UCL AdM 20140303public - mathematik.hu-berlin.de

What can have a 3-sphere as its

boundary, and why should you ask Isaac

Newton?

Σ ∼= S3

Chris Wendl

University College London

Talk for the UCL AdM Maths Society, 3rd March, 2014

Slides available at:

http://www.homepages.ucl.ac.uk/~ucahcwe/publications.html#talks

PART 1: Differential topology

The n-dimensional sphere

Sn :={x ∈ R

n+1 | x21 + . . . x2n+1 = 1}

= boundary of the (n+1)-dimensional ball

Bn+1 :={x ∈ R

n+1 | x21 + . . . x2n+1 ≤ 1}.

S1 = ∂B2x1

S2 = ∂B3x1

Question: What other (n + 1)-dimensional

objects can have Sn as boundary?Some surfa es � with �� = S1:PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MDe�nitionSuppose M � RN is a subset, U �M is open.An n-dimensional oordinate hart on U is aset of fun tions x1; : : : ; xn : U ! R su h thatthe mapping(x1; : : : ; xn) : U ! Rnis bije tive onto some open subset of Rn.

objects can have Sn as boundary?

Some surfaces Σ with ∂Σ = S1:

replacements

De�nitionSuppose M � RN is a subset, U �M is open.An n-dimensional oordinate hart on U is aset of fun tions x1; : : : ; xn : U ! R su h thatthe mapping(x1; : : : ; xn) : U ! Rnis bije tive onto some open subset of Rn.2

replacements

Definition

Suppose M ⊂ RN is a subset, U ⊂ M is open.

An n-dimensional coordinate chart on U is a

set of functions x1, . . . , xn : U → R such that

the mapping

(x1, . . . , xn) : U → Rn

is bijective onto some open subset of Rn.

M is a (smooth, n-dimensional) manifold if:

� Every point p 2M is ontained in an opensubset U �M admitting an n-dimensional oordinate hart;� Wherever two oordinate harts overlap,the resulting oordinate transformationmaps are in�nitely di�erentiable.Two manifolds M and M 0 are di�eomorphi (M �=M 0) if there exists a bije tionf :M !M 0su h that both f and f�1 are everywhere in-�nitely di�erentiable when expressed in oor-dinate harts.M is ompa t if it is a losed and boundedsubset of RN . (Equivalently: every sequen ein M has a onvergent subsequen e!)PropositionIf M �= M 0, then they have the same dimen-sion, and M ompa t , M 0 ompa t.

• Every point p ∈ M is contained in an open

subset U ⊂ M admitting an n-dimensional

coordinate chart;� Wherever two oordinate harts overlap,the resulting oordinate transformationmaps are in�nitely di�erentiable.Two manifolds M and M 0 are di�eomorphi (M �=M 0) if there exists a bije tionf :M !M 0su h that both f and f�1 are everywhere in-�nitely di�erentiable when expressed in oor-dinate harts.M is ompa t if it is a losed and boundedsubset of RN . (Equivalently: every sequen ein M has a onvergent subsequen e!)PropositionIf M �= M 0, then they have the same dimen-sion, and M ompa t , M 0 ompa t.

coordinate chart;

• Wherever two coordinate charts overlap,

the resulting coordinate transformation

maps are infinitely differentiable.

Two manifolds M and M 0 are di�eomorphi (M �=M 0) if there exists a bije tionf :M !M 0su h that both f and f�1 are everywhere in-�nitely di�erentiable when expressed in oor-dinate harts.M is ompa t if it is a losed and boundedsubset of RN . (Equivalently: every sequen ein M has a onvergent subsequen e!)PropositionIf M �= M 0, then they have the same dimen-sion, and M ompa t , M 0 ompa t.3

coordinate chart;

Two manifolds M and M ′ are diffeomorphic

(M ∼= M ′) if there exists a bijection

f : M → M ′

such that both f and f−1 are everywhere in-

finitely differentiable when expressed in coor-

dinate charts.M is ompa t if it is a losed and boundedsubset of RN . (Equivalently: every sequen ein M has a onvergent subsequen e!)PropositionIf M �= M 0, then they have the same dimen-sion, and M ompa t , M 0 ompa t.3

coordinate chart;

f : M → M ′

dinate charts.

