If Y is a lagrangian submanifold of (M1 — M2 , ω), then its “twist” Y σ := σ(Y )

is a lagrangian submanifold of (M1 — M2 , ω).

Recipe for producing symplectomorphisms M1 = T — X1 ’ M2 = T — X2 :

1. Start with a lagrangian submanifold Y of (M1 — M2 , ω).

2. Twist it to obtain a lagrangian submanifold Y σ of (M1 — M2 , ω).

3. Check whether Y σ is the graph of some di¬eomorphism • : M1 ’ M2 .

4. If it is, then • is a symplectomorphism.

Let i : Y σ ’ M1 — M2 be the inclusion map

Yσ

d

pr1 —¦ i d pr2 —¦ i

d

d

© d

‚

•? E M2

M1

Step 3 amounts to checking whether pr1 —¦ i and pr2 —¦ i are di¬eomorphisms. If

yes, then • := (pr2 —¦ i) —¦ (pr1 —¦ i)’1 is a di¬eomorphism.

T — (X1 — X2 ), we

In order to obtain lagrangian submanifolds of M1 — M2

can use the method of generating functions.

22

23

4.2 Method of Generating Functions

4.2 Method of Generating Functions

For any f ∈ C ∞ (X1 — X2 ), df is a closed 1-form on X1 — X2 . The lagrangian

submanifold generated by f is

Yf := {((x, y), (df )(x,y) ) | (x, y) ∈ X1 — X2 } .

We adopt the notation

—

:= (df )(x,y) projected to Tx X1 — {0},

dx f

—

:= (df )(x,y) projected to {0} — Ty X2 ,

dy f

which enables us to write

Yf = {(x, y, dx f, dy f ) | (x, y) ∈ X1 — X2 }

and

Yfσ = {(x, y, dx f, ’dy f ) | (x, y) ∈ X1 — X2 } .

When Yfσ is in fact the graph of a di¬eomorphism • : M1 ’ M2 , we call • the

symplectomorphism generated by f , and call f the generating function,

of • : M1 ’ M2 .

So when is Yfσ the graph of a di¬eomorphism • : M1 ’ M2 ?

Let (U1 , x1 , . . . , xn ), (U2 , y1 , . . . , yn ) be coordinate charts for X1 , X2 , with

associated charts (T — U1 , x1 , . . . , xn , ξ1 , . . . , ξn ), (T — U2 , y1 , . . . , yn , ·1 , . . . , ·n ) for

M1 , M2 . The set

Yfσ = {(x, y, dx f, ’dy f ) | (x, y) ∈ X1 — X2 }

is the graph of • : M1 ’ M2 if and only if, for any (x, ξ) ∈ M1 and (y, ·) ∈ M2 ,

we have

•(x, ξ) = (y, ·) ⇐’ ξ = dx f and · = ’dy f .

Therefore, given a point (x, ξ) ∈ M1 , to ¬nd its image (y, ·) = •(x, ξ) we must

solve the “Hamilton” equations

±

‚f

ξi =

(x, y) ()

‚xi

· = ’ ‚f (x, y)

i ()

‚yi

If there is a solution y = •1 (x, ξ) of ( ), we may feed it to ( ) thus obtaining

· = •2 (x, ξ), so that •(x, ξ) = (•1 (x, ξ), •2 (x, ξ)). Now by the implicit function

theorem, in order to solve ( ) locally for y in terms of x and ξ, we need the

condition n

‚ ‚f

det =0.

‚yj ‚xi i,j=1

24 4 GENERATING FUNCTIONS

This is a necessary local condition for f to generate a symplectomorphism •.

Locally this is also su¬cient, but globally there is the usual bijectivity issue.

2

Rn , and f (x, y) = ’ |x’y| , the

R n , X2 = U 2

Example. Let X1 = U1 2

square of euclidean distance up to a constant.

The “Hamilton” equations are

± ±

‚f

ξi = yi = xi + ξ i

= y i ’ xi

‚xi ⇐’

· = ’ ‚f = y ’ x

i ·i = ξ i

i i

‚yi

The symplectomorphism generated by f is

•(x, ξ) = (x + ξ, ξ) .

If we use the euclidean inner product to identify T — Rn with T Rn , and hence

regard • as • : T Rn ’ T Rn and interpret ξ as the velocity vector, then the sym-

plectomorphism • corresponds to free translational motion in euclidean space.

B

¨

¨¨

r¨

ξ B¨

¨

¨ x+ξ

¨¨

¨

r

x

™¦

4.3 Application to Geodesic Flow

Let V be an n-dimensional vector space. A positive inner product G on V

is a bilinear map G : V — V ’ R which is

symmetric : G(v, w) = G(w, v) , and

positive-de¬nite : G(v, v) > 0 when v=0.

De¬nition 4.1 A riemannian metric on a manifold X is a function g which

assigns to each point x ∈ X a positive inner product gx on Tx X.

A riemannian metric g is smooth if for every smooth vector ¬eld v : X ’

T X the real-valued function x ’ gx (vx , vx ) is a smooth function on X.

De¬nition 4.2 A riemannian manifold (X, g) is a manifold X equipped with

a smooth riemannian metric g.

25

4.3 Application to Geodesic Flow

The arc-length of a piecewise smooth curve γ : [a, b] ’ X on a riemannian

manifold (X, g) is

b

dγ dγ

gγ(t) , dt .

dt dt

a

y = γ(b)

r

γ

E

r

x = γ(a)

De¬nition 4.3 The riemannian distance between two points x and y of a

connected riemannian manifold (X, g) is the in¬mum d(x, y) of the set of all

arc-lengths for piecewise smooth curves joining x to y.

A smooth curve joining x to y is a minimizing geodesic7 if its arc-length

is the riemannian distance d(x, y).

A riemannian manifold (X, g) is geodesically convex if every point x is

joined to every other point y by a unique minimizing geodesic.

Example. On X = Rn with T X Rn — Rn , let gx (v, w) = v, w , gx (v, v) =

|v|2 , where ·, · is the euclidean inner product, and | · | is the euclidean norm.

Then (Rn , ·, · ) is a geodesically convex riemannian manifold, and the rieman-

nian distance is the usual euclidean distance d(x, y) = |x ’ y|. ™¦

Suppose that (X, g) is a geodesically convex riemannian manifold. Consider

the function

d(x, y)2

f : X — X ’’ R , f (x, y) = ’ .

2

What is the symplectomorphism • : T — X ’ T — X generated by f ?

