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(1)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

The H-principle for Isometric Embeddings

Vincent Borrelli

Université Lyon 1

(2)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Isometric Maps

Definition.– A map f : (M n , g) −→ C

1

E q is an isometry if f h., .i = g.

The length of curves is preserved by an isometric map.

In a local coordinate system x = (x 1 , ..., x n ) : 1 ≤ i ≤ j ≤ n, h ∂f

∂x i (x ), ∂f

∂x j (x )i = g ij (x ) The Janet dimension :

s n = n(n + 1)

2 .

(3)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Isometric Maps

Definition.– A map f : (M n , g) −→ C

1

E q is an isometry if f h., .i = g.

The length of curves is preserved by an isometric map.

In a local coordinate system x = (x 1 , ..., x n ) : 1 ≤ i ≤ j ≤ n, h ∂f

∂x i (x ), ∂f

∂x j (x )i = g ij (x )

The Janet dimension :

s n = n(n + 1)

2 .

(4)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Isometric Maps

Definition.– A map f : (M n , g) −→ C

1

E q is an isometry if f h., .i = g.

The length of curves is preserved by an isometric map.

In a local coordinate system x = (x 1 , ..., x n ) : 1 ≤ i ≤ j ≤ n, h ∂f

∂x i (x ), ∂f

∂x j (x )i = g ij (x ) The Janet dimension :

s n = n(n + 1)

2 .

(5)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Nash-Kuiper Theorem

Definition.– A map f : (M n , g) −→ C

1

E q is called strictly short if f h., .i < g.

John Forbes Nash Nicolaas Kuiper

Theorem (1954-55-86).– Let M n be a compact Riemannian manifold and f 0 : (M n , g) C

1

−→ E q be a strictly short

embedding. Then, for every > 0, there exists a C 1

isometric embedding f : (M n , g) −→ E q such that

kf − f 0 k C

0

≤ .

(6)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Nash-Kuiper Theorem

Definition.– A map f : (M n , g) −→ C

1

E q is called strictly short if f h., .i < g.

John Forbes Nash Nicolaas Kuiper

Theorem (1954-55-86).– Let M n be a compact Riemannian manifold and f 0 : (M n , g) −→ C

1

E q be a strictly short

embedding. Then, for every > 0, there exists a C 1

isometric embedding f : (M n , g) −→ E q such that

kf − f 0 k C

0

≤ .

(7)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Nash-Kuiper Spheres

?

Nash-Kuiper Spheres.– Let 0 < r < 1. There exists a

C 1 -isometric embedding of the unit sphere of E 3 in a ball of

radius r .

(8)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Sphere Eversion

Sphere Eversion.– The sphere S 2 ⊂ E 3 can be turned

inside out by a regular homotopy of isometric C 1

immersions.

(9)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Flat Tori

Definition.– Any quotient E 2 /Λ of the Euclidean 2-space by a lattice Λ ⊂ E 2 is called a flat torus

Flat Tori.– Any flat torus E 2 /Λ admits a C 1 isometric

embedding in E 3 .

(10)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Flat Tori

Definition.– Any quotient E 2 /Λ of the Euclidean 2-space by a lattice Λ ⊂ E 2 is called a flat torus

Flat Tori.– Any flat torus E 2 /Λ admits a C 1 isometric

embedding in E 3 .

(11)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Our Goal

Goal of this talk.– To recover the Nash-Kuiper result on C 1 isometric embeddings from the machinery of the Gromov Integration Theory.

Our main ingredient.– The Gromov Theorem : Let R ⊂ J 1 (M, N) be an open and ample differential relation. Then R satisfies the parametric h-principle i. e.

J : Sol(R) −→ Γ(R) is a weak homotopy equivalence.

Our main obstacles.–

• The isometric relation is not ample

• The isometric relation is closed

(12)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Decomposition

• For simplicity M n = [0, 1] n .

• The image of the metric distorsion

∆ := g − f 0 h., .i E

q

lies inside the positive cone M of inner products of E n .

• There exist S ≥ n(n+1) 2 linear forms ` 1 , . . . ` S of E n such that

g − f 0 h., .i E

q

=

S

X

j=1

ρ j ` j ⊗ ` j

where ρ j > 0.

(13)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Adapting the Gromov machinery

The strategy is to do the successive convex integrations along the S directions corresponding to the S linear forms

` 1 , ..., ` S .

rather than

along the n directions of the coordinates in [0, 1] n .