M is compact if it is a closed and bounded

subset of RN . (Equivalently: every sequen ein M has a onvergent subsequen e!)PropositionIf M �= M 0, then they have the same dimen-sion, and M ompa t , M 0 ompa t.3

coordinate chart;

f : M → M ′

dinate charts.

subset of RN . (Equivalently: every sequence

in M has a convergent subsequence!)PropositionIf M �= M 0, then they have the same dimen-sion, and M ompa t , M 0 ompa t.3

coordinate chart;

f : M → M ′

dinate charts.

subset of RN . (Equivalently: every sequence

in M has a convergent subsequence!)

Proposition

If M ∼= M ′, then they have the same dimen-

sion, and M compact ⇔ M ′ compact.

Some examples of manifolds

• Rn (dimension = n)� C = fx+ iy j x; y 2 Rg �= R2� Cn �= R2n (dimension = 2n)� Spheres Sn and balls Bn (dimension = n)( ompa t)� Surfa es of genus g (dimension = 2)( ompa t)� Various matrix groups (\Lie groups"):{ GL(n;R) = fA 2 Rn�n j A invertibleg(dimension = n2){ SL(n;R) = fA 2 GL(n;R) j detA= 1g(dimension = n2 � 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• Rn (dimension = n)

• C = {x+ iy | x, y ∈ R} ∼= R2� Cn �= R2n (dimension = 2n)� Spheres Sn and balls Bn (dimension = n)( ompa t)� Surfa es of genus g (dimension = 2)( ompa t)� Various matrix groups (\Lie groups"):{ GL(n;R) = fA 2 Rn�n j A invertibleg(dimension = n2){ SL(n;R) = fA 2 GL(n;R) j detA= 1g(dimension = n2 � 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

• Cn ∼= R2n (dimension = 2n)� Spheres Sn and balls Bn (dimension = n)( ompa t)� Surfa es of genus g (dimension = 2)( ompa t)� Various matrix groups (\Lie groups"):{ GL(n;R) = fA 2 Rn�n j A invertibleg(dimension = n2){ SL(n;R) = fA 2 GL(n;R) j detA= 1g(dimension = n2 � 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

• Cn ∼= R2n (dimension = 2n)

• Spheres Sn and balls Bn (dimension = n)

(compact)� Surfa es of genus g (dimension = 2)( ompa t)� Various matrix groups (\Lie groups"):{ GL(n;R) = fA 2 Rn�n j A invertibleg(dimension = n2){ SL(n;R) = fA 2 GL(n;R) j detA= 1g(dimension = n2 � 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

(compact)

• Surfaces of genus g (dimension = 2)

(compact)� Various matrix groups (\Lie groups"):{ GL(n;R) = fA 2 Rn�n j A invertibleg(dimension = n2){ SL(n;R) = fA 2 GL(n;R) j detA= 1g(dimension = n2 � 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

(compact)

• Various matrix groups (“Lie groups”):

– GL(n,R) = {A ∈ Rn×n | A invertible}

(dimension = n2){ SL(n;R) = fA 2 GL(n;R) j detA= 1g(dimension = n2 � 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

(compact)

(dimension = n2)

– SL(n,R) = {A ∈ GL(n,R) | detA = 1}

(dimension = n2 − 1){ O(n) = fA 2 GL(n;R) j ATA = Ig(dimension = n(n+1)=2) ( ompa t)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

(compact)

(dimension = n2)

– SL(n,R) = {A ∈ GL(n,R) | detA = 1}

(dimension = n2 − 1)

– O(n) = {A ∈ GL(n,R) | ATA = 1}

(dimension = n(n+1)/2) (compact)� The universe?(dimension = 4? 10? 11?) ( ompa t?)4

• C = {x+ iy | x, y ∈ R} ∼= R2

(compact)

(dimension = n2)

– SL(n,R) = {A ∈ GL(n,R) | detA = 1}

(dimension = n2 − 1)

– O(n) = {A ∈ GL(n,R) | ATA = 1}

(dimension = n(n+1)/2) (compact)

• The universe?

(dimension = 4? 10? 11?) (compact?)

