy taking C τ to C τ0. And by Lemma 4.3 there is a ν τ0 > 0 so that C τ is δ τ0 -close to C τ0 for all 0 τ τ 0 < ν τ0. Let ν = min τ0 I{ν τ0 } and let 0 = τ 1 <... < τ n = 1 be a partition of I so that τ i+1 τ i < ν for 1 i < n. Then by Lemma 4.2 there is an area preserving isotopy taking C τi+1 to C τi for i = 1,..., n 1. Composing the inverses of these isotopies we thus get an area preserving isotopy taking C to C. References [1] V. I. Arnold. Mathematical Methods of Classical Mechanics. Springer- Verlag,

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