The metric gx : Tx X — Tx X ’ R induces an identi¬cation

—

’’ Tx X

g x : Tx X

’’ gx (v, ·)

v

Use g to translate • into a map • : T X ’ T X.

7 Inriemannian geometry, a geodesic is a curve which locally minimizes distance and

whose velocity is constant.

26 4 GENERATING FUNCTIONS

We need to solve

gx (v) = ξi = dx f (x, y)

= ’dy f (x, y)

gy (w) = ·i

for (y, ·) in terms of (x, ξ) in order to ¬nd •, or, equivalently, for (y, w) in terms

(x, v) in order to ¬nd •.

Let γ be the geodesic with initial conditions γ(0) = x and dγ (0) = v.

dt

¢

¢ γ

v¢ E

¢

¢

¢

r

x

Then the symplectomorphism • corresponds to the map

’’ T X

•: TX

’’ (γ(1), dγ (1)) .

(x, v) dt

This is called the geodesic ¬‚ow on X (see Homework 4).

Homework 4: Geodesic Flow

This set of problems is adapted from [52].

Let (X, g) be a riemannian manifold. The arc-length of a smooth curve

γ : [a, b] ’ X is

b

dγ dγ dγ dγ

arc-length of γ := dt , where := gγ(t) , .

dt dt dt dt

a

1. Show that the arc-length of γ is independent of the parametrization of

γ, i.e., show that, if we reparametrize γ by „ : [a , b ] ’ [a, b], the new

curve γ = γ —¦ „ : [a , b ] ’ X has the same arc-length. A curve γ is

called a curve of constant velocity when dγ is independent of t. Show

dt

that, given any curve γ : [a, b] ’ X (with dγ never vanishing), there is

dt

a reparametrization „ : [a, b] ’ [a, b] such that γ —¦ „ : [a, b] ’ X is of

constant velocity.

2

b

dγ

2. Given a smooth curve γ : [a, b] ’ X, the action of γ is A(γ) := dt.

dt

a

Show that, among all curves joining x to y, γ minimizes the action if and

only if γ is of constant velocity and γ minimizes arc-length.

Hint: Suppose that γ is of constant velocity, and let „ : [a, b] ’ [a, b] be

a reparametrization. Show that A(γ —¦ „ ) ≥ A(γ), with equality only when

„ = identity.

3. Assume that (X, g) is geodesically convex, that is, any two points x, y ∈ X

are joined by a unique (up to reparametrization) minimizing geodesic; its

arc-length d(x, y) is called the riemannian distance between x and y.

Assume also that (X, g) is geodesically complete, that is, every minimizing

geodesic can be extended inde¬nitely. Given (x, v) ∈ T X, let exp(x, v) :

R ’ X be the unique minimizing geodesic of constant velocity with initial

conditions exp(x, v)(0) = x and d exp(x,v) (0) = v.

dt

1

Consider the function • : X —X ’ R given by •(x, y) = ’ 2 ·d(x, y)2 . Let

—

d•x and d•y be the components of d•(x,y) with respect to T(x,y) (X —X)

— —

Tx X — Ty X. Recall that, if

“σ = {(x, y, d•x , ’d•y ) | (x, y) ∈ X — X}

•

is the graph of a di¬eomorphism f : T — X ’ T — X, then f is the symplec-

tomorphism generated by •. In this case, f (x, ξ) = (y, ·) if and only if

ξ = d•x and · = ’d•y .

Show that, under the identi¬cation of T X with T — X by g, the sym-

plectomorphism generated by • coincides with the map T X ’ T X,

(x, v) ’ exp(x, v)(1).

27

28 HOMEWORK 4

Hint: The metric g provides the identi¬cations Tx Xv ξ(·) = gx (v, ·) ∈

— X. We need to show that, given (x, v) ∈ T X, the unique solution of

Tx

gx (v, ·) = d•x (·)

is (y, w) = (exp(x, v)(1), d exp(x,v) (1)).

()

gy (w, ·) = ’d•y (·) dt

Look up the Gauss lemma in a book on riemannian geometry. It asserts that

geodesics are orthogonal to the level sets of the distance function.

d exp(x,v)

To solve the ¬rst line in ( ) for y, evaluate both sides at v = (0).

dt

Conclude that y = exp(x, v)(1). Check that d•x (v ) = 0 for vectors v ∈ Tx X

orthogonal to v (that is, gx (v, v ) = 0); this is a consequence of •(x, y) being

the arc-length of a minimizing geodesic, and it su¬ces to check locally.

The vector w is obtained from the second line of ( ). Compute

d exp(x,v)

’d•y ( (1)). Then evaluate ’d•y at vectors w ∈ Ty X orthogo-

dt

nal to d exp(x,v) (1); this pairing is again 0 because •(x, y) is the arc-length

dt

of a minimizing geodesic. Conclude, using the nondegeneracy of g, that

w = d exp(x,v) (1).

dt

For both steps, it might be useful to recall that, given a function f : X ’ R

d

and a tangent vector v ∈ Tx X, we have dfx (v) = [f (exp(x, v)(u))] u=0 .

du

5 Recurrence

5.1 Periodic Points

Let X be an n-dimensional manifold. Let M = T — X be its cotangent bundle

with canonical symplectic form ω.

Suppose that we are given a smooth function f : X —X ’ R which generates

a symplectomorphism • : M ’ M , •(x, dx f ) = (y, ’dy f ), by the recipe of the

previous lecture.

What are the ¬xed points of •?

De¬ne ψ : X ’ R by ψ(x) = f (x, x).

Proposition 5.1 There is a one-to-one correspondence between the ¬xed points

of • and the critical points of ψ.

Proof. At x0 ∈ X, dx0 ψ = (dx f + dy f )|(x,y)=(x0 ,x0 ) . Let ξ = dx f |(x,y)=(x0 ,x0 ) .

x0 is a critical point of ψ ⇐’ dx0 ψ = 0 ⇐’ dy f |(x,y)=(x0 ,x0 ) = ’ξ .

Hence, the point in “σ corresponding to (x, y) = (x0 , x0 ) is (x0 , x0 , ξ, ξ). But

f

“σ is the graph of •, so •(x0 , ξ) = (x0 , ξ) is a ¬xed point. This argument also

f

works backwards.