(14)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Adapting the Gromov machinery

• This will produce S intermediary maps f 1 , ..., f S such that

g − f 1 h., .i E

q

≈ρ 2 ` 2 23 ` 2 3 +...+ρ S ` 2 S g − f 2 h., .i E

q

≈ ρ 3 ` 2 3 +...+ρ S ` 2 S

.. . .. .

g − f S−1 h., .i E

q

≈ ρ S ` 2 S g − f S h., .i E

q

≈ 0.

• The map f := f S is then a solution of R e = Op(R).

(15)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Adapting the Gromov machinery

We have

f j h., .i E

q

− f j−1 h., .i E

q

=(g − f j−1 h., .i E

q

) − (g − f j h., .i E

q

)

≈ ρ j ` j ⊗ ` j .

Hence, the fundamental problem is the following : Fundamental Problem.– Given a positive function ρ, a linear form ` 6= 0 and an embedding f 0 how to build an other embedding f such that

f h., .i E

q

≈ µ

where µ := f 0 h., .i E

q

+ ρ ` ⊗ ` ?

(16)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

The one dimensional case

One dimensional fundamental problem.– Let

f 0 : [0, 1] −→ E q be an embedding, ρ a positive function,

` 6= 0 a linear form on R , how to build an other embedding f : [0, 1] −→ E q such that

∀u ∈ [0, 1], kf 0 (u)k 2 ≈ kf 0 0 (u)k 2 + ρ(u)` 2 (∂ u ) ?

• For short, we set r (u) :=

q

kf 0 0 (u)k 2 + ρ(u)` 2 (∂ u ).

(17)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Choosing the Loops

with

∀s ∈ R / Z , h u (s) := r (u)e iα(u) cos(2πs)

and α(u) > 0 is such that Z 1

0

r (u)e iα(u) cos(2πs) ds = f 0 0 (u).

(18)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Choosing the Loops

with

∀s ∈ R / Z , h u (s) := r (u)e iα(u) cos(2πs) and α(u) > 0 is such that

Z 1 0

r (u)e iα(u) cos(2πs) ds = f 0 0 (u).

(19)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Our Convex Integration Process

• The convex integration formula : f (t) := f 0 (0) +

Z t

0

r (u)e iα(u) cos 2π N u du.

where e := cos θ t + sin θ n with t := f

0 0

kf

00

k and n is a unit normal to the curve f 0 .

Lemma.– The map f solves the one dimensional

fundamental problem. Its speed kf 0 k is equal to the given function r = (kf 0 0 k 2 + ρ` 2 (∂ c ))

12

. Moreover

kf − f 0 k C

0

= O 1

N

and if N is large enough f is an embedding.

(20)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Our Convex Integration Process

• The convex integration formula : f (t) := f 0 (0) +

Z t

0

r (u)e iα(u) cos 2π N u du.

where e := cos θ t + sin θ n with t := f

0 0

kf

00

k and n is a unit normal to the curve f 0 .

Lemma.– The map f solves the one dimensional

fundamental problem. Its speed kf 0 k is equal to the given function r = (kf 0 0 k 2 + ρ` 2 (∂ c ))

12

. Moreover

kf − f 0 k C

0

= O 1

N

and if N is large enough f is an embedding.

(21)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Our Convex Integration Process

A short curve f

0

(black) and the curve f obtained with the one

dimensional convex integration formula (grey, N = 9 and N = 20).

(22)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• We assume for simplicity that ker ` = Span(e 2 , ..., e n ) and

`(e 1 ) = 1.

• Let s ∈ [0, 1] and c = (c 2 , ..., c n ) ∈ [0, 1] n−1 , we set f (s, c) := f 0 (0, c) +

Z s

0

r (u, c)e iα(u,c) cos 2πNu du

with r = p

µ(e 1 , e 1 ) = p

kdf 0 (e 1 )k 2 + ρ.

• The map f is not a solution of our Fundamental Problem. We do not have

kf h., .i E

q

− µk C

0

= O 1

N

with µ := f 0 h., .i E

q

+ ρ ` ⊗ `.

(23)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• We assume for simplicity that ker ` = Span(e 2 , ..., e n ) and

`(e 1 ) = 1.

