L1: Nash-Kuiper Theorem

(1)

L1: Nash-Kuiper Theorem

Vincent Borrelli

Institut Camille Jordan - Université Claude Bernard Lyon 1

(2)

Isometric embeddings

Definition.– A mapf : (Mⁿ,g)−→^C¹ E^q= (R^q,h., .i)isisometricif f^∗h., .i=g.

•In coordinates, the conditionf^∗h., .i=greduces to a system of n(n+1)/2 equations

h∂f

∂x_i, ∂f

∂x_ji=g_ij

of thequnknown functionsf : (x₁, ....,x_n)7→(f₁, ...,f_q)with 0≤i ≤j ≤n.

•The number

s_n = n(n+1) 2 is called theJanet dimension.

(3)

Isometric embeddings

h∂f

∂x_i, ∂f

∂x_ji=g_ij

•The number

(4)

Isometric embeddings

h∂f

∂x_i, ∂f

∂x_ji=g_ij

•The number

(5)

Schläfli Conjecture

Ludwig Schläfli

(6)

Historical perpective

Janet-Cartan Theorem (1926-27).–Any n dimensional C^ω

Riemannian manifold admits locally an isometric embedding intoE^q with q=s_n.

Whitney Theorem (1936).–Any n dimensional differentiable manifold admits an embedding intoR²ⁿ.

Nash-KuiperC¹Embedding Theorem (1954-1955).–Statement in a couple of minutes...

(7)

Historical perpective

(8)

Historical perpective

(9)

Historical perpective

NashC^∞Embedding Theorem (1956).–Any C^∞compact

Riemannian manifold admits a C^∞isometric embedding intoE^qwith q =sn+4n.

•Newton Iterative Method +C¹Isometric Embedding Theorem

•In 1960, J. T. Schwartz observes that the Nash’s proof hides an Implicit Function Theorem on Fréchet spaces, the so called Nash-Moser (or

Newton-Nash-Moser) Theorem.

•In 1990, M. Günther provides a proof of the NashC^∞Embedding Theorem by using the "classical" tool of contractions.

Theorem (Gromov, Rokhlin, Greene 1970).–Any C^∞compact Riemannian manifold admits locally a C^∞isometric embedding intoE^q with q=s_n+n.

(10)

Historical perpective

(11)

Historical perpective

(12)

Historical perpective

(13)

Historical perpective

Theorem (Poznyak, 1973).–Any C^∞compact Riemannian surface admits locally a C^∞isometric embedding intoE⁴.

Theorem (Gromov 1986, Gunther 1989).–In the Nash C^∞ Embedding Theorem, we can take

q= max{s_n+2n,s_n+n+5}.

•Remark that ifn=2 thens₂=3 andq= max{7,10}=10.

Theorem (Gromov 1989).–Any C^∞compact Riemannian surface admits a C^∞isometric embedding intoE⁵.

(14)

Historical perpective

Theorem (Gromov 1989).–Any C^∞compact Riemannian surface admits a C^∞isometric embedding intoE⁵.

(15)

Historical perpective

(16)

Nash-Kuiper Theorem

John Nash and Nicolaas Kuiper

Definition.– A mapf : (Mⁿ,g)−→^C¹ E^qis said (strictly)shortif f^∗h., .i ≤Kg with 0<K <1.

(17)

Nash-Kuiper Theorem

Theorem (1954-55).– Let Mⁿbe a compact manifold and

f₀: (Mⁿ,g)−→^C¹ E^q, q>n, be a short embedding. Then, for every >0,there exists a C¹-isometric embedding f : (Mⁿ,g)−→E^q such thatkf −f₀k_C0 ≤.

•The assumption about the compacity is not essential but allows to simplify the statement of the theorem.

•Nash proved the caseq ≥n+2 in 1954 and Kuiper improved the Nash’s proof to the caseq=n+1 in 1955.

•TheC⁰-closeness condition appears latter (Kuiper, 1959).

(18)

Nash-Kuiper Theorem

(19)

Nash-Kuiper Theorem

(20)

Nash-Kuiper Theorem

(21)

Some puzzling corollaries

Corollary (Nash-Kuiper).– Any Riemannian compact manifold Mⁿ admits a C¹isometric embedding intoE²ⁿ.