Consider the iterates of •,

•(N ) = • —¦ • —¦ . . . —¦ • : M ’ M , N = 1, 2, . . . ,

N

each of which is a symplectomorphism of M . According to the previous propo-

sition, if •(N ) : M ’ M is generated by f (N ) , then there is a correspondence

critical points of

1’1

¬xed points of •(N ) ←’ (N )

: X ’ R , ψ (N ) (x) = f (N ) (x, x)

ψ

Knowing that • is generated by f , does •(2) have a generating function?

The answer is a partial yes:

Fix x, y ∈ X. De¬ne a map

’’ R

X

’’ f (x, z) + f (z, y)

z

Suppose that this map has a unique critical point z0 , and that z0 is nondegen-

erate. Let

f (2) (x, y) := f (x, z0 ) + f (z0 , y) .

Theorem 5.2 The function f (2) : X — X ’ R is smooth and is a generating

function for •(2) .

29

30 5 RECURRENCE

Proof. The point z0 is given implicitly by dy f (x, z0 ) + dx f (z0 , y) = 0. The

nondegeneracy condition is

‚ ‚f ‚f

det (x, z) + (z, y) =0.

‚zi ‚yj ‚xj

By the implicit function theorem, z0 = z0 (x, y) is smooth.

As for the second assertion, f (2) (x, y) is a generating function for •(2) if and

only if

•(2) (x, dx f (2) ) = (y, ’dy f (2) )

(assuming that, for each ξ ∈ Tx X, there is a unique y ∈ X for which dx f (2) = ξ).

—

Since • is generated by f , and z0 is critical, we obtain,

= •(•(x, dx f (2) (x, y))

•(2) (x, dx f (2) (x, y)) = •(z0 , ’dy f (x, z0 ))

=dx f (x,z0 )

= (y, ’dy f (z0 , y) ) .

= •(z0 , dx f (z0 , y))

=’dy f (2) (x,y)

Exercise. What is a generating function for •(3) ?

Hint: Suppose that the function

X — X ’’ R

(z, u) ’’ f (x, z) + f (z, u) + f (u, y)

has a unique critical point (z0 , u0 ), and that it is a nondegenerate critical point.

Let ψ (3) (x, y) = f (x, z0 ) + f (z0 , u0 ) + f (u0 , y). ™¦

5.2 Billiards

Let χ : R ’ R2 be a smooth plane curve which is 1-periodic, i.e., χ(s+1) = χ(s),

dχ

and parametrized by arc-length, i.e., = 1. Assume that the region Y

ds

enclosed by χ is convex, i.e., for any s ∈ R, the tangent line {χ(s) + t dχ | t ∈ R}

ds

intersects X := ‚Y (= the image of χ) at only the point χ(s).

31

5.2 Billiards

'

X = ‚Y

r χ(s)

Suppose that we throw a ball into Y rolling with constant velocity and

bouncing o¬ the boundary with the usual law of re¬‚ection. This determines a

map

• : R/Z — (’1, 1) ’’ R/Z — (’1, 1)

(x, v) ’’ (y, w)

by the rule

when the ball bounces o¬ x with angle θ = arccos v, it will next collide with y

and bounce o¬ with angle ν = arccos w.

4

4

%

4

4

4

xr

˜

˜

˜

˜

˜

j ˜

˜ 4

˜ 4

B

˜ 4

˜4

r

y

Let f : R/Z — R/Z ’ R be de¬ned by f (x, y) = ’|x ’ y|; f is smooth o¬

the diagonal. Use χ to identify R/Z with the image curve X.

Suppose that •(x, v) = (y, w), i.e., (x, v) and (y, w) are successive points on

32 5 RECURRENCE

the orbit described by the ball. Then

± df x’y

=’ projected onto Tx X = v

dx

|x ’ y|

df

y’x

=’ = ’w

projected onto Ty X

|x ’ y|

dy

or, equivalently,

±d y ’ x dχ

·

ds f (χ(s), y) = = cos θ = v

|x ’ y| ds

d

x ’ y dχ

· = ’ cos ν = ’w

f (x, χ(s)) =

|x ’ y| ds

ds

We conclude that f is a generating function for •. Similar approaches work

for higher dimensional billiards problems.

Periodic points are obtained by ¬nding critical points of

X —... —X ’’ R , N >1

N

(x1 , . . . , xN ) ’’ f (x1 , x2 ) + f (x2 , x3 ) + . . . + f (xN ’1 , xN ) + f (xN , x1 )

= |x1 ’ x2 | + . . . + |xN ’1 ’ xN | + |xN ’ x1 |

that is, by ¬nding the N -sided (generalized) polygons inscribed in X of critical

perimeter.

Notice that

R/Z — (’1, 1) {(x, v) | x ∈ X, v ∈ Tx X, |v| < 1} A

is the open unit tangent ball bundle of a circle X, that is, an open annulus A.

The map • : A ’ A is area-preserving.

5.3 Poincar´ Recurrence

e

Theorem 5.3 (Poincar´ Recurrence Theorem) Suppose that • : A ’ A

e

is an area-preserving di¬eomorphism of a ¬nite-area manifold A. Let p ∈ A,

and let U be a neighborhood of p. Then there is q ∈ U and a positive integer N

such that •(N ) (q) ∈ U.

Proof. Let U0 = U, U1 = •(U), U2 = •(2) (U), . . .. If all of these sets were

disjoint, then, since Area (Ui ) = Area (U) > 0 for all i, we would have

Area A ≥ Area (U0 ∪ U1 ∪ U2 ∪ . . .) = Area (Ui ) = ∞ .

i

33

5.3 Poincar´ Recurrence

e

To avoid this contradiction we must have •(k) (U) © •(l) (U) = … for some k > l,

which implies •(k’l) (U) © U = ….

Hence, eternal return applies to billiards...

Remark. Theorem 5.3 clearly generalizes to volume-preserving di¬eomor-

™¦

phisms in higher dimensions.

Theorem 5.4 (Poincar´™s Last Geometric Theorem) Suppose • : A ’

e

A is an area-preserving di¬eomorphism of the closed annulus A = R/Z — [’1, 1]

which preserves the two components of the boundary, and twists them in opposite

directions. Then • has at least two ¬xed points.

This theorem was proved in 1925 by Birkho¬, and hence is also called the

Poincar´-Birkho¬ theorem. It has important applications to dynamical sys-

e

tems and celestial mechanics. The Arnold conjecture (1966) on the existence of

¬xed points for symplectomorphisms of compact manifolds (see Lecture 9) may

be regarded as a generalization of the Poincar´-Birkho¬ theorem. This con-

e

jecture has motivated a signi¬cant amount of recent research involving a more

general notion of generating function; see, for instance, [34, 45].