More precise question: What kinds of com-

pact (n+1)-manifolds M can have ∂M ∼= Sn?Answer: Almost any!Let M = any ompa t (n+1)-manifold with-out boundary, pi k a point p 2 M and a o-ordinate hart on some open set U 3 p su hthat p has oordinates (0; : : : ;0) 2 Rn+1. Thenfor � > 0 small, de�neM := M nB�(p);whereB�(p) := nx21+ : : :+ x2n+1 � �o � U :

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MNow �M = fx21+ : : :+ x2n+1 = �g �= Sn.Con lusion: We asked the wrong question.The answer was too easy!5

pact (n+1)-manifolds M can have ∂M ∼= Sn?

Answer: Almost any!Let M = any ompa t (n+1)-manifold with-out boundary, pi k a point p 2 M and a o-ordinate hart on some open set U 3 p su hthat p has oordinates (0; : : : ;0) 2 Rn+1. Thenfor � > 0 small, de�neM := M nB�(p);whereB�(p) := nx21+ : : :+ x2n+1 � �o � U :

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MNow �M = fx21+ : : :+ x2n+1 = �g �= Sn.Con lusion: We asked the wrong question.The answer was too easy!5

Answer: Almost any!

Let M = any compact (n+1)-manifold with-

out boundary, pi k a point p 2 M and a o-ordinate hart on some open set U 3 p su hthat p has oordinates (0; : : : ;0) 2 Rn+1. Thenfor � > 0 small, de�neM := M nB�(p);whereB�(p) := nx21+ : : :+ x2n+1 � �o � U :

Now �M = fx21+ : : :+ x2n+1 = �g �= Sn.Con lusion: We asked the wrong question.The answer was too easy!5

Answer: Almost any!

out boundary, pick a point p ∈ M and a o-ordinate hart on some open set U 3 p su hthat p has oordinates (0; : : : ;0) 2 Rn+1. Thenfor � > 0 small, de�neM := M nB�(p);whereB�(p) := nx21+ : : :+ x2n+1 � �o � U :p

Answer: Almost any!

out boundary, pick a point p ∈ M and a

coordinate chart on some open set U ∋ p

such that p has coordinates (0, . . . ,0) ∈ Rn+1.Then for � > 0 small, de�neM := M nB�(p);whereB�(p) := nx21+ : : :+ x2n+1 � �o � U :p

Answer: Almost any!

such that p has coordinates (0, . . . ,0) ∈ Rn+1.

Then for ǫ > 0 small, define

M := M \Bǫ(p),

Bǫ(p) :={x21 + . . .+ x2n+1 ≤ ǫ

}⊂ U .

Answer: Almost any!

M := M \Bǫ(p),

Bǫ(p) :={x21 + . . .+ x2n+1 ≤ ǫ

}⊂ U .

Bǫ(p)

Answer: Almost any!

M := M \Bǫ(p),

Bǫ(p) :={x21 + . . .+ x2n+1 ≤ ǫ

}⊂ U .

Now ∂M = {x21 + . . .+ x2n+1 = ǫ} ∼= Sn.Con lusion: We asked the wrong question.The answer was too easy!5

Answer: Almost any!

M := M \Bǫ(p),

Bǫ(p) :={x21 + . . .+ x2n+1 ≤ ǫ

}⊂ U .

Now ∂M = {x21 + . . .+ x2n+1 = ǫ} ∼= Sn.

Conclusion: We asked the wrong question.

The answer was too easy!

PART 2: Dynamics

Newton (18th century):

A system of particles moving with n degrees

of freedom is described by a path in Rn,

q(t) := (q1(t), . . . , qn(t)) ∈ Rn.If the system is onservative, its for es arederived from a potential fun tion V (q) byF(q) = �rV (q).Then Newton's se ond law givesmj�qj = ��V�qj�;a system of n se ond-order ordinary di�eren-tial equations (ODE). Its total energyE = nXj=1 �12�mj _q2j + V (q)is onserved, i.e. �dEdt �= 0.

PART 2: Dynamics

q(t) := (q1(t), . . . , qn(t)) ∈ Rn.

If the system is conservative, its forces are

derived from a potential function V (q) by

F(q) = −∇V (q).