Corollary (Nash-Kuiper).– Any flat torus Tⁿ =Eⁿ/Λadmits a C¹ isometric embedding intoEⁿ⁺¹.

Corollary (Nash-Kuiper).– Let x ∈Mⁿbe any point of a Riemannian manifold. There exists a neighborhood V(x)of x which admits C¹ isometric embedding intoEⁿ⁺¹.

Corollary (Existence of reduced spheres, 1959 ?).– Let0<r <1. There exists a C¹isometric embedding of the unit sphereSⁿ⊂Eⁿ⁺¹ inside a ball B(r)⊂Eⁿ⁺¹.

Corollary (Gromov 1989).– It is possible to perform an eversion of the 2-sphere through C¹isometric immersions.

(22)

Some puzzling corollaries

Corollary (Nash-Kuiper).– Any flat torus Tⁿ=Eⁿ/Λadmits a C¹ isometric embedding intoEⁿ⁺¹.

(23)

Some puzzling corollaries

(24)

Some puzzling corollaries

Corollary (Existence of reduced spheres, 1959 ?).– Let0<r <1.

There exists a C¹isometric embedding of the unit sphereSⁿ⊂Eⁿ⁺¹ inside a ball B(r)⊂Eⁿ⁺¹.

(25)

Some puzzling corollaries

Corollary (Existence of reduced spheres, 1959 ?).– Let0<r <1.

There exists a C¹isometric embedding of the unit sphereSⁿ⊂Eⁿ⁺¹ inside a ball B(r)⊂Eⁿ⁺¹.

(26)

Strategy of the proof (compact case)

•Let∆ =g−f₀^∗h., .ibe the isometric default. Sincef₀is strictly short,

∆is a metric.

•We construct a sequence of maps(f_k)_k∈_N^∗ such that kg−f_k^∗h., .ik_C0 ≤ k∆k_C0

2^k

•Eachf_k is built iteratively fromf_k−1. The parameters of the construction allow to insure that for allk,

kf_k₊₁−f_kk_C0 ≤ 1

2^k and kdf_k+1−df_kk_C0 ≤ C 2^k/2 withC>0. Therefore, the limitf∞is aC¹isometric map.

•Eachf_k is an embedding and we show by using theC⁰-closeness property that the limit still is an embedding.

(27)

Strategy of the proof (compact case)

∆is a metric.

2^k

kf_k₊₁−f_kk_C0 ≤ 1

(28)

Strategy of the proof (compact case)

∆is a metric.

2^k

kf_k₊₁−f_kk_C0 ≤ 1

(29)

Strategy of the proof (compact case)

∆is a metric.

2^k

kf_k₊₁−f_kk_C0 ≤ 1

2^k and kdf_k+1−df_kk_C0 ≤ C 2^k/2

1

(30)

Step 1 : Decomposition of ∆

•It is enough to work locally since the use of a partition of unity (U_α, ψα)allows to move from a local construction to a global one.

•We choose theU_α’s so that for eachα,U_αis compact and each chart extends toU_α.

•OnU_α, the isometric default induces a map∆ :U_α → S₂⁺(Rⁿ).

•The spaceS₂⁺(Rⁿ)of positive definite symetric bilinear forms onRⁿ is an open convex cone of dimensions_n.

•The first step is to write the isometric default as a sum of squares of linear forms :

∆(x) =

P0

X

j=1

ρ_j(x)`_j⊗`_j

withρ_j(x)≥0,j ∈ {1, ...,P₀}andx ∈ U_α.

(31)

Step 1 : Decomposition of ∆

∆(x) =

P0

X

j=1

ρ_j(x)`_j⊗`_j

(32)

Step 1 : Decomposition of ∆

∆(x) =

P0

X

j=1

ρ_j(x)`_j⊗`_j

(33)

Step 1 : Decomposition of ∆

∆(x) =

P0

X

j=1

ρ_j(x)`_j⊗`_j

(34)

Step 1 : Decomposition of ∆

∆(x) =

P0

X

j=1

ρ_j(x)`_j⊗`_j

withρ(x)≥0,j ∈ {1, ...,P }andx ∈ U .