Part III

Local Forms

Inspired by the elementary normal form in symplectic linear algebra (Theo-

rem 1.1), we will go on to describe normal neighborhoods of a point (the Dar-

boux theorem) and of a lagrangian submanifold (the Weinstein theorems), inside

a symplectic manifold. The main tool is the Moser trick, explained in Lecture 7,

which leads to the crucial Moser theorems and which is at the heart of many

arguments in symplectic geometry.

In order to prove the normal forms, we need the (non-symplectic) ingredients

discussed in Lecture 6; for more on these topics, see, for instance, [18, 54, 94].

6 Preparation for the Local Theory

6.1 Isotopies and Vector Fields

Let M be a manifold, and ρ : M — R ’ M a map, where we set ρt (p) := ρ(p, t).

De¬nition 6.1 The map ρ is an isotopy if each ρt : M ’ M is a di¬eomor-

phism, and ρ0 = idM .

Given an isotopy ρ, we obtain a time-dependent vector ¬eld, that is, a

family of vector ¬elds vt , t ∈ R, which at p ∈ M satisfy

d

q = ρ’1 (p) ,

vt (p) = ρs (q) where t

ds s=t

i.e.,

dρt

= vt —¦ ρt .

dt

Conversely, given a time-dependent vector ¬eld vt , if M is compact or if the

vt ™s are compactly supported, there exists an isotopy ρ satisfying the previous

ordinary di¬erential equation.

Suppose that M is compact. Then we have a one-to-one correspondence

1’1

{isotopies of M } ←’ {time-dependent vector ¬elds on M }

ρt , t ∈ R ←’ vt , t ∈ R

De¬nition 6.2 When vt = v is independent of t, the associated isotopy is called

the exponential map or the ¬‚ow of v and is denoted exp tv; i.e., {exp tv :

M ’ M | t ∈ R} is the unique smooth family of di¬eomorphisms satisfying

d

exp tv|t=0 = idM and (exp tv)(p) = v(exp tv(p)) .

dt

35

36 6 PREPARATION FOR THE LOCAL THEORY

De¬nition 6.3 The Lie derivative is the operator

d

(exp tv)— ω|t=0 .

Lv : „¦k (M ) ’’ „¦k (M ) Lv ω :=

de¬ned by

dt

When a vector ¬eld vt is time-dependent, its ¬‚ow, that is, the corresponding

isotopy ρ, still locally exists by Picard™s theorem. More precisely, in the neigh-

borhood of any point p and for su¬ciently small time t, there is a one-parameter

family of local di¬eomorphisms ρt satisfying

dρt

= vt —¦ ρt and ρ0 = id .

dt

Hence, we say that the Lie derivative by vt is

d

(ρt )— ω|t=0 .

Lvt : „¦k (M ) ’’ „¦k (M ) Lvt ω :=

de¬ned by

dt

Exercise. Prove the Cartan magic formula,

Lv ω = ±v dω + d±v ω ,

and the formula

d—

ρ t ω = ρ — Lvt ω , ()

t

dt

where ρ is the (local) isotopy generated by vt . A good strategy for each formula

is to follow the steps:

1. Check the formula for 0-forms ω ∈ „¦0 (M ) = C ∞ (M ).

2. Check that both sides commute with d.

3. Check that both sides are derivations of the algebra („¦— (M ), §). For

instance, check that

Lv (ω § ±) = (Lv ω) § ± + ω § (Lv ±) .

4. Notice that, if U is the domain of a coordinate system, then „¦• (U) is gen-

erated as an algebra by „¦0 (U) and d„¦0 (U), i.e., every element in „¦• (U) is

a linear combination of wedge products of elements in „¦0 (U) and elements

in d„¦0 (U).

™¦

We will need the following improved version of formula ( ).

Theorem 6.4 For a smooth family ωt , t ∈ R, of d-forms, we have

d— dωt

ρ t ωt = ρ — L v t ωt + .

t

dt dt

37

6.2 Tubular Neighborhood Theorem

Proof. If f (x, y) is a real function of two variables, by the chain rule we have

d d d

f (t, t) = f (x, t) + f (t, y) .

dt dx dy

x=t y=t

Therefore,

d— d— d—

ρ ωt = ρ ωt + ρ ωy

dt t dx x dy t

x=t y=t

dωy

ρ— L v x ω t by ( ) ρ—

x t dy

x=t y=t

dωt

= ρ — L v t ωt + .

t

dt

6.2 Tubular Neighborhood Theorem

Let M be an n-dimensional manifold, and let X be a k-dimensional submanifold

where k < n and with inclusion map

i:X ’M .

At each x ∈ X, the tangent space to X is viewed as a subspace of the tangent

space to M via the linear inclusion dix : Tx X ’ Tx M , where we denote x =

i(x). The quotient Nx X := Tx M/Tx X is an (n ’ k)-dimensional vector space,

known as the normal space to X at x. The normal bundle of X is

N X = {(x, v) | x ∈ X , v ∈ Nx X} .

The set N X has the structure of a vector bundle over X of rank n ’ k under the

natural projection, hence as a manifold N X is n-dimensional. The zero section

of N X,

i0 : X ’ N X , x ’ (x, 0) ,

embeds X as a closed submanifold of N X. A neighborhood U0 of the zero

section X in N X is called convex if the intersection U0 © Nx X with each ¬ber

is convex.

Theorem 6.5 (Tubular Neighborhood Theorem) There exist a convex

neighborhood U0 of X in N X, a neighborhood U of X in M , and a di¬eomor-

phism • : U0 ’ U such that

• E U ⊆M

N X ⊇ U0

d

s

d

d commutes.

i0 d i

d

X

38 6 PREPARATION FOR THE LOCAL THEORY

Outline of the proof.

• Case of M = Rn , and X is a compact submanifold of Rn .

Theorem 6.6 (µ-Neighborhood Theorem)

Let U µ = {p ∈ Rn : |p ’ q| < µ for some q ∈ X} be the set of points at a

distance less than µ from X. Then, for µ su¬ciently small, each p ∈ U µ

has a unique nearest point q ∈ X (i.e., a unique q ∈ X minimizing |q ’x|).

π

Moreover, setting q = π(p), the map U µ ’ X is a (smooth) submersion

with the property that, for all p ∈ U µ , the line segment (1 ’ t)p + tq,

0 ¤ t ¤ 1, is in U µ .

