L2: From Nash-Kuiper to Gromov Vincent Borrelli

L2: From Nash-Kuiper to Gromov

Vincent Borrelli

Institut Camille Jordan - Université Claude Bernard Lyon 1

An incredible result

«At first, I looked at one of Nash’s papers and thought it was just nonsense [...] It was incredible. It could not be true but it was true ».

An incredible

«I was thinking about this for several years, trying to understand the mechanism behind [the Nash’s proof] »

An inspirational source

The Nash’s proof was an inspirational source for the Gromov’s Convex Integration Theory

Back to the Nash-Kuiper’s proof

The step 2 problem.– Let

•f₀:U ⊂Rⁿ→E^qbe an immersion,

•ρ:U →R≥0

•`:Rⁿ→Rbe a linear form,

• >0

Findf :U →E^q such that : i) f^∗h., .i=f₀^∗h., .i+ρ`⊗` ii) kf−f₀k_C0 <

Back to the Nash-Kuiper’s proof

The step 2 problem (rephrasing+codimension at least 2).– Let

•f₀: [0,1]ⁿ→E^q be an immersion withq≥n+2,

•ρ: [0,1]ⁿ→R≥0

•`=dx₁

• >0

Findf : [0,1]ⁿ→E^q such that : i) f^∗h., .i=f₀^∗h., .i+ρdx₁⊗dx₁ ii) kf−f₀k_C0 <

Solution

•If we apply conditioni)to(∂1, ∂1)we obtain

f^∗h∂₁, ∂1i=f₀^∗h∂₁, ∂1i+ρdx₁(∂1)dx₁(∂1) i.e.

k∂₁f(x)k²=k∂₁f₀(x)k²+ρ(x)

•We first build a map∂1f(x)satisfying the above equation and then we definef to be a primitive of that map :

f(x) =f₀(0,x₂, ...,x_n) + Z x1

0

∂₁f(u,x₁, ...,x_n)du

Solution

•If we apply conditioni)to(∂1, ∂1)we obtain

f^∗h∂₁, ∂1i=f₀^∗h∂₁, ∂1i+ρdx₁(∂1)dx₁(∂1) i.e.

k∂₁f(x)k²=k∂₁f₀(x)k²+ρ(x)

•We first build a map∂1f(x)satisfying the above equation and then we definef to be a primitive of that map :

f(x) =f₀(0,x₂, ...,x_n) + Z x1

0

∂₁f(u,x₁, ...,x_n)du

Solution

•To define∂1f we follow the Nash’s approach. Given two unit normal vectors off₀:

u,v: [0,1]ⁿ→E^q

such thathu,vi=0, we look for a solution∂₁f behaving as a tangent vector to an helix.

Convex Integration

•We put

∂₁f(x) =p

ρ(x)e^iθ(x)+∂₁f₀(x)

wheree^iθ:= cosθu+ sinθvandθ: [0,1]ⁿ−→Rwill be chosen latter.

Convex Integration

•Since(u,v)are unit normal vectors, we have

∂₁f(x) =p

ρ(x)e^iθ(x)+∂₁f₀(x) =⇒ k∂₁f(x)k²=ρ(x) +k∂₁f₀(x)k²

Convex Integration

•A possible choice forθis

θ(x) =2πNx₁

whereN ∈N^∗ is a free parameter (= the number of spirals) .

C

-density

•For everyx ∈[0,1]ⁿ we set f(x):=f₀(0,x₂, ...,x_n) +

Z x1

0

pρ(u,x₂, ...,x_n)e^i2πNu+∂₁f₀(u,x₂, ...,x_n)du

= f₀(x) + Z x1

0

pρ(u,x₂, ...,x_n)e^i2πNudu

•SinceR1

0 e^i2πudu =0, we have (see the lemma below) kf−f₀k_C0 =O(1

N) and k∂_jf −∂_jf₀k_C0 =O(1 N) for everyj≥2.

•Thus, ifN is large enough,f fulfills theC⁰-closeness conditionii).

C

-density

•For everyx ∈[0,1]ⁿ we set f(x):=f₀(0,x₂, ...,x_n) +

Z x1

0

pρ(u,x₂, ...,x_n)e^i2πNu+∂₁f₀(u,x₂, ...,x_n)du

= f₀(x) + Z x1

0

pρ(u,x₂, ...,x_n)e^i2πNudu

•SinceR1

0 e^i2πudu=0, we have (see the lemma below) kf−f₀k_C0 =O(1

N) and k∂_jf −∂_jf₀k_C0 =O(1 N) for everyj≥2.

•Thus, ifN is large enough,f fulfills theC⁰-closeness conditionii).