(35)

Step 1 : Decomposition of ∆

•To do so, we choose a locally finite covering ofS₂⁺(Rⁿ)by open simplices and a partition of unity(ϕ )subordinated to that covering.

(36)

Step 1 : Decomposition of ∆

•Furthermore we require this finite number to be uniformly bounded, say byW (we admit the existence of such a covering).

(37)

Step 1 : Decomposition of ∆

•Each simplexσ hassn+1 verticesV_σ,0, ...,Vσ,sn and each vertex has a decomposition as a sum ofnsquares of linear forms

(38)

Step 1 : Decomposition of ∆

•We then write∆as a sum of squares of linear forms :

∆(x) =X

σ

ϕσ(∆(x))∆(x) =X

σ

ϕσ(∆(x))X

τ

ασ,τ(x)V_σ,τ

=X

ϕσ(∆(x))X

ασ,τ(x)

n

X`σ,τ,i⊗`σ,τ,i

(39)

Step 1 : Decomposition of ∆

•Since∆(U_α)is compact, it intersects a finite number of simplices Q(∆,U_α). Reindexing the above sum, we obtain

P

(40)

Step 1 : Decomposition of ∆

•A crucial observation is that, for eachx ∈ U_α the decomposition

∆(x) =

P0

X

j=1

ρ_j(x)`²_j

has at most(s +1)nW non vanishing coefficientsρ(x).

(41)

Step 2 : Iterations

•We build fromf₀a sequence of maps f_1,1, ...,f_1,P₀ =f₁ such that

g−f_1,i^∗ h., .i ' 1 4

i

X

j=1

ρ_j`²_j +

P0

X

j=i+1

ρ_j`²_j

In particular

g−f_1,P^∗

0h., .i ' 1 4∆

=⇒

1

(42)

Step 2 : Iterations

•The maps build by Nash are given iteratively by the formula f_1,i =f1,i−1+

√3ρ_i

2N1,i cos(N1,i`_i)u+ sin(N1,i`_i)v

whereu=u_1,i andv=v_1,i are two orthogonal unit normal vectors. We have :

f_1,i^∗ h., .i −f_1,i−1^∗ h., .i= 3

4ρ_i`²_i +O(1/N1,i)

(43)

Step 2 : Iterations

•Therefore

g−f_1,P^∗

0h., .i= 1 4∆ +

P

X

j=1

O(1/N_1,j)

and if theN_1,j’s are large enough :

k∆k

(44)

Step 2 : Iterations

« Actually the conditionq ≥n+2 might be replaced byq ≥n+1. This would come from use of a less easily controlled perturbation process needing only one direction normal to the imbedding. » Nash, 1954

(45)

Step 2 : Iterations

•The maps built by Kuiper are given iteratively by the formula f_1,i =f_1,i−1− 3ρ_i

16N1,isin(2N1,i `_i)t+

√3ρ_i

√

2N1,isin N1,i `_i−3ρ_i

16 sin(2N1,i`_i) w

wheret=t_1,i−1is a (convenient) unit tangent vector andw=w_1,i−1a unit normal vector. We have

(46)

Step 2 : Iterations

•The maps built by Kuiper are given iteratively by the formula f_1,i =f_1,i−1− 3ρ_i

16N1,isin(2N1,i `_i)t+

√3ρ_i

√

2N1,isin N1,i `_i−3ρ_i

16 sin(2N1,i`_i) w

wheret=t_1,i−1is a (convenient) unit tangent vector andw=w_1,i−1a unit normal vector. We have

f_1,i^∗ h., .i −f_1,i−1^∗ h., .i= 3

4ρi`²_i +O(ρ²_i) +O(1/N1,i)

(47)

Step 2 : Iterations

« Our proof follows Nash’ proof with the exception of a different kind of one step device : a strain. This strain however requires considerations

(48)

Step 2 : Iterations

•The passing from the codimension 2 to the codimension 1 shows a real technical problem.

•This problem can be settled by substituting aConvex Integrationto the Kuiper formula (we shall see how latter).

•The new mapf_1,i thus defined is such that f_1,i^∗ h., .i −f_1,i−1^∗ h., .i= 3

4ρ_i`²_i +O(1/N_1,i)

(49)

Step 2 : Iterations

•The passing from the codimension 2 to the codimension 1 shows a real technical problem.