The proof is part of Homework 5. Here are some hints.

At any x ∈ X, the normal space Nx X may be regarded as an (n ’ k)-

dimensional subspace of Rn , namely the orthogonal complement in Rn of

the tangent space to X at x:

{v ∈ Rn : v ⊥ w , for all w ∈ Tx X} .

Nx X

We de¬ne the following open neighborhood of X in N X:

N X µ = {(x, v) ∈ N X : |v| < µ} .

Let

’’ Rn

exp : NX

’’ x + v .

(x, v)

Restricted to the zero section, exp is the identity map on X.

Prove that, for µ su¬ciently small, exp maps N X µ di¬eomorphically onto

U µ , and show also that the diagram

exp E Uµ

NXµ

d

d

d commutes.

π0 d π

‚

d

©

X

• Case where X is a compact submanifold of an arbitrary manifold M .

Put a riemannian metric g on M , and let d(p, q) be the riemannian distance

between p, q ∈ M . The µ-neighborhood of a compact submanifold X is

U µ = {p ∈ M | d(p, q) < µ for some q ∈ X} .

Prove the µ-neighborhood theorem in this setting: for µ small enough, the

following assertions hold.

39

6.3 Homotopy Formula

“ Any p ∈ U µ has a unique point q ∈ X with minimal d(p, q). Set q = π(p).

π

“ The map U µ ’ X is a submersion and, for all p ∈ U µ , there is a unique

geodesic curve γ joining p to q = π(p).

“ The normal space to X at x ∈ X is naturally identi¬ed with a subspace

of Tx M :

{v ∈ Tx M | gx (v, w) = 0 , for any w ∈ Tx X} .

Nx X

Let N X µ = {(x, v) ∈ N X | gx (v, v) < µ}.

“ De¬ne exp : N X µ ’ M by exp(x, v) = γ(1), where γ : [0, 1] ’ M

is the geodesic with γ(0) = x and dγ (0) = v. Then exp maps N X µ

dt

di¬eomorphically to U µ .

• General case.

When X is not compact, adapt the previous argument by replacing µ by

an appropriate continuous function µ : X ’ R+ which tends to zero fast

enough as x tends to in¬nity.

Restricting to the subset U 0 ⊆ N X from the tubular neighborhood theorem,

π0 ’1

we obtain a submersion U0 ’’ X with all ¬bers π0 (x) convex. We can carry

this ¬bration to U by setting π = π0 —¦ •’1 :

U0 ⊆ NX U ⊆M

is a ¬bration =’ is a ¬bration

π0 “ π“

X X

This is called the tubular neighborhood ¬bration.

6.3 Homotopy Formula

Let U be a tubular neighborhood of a submanifold X in M . The restriction i— :

d d

HdeRham(U) ’ HdeRham (X) by the inclusion map is surjective. As a corollary

of the tubular neighborhood ¬bration, i— is also injective: this follows from the

homotopy-invariance of de Rham cohomology.

Corollary 6.7 For any degree , HdeRham (U) HdeRham (X).

At the level of forms, this means that, if ω is a closed -form on U and i— ω

is exact on X, then ω is exact. We will need the following related result.

Theorem 6.8 If a closed -form ω on U has restriction i— ω = 0, then ω is

exact, i.e., ω = dµ for some µ ∈ „¦d’1 (U). Moreover, we can choose µ such that

µx = 0 at all x ∈ X.

40 6 PREPARATION FOR THE LOCAL THEORY

Proof. Via • : U0 ’’ U, it is equivalent to work over U0 . De¬ne for every

0 ¤ t ¤ 1 a map

U0 ’’ U0

ρt :

(x, v) ’’ (x, tv) .

This is well-de¬ned since U0 is convex. The map ρ1 is the identity, ρ0 = i0 —¦

π0 , and each ρt ¬xes X, that is, ρt —¦ i0 = i0 . We hence say that the family

{ρt | 0 ¤ t ¤ 1} is a homotopy from i0 —¦ π0 to the identity ¬xing X. The

map π0 : U0 ’ X is called a retraction because π0 —¦ i0 is the identity. The

submanifold X is then called a deformation retract of U.

A (de Rham) homotopy operator between ρ0 = i0 —¦ π0 and ρ1 = id is a

linear map

Q : „¦d (U0 ) ’’ „¦d’1 (U0 )

satisfying the homotopy formula

Id ’ (i0 —¦ π0 )— = dQ + Qd .

When dω = 0 and i— ω = 0, the operator Q gives ω = dQω, so that we can take

0

µ = Qω. A concrete operator Q is given by the formula:

1

ρ— (±vt ω) dt ,

Qω = t

0

where vt , at the point q = ρt (p), is the vector tangent to the curve ρs (p) at

s = t. The proof that Q satis¬es the homotopy formula is below.

In our case, for x ∈ X, ρt (x) = x (all t) is the constant curve, so vt vanishes

at all x for all t, hence µx = 0.

To check that Q above satis¬es the homotopy formula, we compute

1 1

ρ— (±vt dω)dt ρ— (±vt ω)dt

Qdω + dQω = +d

t t

0 0

1

ρ— (±vt dω + d±vt ω)dt ,

= t

0

Lvt ω

where Lv denotes the Lie derivative along v (reviewed in the next section), and

we used the Cartan magic formula: Lv ω = ±v dω + d±v ω. The result now follows

from

d—

ρ t ω = ρ — Lvt ω

t

dt

and from the fundamental theorem of calculus:

1

d—

ρ ω dt = ρ— ω ’ ρ— ω .

Qdω + dQω = 1 0

dt t

0

Homework 5: Tubular Neighborhoods in Rn

1. Let X be a k-dimensional submanifold of an n-dimensional manifold M .

Let x be a point in X. The normal space to X at x is the quotient space

Nx X = Tx M/Tx X ,

and the normal bundle of X in M is the vector bundle N X over X

whose ¬ber at x is Nx X.

(a) Prove that N X is indeed a vector bundle.

(b) If M is Rn , show that Nx X can be identi¬ed with the usual “normal

space” to X in Rn , that is, the orthogonal complement in Rn of the

tangent space to X at x.

2. Let X be a k-dimensional compact submanifold of Rn . Prove the tubular

neighborhood theorem in the following form.

(a) Given µ > 0 let Uµ be the set of all points in Rn which are at a distance

less than µ from X. Show that, for µ su¬ciently small, every point

p ∈ Uµ has a unique nearest point π(p) ∈ X.