• Let s ∈ [0, 1] and c = (c 2 , ..., c n ) ∈ [0, 1] n−1 , we set f (s, c) := f 0 (0, c) +

Z s

0

r (u, c)e iα(u,c) cos 2πNu du

with r = p

µ(e 1 , e 1 ) = p

kdf 0 (e 1 )k 2 + ρ.

• The map f is not a solution of our Fundamental Problem.

We do not have

kf h., .i E

q

− µk C

0

= O 1

N

with µ := f 0 h., .i E

q

+ ρ ` ⊗ `.

(24)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• By C 1, ˆ 1 -density we have

df (e j ) = df 0 (e j ) + O( 1 N )

• Thus, for all 1 < i , 1 < j, we have

(f h., .i E

q

)(e i , e j ) = hdf (e i ), df (e j )i E

q

= hdf 0 (e i ), df 0 (e j )i E

q

+ O 1

N

= µ(e i , e j ) + O 1

N

• Therefore

k(f h., .i E

q

− µ)(e i , e j )k C

0

= O 1

N

for all 1 < i, 1 < j.

(25)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• By C 1, ˆ 1 -density we have

df (e j ) = df 0 (e j ) + O( 1 N )

• Thus, for all 1 < i, 1 < j , we have

(f h., .i E

q

)(e i , e j ) = hdf (e i ), df (e j )i E

q

= hdf 0 (e i ), df 0 (e j )i E

q

+ O 1

N

= µ(e i , e j ) + O 1

N

• Therefore

k(f h., .i E

q

− µ)(e i , e j )k C

0

= O 1

N

for all 1 < i, 1 < j.

(26)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• By C 1, ˆ 1 -density we have

df (e j ) = df 0 (e j ) + O( 1 N )

• Thus, for all 1 < i, 1 < j , we have

(f h., .i E

q

)(e i , e j ) = hdf (e i ), df (e j )i E

q

= hdf 0 (e i ), df 0 (e j )i E

q

+ O 1

N

= µ(e i , e j ) + O 1

N

• Therefore

k(f h., .i E

q

− µ)(e i , e j )k C

0

= O 1

N

for all 1 < i , 1 < j.

(27)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• By definition of f we have

df (s,c) (e 1 ) = r (s, c)e iα(s,c) cos 2πNs .

• Thus

(f h., .i E

q

)(e 1 , e 1 ) = hdf (e 1 ), df (e 1 )i E

q

= r 2 (s, c)

= µ(e 1 , e 1 )

• Therefore

k(f h., .i E

q

− µ)(e 1 , e 1 )k C

0

= 0.

(28)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• By definition of f we have

df (s,c) (e 1 ) = r (s, c)e iα(s,c) cos 2πNs .

• Thus

(f h., .i E

q

)(e 1 , e 1 ) = hdf (e 1 ), df (e 1 )i E

q

= r 2 (s, c)

= µ(e 1 , e 1 )

• Therefore

k(f h., .i E

q

− µ)(e 1 , e 1 )k C

0

= 0.

(29)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• By definition of f we have

df (s,c) (e 1 ) = r (s, c)e iα(s,c) cos 2πNs .

• Thus

(f h., .i E

q

)(e 1 , e 1 ) = hdf (e 1 ), df (e 1 )i E

q

= r 2 (s, c)

= µ(e 1 , e 1 )

• Therefore

k(f h., .i E

q

− µ)(e 1 , e 1 )k C

0

= 0.

(30)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• The problem arises with the mixted term hdf (e 1 ), df (e j )i E

q

, j > 1.

• Indeed, in the one hand

(f h., .i E

q

)(e 1 , e j ) = hdf (e 1 ), df (e j )i E

q

= hdf (e 1 ), df 0 (e j )i E

q

+ O 1

N

= hr e iF , df 0 (e j )i E

q

+ O 1

N

= hr cos( F )t, df 0 (e j )i E

q

+hr sin( F )n, df 0 (e j )i E

q

+ O

1 N

= r cos( F )

kdf 0 (e 1 )k hdf 0 (e 1 ), df 0 (e j )i E

q

+O

1 N

(31)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• The problem arises with the mixted term hdf (e 1 ), df (e j )i E

q

, j > 1.