Then Newton’s second law gives

mj qj = −∂V

∂qj,a system of n se ond-order ordinary di�eren-tial equations (ODE). Its total energyE = nXj=1 �12�mj _q2j + V (q)is onserved, i.e. �dEdt �= 0.

PART 2: Dynamics

q(t) := (q1(t), . . . , qn(t)) ∈ Rn.

F(q) = −∇V (q).

mj qj = −∂V

∂qj,

a system of n second-order ordinary differen-

tial equations (ODE). Its total energyE = nXj=1 �12�mj _q2j + V (q)is onserved, i.e. �dEdt �= 0.6

PART 2: Dynamics

q(t) := (q1(t), . . . , qn(t)) ∈ Rn.

F(q) = −∇V (q).

mj qj = −∂V

∂qj,

a system of n second-order ordinary differen-

tial equations (ODE). Its total energy

E =n∑

2j + V (q)

is conserved, i.e. dEdt = 0.

Hamilton (19th century):

Pretend qi and pj := mj qj (momentum) are

independent variables moving in the “phase

space” R2n. The total energy de�nes theHamiltonian fun tion:H : R2n ! R : (q; p) 7! nXj=1 � p2j2mj�+ V (q);and Newton's se ond-order system be omesHamilton's (�rst-order!) equations:_qj = ��H�pj�; _pj = ��H�qj�; j = 1; : : : ; n: (�)Idea: To study motion of systems satisfy-ing onstraints, we an treat (q; p) as lo al oordinates of a point moving in a manifold.

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4M7

space” R2n. The total energy defines the

Hamiltonian function:

H : R2n → R : (q,p) 7→n∑

2mj+ V (q),

and Newton’s second-order system becomes

Hamilton’s (first-order!) equations:

qj =∂H

∂pj, pj = −

∂qj, j = 1, . . . , n. (∗)Idea: To study motion of systems satisfy-ing onstraints, we an treat (q; p) as lo al oordinates of a point moving in a manifold.

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4M7

space” R2n. The total energy defines the

Hamiltonian function:

H : R2n → R : (q,p) 7→n∑

2mj+ V (q),

and Newton’s second-order system becomes

Hamilton’s (first-order!) equations:

qj =∂H

∂pj, pj = −

∂qj, j = 1, . . . , n. (∗)

Idea: To study motion of systems satisfy-

ing constraints, we can treat (q,p) as local

coordinates of a point moving in a manifold.

qj =∂H

∂pj, pj = −

∂qj(∗)

Complication: A system that satisfies (∗)

for one particular choice of coordinates might

not satisfy it for all other choices.De�nitionA 2n-dimensional manifold M has a symple -ti stru ture if it is overed by spe ial oordi-nate harts of the form (q1; : : : ; qn; p1; : : : ; pn)su h that for any smooth fun tion H : M !R, all oordinate transformations preserve theform of Hamilton's equations (�).Exer ise: A transformation on R2 preserves (�), it is area and orientation preserving.Simple examples� Symple ti : R2n, all orientable surfa es� Not symple ti : S2n for n > 1( an prove using de Rham ohomology)

qj =∂H

∂pj, pj = −

∂qj(∗)

not satisfy it for all other choices.

Definition

A 2n-dimensional manifold M has a symplec-

tic structure if it is covered by special coordi-

nate charts of the form (q1, . . . , qn, p1, . . . , pn)

such that for any smooth function H : M →

R, all coordinate transformations preserve the

form of Hamilton’s equations (∗).Exer ise: A transformation on R2 preserves (�), it is area and orientation preserving.Simple examples� Symple ti : R2n, all orientable surfa es� Not symple ti : S2n for n > 1( an prove using de Rham ohomology)

qj =∂H

∂pj, pj = −

∂qj(∗)

Definition

form of Hamilton’s equations (∗).

Exercise: A transformation on R2 preserves (∗)

⇔ it is area and orientation preserving.Simple examples� Symple ti : R2n, all orientable surfa es� Not symple ti : S2n for n > 1( an prove using de Rham ohomology)

qj =∂H

∂pj, pj = −

∂qj(∗)

Definition

⇔ it is area and orientation preserving.