C

-density

•For everyx ∈[0,1]ⁿ we set f(x):=f₀(0,x₂, ...,x_n) +

Z x1

0

pρ(u,x₂, ...,x_n)e^i2πNu+∂₁f₀(u,x₂, ...,x_n)du

= f₀(x) + Z x1

0

pρ(u,x₂, ...,x_n)e^i2πNudu

•SinceR1

0 e^i2πudu=0, we have (see the lemma below) kf−f₀k_C0 =O(1

N) and k∂_jf −∂_jf₀k_C0 =O(1 N) for everyj≥2.

•Thus, ifN is large enough,f fulfills theC⁰-closeness conditionii).

A useful lemma

Lemma.– Let f : [a,b]→E^qbe a C¹function and h:R→Rbe a continuous T -periodic function then

Z b a

f(s)h(Ns)ds= Z b

a

f(s)ds

! 1 T

Z T 0

h(s)ds

! +O

1 N

In particular, if h=0, then Z b

a

f(s)h(Ns)ds=O 1

N

A useful lemma

Proof.– Letg=h−h. We have Z b

a

f(s) h(Ns)−h ds =

Z b a

f(s)g(Ns)ds

=

f(s)G(Ns) N

b a

− Z b

a

f⁰(s)G(Ns) N ds whereGis the primitive ofg given byG(t) =Rt

0g(s)ds.

Since

Z T 0

g(s)ds= Z T

0

(h(s)−h)ds=0 the primitiveGisT-periodic thus bounded. We conclude

Z b a

f(s) h(Ns)−h

ds=O 1

N

A useful lemma

Proof.– Letg=h−h. We have Z b

a

f(s) h(Ns)−h ds =

Z b a

f(s)g(Ns)ds

=

f(s)G(Ns) N

b a

− Z b

a

f⁰(s)G(Ns) N ds whereGis the primitive ofg given byG(t) =Rt

0g(s)ds.

Since

Z T 0

g(s)ds= Z T

0

(h(s)−h)ds=0 the primitiveGisT-periodic thus bounded.

We conclude Z b

a

f(s) h(Ns)−h

ds=O 1

N

A useful lemma

Proof.– Letg=h−h. We have Z b

a

f(s) h(Ns)−h ds =

Z b a

f(s)g(Ns)ds

=

f(s)G(Ns) N

b a

− Z b

a

f⁰(s)G(Ns) N ds whereGis the primitive ofg given byG(t) =Rt

0g(s)ds.

Since

Z T 0

g(s)ds= Z T

0

(h(s)−h)ds=0 the primitiveGisT-periodic thus bounded. We conclude

Z b a

f(s) h(Ns)−h

ds=O 1

N

The isometric condition

•By construction, forj =1, we have

k∂₁f(x)k²=k∂₁f₀k²+ρ(x)

for everyx ∈[0,1]ⁿ. Thus,f fulfills Conditioni)for the couple(∂1, ∂1).

•However, for the other couples(∂_i, ∂_j),(i,j)6= (1,1),we have h∂_if(x), ∂_jf(x)i = h∂_if₀(x), ∂_jf₀(x)i+O(1

N)

= h∂_if₀(x), ∂_jf₀(x)i+ρ(x)dx₁(∂_i)dx₁(∂_j) +O(1 N)

•Finally, we have solved conditioni)approximately f^∗h., .i=f₀^∗h., .i+ρdx₁⊗dx₁+O(1

N)

•Note that Nash also solved this condition approximately.

The isometric condition

•By construction, forj =1, we have

k∂₁f(x)k²=k∂₁f₀k²+ρ(x)

for everyx ∈[0,1]ⁿ. Thus,f fulfills Conditioni)for the couple(∂1, ∂1).

•However, for the other couples(∂_i, ∂_j),(i,j)6= (1,1),we have h∂_if(x), ∂_jf(x)i = h∂_if₀(x), ∂_jf₀(x)i+O(1

N)

= h∂_if₀(x), ∂_jf₀(x)i+ρ(x)dx₁(∂_i)dx₁(∂_j) +O(1 N)

•Finally, we have solved conditioni)approximately f^∗h., .i=f₀^∗h., .i+ρdx₁⊗dx₁+O(1

N)

•Note that Nash also solved this condition approximately.

The isometric condition

•By construction, forj =1, we have

k∂₁f(x)k²=k∂₁f₀k²+ρ(x)

for everyx ∈[0,1]ⁿ. Thus,f fulfills Conditioni)for the couple(∂1, ∂1).