(b) Let π : Uµ ’ X be the map de¬ned in (a) for µ su¬ciently small.

Show that, if p ∈ Uµ , then the line segment (1 ’ t) · p + t · π(p),

0 ¤ t ¤ 1, joining p to π(p) lies in Uµ .

(c) Let N Xµ = {(x, v) ∈ N X such that |v| < µ}. Let exp : N X ’ Rn

be the map (x, v) ’ x + v, and let ν : N Xµ ’ X be the map

(x, v) ’ x. Show that, for µ su¬ciently small, exp maps N Xµ dif-

feomorphically onto Uµ , and show also that the following diagram

commutes:

exp E Uµ

N Xµ

d

d

d

νd π

‚

d

©

X

3. Suppose that the manifold X in the previous exercise is not compact.

Prove that the assertion about exp is still true provided we replace µ by a

continuous function

µ : X ’ R+

which tends to zero fast enough as x tends to in¬nity.

41

7 Moser Theorems

7.1 Notions of Equivalence for Symplectic Structures

Let M be a 2n-dimensional manifold with two symplectic forms ω0 and ω1 , so

that (M, ω0 ) and (M, ω1 ) are two symplectic manifolds.

De¬nition 7.1 We say that

• (M, ω0 ) and (M, ω1 ) are symplectomorphic if there is a di¬eomorphism

• : M ’ M with •— ω1 = ω0 ;

• (M, ω0 ) and (M, ω1 ) are strongly isotopic if there is an isotopy ρt :

M ’ M such that ρ— ω1 = ω0 ;

1

• (M, ω0 ) and (M, ω1 ) are deformation-equivalent if there is a smooth

family ωt of symplectic forms joining ω0 to ω1 ;

• (M, ω0 ) and (M, ω1 ) are isotopic if they are deformation-equivalent with

[ωt ] independent of t.

Clearly, we have

strongly isotopic =’ symplectomorphic , and

isotopic =’ deformation-equivalent .

We also have

strongly isotopic =’ isotopic

because, if ρt : M ’ M is an isotopy such that ρ— ω1 = ω0 , then the set ωt :=

1

—

ρt ω1 is a smooth family of symplectic forms joining ω1 to ω0 and [ωt ] = [ω1 ],

∀t, by the homotopy invariance of de Rham cohomology. As we will see below,

the Moser theorem states that, on a compact manifold,

isotopic =’ strongly isotopic .

7.2 Moser Trick

Problem. Given a 2n-dimensional manifold M , a k-dimensional submanifold

X, neighborhoods U0 , U1 of X, and symplectic forms ω0 , ω1 on U0 , U1 , does

there exist a symplectomorphism preserving X? More precisely, does there

exist a di¬eomorphism • : U0 ’ U1 with •— ω1 = ω0 and •(X) = X?

At the two extremes, we have:

Case X = point: Darboux theorem “ see Lecture 8.

Case X = M : Moser theorem “ discussed here:

Let M be a compact manifold with symplectic forms ω0 and ω1 .

42

43

7.2 Moser Trick

“ Are (M, ω0 ) and (M, ω1 ) symplectomorphic?

I.e., does there exist a di¬eomorphism • : M ’ M such that •— ω0 = ω1 ?

1

Moser asked whether we can ¬nd such an • which is homotopic to id M . A

necessary condition is [ω0 ] = [ω1 ] ∈ H 2 (M ; R) because: if • ∼ idM , then, by

the homotopy formula, there exists a homotopy operator Q such that

id— ω1 ’ •— ω1 = dQω1 + Q dω1

M

0

—

=’ ω1 = • ω1 + d(Qω1 )

[ω1 ] = [•— ω1 ] = [ω0 ] .

=’

“ If [ω0 ] = [ω1 ], does there exist a di¬eomorphism • homotopic to idM such

that •— ω1 = ω0 ?

Moser [86] proved that the answer is yes, with a further hypothesis as in

Theorem 7.2. McDu¬ showed that, in general, the answer is no; for a coun-

terexample, see Example 7.23 in [82].

Theorem 7.2 (Moser Theorem “ Version I) Suppose that [ω0 ] = [ω1 ] and

that the 2-form ωt = (1 ’ t)ω0 + tω1 is symplectic for each t ∈ [0, 1]. Then there

exists an isotopy ρ : M — R ’ M such that ρ— ωt = ω0 for all t ∈ [0, 1].

t

In particular, • = ρ1 : M ’’ M , satis¬es •— ω1 = ω0 .

The following argument, due to Moser, is extremely useful; it is known as

the Moser trick.

Proof. Suppose that there exists an isotopy ρ : M — R ’ M such that ρ— ωt =

t

ω0 , 0 ¤ t ¤ 1. Let

dρt

—¦ ρ’1 , t∈R.

vt = t

dt

Then

d dωt

0 = (ρ— ωt ) = ρ— Lvt ωt +

t t

dt dt

dωt

⇐’ L v t ωt + =0. ()

dt

Suppose conversely that we can ¬nd a smooth time-dependent vector ¬eld

vt , t ∈ R, such that ( ) holds for 0 ¤ t ¤ 1. Since M is compact, we can

integrate vt to an isotopy ρ : M — R ’ M with

d—

ρ — ωt = ρ — ω0 = ω 0 .

(ρ ωt ) = 0 =’ 0

dt t t

So everything boils down to solving ( ) for vt .

First, from ωt = (1 ’ t)ω0 + tω1 , we conclude that

dωt

= ω1 ’ ω0 .

dt

44 7 MOSER THEOREMS

Second, since [ω0 ] = [ω1 ], there exists a 1-form µ such that

ω1 ’ ω0 = dµ .

Third, by the Cartan magic formula, we have

Lvt ωt = d±vt ωt + ±vt dωt .

0

Putting everything together, we must ¬nd vt such that

d±vt ωt + dµ = 0 .

It is su¬cient to solve ±vt ωt + µ = 0. By the nondegeneracy of ωt , we can solve

this pointwise, to obtain a unique (smooth) vt .

Theorem 7.3 (Moser Theorem “ Version II) Let M be a compact man-

ifold with symplectic forms ω0 and ω1 . Suppose that ωt , 0 ¤ t ¤ 1, is a smooth

family of closed 2-forms joining ω0 to ω1 and satisfying:

d d

(1) cohomology assumption: [ωt ] is independent of t, i.e., dt [ωt ] = dt ωt = 0,

(2) nondegeneracy assumption: ωt is nondegenerate for 0 ¤ t ¤ 1.