Simple examples

• Symplectic: R2n, all orientable surfaces� Not symple ti : S2n for n > 1( an prove using de Rham ohomology)8

qj =∂H

∂pj, pj = −

∂qj(∗)

Definition

⇔ it is area and orientation preserving.

Simple examples

• Symplectic: R2n, all orientable surfaces

• Not symplectic: S2n for n > 1

(can prove using de Rham cohomology)

Assume M is symplectic, H : M → R a smoothfunction. Then any path γ : R → M satisfyingHamilton’s equations “conserves energy”:

dtH(γ(t)) = 0,) orbits are on�ned to level sets H�1( ).Question: Given H : M ! R and , mustthere exist a periodi orbit in H�1( )?Theorem (Rabinowitz-Weinstein '78)Given H : R2n ! R, any star-shaped levelset H�1( ) � R2n admits a periodi orbit.PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4M

dtH(γ(t)) = 0,

⇒ orbits are confined to level sets H−1(c).Question: Given H : M ! R and , mustthere exist a periodi orbit in H�1( )?Theorem (Rabinowitz-Weinstein '78)Given H : R2n ! R, any star-shaped levelset H�1( ) � R2n admits a periodi orbit.PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4M9

dtH(γ(t)) = 0,

⇒ orbits are confined to level sets H−1(c).

Question: Given H : M → R and c, mustthere exist a periodic orbit in H−1(c)?Theorem (Rabinowitz-Weinstein '78)Given H : R2n ! R, any star-shaped levelset H�1( ) � R2n admits a periodi orbit.PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4M

dtH(γ(t)) = 0,

Question: Given H : M → R and c, mustthere exist a periodic orbit in H−1(c)?

Theorem (Rabinowitz-Weinstein ’78)Given H : R2n → R, any star-shaped levelset H−1(c) ⊂ R2n admits a periodic orbit.

dtH(γ(t)) = 0,

Definitions

A submanifold N of a manifold M is a subset

N ⊂ M such that the natural inclusion map

N → M is infinitely differentiable.A hypersurfa e N �M is a submanifold withdimN = dimM � 1.A hypersurfa e N � R2n is star-shaped if itinterse ts every ray from the origin exa tlyon e, transversely.Exer iseAny star-shaped hypersurfa e in R2n is dif-feomorphi to S2n�1.10

Definitions

N → M is infinitely differentiable.

A hypersurface N ⊂ M is a submanifold with

dimN = dimM − 1.A hypersurfa e N � R2n is star-shaped if itinterse ts every ray from the origin exa tlyon e, transversely.Exer iseAny star-shaped hypersurfa e in R2n is dif-feomorphi to S2n�1.10

Definitions

dimN = dimM − 1.

A hypersurface N ⊂ R2n is star-shaped if it

intersects every ray from the origin exactly

once, transversely.Exer iseAny star-shaped hypersurfa e in R2n is dif-feomorphi to S2n�1.10

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

Any star-shaped hypersurface in R2n is dif-

feomorphic to S2n−1.

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

Definitions

dimN = dimM − 1.

once, transversely.

Exercise

PART 3: Symplectic topology

In 1985, Mikhail Gromov

published a paper called

Pseudoholomorphic curves

in symplectic manifolds.Among other remarkableresults, it proved:Our main theoremSuppose M is a ompa t 4-manifold withan exa t symple ti stru ture whi h, at itsboundary, looks like a star-shaped hypersur-fa e in R4. Then M �= B4.A generalisation to all dimensions � 4 waspublished in 1991, due to

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MYashaEliashberg

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MAndreasFloer

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MDusaM Du�11

in symplectic manifolds.

Among other remarkable

results, it proved:

Our main theorem

Suppose M is a compact 4-manifold withan exact symplectic structure which, at itsboundary, looks like a star-shaped hypersur-face in R4. Then M ∼= B4.A generalisation to all dimensions � 4 waspublished in 1991, due to

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MYashaEliashberg

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MAndreasFloer

PSfrag repla ements� �= S3S2 = �B3S1 = �B2x1x2x3pB�(p)S1uwvw�B4MDusaM Du�11

results, it proved:

Our main theorem

Suppose M is a compact 4-manifold withan exact symplectic structure which, at itsboundary, looks like a star-shaped hypersur-face in R4. Then M ∼= B4.