• Indeed, in the one hand

(f h., .i E

q

)(e 1 , e j ) = hdf (e 1 ), df (e j )i E

q

= hdf (e 1 ), df 0 (e j )i E

q

+ O 1

N

= hr e iF , df 0 (e j )i E

q

+ O 1

N

= hr cos( F )t, df 0 (e j )i E

q

+hr sin( F )n, df 0 (e j )i E

q

+ O

1 N

= r cos( F )

kdf 0 (e 1 )k hdf 0 (e 1 ), df 0 (e j )i E

q

+O

1 N

(32)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• In the other hand, since `(e j ) = 0, we have µ(e 1 , e j ) = hdf 0 (e 1 ), df 0 (e j )i E

q

.

• Therefore

k(f h., .i E

q

− µ)(e 1 , e j )k C

0

6= O 1

N

unless hdf 0 (e i ), df 0 (e j )i E

q

≡ 0.

Claim.– This difficulty vanishes if the convex integration is done along the integral lines of a vector field W such that

∀j ∈ {2, ..., m}, µ(W , e j ) = 0

i. e. W is µ-orthogonal to ker `.

(33)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• In the other hand, since `(e j ) = 0, we have µ(e 1 , e j ) = hdf 0 (e 1 ), df 0 (e j )i E

q

.

• Therefore

k(f h., .i E

q

− µ)(e 1 , e j )k C

0

6= O 1

N

unless hdf 0 (e i ), df 0 (e j )i E

q

≡ 0.

Claim.– This difficulty vanishes if the convex integration is done along the integral lines of a vector field W such that

∀j ∈ {2, ..., m}, µ(W , e j ) = 0

i. e. W is µ-orthogonal to ker `.

(34)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

A technical difficulty

• In the other hand, since `(e j ) = 0, we have µ(e 1 , e j ) = hdf 0 (e 1 ), df 0 (e j )i E

q

.

• Therefore

k(f h., .i E

q

− µ)(e 1 , e j )k C

0

6= O 1

N

unless hdf 0 (e i ), df 0 (e j )i E

q

≡ 0.

Claim.– This difficulty vanishes if the convex integration is done along the integral lines of a vector field W such that

∀j ∈ {2, ..., m}, µ(W , e j ) = 0

i. e. W is µ-orthogonal to ker `.

(35)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Adjusting the convex integration formula

Proposition.– The resulting map f solves the fundamental problem. Precisely

kf h., .i E

q

− µk = O 1

N

where µ = f 0 h., .i E

q

+ ρ ` ⊗ `. Moreover 1) kf − f 0 k C

0

= O N 1

, 2) kdf − df 0 k C

0

Cte N + √

12

|`(W )|,

and if N is large enough, f is an embedding.

(36)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential

Relation

(37)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential

Relation

(38)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential

Relation

(39)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential Relation

Question.– Is the limit

−→0 lim f an isometric map (if it exists) ?

Answer.– No !

−→0 lim f = f 0 .

(40)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential Relation

Question.– Is the limit

−→0 lim f an isometric map (if it exists) ? Answer.– No !

−→0 lim f = f 0 .

(41)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Approximating the Differential

Relation

(42)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Approximating the Differential

Relation

(43)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Iterated Convex Integrations

(44)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Iterated Convex Integrations

(45)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Iterated Convex Integrations

(46)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential Relation

Question.– Is the limit

k−→+∞ lim f k an isometric map (if it exists) ?

Answer.– Yes ! Let us see why

i) it is C 0 converging, ii) it is C 1 converging. Consequently

f ∞ := lim

k−→+∞ f k is a C 1 isometric map.

(47)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

Thickening the Differential Relation

Question.– Is the limit

k−→+∞ lim f k an isometric map (if it exists) ? Answer.– Yes !

Let us see why

i) it is C 0 converging, ii) it is C 1 converging.

Consequently

f ∞ := lim

k−→+∞ f k is a C 1 isometric map.

(48)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

C 0 Convergence

It is enough to control the difference kf k − f k−1 k C

0

.

We set

f = lim

k→+∞ f k .

(49)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

C 1 convergence

It is enough to control the difference kdf k − df k−1 k C

0

.

kdf k − df k−1 k C

0

≤ C te q

dist ( R e k −1 , R e k )

(50)

H-principle for Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

C 1 convergence

It is enough to control the difference kdf k − df k−1 k C

0

.

kdf k − df k−1 k C

0

≤ C te q

dist ( R e k−1 , R e k )

(51)

The H-principle for

Isometric Embeddings

V.Borrelli

Isometric Maps How to deal with the non- ampleness ? The one dimensional case How to deal with a closed relation ?

John Nash

Références

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