•However, for the other couples(∂_i, ∂_j),(i,j)6= (1,1),we have h∂_if(x), ∂_jf(x)i = h∂_if₀(x), ∂_jf₀(x)i+O(1

N)

= h∂_if₀(x), ∂_jf₀(x)i+ρ(x)dx₁(∂_i)dx₁(∂_j) +O(1 N)

•Finally, we have solved conditioni)approximately f^∗h., .i=f₀^∗h., .i+ρdx₁⊗dx₁+O(1

N)

•Note that Nash also solved this condition approximately.

The isometric condition

•By construction, forj =1, we have

k∂₁f(x)k²=k∂₁f₀k²+ρ(x)

for everyx ∈[0,1]ⁿ. Thus,f fulfills Conditioni)for the couple(∂1, ∂1).

•However, for the other couples(∂_i, ∂_j),(i,j)6= (1,1),we have h∂_if(x), ∂_jf(x)i = h∂_if₀(x), ∂_jf₀(x)i+O(1

N)

= h∂_if₀(x), ∂_jf₀(x)i+ρ(x)dx₁(∂_i)dx₁(∂_j) +O(1 N)

•Finally, we have solved conditioni)approximately f^∗h., .i=f₀^∗h., .i+ρdx₁⊗dx₁+O(1

N)

Convex Integration Formula

Definition.– Let

•γ : [0,1]ⁿ×R/Z→R^qbe a family of loops

•f₀: [0,1]ⁿ→R^q a map

•N>0

We define a new mapF : [0,1]ⁿ→R^qby setting F(x) :=f₀(0,x₂, ...,x_n) +

Z x1

0

γ(u,x₂, ...,x_n;Nu)du for everyx ∈[0,1]ⁿ. The mapF is said to be obtained fromf₀by Convex Integration. It is denoted byF =CI_γ(f₀, ∂₁,N).

Convex Integration Formula

Example.– The mapf previously built is obtained by convex integration fromf₀and with the following choice forγ :

γ(x,t) =p

ρ(x)e^2iπt +∂₁f₀(x)

•Observe that

Z 1 0

γ(x,t)dt =∂1f₀(x).

In average, the effect ofγ and∂₁f₀are the same. This is the reason whyf isC⁰-close tof₀.

Definition.– A family of loopsγ : [0,1]ⁿ×R/Z→R^q satisfies the average condition with respect to f₀and in the direction of∂₁if

∀x ∈ [0,1]ⁿ, Z 1

0

γ(x,t)dt=∂₁f₀(x).

Convex Integration Formula

Example.– The mapf previously built is obtained by convex integration fromf₀and with the following choice forγ :

γ(x,t) =p

ρ(x)e^2iπt +∂₁f₀(x)

•Observe that

Z 1 0

γ(x,t)dt =∂₁f₀(x).

In average, the effect ofγ and∂₁f₀are the same. This is the reason whyf isC⁰-close tof₀.

Definition.– A family of loopsγ : [0,1]ⁿ×R/Z→R^q satisfies the average condition with respect to f₀and in the direction of∂₁if

∀x ∈ [0,1]ⁿ, Z 1

0

γ(x,t)dt=∂₁f₀(x).

Convex Integration Formula

Example.– The mapf previously built is obtained by convex integration fromf₀and with the following choice forγ :

γ(x,t) =p

ρ(x)e^2iπt +∂₁f₀(x)

•Observe that

Z 1 0

γ(x,t)dt =∂₁f₀(x).

In average, the effect ofγ and∂₁f₀are the same. This is the reason whyf isC⁰-close tof₀.

Definition.– A family of loopsγ : [0,1]ⁿ×R/Z→R^q satisfies the average condition with respect to f₀and in the direction of∂₁if

∀x ∈ [0,1]ⁿ, Z 1

0

γ(x,t)dt=∂₁f₀(x).

Convex Integration Formula

Proposition.–Ifγ satisfies the average condition with respect to f₀ and in the direction∂₁then the following properties hold for

F =CI_γ(f₀, ∂1,N):

(P₁) kf₀−Fk_C0 =O(1/N),

(P₂) k∂_if₀−∂_iFk_C0 =O(1/N)for every i 6=1, (P₃) ∀x ∈[0,1]ⁿ, ∂₁F(x) =γ(x,Nx₁).

Proof.– Postponed to the lecture devoted to the 1D Convex Integration.

Convex Integration Formula

Proposition.–Ifγ satisfies the average condition with respect to f₀ and in the direction∂₁then the following properties hold for

F =CI_γ(f₀, ∂1,N):

(P₁) kf₀−Fk_C0 =O(1/N),

(P₂) k∂_if₀−∂_iFk_C0 =O(1/N)for every i 6=1, (P₃) ∀x ∈[0,1]ⁿ, ∂₁F(x) =γ(x,Nx₁).