Then there exists an isotopy ρ : M — R ’ M such that ρ— ωt = ω0 , 0 ¤ t ¤ 1.

t

Proof. (Moser trick) We have the following implications from the hypotheses:

(1) =’ ∃ family of 1-forms µt such that

dωt

0¤t¤1.

= dµt ,

dt

We can indeed ¬nd a smooth family of 1-forms µt such that dωt = dµt . The

dt

argument involves the Poincar´ lemma for compactly-supported forms,

e

together with the Mayer-Vietoris sequence in order to use induction on

the number of charts in a good cover of M . For a sketch of the argument,

see page 95 in [82].

(2) =’ ∃ unique family of vector ¬elds vt such that

±v t ω t + µ t = 0 (Moser equation) .

Extend vt to all t ∈ R. Let ρ be the isotopy generated by vt (ρ exists by

compactness of M ). Then we indeed have

d— dωt

(ρt ωt ) = ρ— (Lvt ωt + ) = ρ— (d±vt ωt + dµt ) = 0 .

t t

dt dt

45

7.3 Moser Local Theorem

The compactness of M was used to be able to integrate vt for all t ∈ R.

If M is not compact, we need to check the existence of a solution ρt for the

di¬erential equation dρt = vt —¦ ρt for 0 ¤ t ¤ 1.

dt

Picture. Fix c ∈ H 2 (M ). De¬ne Sc = {symplectic forms ω in M with [ω] = c}.

The Moser theorem implies that, on a compact manifold, all symplectic forms

on the same path-connected component of Sc are symplectomorphic.

7.3 Moser Local Theorem

Theorem 7.4 (Moser Theorem “ Local Version) Let M be a manifold,

X a submanifold of M , i : X ’ M the inclusion map, ω0 and ω1 symplectic

forms in M .

Hypothesis: ω0 |p = ω1 |p , ∀p ∈ X .

Conclusion: There exist neighborhoods U0 , U1 of X in M ,

and a di¬eomorphism • : U0 ’ U1 such that

• E U1

U0

d

s

d

d commutes

id i

d

X

and •— ω1 = ω0 .

Proof.

1. Pick a tubular neighborhood U0 of X. The 2-form ω1 ’ ω0 is closed on U0 ,

and (ω1 ’ ω0 )p = 0 at all p ∈ X. By the homotopy formula on the tubular

neighborhood, there exists a 1-form µ on U0 such that ω1 ’ ω0 = dµ and

µp = 0 at all p ∈ X.

2. Consider the family ωt = (1 ’ t)ω0 + tω1 = ω0 + tdµ of closed 2-forms

on U0 . Shrinking U0 if necessary, we can assume that ωt is symplectic for

0 ¤ t ¤ 1.

3. Solve the Moser equation: ±vt ωt = ’µ. Notice that vt = 0 on X.

4. Integrate vt . Shrinking U0 again if necessary, there exists an isotopy ρ :

U0 — [0, 1] ’ M with ρ— ωt = ω0 , for all t ∈ [0, 1]. Since vt |X = 0, we have

t

ρt |X = idX .

Set • = ρ1 , U1 = ρ1 (U0 ).

Exercise. Prove the Darboux theorem. (Hint: apply the local version of the

Moser theorem to X = {p}, as in the next lecture.) ™¦

8 Darboux-Moser-Weinstein Theory

8.1 Classical Darboux Theorem

Theorem 8.1 (Darboux) Let (M, ω) be a symplectic manifold, and let p be

any point in M . Then we can ¬nd a coordinate system (U, x1 , . . . , xn , y1 , . . . yn )

centered at p such that on U

n

dxi § dyi .

ω=

i=1

As a consequence of Theorem 8.1, if we prove for (R2n , dxi § dyi ) a lo-

cal assertion which is invariant under symplectomorphisms, then that assertion

holds for any symplectic manifold.

Proof. Apply the Moser local theorem (Theorem 7.4) to X = {p}:

Use any symplectic basis for Tp M to construct coordinates (x1 , . . . , xn , y1 , . . . yn )

centered at p and valid on some neighborhood U , so that

dxi § dyi

ωp = .

p

There are two symplectic forms on U : the given ω0 = ω and ω1 = dxi §

dyi . By the Moser theorem, there are neighborhoods U0 and U1 of p, and a

di¬eomorphism • : U0 ’ U1 such that

•— ( dxi § dyi ) = ω .

•(p) = p and

Since •— ( dxi § dyi ) = d(xi —¦ •) § d(yi —¦ •), we only need to set new

coordinates xi = xi —¦ • and yi = yi —¦ •.

If in the Moser local theorem (Theorem 7.4) we assume instead

Hypothesis: X is an n-dimensional submanifold with

i— ω0 = i— ω1 = 0 where i : X ’ M is inclusion, i.e.,

X is a submanifold lagrangian for ω0 and ω1 ,

then Weinstein [102] proved that the conclusion still holds. We need some

algebra for the Weinstein theorem.

8.2 Lagrangian Subspaces

Suppose that U, W are n-dimensional vector spaces, and „¦ : U — W ’ R

is a bilinear pairing; the map „¦ gives rise to a linear map „¦ : U ’ W — ,

„¦(u) = „¦(u, ·). Then „¦ is nondegenerate if and only if „¦ is bijective.

46

47

8.2 Lagrangian Subspaces

Proposition 8.2 Suppose that V is a 2n-dimensional vector space and „¦ :

V — V ’ R is a nondegenerate skew-symmetric bilinear pairing. Let U be a

lagrangian subspace of (V, „¦) (i.e., „¦|U —U = 0 and U is n-dimensional). Let W

be any vector space complement to U , not necessarily lagrangian.

Then from W we can canonically build a lagrangian complement to U .

„¦

Proof. The pairing „¦ gives a nondegenerate pairing U — W ’ R. Therefore,

„¦ : U ’ W — is bijective. We look for a lagrangian complement to U of the

form

W = {w + Aw | w ∈ W } ,

A : W ’ U being a linear map. For W to be lagrangian we need

∀ w1 , w2 ∈ W , „¦(w1 + Aw1 , w2 + Aw2 ) = 0

=’ „¦(w1 , w2 ) + „¦(w1 , Aw2 ) + „¦(Aw1 , w2 ) + „¦(Aw1 , Aw2 ) = 0

∈U

0

=’ „¦(w1 , w2 ) = „¦(Aw2 , w1 ) ’ „¦(Aw1 , w2 )

= „¦ (Aw2 )(w1 ) ’ „¦ (Aw1 )(w2 ) .