•This problem can be settled by substituting aConvex Integrationto the Kuiper formula (we shall see how latter).

•The new mapf_1,i thus defined is such that

(50)

Step 3 : Convergence

•We re-do all the process starting withf₁and decomposing the new isometric default as a sum ofP₁squares of linear forms

∆1(x) =g−f₁^∗h., .i=

P₁

X

j=1

ρ1,j(x)`²_1,j

and then redoingP₁iterations to obtainf₂:=f_1,P₁.And so on...

•If theN_k,i’s are large enough, the resulting sequence of maps satisfies :

kg−f_k^∗h., .ik_C0 ≤ k∆k_C0

2^k

(51)

Step 3 : Convergence

•We re-do all the process starting withf₁and decomposing the new isometric default as a sum ofP₁squares of linear forms

∆1(x) =g−f₁^∗h., .i=

P₁

X

j=1

ρ1,j(x)`²_1,j

and then redoingP₁iterations to obtainf₂:=f_1,P₁.And so on...

•If theN_k,i’s are large enough, the resulting sequence of maps satisfies :

kg−f^∗h., .ik 0 ≤ k∆k_C0

(52)

Step 3 : Convergence

•A direct computation shows that the sequence(f_k)isC¹converging.

Nash :

f_k,i =f_k,i−1+

p3ρ_k,i

2Nk,i cos(Nk,i `k,i)u+ sin(Nk,i`k,i)v Kuiper :

f_k_,i = fk,i−1− 3ρ_k,i

16Nk,i sin(2Nk,i`_k,i)t +

p3ρ_k,i

√

2Nk,i sin Nk,i`_k,i−3ρ_k,i

16 sin(2Nk,i `_k,i) w

(here againu=u_1,i andv=v_1,i are two orthogonal unit normal vectors off_k,i−1).

•Thus the limit mapf∞isC¹isometric.

(53)

Step 3 : Convergence

•A direct computation shows that the sequence(f_k)isC¹converging.

Nash :

f_k,i =f_k,i−1+

p3ρ_k,i

2Nk,i cos(Nk,i `k,i)u+ sin(Nk,i`k,i)v Kuiper :

f_k_,i = fk,i−1− 3ρ_k,i

16Nk,i sin(2Nk,i`_k,i)t +

p3ρ_k,i

√

2Nk,i sin Nk,i`_k,i−3ρ_k,i

16 sin(2Nk,i `_k,i) w

(here againu=u andv=v are two orthogonal unit normal

(54)

Step 4 : The limit map f

∞

is an embedding

•The image off_1,1is a graph abovef₀(lying in a normal neiborhood of f₀), thereforef_1,1is an embedding. For the same reason, eachf_k is an embedding.

•Letx₁andx₂be two distinct points ofMand letk >0, we put d_k(x₁,x₂) :=dist(f_k(x₁),f_k(x₂)).

Sincekf_k+1−f_kk_C

0 ≤ ¹

2^k we have

dist(f∞(x_i)−f_k(x_i))≤ 1 2^k−1 and thus

d_k(x₁,x₂)− 1

2^k−2 ≤dist(f∞(x₁),f∞(x₂)) =d∞(x₁,x₂)

(55)

Step 4 : The limit map f

∞

is an embedding

•The image off_1,1is a graph abovef₀(lying in a normal neiborhood of f₀), thereforef_1,1is an embedding. For the same reason, eachf_k is an embedding.

•Letx₁andx₂be two distinct points ofMand letk >0, we put d_k(x₁,x₂) :=dist(f_k(x₁),f_k(x₂)).

Sincekf_k+1−f_kk_C

0 ≤ ¹

2^k we have

dist(f∞(x_i)−f_k(x_i))≤ 1 2^k−1 and thus

(56)

Step 4 : The limit map f

∞

is an embedding

•We shall show that, for every couple of distinct points(x₁,x₂), there existsk such thatd_k(x₁,x₂)− ₂_k¹₋₂ >0. This will imply thatf_∞is an embedding.