A generalisation to all dimensions ≥ 4 waspublished in 1991, due to

Eliashberg

AndreasFloer DusaM Du�11

results, it proved:

Our main theorem

Eliashberg

Andreas

DusaM Du�11

results, it proved:

Our main theorem

Eliashberg

Andreas

McDuff11

Some preparation from complex analysis

A function f = u + iv : C → C is analytic /

holomorphic if it satisfies the Cauchy-Riemann

equations:

∂su(s+ it) = ∂tv(s+ it),

∂tu(s+ it) = −∂sv(s+ it).

Equivalently: �sf + i �tf = 0 : (��)A map f : C ! Cn satisfying this equation is alled a holomorphi urve in Cn.A 2n-dimensional manifold M has a omplexstru ture if it is overed by spe ial ( omplex) oordinate harts of the form (z1; : : : ; zn) :U ! Cn su h that all oordinate transfor-mations preserve the form of the Cau hy-Riemann equation (��).Thus one an speak of holomorphi urvesin any omplex manifold.Examples: Cn, SL(n;C), C [ f1g �= S212

equations:

Equivalently: ∂sf + i ∂tf = 0 . (∗∗)

A map f : C → Cn satisfying this equation is

called a holomorphic curve in Cn.A 2n-dimensional manifold M has a omplexstru ture if it is overed by spe ial ( omplex) oordinate harts of the form (z1; : : : ; zn) :U ! Cn su h that all oordinate transfor-mations preserve the form of the Cau hy-Riemann equation (��).Thus one an speak of holomorphi urvesin any omplex manifold.Examples: Cn, SL(n;C), C [ f1g �= S212

equations:

called a holomorphic curve in Cn.

A 2n-dimensional manifold M has a complex

structure if it is covered by special (complex)

coordinate charts of the form (z1, . . . , zn) :

U → Cn such that all coordinate transfor-

mations preserve the form of the Cauchy-

Riemann equation (∗∗).

Thus one can speak of holomorphic curves

in any complex manifold.Examples: Cn, SL(n;C), C [ f1g �= S212

equations:

in any complex manifold.

Examples: Cn, SL(n;C), C [ f1g �= S212

equations:

Examples: Cn, SL(n,C), C [ f1g �= S212

equations:

Examples: Cn, SL(n,C), C ∪ {∞} ∼= S2

Unfortunately, symplectic manifolds are not

always complex, so one cannot generally make

sense of holomorphic curves in them.The next best thing. . .An almost omplex stru ture on Cn is a smoothfun tionJ : Cn ! freal-linear maps Cn ! Cng �= R2n�2nsu h that for all p 2 Cn, [J(p)℄2 = �1.A map f : C ! Cn is then alled a pseudo-holomorphi urve if it satis�es the nonlinearCau hy-Riemann equation:�sf + J(f) �tf = 0 : (� � �)This is a nonlinear �rst-order ellipti partialdi�erential equation (PDE).Fundamental lemma:Every symple ti manifold admits a spe ial lass of ompatible almost omplex stru -tures.13

sense of holomorphic curves in them.

The next best thing. . .

An almost complex structure on Cn is a smooth

function

J : Cn → {real-linear maps Cn → C

n} ∼= R2n×2n

such that for all p ∈ Cn, [J(p)]2 = −1.A map f : C ! Cn is then alled a pseudo-holomorphi urve if it satis�es the nonlinearCau hy-Riemann equation:�sf + J(f) �tf = 0 : (� � �)This is a nonlinear �rst-order ellipti partialdi�erential equation (PDE).Fundamental lemma:Every symple ti manifold admits a spe ial lass of ompatible almost omplex stru -tures.13

function

n} ∼= R2n×2n

such that for all p ∈ Cn, [J(p)]2 = −1.

A map f : C → Cn is then called a pseudo-

holomorphic curve if it satisfies the nonlinear

Cauchy-Riemann equation:

∂sf + J(f) ∂tf = 0 . (∗ ∗ ∗)

This is a nonlinear first-order elliptic partial

differential equation (PDE).Fundamental lemma:Every symple ti manifold admits a spe ial lass of ompatible almost omplex stru -tures.13

function

n} ∼= R2n×2n

such that for all p ∈ Cn, [J(p)]2 = −1.