Let A = „¦ —¦ A : W ’ W — , and look for A such that

∀ w1 , w2 ∈ W , „¦(w1 , w2 ) = A (w2 )(w1 ) ’ A (w1 )(w2 ) .

The canonical choice is A (w) = ’ 1 „¦(w, ·). Then set A = („¦ )’1 —¦ A .

2

Proposition 8.3 Let V be a 2n-dimensional vector space, let „¦0 and „¦1 be

symplectic forms in V , let U be a subspace of V lagrangian for „¦0 and „¦1 , and

let W be any complement to U in V . Then from W we can canonically construct

a linear isomorphism L : V ’ V such that L|U = IdU and L— „¦1 = „¦0 .

Proof. From W we canonically obtain complements W0 and W1 to U in V such

that W0 is lagrangian for „¦0 and W1 is lagrangian for „¦1 . The nondegenerate

bilinear pairings

„¦

„¦0 : W0 ’’ U —

0

W0 — U ’’ R

give isomorphisms

„¦

„¦1 : W1 ’’ U — .

1

W1 — U ’’ R

Consider the diagram

„¦

U—

0

’’

W0

B“ “ id

„¦

U—

1

’’

W1

48 8 DARBOUX-MOSER-WEINSTEIN THEORY

where the linear map B satis¬es „¦1 —¦ B = „¦0 , i.e., „¦0 (ω0 , u) = „¦1 (Bω0 , u),

∀ω0 ∈ W0 , ∀u ∈ U . Extend B to the rest of V by setting it to be the identity

on U :

L := IdU • B : U • W0 ’’ U • W1 .

Finally, we check that L— „¦1 = „¦0 .

(L— „¦1 )(u • w0 , u • w0 ) = „¦1 (u • Bω0 , u • Bω0 )

= „¦1 (u, Bω0 ) + „¦1 (Bω0 , u )

= „¦0 (u, ω0 ) + „¦0 (ω0 , u )

„¦0 (u • w0 , u • w0 ) .

=

8.3 Weinstein Lagrangian Neighborhood Theorem

Theorem 8.4 (Weinstein Lagrangian Neighborhood Theorem [102])

Let M be a 2n-dimensional manifold, X an n-dimensional submanifold, i : X ’

M the inclusion map, and ω0 and ω1 symplectic forms on M such that i— ω0 =

i— ω1 = 0, i.e., X is a lagrangian submanifold of both (M, ω0 ) and (M, ω1 ).

Then there exist neighborhoods U0 and U1 of X in M and a di¬eomorphism

• : U0 ’ U1 such that

• E U1

U0

d

s

d

• — ω1 = ω 0 .

d commutes and

id i

d

X

The proof of the Weinstein theorem uses the Whitney extension theorem.

Theorem 8.5 (Whitney Extension Theorem) Let M be an n-dimensional

manifold and X a k-dimensional submanifold with k < n. Suppose that at

each p ∈ X we are given a linear isomorphism Lp : Tp M ’ Tp M such that

Lp |Tp X = IdTp X and Lp depends smoothly on p. Then there exists an embedding

h : N ’ M of some neighborhood N of X in M such that h|X = idX and

dhp = Lp for all p ∈ X.

The linear maps L serve as “germs” for the embedding.

Proof of the Weinstein theorem. Put a riemannian metric g on M ; at each

p ∈ M , gp (·, ·) is a positive-de¬nite inner product. Fix p ∈ X, and let V = Tp M ,

U = Tp X and W = U ⊥ = orthocomplement of U in V relative to gp (·, ·).

Since i— ω0 = i— ω1 = 0, U is a lagrangian subspace of both (V, ω0 |p ) and

(V, ω1 |p ). By symplectic linear algebra, we canonically get from U ⊥ a linear

49

8.3 Weinstein Lagrangian Neighborhood Theorem

isomorphism Lp : Tp M ’ Tp M , such that Lp |Tp X = IdTp X and L— ω1 |p = ω0 |p .

p

Lp varies smoothly with respect to p since our recipe is canonical!

By the Whitney theorem, there are a neighborhood N of X and an embed-

ding h : N ’ M with h|X = idX and dhp = Lp for p ∈ X. Hence, at any

p ∈ X,

(h— ω1 )p = (dhp )— ω1 |p = L— ω1 |p = ω0 |p .

p

Applying the Moser local theorem (Theorem 7.4) to ω0 and h— ω1 , we ¬nd a

neighborhood U0 of X and an embedding f : U0 ’ N such that f |X = idX and

f — (h— ω1 ) = ω0 on Uo . Set • = h —¦ f .

Sketch of proof for the Whitney theorem.

Case M = Rn :

For a compact k-dimensional submanifold X, take a neighborhood of the

form

U µ = {p ∈ M | distance (p, X) ¤ µ}

For µ su¬ciently small so that any p ∈ U µ has a unique nearest point in X,

de¬ne a projection π : U µ ’ X, p ’ point on X closest to p. If π(p) = q, then

p = q + v for some v ∈ Nq X where Nq X = (Tq X)⊥ is the normal space at q;

see Homework 5. Let

h : U µ ’’ Rn

p ’’ q + Lq v

where q = π(p) and v = p ’ π(p) ∈ Nq X. Then hX = idX and dhp = Lp for

p ∈ X. If X is not compact, replace µ by a continuous function µ : X ’ R+

which tends to zero fast enough as x tends to in¬nity.

General case:

Choose a riemannian metric on M . Replace distance by riemannian distance,

replace straight lines q + tv by geodesics exp(q, v)(t) and replace q + Lq v by the

value at t = 1 of the geodesic with initial value q and initial velocity Lq v.

In Lecture 30 we will need the following generalization of Theorem 8.4. For

a proof see, for instance, either of [47, 57, 105].

Theorem 8.6 (Coisotropic Embedding Theorem) Let M be a manifold

of dimension 2n, X a submanifold of dimension k ≥ n, i : X ’ M the inclusion

map, and ω0 and ω1 symplectic forms on M , such that i— ω0 = i— ω1 and X is

coisotropic for both (M, ω0 ) and (M, ω1 ). Then there exist neighborhoods U0 and

U1 of X in M and a di¬eomorphism • : U0 ’ U1 such that

• E U1

U0

d

s