•Let(f_k_,i)_i∈{1,...,P_k_}be the sequence joiningf_k tof_k+1. We first observe that

N_k,i+1lim→+∞kf_k,i+1−f_k,ik_C0 =0 implies

N_k,i+1lim→+∞kd_k,i+1(., .)−d_k,i(., .)k_C0 =0 Thus, for everyk and everyi, there existsN_k,i+1such that

d_k,i+1≥ 2

3 1/P_k

d_k,i

(57)

Step 4 : The limit map f

∞

is an embedding

•We shall show that, for every couple of distinct points(x₁,x₂), there existsk such thatd_k(x₁,x₂)− ₂_k¹₋₂ >0. This will imply thatf_∞is an embedding.

•Let(f_k,i)_i∈{1,...,P_k_}be the sequence joiningf_k tof_k₊₁. We first observe that

N_k,i+1lim→+∞kf_k,i+1−f_k,ik_C0 =0 implies

N_k,i+1lim→+∞kd_k,i+1(., .)−d_k,i(., .)k_C0 =0 Thus, for everyk and everyi, there existsN_k,i+1such that

(58)

Step 4 : The limit map f

∞

is an embedding

•As a consequenced_k₊₁≥ ²₃d_k for everyk and

d_k ≥ 2

3 k

d₀

•We deduce

d_k(x₁,x₂)− 1

2^k−2 ≥d₀(x₁,x₂) 2

3 k

−4 1

2 k

•Ifk is large enough, the right term is positive, henced∞(x₁,x₂)>0. Thus, the mapf∞is an embedding.

•We have proved the Nash-Kuiper Theorem

(59)

Step 4 : The limit map f

∞

is an embedding

d_k ≥ 2

3 k

d₀

•We deduce

d_k(x₁,x₂)− 1

2^k−2 ≥d₀(x₁,x₂) 2

3 k

−4 1

2 k

•Ifk is large enough, the right term is positive, henced∞(x₁,x₂)>0. Thus, the mapf∞is an embedding.

(60)

Step 4 : The limit map f

∞

is an embedding

d_k ≥ 2

3 k

d₀

•We deduce

d_k(x₁,x₂)− 1

2^k−2 ≥d₀(x₁,x₂) 2

3 k

−4 1

2 k

•Ifk is large enough, the right term is positive, henced∞(x₁,x₂)>0.

Thus, the mapf∞is an embedding.

(61)

Step 4 : The limit map f

∞

is an embedding

d_k ≥ 2

3 k

d₀

•We deduce

d_k(x₁,x₂)− 1

2^k−2 ≥d₀(x₁,x₂) 2

3 k

−4 1

2 k

•Ifk is large enough, the right term is positive, henced∞(x₁,x₂)>0.

Thus, the mapf∞is an embedding.

(62)

The outrageously simple idea

John Nash

The isometric default must be reduced iteratively and not all at once.

(63)

The outrageously simple idea

By considering f_1,i =f_1,i−1+

p1ρ_1,i

N_1,i cosN_k,i `1,i)u+ sin(N_1,i `1,i)v instead of

f_1,i =f_1,i−1+

p3ρ_1,i

2N_1,i cosN_1,i `_1,i)u+ sin(N_1,i `_1,i)v it is obviously possible to kill the whole isometric default in each direction`1,i up to aO(1/N_1,i):

f_1,i^∗ h., .i −f_1,i−1^∗ h., .i=1×ρ_i`²_i +O(1/N_1,i) to get a mapf₁approximately isometric

(64)

The outrageously simple idea

BUT

This leads to a dead end. Indeed there is no control on the sign of O(1/N_1,i)and consequentlyf₁is no longer a short map in general. It lengthens some curves and the helix deformation can not reduce their length.

NASH

bypasses this difficulty with an iterative approach, dividing the isometric default by 2 at each step rather than trying to reduce it to zero all at once.

(65)

The outrageously simple idea

BUT

This leads to a dead end. Indeed there is no control on the sign of O(1/N_1,i)and consequentlyf₁is no longer a short map in general. It lengthens some curves and the helix deformation can not reduce their length.

NASH

bypasses this difficulty with an iterative approach, dividing the isometric default by 2 at each step rather than trying to reduce it to

(66)

That’s all folks !

John Nash