A map f : C → Cn is then called a pseudo-

holomorphic curve if it satisfies the nonlinear

Cauchy-Riemann equation:

∂sf + J(f) ∂tf = 0 . (∗ ∗ ∗)

This is a nonlinear first-order elliptic partial

differential equation (PDE).

Fundamental lemma:

Every symplectic manifold admits a special

class of compatible almost complex struc-

tures.

A decomposition of the standard B4 ⊂ R4

Identify R4 = C2 and define

J0(p) := i for all p ∈ R4.

We now see two obvious 2-dimensional fam-

ilies of pseudoholomorphic curves:

uw : C → C2 : z 7→ (z, w) for w ∈ C,

vw : C → C2 : z 7→ (w, z) for w ∈ C.

They form two transverse foliations of C2:

Proof of the main theorem

Given ∂M = Σ ⊂ R4 star-shaped, construct

a symplectic manifold W by surgery :

(1) Remove from R4 = C2 the interior of Σ;

(2) Attach M along its boundary to Σ.

Σ ∼= S3

Choose J mat hing J0 outside a large ball.Then for large jwj, the pseudoholomorphi urves uw and vw also exist in W .15

Proof of the main theorem

Given ∂M = Σ ⊂ R4 star-shaped, construct

a symplectic manifold W by surgery :

(1) Remove from R4 = C2 the interior of Σ;

(2) Attach M along its boundary to Σ.

Σ ∼= S3

Choose J matching J0 outside a large ball.

Then for large |w|, the pseudoholomorphic

curves uw and vw also exist in W .

Let Mu and Mv denote the families of pseu-

doholomorphic curves in W containing the

curves uw and vw respectively. Using fun -tional analysis and PDE theory, one an show:Lemma 1 (smoothness):One an hoose J su h that Mu and Mvare ea h parametrized by smooth, oriented 2-dimensional manifolds, and within ea h fam-ily, any two distin t urves are disjoint. More-over, every urve inMu interse ts every urvein Mv exa tly on e, transversely.Lemma 2 ( ompa tness):Any bounded sequen e of urves in Mu orMv has a onvergent subsequen e.These lemmas on ern general properties ofsolution spa es.One an prove them without knowing how tosolve the PDE, and without knowing whatMa tually is!16

curves uw and vw respectively. Using func-

tional analysis and PDE theory, one can show:

Lemma 1 (smoothness):

One can choose J such that Mu and Mv

are each parametrized by smooth, oriented 2-

dimensional manifolds, and within each fam-

ily, any two distinct curves are disjoint. More-

over, every curve in Mu intersects every curve

in Mv exactly once, transversely.Lemma 2 ( ompa tness):Any bounded sequen e of urves in Mu orMv has a onvergent subsequen e.These lemmas on ern general properties ofsolution spa es.One an prove them without knowing how tosolve the PDE, and without knowing whatMa tually is!16

in Mv exactly once, transversely.

Lemma 2 (compactness):

Any bounded sequence of curves in Mu or

Mv has a convergent subsequence.These lemmas on ern general properties ofsolution spa es.One an prove them without knowing how tosolve the PDE, and without knowing whatMa tually is!16

in Mv exactly once, transversely.

Lemma 2 (compactness):

Any bounded sequence of curves in Mu or

Mv has a convergent subsequence.

These lemmas concern general properties of

solution spaces.

One can prove them without knowing how to

solve the PDE, and without knowing what M

actually is!

Final step: “turn on the machine. . . ”

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

Σ ∼= S3

) W �= C2.17

⇒ W ∼= C2.

That was nearly 30 years ago.

Here is a more recent but similar result. . .

Theorem (W. 2010)

The only exact symplectic fillings of a 3-

dimensional torus

T3 := S1 × S1 × S1

are star-shaped domains in the cotangent bun-

dle of T2.

Question:

For a surface Σ of genus g ≥ 2, does the unit

cotangent bundle have more than one exact

symplectic filling?

No one has any idea.

UCL AdM 20140303public - mathematik.hu-berlin.de

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