Proof.– Postponed to the lecture devoted to the 1D Convex Integration.

Improving the Kuiper Formula

The step 2 problem (codimension 1).– We assumeq=n+1. Given >0 we want to constructf : [0,1]ⁿ →Eⁿ⁺¹such that

i) kf^∗h., .i −(f₀^∗h., .i+ρdx₁⊗dx₁)k_C0 <

ii) kf−f₀k_C0 <

Solution.–We are going to buildf by a convex integration fromf₀in the direction∂₁. Any such map will satisfy property(P2):

k∂_if−∂_if₀k_C0 =O(1/N)pour touti6=1 which implies that

h∂_if, ∂_jfi=h∂_if₀, ∂_jf₀i+O(1/N) for everyi6=1,j6=1.

Improving the Kuiper Formula

The step 2 problem (codimension 1).– We assumeq=n+1. Given >0 we want to constructf : [0,1]ⁿ →Eⁿ⁺¹such that

i) kf^∗h., .i −(f₀^∗h., .i+ρdx₁⊗dx₁)k_C0 <

ii) kf−f₀k_C0 <

Solution.–We are going to buildf by a convex integration fromf₀in the direction∂₁. Any such map will satisfy property(P2):

k∂_if−∂_if₀k_C0 =O(1/N)pour touti6=1 which implies that

h∂_if, ∂_jfi=h∂_if₀, ∂_jf₀i+O(1/N) for everyi6=1,j6=1.

Improving the Kuiper Formula

•It remains to solve(i)for the couples(1,i),i ∈ {1, ...,n}, i. e.

( h∂₁f, ∂_ifi=h∂₁f₀, ∂_if₀i+O(1/N)for everyi 6=1 k∂₁fk²=k∂₁f₀k²+ρ+O(1/N)

Or equivalently

( h∂₁f, ∂_if0i=h∂₁f₀, ∂_if₀i+O(1/N)for everyi6=1 k∂₁fk²=k∂₁f₀k²+ρ+O(1/N)

since

k∂_if−∂_if₀k_C0 =O(1/N)pour touti6=1

•For everyx ∈[0,1]ⁿ, we put

R_x = (

v ∈Rⁿ⁺¹| hv, ∂_if₀(x)i=h∂₁f₀(x), ∂_if₀(x)ifor everyi 6=1 kvk²=k∂₁f₀(x)k²+ρ(x)

)

Improving the Kuiper Formula

•It remains to solve(i)for the couples(1,i),i ∈ {1, ...,n}, i. e.

( h∂₁f, ∂_ifi=h∂₁f₀, ∂_if₀i+O(1/N)for everyi 6=1 k∂₁fk²=k∂₁f₀k²+ρ+O(1/N)

Or equivalently

( h∂₁f, ∂_if0i=h∂₁f₀, ∂_if₀i+O(1/N)for everyi6=1 k∂₁fk²=k∂₁f₀k²+ρ+O(1/N)

since

k∂_if−∂_if₀k_C0 =O(1/N)pour touti6=1

•For everyx ∈[0,1]ⁿ, we put

R_x = (

v ∈Rⁿ⁺¹| hv, ∂_if₀(x)i=h∂₁f₀(x), ∂_if₀(x)ifor everyi 6=1 kvk²=k∂ f (x)k²+ρ(x)

)

Improving the Kuiper Formula

R_x = (

v ∈Rⁿ⁺¹| hv, ∂_if₀(x)i=h∂₁f₀(x), ∂_if₀(x)ifor everyi 6=1 kvk²=k∂₁f₀(x)k²+ρ(x)

)

•The setR_x is the intersection of a hypersphereSⁿ(R)of radius

R= q

k∂₁f₀(x)k²+ρ(x) and of an affine 2-plane

W ={v ∈Rⁿ⁺¹| hv, ∂if₀(x)i=h∂₁f₀(x), ∂if₀(x)ifor everyi 6=1}

Improving the Kuiper Formula

•It is easily seen thatR_x is a circle whose center is given by the projectionπ(∂1f₀(x))of∂1f₀(x)onP =Span(∂₂f₀(x), ..., ∂_nf₀(x))and whose radius is

r(x) = q

k∂₁f₀(x)k²+ρ(x)− kπ(∂₁f₀(x))k²

Improving the Kuiper Formula

•We have to choose a family of loopsγ : [0,1]ⁿ×R/Z→Rⁿ⁺¹such that

1)t7→γ(x,t)∈ R_x 2)

Z 1 0

γ(x,t)dt =∂1f₀(x)

Improving the Kuiper Formula

•We set

t= ∂1f₀−π(∂1f₀)

k∂₁f₀−π(∂₁f₀)k and n= ∂1f₀∧...∧∂nf₀ k∂₁f₀∧...∧∂nf₀k

Improving the Kuiper Formula

•We defineγ to be

γ(x,t) =π(∂₁f₀(x)) +r(x)(cosθt+ sinθn)

withθ(x,t) =α(x) cos2πtandα(x)is to be determined.

Improving the Kuiper Formula

•We defineγ to be

γ(x,t) =π(∂₁f₀(x)) +r(x)(cosθt+ sinθn) withθ(x,t) =α(x) cos2πtandα(x)is to be determined.

Improving the Kuiper Formula

•We then have Z 1

0

γ(x,t)dt=r(x)J₀(α(x))t+π(∂1f₀(x)) whereJ₀the Bessel function.

The Bessel Function J

J (α) = 1Z π

cos(αsinu)du

Improving the Kuiper Formula

•We then have Z 1

0

γ(x,t)dt=r(x)J₀(α(x))t+π(∂₁f₀(x))

whereJ₀the Bessel function.

•To ensure the average to be equal to∂₁f₀, it is enough to choose α(x) =J₀⁻¹

k∂₁f₀(x)−π(∂₁f₀(x))k r(x)

•SincerJ₀(α)t+π(∂₁f₀) =∂₁f₀, we can write

γ(x,t) =r(cos(αcos2πt)−J₀(α))t+rsin(αcos2πt)n+∂1f₀

Improving the Kuiper Formula

•We then have Z 1

0

γ(x,t)dt=r(x)J₀(α(x))t+π(∂₁f₀(x))

whereJ₀the Bessel function.

•To ensure the average to be equal to∂₁f₀, it is enough to choose α(x) =J₀⁻¹

k∂₁f₀(x)−π(∂₁f₀(x))k r(x)

•SincerJ₀(α)t+π(∂₁f₀) =∂₁f₀, we can write

γ(x,t) =r(cos(αcos2πt)−J₀(α))t+rsin(αcos2πt)n+∂1f₀

Improving the Kuiper Formula

To sum up.– The mapf =CI_γ(f₀, ∂₁,N)with

γ(x,t) =r(cos(αcos2πt)−J₀(α))t+rsin(αcos2πt)n+∂1f₀ and

r = q

k∂₁f₀k²+ρ− kπ(∂₁f₀)k², α=J₀⁻¹

k∂₁f₀−π(∂₁f₀)k r

satisfies the following properties

i) f^∗h., .i=f₀^∗h., .i+ρdx₁⊗dx₁+O(1/N) ii) kf−f₀k_C0 =O(1/N)

iii) k∂_if−∂if₀k_C0 =O(1/N)for everyi6=1.

Improving the Kuiper Formula

Analytical expression.– The mapf =CI_γ(f₀, ∂₁,N)has the following expression

f(x) =f₀(0,x₂, ...,x_m) + Z x1

0

γ(u,x₂, ...,x_m;Nu)du

with

γ(x,t) =π(∂₁f₀(x))+r(x)(cos(α(x) cos2πt)t(x)+sin(α(x) cos2πt)n(x)).

•By comparison the Kuiper formula is :

f(x)=f₀(x)−3ρ(x)

16N sin(2Nx₁)t(x)+

p3ρ(x)

√

2N sin Nx₁−3ρ(x)

16 sin(2Nx₁) n(x)

Improving the Kuiper Formula

Analytical expression.– The mapf =CI_γ(f₀, ∂₁,N)has the following expression

f(x) =f₀(0,x₂, ...,x_m) + Z x1

0

γ(u,x₂, ...,x_m;Nu)du

with

γ(x,t) =π(∂₁f₀(x))+r(x)(cos(α(x) cos2πt)t(x)+sin(α(x) cos2πt)n(x)).

•By comparison the Kuiper formula is :

f(x)=f₀(x)−3ρ(x)

16N sin(2Nx₁)t(x)+

p3ρ(x)

√

2N sin Nx₁−3ρ(x)

16 sin(2Nx₁) n(x)

The outrageously simple idea

Mikhaïl Gromov

•TheO(ρ²)default in the Kuiper process deserves to be corrected

•This can be done by combining a geometrical approach with a simple integral formula.

The outrageously simple idea

Mikhaïl Gromov

•TheO(ρ²)default in the Kuiper process deserves to be corrected

•This can be done by combining a geometrical approach with a simple integral formula.

Mikhaïl Gromov